The materials to prepare students for

The materials to prepare students for practical lessons of inorganic chemistry

LESSON № 2.

Theme. Basic laws of chemistry. The notion about equivalent of substance. Experimental determination of metal’s equivalent

PLAN

1. Basic laws of chemistry:

a) Law of constant composition;

b) Law of multiple proportions;

c) Law of combining volumes;

c) Law of combining volumes;

d) Avogadro’s law;

2. Molar volume of gas. Relation between density of gas and it molecular mass.  Normal state of gas. Clapeyron’s – Mendeleyev’s equation

a) Ideal gas law;

 b) General gas law.

3. Equivalent. Law equivalents

4. Chemical formulas, their types.

 

1. BASIC LAWS OF CHEMISTRY.\

Chemical laws are those laws of nature relevant to chemistry. The most fundamental concept in chemistry is the law of conservation of mass, which states that there is no detectable change in the quantity of matter during an ordinary chemical reaction. Modern physics shows that it is actually energy that is conserved, and that energy and mass are related; a concept which becomes important iuclear chemistry. Conservation of energy leads to the important concepts of equilibrium, thermodynamics, and kinetics.

Additional laws of chemistry elaborate on the law of conservation of mass. Joseph Proust’s law of definite composition says that pure chemicals are composed of elements in a definite formulation; we now know that the structural arrangement of these elements is also important.

Dalton’s law of multiple proportions says that these chemicals will present themselves in proportions that are small whole numbers (i.e. 1:2 O:H in water); although in many systems (notably biomacromolecules and minerals) the ratios tend to require large numbers, and are frequently represented as a fraction. Such compounds are known as non-stoichiometric compounds

More modern laws of chemistry define the relationship between energy and transformations.

·        In equilibrium, molecules exist in mixture defined by the transformations possible on the timescale of the equilibrium, and are in a ratio defined by the intrinsic energy of the molecules—the lower the intrinsic energy, the more abundant the molecule.

·        Transforming one structure to another requires the input of energy to cross an energy barrier; this can come from the intrinsic energy of the molecules themselves, or from an external source which will generally accelerate transformations. The higher the energy barrier, the slower the transformation occurs.

·        There is a hypothetical intermediate, or transition structure, that corresponds to the structure at the top of the energy barrier. The Hammond-Leffler Postulate states that this structure looks most similar to the product or starting material which has intrinsic energy closest to that of the energy barrier. Stabilizing this hypothetical intermediate through chemical interaction is one way to achieve catalysis.

·        All chemical processes are reversible (law of microscopic reversibility) although some processes have such an energy bias, they are essentially irreversible.

a) Law of constant composition

Law of constant composition

For the first time has formulated by G. Prust (1808).

“All individual chemical substances have constant quality and quantity composition and definite chemical structure and does not depend on how this substance was prepared.”

From the law of constant composition follows that at complex substance formation the elements combine with each other in definite mass proportions.

An equivalent statement is the law of constant composition, which states that all samples of a given chemical compound have the same elemental composition. For example, oxygen makes up ⁸/₉ of the mass of any sample of pure water, while hydrogen makes up the remaining ¹/₉ of the mass. Along with the law of multiple proportions, the law of definite proportions forms the basis of stoichiometry.

This observation was first made by the French chemist Joseph Proust, based on several experiments conducted between 1798 and 1804. Based on such observations, Proust made statements like this one, in 1806:

“I shall conclude by deducing from these experiments the principle I have established at the commencement of this memoir, viz. that iron like many other metals is subject to the law of nature which presides at every true combination, that is to say, that it unites with two constant proportions of oxygen. In this respect it does not differ from tin, mercury, and lead, and, in a word, almost every known combustible.”

The law of definite proportions might seem obvious to the modern chemist, inherent in the very definition of a chemical compound. At the end of the 18th century, however, when the concept of a chemical compound had not yet been fully developed, the law was novel. In fact, when first proposed, it was a controversial statement and was opposed by other chemists, most notably Proust’s fellow Frenchman Claude Louis Berthollet, who argued that the elements could combine in any proportion. The very existence of this debate underscores that at the time, the distinction between pure chemical compounds and mixtures had not yet been fully developed.

The law of definite proportions contributed to, and was placed on a firm theoretical basis by, the atomic theory that John Dalton promoted beginning in 1803, which explained matter as consisting of discrete atoms, that there was one type of atom for each element, and that the compounds were made of combinations of different types of atoms in fixed proportions. A related early idea was Prout’s hypothesis, which supposed that hydrogen was the only functional unit, and was related to the whole number rule, which was the rule of thumb that atomic masses were whole number multiples of the mass of hydrogen. This was later rejected in the 1820s and 30s following more refined measurements of atomic mass, notably by Jöns Jacob Berzelius, which revealed in particular that the atomic mass of chlorine was 35.45, which was incompatible with the hypothesis. Today however this discrepancy is understood (since the 1920s) to be due to the presence of isotopes, and the mass of given isotopes is very close to satisfying the whole number rule, with the mass defect caused by differing binding energies being significantly smaller.

It may be noted that although very useful in the foundation of modern chemistry, the law of definite proportions is not universally true. There exist non-stoichiometric compounds whose elemental composition can vary from sample to sample. An example is the iron oxide wüstite, which can contain between 0.83 and 0.95 iron atoms for every oxygen atom, and thus contain anywhere between 23% and 25% oxygen. In general, Proust’s measurements were not accurate enough to detect such variations. In addition, the isotopic composition of an element can vary depending on its source, hence its weight in a pure stoichiometric compound may vary. This fact is used in geochemical dating since astronomical, atmospheric, oceanic, crustal and deep Earth processes may concentrate lighter or heavier isotopes preferentially. With the exception of hydrogen and its isotopes, the effect is usually small but measurable with modern instrumentation. An additional note: many natural polymers vary in composition (for instance RNA, proteins, carbohydrates) even when “pure”. Polymers are generally not considered “pure chemical compounds” except when their molecular weight is uniform (monodisperse) and their stoichiometry is constant. In this unusual case, they still may violate the law due to isotopic variations.

From the law of constant composition follows that at complex substance formation the elements combine with each other in definite mass proportions.

S +O₂ ®SO₂

Cu + 2H₂SO_4(конц.)® CuSO₄ + SO₂ + 2H₂O

2H₂S + 3O₂® 2SO₂ + 2H₂O

 

Example:

CuS ‑ copper sulphide. m (Cu) : m (S) = A_r (Cu) : A_r (S) = 64 : 32 = 2 : 1

To get copper sulphide (CuS) it is necessary to mix up the powders of copper and sulphur in mass relations 2:1.

If taken amounts of source substances do not correspond their correlation in the chemical formula of compound one of them stay in the excess.

For instance, if take 3 g. copper and 1 g. sulphur than after the reaction 1 g. copper, which did not enter in the chemical reaction will stay.

Substances with non-molecular structure do not possess strictly constant composition. Their composition depends on conditions of preparation.

The mass share of element W_(e) shows what part forms the mass of given element from the whole mass of substance: where– a number of atoms; A_r(e) – relative atomic mass of element; M_r – relative molecular mass of substance.

		n * Ar(e)
W(e))	=	———
		Mr

 

Knowing the quantitative element composition of substance its simplest molecular formula is possible to determine:

1. Mark formula of compounds A_x B_y C_z.

2. Calculate an attitude X: Y: Z through the mass shares of elements:

3. Obtained numerals divide on the least for getting total numbers.

4. Write a formula of compound.

b) Law of multiple proportions (D. Dalton, 1803)

The law of multiple proportions is one of the fundamental laws of stoichiometry and was first discovered by the English chemist John Dalton in 1803. The law states that when chemical elements combine, they do so in a ratio of small whole numbers. For example, carbon and oxygen react to form carbon monoxide (CO) or carbon dioxide (CO₂), but not CO_1.3. Further, it states that if two elements form more than one compound between them, the ratios of the masses of the second element to a fixed mass of the first element will also be in small whole numbers.

If two elements form several chemical compounds with each other, then the masses of one of the elements corresponding to the same mass of the other element in these compounds are in a simple integral proportion.

N₂O                 N₂O₃               NO₂(N₂O₄)                  N₂O₅

A number of oxygen atoms in molecules of such compounds corresponding to the two nitrogen atoms are in a proportion 1:3:4:5.

 

Law of combining volumes (Gay-Lussac, 1808)

“When gases react, the volumes consumed and produced, measured at the same temperature and pressure, are in ratios of small whole numbers”.

The expression Gay-Lussac’s law is used for each of the two relationships named after the French chemist Joseph Louis Gay-Lussac and which concern the properties of gases, though it is more usually applied to his law of combining volumes, the first listed here. The first law relates to volumes before and after a chemical reaction while the second concerns the pressure and temperature relationship for a sample of gas often known as Amontons’ Law. The law of combining volumes states that, when gases react together to form other gases, and all volumes are measured at the same temperature and pressure:

The ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.

For example, Gay-Lussac found that 2 volumes of Hydrogen and 1 volume of Oxygen would react to form 2 volume of gaseous water. In addition to Gay-Lussac’s results, Amedeo Avogadro theorized that, at the same temperature and pressure, equal volumes of gas contain equal numbers of molecules (Avogadro’s law). This hypothesis meant that the previously stated result

2 volumes of Hydrogen + 1 volume of Oxygen = 2 volumes of gaseous water could also be expressed as

2 molecules of Hydrogen + 1 molecule of Oxygen = 2 molecules of water.

The law of combining gases was made public by Joseph Louis Gay-Lussac in 1808. Avogadro’s hypothesis, however, was not initially accepted by chemists until the Italian chemist Stanislao Cannizzaro was able to convince the First International Chemical Congress in 1860.

Pressure-temperature law

This law is often referred to as Amontons’ Law of Pressure-Temperature after Guillaume Amontons, who, between 1700 and 1702, discovered the relationship between the pressure and temperature of a fixed mass of gas kept at a constant volume. Amontons discovered this while building an “air thermometer”.

The pressure of a gas of fixed mass and fixed volume is directly proportional to the gas’ absolute temperature.

Simply put, if a gas’ temperature increases, then so does its pressure if the mass and volume of the gas are held constant. The law has a particularly simple mathematical form if the temperature is measured on an absolute scale, such as in kelvins. The law can then be expressed mathematically as:

or

where:

P is the pressure of the gas

T is the temperature of the gas (measured in Kelvin).

k is a constant.

This law holds true because temperature is a measure of the average kinetic energy of a substance; as the kinetic energy of a gas increases, its particles collide with the container walls more rapidly, thereby exerting increased pressure.

For comparing the same substance under two different sets of conditions, the law can be written as:

Because Amontons discovered the law beforehand, Gay-Lussac’s name is now generally associated within chemistry with the law of combining volumes discussed in the section above. Some introductory physics textbooks still define the pressure-temperature relationship as Gay-Lussac’s law. Gay-Lussac primarily investigated the relationship between volume and temperature and published it in 1802, but his work did cover some comparison between pressure and temperature. Given the relative technology to both men, Amonton was only able to work with air as a gas, where Gay-Lussac was able to experiment with multiple types of common gases, such as oxygen, nitrogen, and hydrogen. Gay-Lussac did attribute his findings to Jacques Charles because he used much of Charles’s unpublished data from 1787 – hence, the law became known as Charles’s law or the Law of Charles and Gay-Lussac. However, in recent years the term has fallen out of favor.

Gay-Lussac’s (Amontons’) Law, Charles’ Law, and Boyle’s law form the combined gas law. These three gas laws in combination with Avogadro’s Law can be generalized by the ideal gas law.

Consequence. Stoichiometric coefficients in equations of chemical reactions for molecules of gaseous substances shows in which volume combinations gaseous substances are got or react.

Examples

a) 2CO + O₂ → 2CO₂

In the time of oxidizing of two volumes of carbon (II) oxide by one volume of oxygen forms 2 volume of carbon (IV) oxide, i.e. the volume of source reaction mixture decrease on 1 volume.

b) In the time of synthesis of ammonia from elements:

N₂ + 3H₂ → 2NH₃

One volume of nitrogen is reacting with three volumes of hydrogen: 2 volumes of ammonia are forming – the volume of source reaction mixture will be decreased in 2 times.

 

The Three Laws

Law 1: Law of Conservation of Mass (M.V. Lomonosov, 1748, A. Lavoisier 1789)

The total mass of all products of a chemical reaction is equal to the total mass of all reactants of that reaction. These statements are summaries of many observations, which required a tremendous amount of experimentation to achieve and even more creative thinking to systematize as we have written them here. By making these assumptions, we can proceed directly with the experiments which led to the development of the atomic-molecular theory. It is also called hard law theory. Dalton believed in this theory.

Goals: The statements above, though correct, are actually more vague than they might first appear. For example, exactly what do we mean when we say that all materials are made from elements? Why is it that the elements cannot be decomposed? What does it mean to combine elements into a compound? We want to understand more about the nature of elements and compounds so we can describe the processes by which elements combine to form compounds, by which compounds are decomposed into elements, and by which compounds are converted from one to another during chemical reactions. One possibility for answering these questions is to assume that a compound is formed when indestructible elements are simply mixed together, as for example, if we imagine stirring together a mixture of sugar and sand. Neither the sand nor the sugar is decomposed in the process. And the mixture can be decomposed back into the original components. In this case, though, the resultant mixture exhibits the properties of both components: for example, the mixture would taste sweet, owing to the sugar component, but gritty, characteristic of the sand component. In contrast, the compound we call iron rust bears little resemblance to elemental iron: iron rust does not exhibit elemental iron’s color, density, hardness, magnetism, etc. Since the properties of the elements are not maintained by the compound, then the compound must not be a simple mixture of the elements.

We could, of course, jump directly to the answers to these questions by stating that the elements themselves are composed of atoms: indivisible, identical particles distinctive of that element. Then a compound is formed by combining the atoms of the composite elements. Certainly, the Law of Conservation of Mass would be easily explained by the existence of immutable atoms of fixed mass. However, if we do decide to jump to conclusions and assume the existence of atoms without further evidence (as did the leading chemists of the seventeenth and eighteenth centuries), it does not lead us anywhere. What happens to iron when, after prolonged heating in air, it converts to iron rust? Why is it that the resultant combination of iron and air does not maintain the properties of either, as we would expect if the atoms of each are mixed together? An atomic view of nature would not yet provide any understanding of how the air and the iron have interacted or combined to form the new compound, and we can’t make any predictions about how much iron will produce how much iron rust. There is no basis for making any statements about the properties of these atoms. We need further observations.

Atomic-molecular concept explains this law by following manner: as a result of chemical reactions atoms do not appear and do not disappear, but occurs their rearrangement (i.e. chemical conversion-process of bond breakup between atoms with another bond formation, as a result of such conversation the molecules of source substances transform into the molecules of products of reaction). As far as a number of atoms before and after reaction stays unchangeable their general mass also must not change. At term «mass» has understood a value-characterizing amount of matter.

At the beginning of 20 century a formulation of mass conversation law was revising in connection with the appearance of relativity theory (A. Einstein, 1905), according to which the mass of substance depends on its velocities, and consequently characterizes not only matter amount, as well as its motion. An Energy DE, obtained by substance, is connected with increasing its mass Dm by the correlation DE = Dm _* c², where c is a light velocity. This correlation is not used in chemical reactions, since 1kJ of energy is corresponding ~ 10^-11g mass changing and practically cannot be measured.

Iucleus reactions, where DE greater in 10⁶ times than in chemical reactions, Dm should take into account.

Originating from the mass conservation law equations of chemical reactions is possible to form with the following calculations. It is a basis of quantitative chemical analysis.

When elements and compounds react to form new products, mass cannot be lost or gained.

“The Law of Conservation of Mass” definition states that mass cannot be created or destroyed, but changed into different forms.

So, in a chemical change, the total mass of reactants must equal the total mass of products.

By using this law, together with atomic and formula masses, you can calculate the quantities of reactants and products involved in a reaction and the simplest formula of a compound.

NOTE: (1) the symbol equation must be correctly balanced to get the right answer! (2) There are good reasons why, when doing a real chemical preparation-reaction to make a substance you will not get 100% of what you theoretically calculate

Example 1: Magnesium + Oxygen ==> Magnesium oxide

2Mg + O₂ ==> 2MgO

(atomic masses required: Mg=24 and O=16) think of the ==> as an = sign, so the mass changes in the reaction are:

(2 x 24) + (2 x 16) = 2 x (24 + 16)

48 + 32 = 2 x 40 and so 80 mass units of reactants = or produces 80 mass units of products (you can work with any mass units such as g, kg or tonne (1 tonne = 1000 kg)

Example 2: iron + sulphur ==> iron sulphide

Fe + S ==> FeS

(atomic masses: Fe = 56, S = 32)

If 59g of iron is heated with 32g of sulphur to form iron sulphide, how much iron is left unreacted? (assuming all the sulphur reacted)

From the atomic masses, 56g of Fe combines with 32g of S to give 88g FeS.

This means 59 – 56 = 3g Fe unreacted.

Example 3: When limestone (calcium carbonate) is strongly heated, it undergoes thermal decomposition to form lime (calcium oxide) and carbon dioxide gas.

CaCO₃ ==> CaO + CO₂

(relative atomic masses: Ca = 40, C = 12 and O = 16)

Calculate the mass of calcium oxide and the mass of carbon dioxide formed by decomposing 50 tonnes of calcium carbonate.

(40 + 12 + 3×16) ==> (40 + 16) + (12 + 2×16)

100 ==> 56 + 44, scaling down by a factor of two, 50 ==> 28 + 22

so decomposing 50 tonnes of limestone produces 28 tonnes of lime and 22 tonnes of carbon dioxide gas.

CHEMICAL EQUATIONS

Chemical transformations take place according to strict ‘rules’ because of the nature of molecular structure. Chemical reactions are really the rearrangement of the connectivity between the atoms in the reagenmts to produce products. The Ancients saw that chemical reactions obeyed the law of multiple proportions very precisely and correctly concluded that this was significant. Because the creation or destruction of nuclei is very unlikely without a nuclear reactor, the same number of atoms of each type (element) must exist both before and after any chemical transformation. This means that molecules must react in simple, whole-number ratios (i.e. the law of multiple proportions), as can be seen by the following example:

Even though we could try to write down the chemical reaction as

X1 * CH₄  +  X2 * O₂  =  X3 * CO₂  +  X4 * H₂O

but this equation is not useful because it is not ‘balanced’. Not balanced means it does not have the correct (or determined) stoichiometric coefficients. The process of ‘balancing’ a chemical equation is simply determining a set of coefficients {Xi}, which represent the proportions of whole molecules in the reaction that balance the number of atoms of each element on both sides of the equal sign. Note: an equal sign, as well as single and double headed arrows, is sometimes used to separate reactants and products in a chemical equation. Nomatter what is used as a separator, a chemical reaction must be written as an equation, which means that the same number of each type of atom exist each side. (Sometimes unscrupulous chemistry intructors provide unbalanced chemical equations (which are really not equations at all) in problems given to students, but they argue that these are the cards that life deals us. Nonetheless, the equations always must be checked before proceeding further with the use of said equation.) How do you balance a chemical equation?

Identify all the different elements in the chemical equation.

Count the number of atoms on both sides of the equal sign of the element that appears in the fewest different molecules in the equation. Make sure to use the correct molecular formula (structure) in this arithmetic

Adjust the Stoichiometric Coefficents of the species that contain this element to balance the count of this element in the equation.

Repeat the last two steps for each element identified in step 1. You must maintain the ratio of coefficients of every molecule that has been balanced for a previous element.

When all the elements have been balanced, multiply the coefficients in the entire equation by any number you wish. This is usually done to obtain the smallest whole number (integer) coefficients, {Xi}

For the above equation:

Carbon:   X1 = X3

Hydrogen: X4 = (2)*X1

Oxygen:   X2 = X3 + (1/2)*X4

If one chooses X1 =1, then X3 = 1, then X4 = 2, then X2 = 2, and the equation is balanced. Practice some more equation balancing on your own.

 

 Arranging of chemical equations

Include three stages:

1. Record formulas of substances: entered in the reaction (on the left) and products of reaction (on the right), having connected them on the sense by signs «+», «→»:

HgO → Hg + O₂

2. Selection the coefficients for each substance so that amount of atoms of each element in left and right part of equation will be equally:

2HgO → 2Hg + O₂

3. Checking a number of atoms of each element in left and right parts of equation.

Calculations on chemical equations

Calculations on chemical equations (stoichiometric calculations) are based on the mass conservation law of substances. In real chemical processes because of incomplete running chemical reactions and losses a products mass usually less theoretically calculated.

Yield of reaction (h) is a ratio of real product mass (m_r) to theoretically possible (m_t), expressed in shares of units or percent.

If in conditions of problems a yield of reactions products does not specified, its take in calculations as 100% (quantitative output).

Example 1.

Calculated the mass of copper formed when reduction 8 g copper oxide by hydrogen. The yield of reaction – 82% from theoretical.

Solution

CuO + H₂ → Cu + H₂O

1. Calculate a theoretical yield of copper on the equation of reaction: 80 g (1 g-mol) CuO when reducing can form 64 g (1 g-mol) Cu. 8 g CuO – X g Cu.

2.  Define how much grams of copper will be formed at 82% yield of product

Example 2.

Calculate the yield of reaction of tungsten preparation by aluminizing, if from 33.14 g of ore concentrate, containing WO₃ and non-reduce admixtures (mass share of admixtures 0.3), were obtained 12.72 g of metal.

Solution

1. Define the WO₃ mass (g) in 33.14 g of ore concentrate.

w (WO₃) = 1.0 – 0.3 = 0.7

m (WO₃) = w(WO₃) _* m_ore = 0.7 _* 33.14 = 23.2 g

2. Define the theoretical yield of tungsten as a result of reduction 23.2 g by aluminum powder.

WO₃ + 2Al → Al₂ + O₃ + W

When reducing 232 g (1g-mol) WO₃ form 187 g (1g-mol) W, but at 23.2 g WO₃ – X g W

3. Calculate the practical yield of tungsten

Example 3

How much grams of barium sulphate will be formed at pouring together the solutions containing 20.8 g of barium chloride and 18.0 g of sodium sulphate?

Solution

BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl

Calculation of product amount carries on the source substance, which is taken in the deficit.

a) Beforehand define which substance stand in the deficit.

Mark the amount of Na₂SO₄ – X

208 g (1 g-mol) BaCl₂ react with 132 g (1 g-mol) Na₂SO₄

20.8 g with X g

We found, that on the reaction with 20.8 g BaCl₂ will be spend 13.2 g Na₂SO₄, but we have 18.0 g. Thus, sodium sulphate in the reaction is taken in the excess and further calculation is necessary to carry on the BaCl₂, which is taken in deficit.

b) Define quantity of BaSO₄ precipitate. 208 g (1 g-mol) BaCl₂ form 233 g BaSO₄.

20.8 g –– Y g.

 

Observation 1: Mass relationships during chemical reactions

The Law of Conservation of Mass, by itself alone, does not require an atomic view of the elements. Mass could be conserved even if matter were not atomic. The importance of the Law of Conservation of Mass is that it reveals that we can usefully measure the masses of the elements which are contained in a fixed mass of a compound. As an example, we can decompose copper carbonate into its constituent elements, copper, oxygen, and carbon, weighing each and taking the ratios of these masses. The result is that every sample of copper carbonate is 51.5% copper, 38.8% oxygen, and 9.7% carbon. Stated differently, the masses of copper, oxygen, and carbon are in the ratio of 5.3 : 4 : 1, for every measurement of every sample of copper carbonate. Similarly, lead sulfide is 86.7% lead and 13.3% sulfur, so that the mass ratio for lead to sulfur in lead sulfide is always 6.5 : 1. Every sample of copper carbonate and every sample of lead sulfide will produce these elemental proportions, regardless of how much material we decompose or where the material came from. These results are examples of a general principle known as the Law of Definite Proportions.

Law 2: Law of Definite Proportions

When two or more elements combine to form a compound, their masses in that compound are in a fixed and definite ratio.

The idea that compounds have defined chemical formulas was first proposed in the late 1700s by the French chemist Joseph Proust. Proust performed a number of experiments and observed that no matter how he caused different elements to react with oxygen, they always reacted in defined proportions. For example, two parts of hydrogen always reacts with one part oxygen when forming water; one part mercury always reacts with one part oxygen when forming mercury calx. Dalton used Proust’s Law of Definite Proportions in developing his atomic theory.

 

The law also applies to multiples of the fundamental proportion, for example:

In both of these examples, the ratio of hydrogen to oxygen to water is 2 to 1 to 1. When reactants are present in excess of the fundamental proportions, some reactants will remain unchanged after the chemical reaction has occurred.

The story of the development of modern atomic theory is one in which scientists built upon the work of others to produce a more accurate explanation of the world around them. This process is common in science, and even incorrect theories can contribute to important scientific discoveries. Dalton, Priestley, and others laid the foundation of atomic theory, and many of their hypotheses are still useful. However, in the decades after their work, other scientists would show that atoms are not solid billiard balls, but complex systems of particles. Thus they would smash apart a bit of Dalton’s atomic theory in an effort to build a more complete view of the world around us.

These data help justify an atomic view of matter. We can simply argue that, for example, lead sulfide is formed by taking one lead atom and combining it with one sulfur atom. If this were true, then we also must conclude that the ratio of the mass of a lead atom to that of a sulfur atom is the same as the 6.5 : 1 lead to sulfur mass ratio we found for the bulk lead sulfide. This atomic explanation looks like the definitive answer to the question of what it means to combine two elements to make a compound, and it should even permit prediction of what quantity of lead sulfide will be produced by a given amount of lead. For example, 6.5g of lead will produce exactly 7.5g of lead sulfide, 50g of lead will produce 57.7g of lead sulfide, etc. There is a problem, however, which we can illustrate with three compounds formed from hydrogen, oxygen, and nitrogen. The three mass proportion measurements are given in Table 1.

First we examine nitric oxide, to find that the mass proportion is 8 : 7 oxygen to nitrogen. If this is one nitrogen atom combined with one oxygen atom, we would expect that the mass of an oxygen atom is 8/7=1.14 times that of a nitrogen atom.

Second we examine ammonia, which is a combination of nitrogen and hydrogen with the mass proportion of 7 : 1.5 nitrogen to hydrogen. If this is one nitrogen combined with one hydrogen, we would expect that a nitrogen atom mass is 4.67 times that of a hydrogen atom mass. These two expectations predict a relationship between the mass of an oxygen atom and the mass of a hydrogen atom. If the mass of an oxygen atom is 1.14 times the mass of a nitrogen atom and if the mass of a nitrogen atom is 4.67 times the mass of a hydrogen atom, then we must conclude that an oxygen atom has a mass which is 1.14 × 4.67 = 5.34 times that of a hydrogen atom.

But there is a problem with this calculation. The third line of Table 1 shows that the compound formed from hydrogen and oxygen is water, which is found to have mass proportion 8:1 oxygen to hydrogen.

Our expectation should then be that an oxygen atom mass is 8.0 times a hydrogen atom mass. Thus the three measurements in Table 1 appear to lead to contradictory expectations of atomic mass ratios. How are we to reconcile these results?

One possibility is that we were mistaken in assuming that there are atoms of the elements which combine to form the different compounds. If so, then we would not be surprised to see variations in relative masses of materials which combine. Another possibility is that we have erred in our reasoning. Looking back, we see that we have to assume how many atoms of each type are contained in each compound to find the relative masses of the atoms. In each of the above examples, we assumed the ratio of atoms to be 1:1 in each compound. If there are atoms of the elements, then this assumption must be wrong, since it gives relative atomic masses which differ from compound to compound. How could we find the correct atomic ratios? It would help if we knew the ratio of the atomic masses: for example, if we knew that the oxygen to hydrogen mass ratio were 8:1, then we could conclude that the atomic ratio in water would be 1 oxygen and 1 hydrogen. Our reasoning seems to circular: to know the atomic masses, we must know the formula of the compound (the numbers of atoms of each type), but to know the formula we must know the masses. Which of these possibilities is correct? Without further observations, we cannot say for certain whether matter is composed of atoms or not.

Observation 2: Multiple Mass Ratios

Significant insight into the above problem is found by studying different compounds formed from the same elements. For example, there are actually three oxides of nitrogen, that is, compounds composed only of nitrogen and oxygen. For now, we will call them oxide A, oxide B, and oxide C. Oxide A has oxygen to nitrogen mass ratio 2.28 : 1. Oxide B has oxygen to nitrogen mass ratio 1.14 : 1, and oxide C has oxygen to nitrogen mass ratio 0.57 : 1.

The fact that there are three mass ratios might seem to contradict the Law of Definite Proportions, which on the surface seems to say that there should be just one ratio. However, each mass combination gives rise to a completely unique chemical compound with very different chemical properties. For example, oxide A is very toxic, whereas oxide C is used as an anesthetic. It is also true that the mass ratio is neither arbitrary nor continuously variable: we cannot pick just any combination of masses in combining oxygen and nitrogen, rather we must obey one of only three. So there is no contradiction: we simply need to be careful with the Law of Definite Proportions to say that each unique compound has a definite mass ratio of combining elements. These new mass ratio numbers are highly suggestive in the following way. Notice that, in each case, we took the ratio of oxygen mass to a nitrogen mass of 1, and that the resultant ratios have a very simple relationship: 2.28 : 1.14 : 0.57 = 2 : 1 : 0.5 = 4 : 2 : 1

The masses of oxygen appearing in these compounds are in simple whole number ratios when we take a fixed amount of nitrogen. The appearance of these simple whole numbers is very significant. These integers suggest that the compounds contain a multiple of a fixed unit of mass of oxygen. The simplest explanation for this fixed unit of mass is that oxygen is particulate. We call the fixed unit of mass an atom. We now assume that the compounds have been formed from combinations of atoms with fixed masses, and that different compounds have differing numbers of atoms. The mass ratios make it clear that oxide B contains twice as many oxygen atoms (per nitrogen atom) as does oxide C and half as many oxygen atoms (per nitrogen atom) as does oxide A. The simple mass ratios must be the result of the simple ratios in which atoms combine into molecules. If, for example, oxide C has the molecular formula NO, then oxide B has the formula NO₂, and oxide A has the formula NO₄. There are other possibilities: if oxide B has molecular formula NO, then oxide A has formula NO₂, and oxide C has formula N₂O. Or if oxide A has formula NO, then oxide B has formula N₂O and oxide C has formula N₄O.

N₂O                 N₂O₃               NO₂  (N₂O₄)                  N₂O₅

Similar data are found for any set of compounds formed from common elements. For example, there are two oxides of carbon, one with oxygen-to-carbon mass ratio 1.33 : 1 and the other with mass ratio 2.66 : 1. The second oxide must have twice as many oxygen atoms, per carbon atom, as does the first. The general statement of this observation is the Law of Multiple Proportions.

Law 3: Law of Multiple Proportions

In chemistry, the law of multiple proportions is one of the basic laws of stoichiometry used to establish the atomic theory, alongside the law of conservation of mass (matter) and the law of definite proportions. It is sometimes called Dalton’s Law after its discoverer, the English chemist John Dalton, who published it in the first part of the first volume of his “New System of Chemical Philosophy” (1808). The statement of the law is:

If two elements form more than one compound between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers.

For example, Dalton knew that the element carbon forms two oxides by combining with oxygen in different proportions. A fixed mass of carbon, say 100 grams, may react with 133 grams of oxygen to produce one oxide, or with 266 grams of oxygen to produce the other. The ratio of the masses of oxygen that can react with 100 grams of carbon is 266:133 ≈ 2:1, a ratio of small whole numbers. Dalton interpreted this result in his atomic theory by proposing (correctly in this case) that the two oxides have one and two oxygen atoms respectively for each carbon atom. In moderotation the first is CO (carbon monoxide) and the second is CO₂ (carbon dioxide).

John Dalton first expressed this observation in 1804. A few years previously, the French chemist Joseph Proust had proposed the law of definite proportions, which expressed that the elements combined to form compounds in certain well-defined proportions, rather than mixing in just any proportion; and Antoine Lavoisier proved the law of conservation of mass, which helped out Dalton. Careful study of the actual numerical values of these proportions led Dalton to propose his law of multiple proportions. This was an important step toward the atomic theory that he would propose later that year, and it laid the basis for chemical formulas for compounds.

Limitations

The law of multiple proportions is best demonstrated using simple compounds. For example, if one tried to demonstrate it using the hydrocarbons decane (chemical formula C₁₀H₂₂) and undecane (C₁₁H₂₄), one would find that 100 grams of carbon could react with 18.46 grams of hydrogen to produce decane or with 18.31 grams of hydrogen to produce undecane, for a ratio of hydrogen masses of 121:120, which is hardly a ratio of “small” whole numbers.

The law fails with non-stoichiometric compounds and also doesn’t work well with polymers and oligomers.

Another example of the law can be seen by comparing ethane (C₂H₆) with propane (C₃H₈). The weight of hydrogen which combines with 1 g carbon is 0.252 g in ethane and 0.224 g in propane. The ratio of those weights is 1.125, which can be expressed as the ratio of two small numbers 9:8.When two elements are forced together to form more than one compound, the mass of element A that combines in the first compound with a specific amount of element B has a simple whole number ratio with the mass of element A that combines in the second compound with the same given mass of element B.

In other words, when two elements can combine to form more than one compound and the same amount of the first element is used in each, then the ratio of the amounts of the other element will be a whole number.

This sounds confusing, but an example clarifies this statement. Consider the carbon oxides, and let carbon be element B and oxygen be element A. Take a fixed given mass of carbon (element B), say 1 gram. The mass of oxygen which combines with 1 gram of carbon to form the first oxide is 1.33 grams. The mass of oxygen which combines with 1 gram of carbon to form the second oxide is 2.66. These masses are in ratio 2.66:1.33=2:1, a simple whole number ratio. In explaining our observations of the Law of Multiple Proportions for the carbon oxides and the nitrogen oxides, we have concluded that the simple mass ratio arises from the simple ratio of atoms contained in the individual molecules. Thus, we have established the above postulates of the Atomic Molecular Theory.

In short, the law of multiple proportions states:

“If two elements combine to form more than one compounds, the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers or simple multiple of it.”

The law of multiple proportions demonstrated with oxygen and 1.00 gram of nitrogen

c) Law of combining volumes (Gay-Lussac, 1808)

“When gases react, the volumes consumed and produced, measured at the same temperature and pressure, are in ratios of small whole numbers”.

Consequence. Stoichiometric coefficients in equations of chemical reactions for molecules of gaseous substances shows in which volume combinations gaseous substances are got or react.

Examples

1. 1 litre of nitrogen gas reacts with 3 litres of hydrogen gas to produce 2 litres of ammonia gas

N₂(g) + 3H₂(g) —–> 2NH₃(g)

Since all the reactants and products are gases, the mole ratio of N₂(g):H₂(g):NH₃(g) of 1:3:2 is also the ratio of the volumes of gases

So, 10mL of nitrogen gas would react with 10 x 3 = 30mL of hydrogen gas to produce 10 x 2 = 20mL ammonia gas

2. 2 litres of hydrogen gas react with 1 litre of oxygen gas to produce 2 litres of water vapour

2H₂(g) + O₂(g) —–> 2H₂O(g)

Since all the reactants and products are gases, the mole ratio of H₂(g):O₂(g):H₂O(g) of 2:1:2 is also the ratio of the volumes of gases

So, 50mL of hydrogen gas would react with 50 x ½ = 25mL oxygen gas to produce 50mL of water vapour

When liquid water undergoes electrolysis to produce hydrogen gas and oxygen gas, the volumes of hydrogen gas and oxygen gas are produced in the ratio of 2:1 but the volume of liquid water required does not follow this relationship since the liquid water is not a gas

2H₂O(l) —-> 2H₂(g) + O₂(g)

Under Standard Laboratory Conditions (S.L.C.) of 25^oC (298K) and 101.3kPa (1 atm), 1 mole of liquid water has a volume of 18mL and will undergo electrolysis to produce 2 moles of hydrogen gas with a volume of 48.94L and 1 mole of oxygen gas with a volume of 24.47L

3.  2CO + O₂ → 2CO₂

In the time of oxidizing of two volumes of carbon (II) oxide by one volume of oxygen forms 2 volume of carbon (IV) oxide, i.e. the volume of source reaction mixture decrease on 1 volume.

d) Avogadro’s law (1811)

Equal volumes of all gases at the same conditions (temperature, pressure) contain the same number of molecules.

This law truth for gaseous substances only.

Consequences:

1. On mole of any substance in the gaseous state occupies the same volume at the same temperature and pressure.

2. One mole of any gas in standard conditions (0°C = 273K, 1 atm = 101.3 kPa) occupies a volume of 22.4 litres.

Avogadro’s law (sometimes referred to as Avogadro’s hypothesis or Avogadro’s principle) is a gas law named after Amedeo Avogadro who, in 1811,hypothesized that two given samples of an ideal gas, at the same temperature, pressure and volume, contain the same number of molecules. Thus, the number of molecules or atoms in a specific volume of gas is independent of their size or the molar mass of the gas.

In the same 1808 article in which Gay-Lussac published his observations on the thermal expansion of gases, he pointed out that when two gases react, they do so in volume ratios that can always be expressed as small whole numbers. This came to be known as the Law of combining volumes.

These “small whole numbers” are of course the same ones that describe the “combining weights” of elements to form simple compounds, as described in the lesson dealing with simplest formulas from experimental data.

The Italian scientist Amedeo Avogadro (1776-1856) drew the crucial conclusion: these volume ratios must be related to the relative numbers of molecules that react, and thus the famous “E.V.E.N principle”:

Avogadro’s law thus predicts a directly proportional relation between the number of moles of a gas and its volume.

This relationship, originally known as Avogadro’s Hypothesis, was crucial in establishing the formulas of simple molecules at a time (around 1811) when the distinction between atoms and molecules was not clearly understood. In particular, the existence of diatomic molecules of elements such as H2, O2, and Cl2 was not recognized until the results of combining-volume experiments such as those depicted below could be interpreted in terms of the E.V.E.N. principle.

How the E.V.E.N. principle led to the correct formula of water

Early chemists made the mistake of assuming that the formula of water is HO. This led them to miscalculate the molecular weight of oxygen as 8 (instead of 16). If this were true, the reaction H + O → HO would correspond to the following combining volumes results according to the E.V.E.N principle:

But a similar experiment on the formation of hydrogen chloride from hydrogen and chlorine yielded twice the volume of HCl that was predicted by the the assumed reaction H + Cl → HCl. This could be explained only if hydrogen and chlorine were diatomic molecules:

This made it necessary to re-visit the question of the formula of water. The experiment immediately confirmed that the correct formula of water is H2O:

This conclusion was also seen to be consistent with the observation, made a few years earlier by the English chemists Nicholson and Carlisle that the reverse of the above reaction, brought about by the electrolytic decomposition of water, yields hydrogen and oxygen in a 2:1 volume ratio.

As an example, equal volumes of molecular hydrogen and nitrogen contain the same number of molecules when they are at the same temperature and pressure, and observe ideal gas behavior. In practice, real gases show small deviations from the ideal behavior and the law holds only approximately, but is still a useful approximation.

 

 

 

Mathematical definition

Avogadro’s law is stated mathematically as:

Where:

V ‑ is the volume of the gas.

n ‑ is the amount of substance of the gas.

k ‑ is a proportionality constant.

The most significant consequence of Avogadro’s law is that the ideal gas constant has the same value for all gases. This means that:

Where:

p ‑ is the pressure of the gas;

T ‑ is the temperature in kelvin of the gas;

Example 1

What volume of hydrogen at s.t.p. would be evolved at dissolution 4.8 g magnesium in excess of hydrochloric acid?

Solution

Mg + 2HCl → MgCl₂ + H₂­

At dissolution of 24 g (1 g-mol) magnesium in HCl –– 22.4 l (1 g-mol) of hydrogen is evaluated; at dissolution of 4.8 g of magnesium –– X l of hydrogen.

Example.

 3.17 g of chlorine is borrowing volume, which equal 1 l (at s.t.p.). Calculate the molecular mass of chlorine.

Solution

Find mass of 22.4 l chlorine.

X = 3.17 _* 22.4 = 71 g

Example.

Calculate the volume that 0.881 moles of oxygen gas at STP will occupy.

Solution

   If the pressure by a gas at 30^oC in a volume of .05 L is 3.52 atm, how many moles of the gas is present?

Example.

If the pressure by a gas at 30^oC in a volume of .05 L is 3.52 atm, how many moles of the gas is present?

Solution

 

 

2. MOLAR VOLUME OF GAS. RELATION BETWEEN DENSITY OF GAS AND IT MOLECULAR MASS. NORMAL STATE OF GAS.

a) Ideal gas law

The ideal gas law is the equation of state of a hypothetical ideal gas. It is a good approximation to the behaviour of many gases under many conditions, although it has several limitations. It was first stated by Émile Clapeyron in 1834 as a combination of Boyle’s law and Charles’s law. The ideal gas law is often introduced in its common form:

PV=nRT,

where P is the absolute pressure of the gas, V is the volume of the gas,is the amount of substance of gas (measured in moles), T is the absolute temperature of the gas and R is the ideal, or universal, gas constant.

It can also be derived from kinetic theory, as was achieved (apparently independently) by August Krönig in 1856 and Rudolf Clausius in 1857. Universal gas constant was discovered and first introduced into the ideal gas law instead of a large number of specific gas constants by Dmitri Mendeleev in 1874.

This equation is known as the ideal gas law.

An Ideal Gas is modelled on the Kinetic Theory of Gases which has 4 basic postulates:

1. Gases consist of small particles (molecules) which are in continuous random motion.

2. The volume of the molecules present is negligible compared to the total volume occupied by the gas.

3. Intermolecular forces are negligible.

4. Pressure is due to the gas molecules colliding with the walls of the container.

Real Gases deviate from Ideal Gas Behaviour because:

·        at low temperatures the gas molecules have less kinetic energy (move around less) so they do attract each other

·        at high pressures the gas molecules are forced closer together so that the volume of the gas molecules becomes significant compared to the volume the gas occupies.

Under ordinary conditions, deviations from Ideal Gas behaviour are so slight that they can be neglected. A gas which deviates from Ideal Gas behaviour is called a non-ideal gas.

The table below essentially simplifies the ideal gas equation for a particular processes, thus making this equation easier to solve using numerical methods.

A thermodynamic process is defined as a system that moves from state 1 to state 2, where the state number is denoted by subscript. As shown in the first column of the table, basic thermodynamic processes are defined such that one of the gas properties (P, V, T, or S) is constant throughout the process.

For a given thermodynamics process, in order to specify the extent of a particular process, one of the properties ratios (listed under the column labeled “known ratio”) must be specified (either directly or indirectly). Also, the property for which the ratio is known must be distinct from the property held constant in the previous column (otherwise the ratio would be unity, and not enough information would be available to simplify the gas law equation).

In the final three columns, the properties (P, V, or T) at state 2 can be calculated from the properties at state 1 using the equations listed.

The equation of state given here applies only to an ideal gas, or as an approximation to a real gas that behaves sufficiently like an ideal gas. There are in fact many different forms of the equation of state. Since the ideal gas law neglects both molecular size and intermolecular attractions, it is most accurate for monatomic gases at high temperatures and low pressures. The neglect of molecular size becomes less important for lower densities, i.e. for larger volumes at lower pressures, because the average distance between adjacent molecules becomes much larger than the molecular size. The relative importance of intermolecular attractions diminishes with increasing thermal kinetic energy, i.e., with increasing temperatures. More detailed equations of state, such as the van der Waals equation, account for deviations from ideality caused by molecular size and intermolecular forces.

A residual property is defined as the difference between a real gas property and an ideal gas property, both considered at the same pressure, temperature, and composition.

Molar volume

Taking STP to be 101.325 kPa and 293.15 K, we can find the volume of one mole of a gas:

For 100.000 kPa and 273.15 K, the molar volume of an ideal gas is 22.71 dm³mol^-1.

Example 1.

What volume of hydrogen at s.t.p. would be evolved at dissolution 4.8 g magnesium in excess of hydrochloric acid?

Solution:

Mg + 2HCl = MgCl₂ + H₂

At dissolution of 24 g (1 g-mol) magnesium in HCl –– 22.4 l (1 g-mol) of hydrogen is evaluated; at dissolution of 4.8 g of magnesium –– X l of hydrogen.

Example 2.

3.17 g of chlorine is borrowing volume, which equal 1 l (at s.t.p. at the same temperature and pressure.). Calculate the molecular mass of chlorine.

Solution

Find mass of 22.4 l chlorine.

 X = 3.17 _* 22.4 = 71 g

b) General gas law

All matter expands when heated, but gases are special in that their degree of expansion is independent of their composition. The French scientists Jacques Charles (1746-1823) and Joseph Gay-Lussac (1778-1850) independently found that if the pressure is held constant, the volume of any gas changes by the same fractional amount (1/273 of its value) for each C° change in temperature.

The volume of a gas confined against a constant pressure is directly proportional to the absolute temperature.

A graphical expression of the law of Charles and Gay-Lussac can be seen in these plots of the volume of one mole of an ideal gas as a function of its temperature at various constant pressures.

What do these plots show?

The straight-line plots show that the ratio V/T (and thus dV/dT) is a constant at any given pressure. Thus we can express the law algebraically as      V/T = constant   or   V₁/T₁ = V₂/T₂

What is the significance of the extrapolation to zero volume?

If a gas contracts by 1/273 of its volume for each degree of cooling, it should contract to zero volume at a temperature of –273°C. This, of course, is theabsolute zero of temperature, and this extrapolation of Charles’ law is the first evidence of the special significance of this temperature.

Why do the plots for different pressures have different slopes?

The lower the pressure, the greater the volume (Boyle’s law), so at low pressures the fraction (V/273) will have a larger value. You might say that the gas must “contract faster” to reach zero volume when its starting volume is larger.

Robert Boyle (1627-91) showed that the volume of air trapped by a liquid in the closed short limb of a J-shaped tube decreased in exact proportion to the pressure produced by the liquid in the long part of the tube. The trapped air acted much like a spring, exerting a force opposing its compression. Boyle called this effect “the spring of the air”, and published his results in a pamphlet of that title.

The difference between the heights of the two mercury columns gives the pressure (76 cm = 1 atm), and the volume of the air is calculated from the length of the air column and the tubing diameter.

Boyle’s law can be expressed as

PV = constant

or, equivalently,

P₁V₁ = P₂V₂

These relations hold true only if the number of moleculesand the temperature are constant. This is a relation of inverse proportionality; any change in the pressure is exactly compensated by an opposing change in the volume. As the pressure decreases toward zero, the volume will increase without limit. Conversely, as the pressure is increased, the volume decreases, but caever reach zero. There will be a separate P-V plot for each temperature; a single P-V plot is therefore called an isotherm

Shown here are some isotherms for one mole of an ideal gas at several different temperatures. Each plot has the shape of a hyperbola— the locus of all points having the property x y = a, where a is a constant. You will see later how the value of this constant (PV=25 for the 300K isotherm shown here) is determined.

We have learned that Boyle’s law pertains to situations in which the temperature remains constant (Fig. 1A), and that Charles’s law pertains to situations in which pressure remains constant (Fig. 1B). It is usually not possible to control pressure or temperature in tanks or bottles of gas subject to the weather and shipboard demands. Boyle’s and Charles’s laws are combined to form the general gas law. This law states: The product of the initial pressure, initial volume, and new temperature (absolute scale) of an enclosed gas is equal to the product of the new pressure, new volume, and initial temperature.

Fig. 1. The general gas law

General gas law is association of three independent private gas laws: Gay-Lussac’s, Charle’s, Boyl’s- Mariott’s, equation which possible write like this:

Conversely, from general gas law under P= const. (P₁=P₂), possible to get:

Gay-Lussac’s law

On the T=const. (T₁=T₂):       P₁V₁= P₂V₂

Gay-Lussac published his observations on the thermal expansion of gases, he pointed out that when two gases react, they do so in volume ratios that can always be expressed as small whole numbers. This came to be known as the Law of combining volumes.

These “small whole numbers” are of course the same ones that describe the “combining weights” of elements to form simple compounds, as described in the lesson dealing with simplest formulas from experimental data.

The Italian scientist Amedeo Avogadro (1776-1856) drew the crucial conclusion: these volume ratios must be related to the relative numbers of molecules that react, and thus the famous “E.V.E.N principle”:

Equal volumes of gases, measured at the same temperature and pressure, contain equal numbers of molecules

(Boyl’s-Mariott’s law).

 On the V=const.

(Charle’s law).

Example

The air pressure in a car tire is 30 psi (pounds per square inch) at 10°C. What will be pressure be after driving has raised its temperature to 45°C ? (Assume that the volume remains unchanged.)

Solution

The gas expands in direct proportion to the ratio of the absolute temperatures:

P₂ = (30 psi) × (318K ÷ 283K) = 33.7 psi

Example

In an industrial process, a gas confined to a volume of 1 L at a pressure of 20 atm is allowed to flow into a 12-L container by opening the valve that connects the two containers. What will be the final pressure of the gas?

Solution

 The final volume of the gas is (1 + 12)L = 13 L. The gas expands in inverse proportion two volumes

P₂ = (20 atm) × (1 L ÷ 13 L) = 1.5 atm

Example

A biscuit made with baking powder has a volume of 20 mL, of which one-fourth consists of empty space created by gas bubbles produced when the baking powder decomposed to CO₂. What weight of NaHCO₃ was present in the baking powder in the biscuit? Assume that the gas reached its final volume during the baking process when the temperature was 400°C.

 (Baking powder consists of sodium bicarbonate mixed with some other solid that produces an acidic solution on addition of water, initiating the reaction

NaHCO₃(s) + H⁺ → Na⁺ + H₂O + CO₂

Solution

Use the ideal gas equation to find the number of moles of CO2 gas; this will be the same as the number of moles of NaHCO3 (84 g mol–1) consumed :

Clapeyron’s – Mendeleyev’s equation

If write general gas law for any mass of any gas than would be Clapeyron’s – Mendeleyev’s equation:

 where, m – gas mass; M – molecular mass; p – pressure; V – volume; T – absolute temperature (K); R – molar gas constant (8.314 J/(mol _* K) or 0.082 l atm/(mol _* K).

For given mass of concrete gas the ratio m/M is a constant, therefore general gas law is obtained from Clapeyron’s – Mendeleyev’s equation.

Example

What volume would be reserved by carbon (II) oxide with mass 84 g at temperature 17°C and pressure 250 kPa?

Solution

Quantity g-mol of CO is:

CO volume at s.t.p is 3 _* 22.4 l = 67.2 l

From general gas law of Boyl’s-Mariott’s and Gay-Lussac’s:

P * V		P0 * V0
——	=	——
T		T2

 

 

follows:

f) Relative density of gases shows in how many times 1 mol of one gas more heavy (or easier) than 1 mol of another.

 

The average molecular mass of gases mixture is peer to a common mass of mixture bisected on a total number of moles:

Example 1

The density of certain gaseous substance on hydrogen is 17. Calculate its density on air (M_av=29).

Solution

М_sub = 2DH₂ = 34

Example 2

Define the density of nitrogen, argon, and carbon (II) oxide mixture on air, if the mass parts of components are 15, 50 and 35% respectively.

Solution

 

 

Relative density, or specific gravity, is the ratio of the density (mass of a unit volume) of a substance to the density of a given reference material. Specific gravity usually means relative density with respect to water. The term “relative density” is often preferred in modern scientific usage. If a substance’s relative density is less than one then it is less dense than the reference; if greater than 1 then it is denser than the reference. If the relative density is exactly 1 then the densities are equal; that is, equal volumes of the two substances have the same mass. If the reference material is water then a substance with a relative density (or specific gravity) less than 1 will float in water. For example, an ice cube, with a relative density of about 0.91, will float. A substance with a relative density greater than 1 will sink.

Temperature and pressure must be specified for both the sample and the reference. Pressure is nearly always 1 atm equal to 101.325 kPa. Where it is not, it is more usual to specify the density directly. Temperatures for both sample and reference vary from industry to industry. In British brewing practice the specific gravity as specified above is multiplied by 1000. Specific gravity is commonly used in industry as a simple means of obtaining information about the concentration of solutions of various materials such as brines, sugar solutions (syrups, juices, honeys, brewers wort, must, etc.) and acids.

Relative density can also help quantify the buoyancy of a substance in a fluid, or determine the density of an unknown substance from the known density of another. Relative density is often used by geologists and mineralogists to help determine the mineral content of a rock or other sample. Gemologists use it as an aid in the identification of gemstones. Water is preferred as the reference because measurements are then easy to carry out in the field (see below for examples of measurement methods). As the principal use of specific gravity measurements in industry is determination of the concentrations of substances in aqueous solutions and these are found in tables of SG vs concentration it is extremely important that the analyst enter the table with the correct form of specific gravity. For example, in the brewing industry, the Plato table, which lists sucrose concentration by weight against true SG, were originally (20 °C/4 °C)  that is based on measurements of the density of sucrose solutions made at laboratory temperature (20 °C) but referenced to the density of water at 4 °C which is very close to the temperature at which water has its maximum density of ρ(H₂O) equal to 0.999972 g/cm³ (or 62.43 lb_m·ft⁻³). The ASBC table in use today in North America, while it is derived from the original Plato table is for apparent specific gravity measurements at (20 °C/20 °C) on the IPTS-68 scale where the density of water is 0.9982071 g/cm³. In the sugar, soft drink, honey, fruit juice and related industries sucrose concentration by weight is taken from this work which uses SG (17.5 °C/17.5 °C). As a final example, the British SG units are based on reference and sample temperatures of 60°F and are thus (15.56°C/15.56°C).

The relative density of a liquid can be measured using a hydrometer. This consists of a bulb attached to a stalk of constant cross-sectional area, as shown in the diagram to the right.

First the hydrometer is floated in the reference liquid (shown in light blue), and the displacement (the level of the liquid on the stalk) is marked (blue line). The reference could be any liquid, but in practice it is usually water. The hydrometer is then floated in a liquid of unknown density (shown in green). The change in displacement, Δx, is noted. In the example depicted, the hydrometer has dropped slightly in the green liquid; hence its density is lower than that of the reference liquid. It is, of course, necessary that the hydrometer floats in both liquids. The application of simple physical principles allows the relative density of the unknown liquid to be calculated from the change in displacement. (In practice the stalk of the hydrometer is pre-marked with graduations to facilitate this measurement.)

In the explanation that follows,

ρ_ref is the known density (mass per unit volume) of the reference liquid (typically water).

ρ_new is the unknown density of the new (green) liquid.

RD_new/ref is the relative density of the new liquid with respect to the reference.

V is the volume of reference liquid displaced, i.e. the red volume in the diagram.

m is the mass of the entire hydrometer.

g is the local gravitational constant.

Δx is the change in displacement. In accordance with the way in which hydrometers are usually graduated, Δx is here taken to be negative if the displacement line rises on the stalk of the hydrometer, and positive if it falls. In the example depicted, Δx is negative.

A is the cross sectional area of the shaft.

Since the floating hydrometer is in static equilibrium, the downward gravitational force acting upon it must exactly balance the upward buoyancy force. The gravitational force acting on the hydrometer is simply its weight, mg. From the Archimedes buoyancy principle, the buoyancy force acting on the hydrometer is equal to the weight of liquid displaced. This weight is equal to the mass of liquid displaced multiplied by g, which in the case of the reference liquid is ρ_refVg. Setting these equal, we have:

(1)

or just:

(2)

Exactly the same equation applies when the hydrometer is floating in the liquid being measured, except that the new volume is V – AΔx (see note above about the sign of Δx). Thus,

Combining (1) and (2) yields

(3)

But from (1) we have V = m/ρ_ref. Substituting into (3) gives

(4)

This equation allows the relative density to be calculated from the change in displacement, the known density of the reference liquid, and the known properties of the hydrometer. If Δx is small then, as a first-order approximation of the geometric series equation (4) can be written as:

This shows that, for small Δx, changes in displacement are approximately proportional to changes in relative density.

Example 1.

The density of certain gaseous substance on hydrogen is 17. Calculate its density on air (M_av=29).

Solution

 М_sub = 2DH₂ = 34

Example 2.

Define the density of nitrogen, argon, and carbon (II) oxide mixture on air, if the mass parts of components are 15, 50 and 35% respectively.

Solution

 

 

3. EQUIVALENT. LAW EQUIVALENTS

The equivalent (Eq or eq) unit- Eq/L is a reasonably common measurement unit used in chemistry. It is a measure of a substance’s ability to combine with other substances. It is frequently used in the context of normality.

The equivalent is formally defined as the mass in grams of a substance which will react with 6.022 x 10²³ electrons. (This is Avogadro’s Number, which is the number of particles in a mole). Another, slightly less precise, definition describes the equivalent as the number of grams of a substance that will react with a gram of free hydrogen. (This is practically true, since a gram of hydrogen is very close to a mole of hydrogen, and free hydrogen has one spare electron; hence one gram of hydrogen is effectively equivalent to 6.022 x 10²³ electrons.)

• For monovalent ions, 1 equivalent = 1 mole

• For divalent ions, 1 Eq = 0.5 mol

• For trivalent ions, 1 Eq = 0.333 mol

Equivalent is equal to 1/V (mol), where V – valence. For example, some compounds HCl, H₂S, NH₃, CH₄, the equivalent of Cl, S, N, C respectively are equal 1mol, 1/2mol, 1/3mol, 1/4mol.

Weight of 1 Equivalent of chemical element is called its equivalent weight. m_e= 35,45 g/mol, m_e=32× ½ =16 g/mol, m_e(N)= 14/3=4,67, m_e(c)=12/4=3 g/mol. Equivalent mass of element is the proportion of m_e (E) = Ar (E) / V = Ar×E, where V-valence, E – equivalent.

This is conceptually easier because the molarity refers to the number of dissolved particles, not the number of available charges. This simplifies understanding the composition of physiological solutions which contain the divalent ions Mg²⁺ and Ca²⁺. An equivalent of a substance is defined as the amount of it which combines with one mole of hydrogen atoms.

 

Experimental determination of the metal equivalent.

To take piece of metal with known mass. At flask add 10-12 ml of 4 mole-eq/l hydrochloric acid and  piece of metal. To measure volume of hydrogen, which is formed in reaction

Mass of metal, g.

Volume of hydrogen, which is formed V, ml.

Temperature  t, ⁰С.

Absolute temperature  Т, К.

Atmospheric pressure of Р, mm. Hg (Pa).

Vapor – pressure of  water at some temperature  Р (Н₂О).

Partial pressure of hydrogen of Р(Н₂) = Р – Р (Н₂О).

To calculate the equivalent of metal by two methods:

1. Mendeleyev-Clapeyron equation:  Р(Н₂) × V = (m/M) × RT

2. Clapeyron equation: V₀P₀/T₀ = VP/T

Knowing the volume of hydrogen equivalent, to calculate the equivalent of metal.

Equivalent Weight of an Element (E): It is defined as the number of parts by weight of the element which combine with or displace from a compund 1 part by weight of Hydrogen, 8 parts by weight of Oxygen or 35.5 parts by weight of Chlorine.

E = M ( Molar Mass) / n_f (n factor)

where:

n_f = Valency in case of an atom

= Total positive or negative oxidatioumber of an atom in a molecule

= Basicity or Acidity.

= Change in oxidatioumber in case of a redox reacton.

Equivalent weight of an element

In the case of an element, the equivalent weight is defined as :

 

Equivalent weight, E=Atomic weight          Valency  = A       x

Note that atomic weight substitutes molecular weight and valency substitutes valence factor in the definition. Valencies of hydrogen, calcium and oxygen are 1,2 and 2 respectively. Hence, their equivalent weights are 1/1 =1, 40/2 = 20 and 16/2 = 8 respectively.

 

Equivalent weight of an acid

The valence factor of an acid is equal to its basicity. The basicity of an acid is equal to furnishable hydrogen ion (proton) in its aqueous solution. Importantly, basicity is not same as the number of hydrogen atoms in acid molecule. Consider acetic acid (CH₃COOH). It contains 4 hydrogen atoms in it, but only 1 furnishable hydrogen ion. As such, basicity of acetic acid is 1. With this background, we define equivalent weight of an acid as :

Equivalent weight, E= Molecular weight of acid / Basicity

Basicity of sulphuric acid is 2. Hence, equivalent weight of sulphuric acid (H₂SO₄) is (2x1 + 32 + 4x16)/2 = 98/2 = 49. Similarly, basicity of oxalic acid is 2. Hence, equivalent weight of oxalic acid (H₂C₂O₄) is (2x1 + 2x12 + 4x16)/2 = 90/2=45.

 

Phosphorous based acids like phosphoric acid (H₃PO₄), phosphorous acid (H₃PO₃) and hypo-phosphorous acid (H₃PO₂) need special mention here to understand their basicity. The structures of three acids are shown here. From the structure, it appears that these compounds may furnish OH ions, but bond strengths between phosphorous and oxygen (P-O) and phosphorous and hydrogen (P-H) are stronger than between oxygen and hydrogen (O-H) in –OH group. As such, these molecules release hydrogen ions from –OH group and behave as acid. Clearly, basicities of phosphoric acid (H₃PO₄), phosphorous acid (H₃PO₃) and hypo-phosphorous acid (H₃PO₂) are 3, 2 and 1 respectively.

Phosphorous based acids

 

Figure 1: Furnishable hydrogen ions of acids

Equivalent weight of a base

The valence factor of a base is equal to its acidity. The acidity of a base is equal to furnishable hydroxyl ion (OH^–) in its aqueous solution. With this background, we define equivalent weight of a base as :

 

Equivalent weight, E=  Molecular weight of base / Acidity

Acidity of KOH is 1, whereas acidity of Ca(OH)₂ is 2. Hence, equivalent weight of KOH is (39 + 16 + 1)/1 = 56/1 = 56. Similarly, equivalent weight of Ca(OH)₂ is {40 + 2x(16+1)}/2 = 74/2=37.

 

Equivalent weight of a compound

The valence factor of a compound depends on the manner a compound is involved in a reaction. The compounds of alkali metal salts and alkaline earth metal salts are, however, constant. These compounds are ionic and they dissociate in ionic components in aqueous solution. In this case, valence factor is equal to numbers of electronic charge on either cation or anion.

Equivalent weight, E= Molecular weight of compound / Numbers of electronic charge on cation or anion

The numbers of electronic charge on cation of NaHCO₃ is 1. Hence, equivalent weight of NaHCO₃ is (23 + 1 + 12 + 3x16)/1 = 84.

If we look at the defining ratio of equivalent weight of a compound (AB) formed of two radicals (say A and B), then we can rearrange the ratio as :

Equivalent weight, E= Molecular weight of Radical A / Numbers of electronic charge  + Molecular weight of Radical B / Numbers of electronic charge

Thus,

⇒Equivalent weight of AB=Equivalent weight of A+Equivalent weight of B

 

Equivalent weight of an ion

The valence factor of an ion is equal to numbers of electronic charge on the ion. Therefore, we define equivalent weight of an ion as :

Equivalent weight, E=  Molecular weight of ion / Numbers of electronic charge

The numbers of electronic charge on carbonate ion (CO32− ) is 2. Hence, equivalent weight of carbonate ion is (12 + 3x16)/1 = 60/2 = 30. Similarly, equivalent weight of aluminum ion (Al³⁺) is 27/3 = 9.

 

Equivalent weight of an oxidizing or reducing agent

In a redox reaction, one of the reacting entities is oxidizing agent (OA). The other entity is reducing agent (RA). The oxidizer is recipient of electrons, whereas reducer is releaser of electrons. The valence factor for either an oxidizing or reducing agent is equal to the numbers of electrons transferred from one entity to another.

Equivalent weight, E = Molecular weight of compound / Numbers of electrons transferred in redox reaction

Alternatively,

Equivalent weight, E = Molecular weight of compound / Change in oxidatioumber in redox reaction

Potassium dichromate in acidic medium is a strong oxidizer. It means it gains electrons during redox reaction. Potassium dichromate in acidic solution results in :

 

K₂Cr₂O₇+14H⁺ +6e−→2K⁺+2Cr³⁺+7H₂O

294.2
‑‑	=	49
6

 

Equivalent weight of K₂Cr₂O₇ =

 

 

Study of redox reaction is in itself an exclusive and extensive topic. We shall, therefore, discuss redox reaction separately.

 

4. CHEMICAL FORMULAS, THEIR TYPES

 

Structural formula for butane. This is not a chemical formula. Examples of chemical formulas for butane are the empirical formula C₂H₅, the molecular formula C₄H₁₀, and the condensed (or semi-structural) formula CH₃CH₂CH₂CH₃

A chemical formula is a way of expressing information about the proportions of atoms that constitute a particular chemical compound, using a single line of chemical element symbols, numbers, and sometimes also other symbols, such as parentheses, dashes, brackets, and plus (+) and minus (−) signs. These are limited to a single typographic line of symbols, which may include subscripts and superscripts. A chemical formula is not a chemical name, and it contains no words. Although a chemical formula may imply certain simple chemical structures, it is not the same as a full chemical structural formula. Chemical formulas are more limiting than chemical names and structural formulas.

The simplest types of chemical formulas are called empirical formulas, which use only letters and numbers indicating atomic proportional ratios (the numerical proportions of atoms of one type to those of other types). Molecular formulas indicate the simple numbers of each type of atom in a molecule of a molecular substance, and are thus sometimes the same as empirical formulas (for molecules that only have one atom of a particular type), and at other times require larger numbers than do empirical formulas. An example of the difference is the empirical formula for glucose, which is CH₂O, while its molecular formula requires all numbers to be increased by a factor of six, giving C₆H₁₂O₆.

Sometimes a chemical formula is complicated by being written as a condensed formula (or condensed molecular formula, occasionally called a “semi-structural formula”), which conveys additional information about the particular ways in which the atoms are chemically bonded together, either in covalent bonds, ionic bonds, or various combinations of these types. This is possible if the relevant bonding is easy to show in one dimension. An example is the condensed molecular/chemical formula for ethanol, which is CH₃-CH₂-OH or CH₃CH₂OH. However, even a condensed chemical formula is necessarily limited in its ability to show complex bonding relationships between atoms, especially atoms that have bonds to four or more different substituents.

Since a chemical formula must be expressed as a single line of chemical element symbols, it often cannot be as informative as a true structural formula, which is a graphical representation of the spacial relationship between atoms in chemical compounds (see for example the figure for butane structural and chemical formulas, at right). For reasons of structural complexity, there is no condensed chemical formula (or semi-structural formula) that specifies glucose (and there exist many different molecules, for example fructose and mannose, have the same molecular formula C₆H₁₂O₆ as glucose). Linear equivalent chemical names exist that can and do specify any complex structural formula, but these names must use many terms (words), rather than the simple element symbols, numbers, and simple typographical symbols that define a chemical formula.

Chemical formulas may be used in chemical equations to describe chemical reactions and other chemical transformations, such as the dissolving of ionic compounds into solution. While, as noted, chemical formulas do not have the full power of structural formulas to show chemical relationships between atoms, they are sufficient to keep track of numbers of atoms and numbers of electical charges in chemical reactions, thus balancing chemical equations so that these equations can be used in chemical problems involving conservation of atoms, and conservation of electric charge.

There are several types of chemical formulas that you can use to represent chemical bonds. These include empirical formulas, molecular (or true) formulas, and structural formulas.

You can predict the formula of an ionic compound based on the loss and gain of electrons, to reach a noble gas configuration. However, you really can’t make that type of prediction with covalent compounds, because they can combine in many ways, and many different possible covalent compounds may result.

Most of the time, you have to know the formula of the molecule you’re studying. But you may have several different types of formulas, and each gives a slightly different amount of information.

 

A chemical formula identifies each constituent element by its chemical symbol and indicates the proportionate number of atoms of each element. In empirical formulas, these proportions begin with a key element and then assigumbers of atoms of the other elements in the compound, as ratios to the key element. For molecular compounds, these ratio numbers can all be expressed as whole numbers. For example, the empirical formula of ethanol may be written C₂H₆O because the molecules of ethanol all contain two carbon atoms, six hydrogen atoms, and one oxygen atom. Some types of ionic compounds, however, cannot be written with entirely whole-number empirical formulas. An example is boron carbide, whose formula of CB_n is a variable non-whole number ratio with ranging from over 4 to more than 6.5.

When the chemical compound of the formula consists of simple molecules, chemical formulas often employ ways to suggest the structure of the molecule. These types of formulas are variously known as molecular formulas and condensed formulas. A molecular formula enumerates the number of atoms to reflect those in the molecule, so that the molecular formula for glucose is C₆H₁₂O₆ rather than the glucose empirical formula, which is CH₂O. However, except for very simple substances, molecular chemical formulas lack needed structural information, and are ambiguous.

For simple molecules, a condensed (or semi-structural) formula is a type of chemical formula that may fully imply a correct structural formula. For example, ethanol may be represented by the condensed chemical formula CH₃CH₂OH, and dimethyl ether by the condensed formula CH₃OCH₃. These two molecules have the same empirical and molecular formulas (C₂H₆O), but may be differentiated by the condensed formulas shown, which are sufficient to represent the full structure of these simple organic compounds.

Condensed chemical formulas may also be used to represent ionic compounds that do not exist as discrete molecules, but nonetheless do contain covalently bound clusters within them. These polyatomic ions are groups of atoms that are covalently bound together and have an overall ionic charge, such as the sulfate [SO₄]²− ion. Each polyatomic ion in a compound is written individually in order to illustrate the separate groupings. For example, the compound dichlorine hexoxide has an empirical formula ClO₃, and molecular formula Cl₂O₆, but in liquid or solid forms, this compound is more correctly shown by an ionic condensed formula [ClO₂]+[ClO₄]−, which illustrates that this compound consists of [ClO₂]+ ions and [ClO₄]− ions. In such cases, the condensed formula only need be complex enough to show at least one of each ionic species.

Chemical formulas must be differentiated from the far more complex chemical systematic names that are used in various systems of chemical nomenclature. For example, one systematic name for glucose is (2R,3S,4R,5R)-2,3,4,5,6-pentahydroxyhexanal. This name, and the rules behind it, fully specify glucose’s structural formula, but the name is not a chemical formula, as it uses many extra terms and words that chemical formulas do not permit. Such chemical names may be able to represent full structural formulas without graphs, but in order to do so, they require word terms that are not part of chemical formulas.

In chemistry, the empirical formula of a chemical is a simple expression of the relative number of each type of atom or ratio of the elements in the compound. Empirical formulas are the standard for ionic compounds, such as CaCl2, and for macromolecules, such as SiO₂. An empirical formula makes no reference to isomerism, structure, or absolute number of atoms. The term empiricalrefers to the process of elemental analysis, a technique of analytical chemistry used to determine the relative percent composition of a pure chemical substance by element.

For example hexane has a molecular formula of C₆H₁₄, or structurally CH₃CH₂CH₂CH₂CH₂CH₃, implying that it has a chain structure of 6 carbon atoms, and 14 hydrogen atoms. However, the empirical formula for hexane is C₃H₇. Likewise the empirical formula for hydrogen peroxide, H₂O₂, is simply HO expressing the 1:1 ratio of component elements. Formaldehyde and acetic acidhave the same empirical formula, CH₂O. This is the actual chemical formula for formaldehyde, but acetic acid has double the number of atoms.

The connectivity of a molecule often has a strong influence on its physical and chemical properties and behavior. Two molecules composed of the same numbers of the same types of atoms (i.e. a pair of isomers) might have completely different chemical and/or physical properties if the atoms are connected differently or in different positions. In such cases, a structural formula is useful, as it illustrates which atoms are bonded to which other ones. From the connectivity, it is often possible to deduce the approximate shape of the molecule.

A condensed chemical formula may represent the types and spatial arrangement of bonds in a simple chemical substance, though it does not necessarily specifyisomers or complex structures. For example ethane consists of two carbon atoms single-bonded to each other, with each carbon atom having three hydrogen atoms bonded to it. Its chemical formula can be rendered as CH₃CH₃. In ethylene there is a double bond between the carbon atoms (and thus each carbon only has two hydrogens), therefore the chemical formula may be written: CH₂CH₂, and the fact that there is a double bond between the carbons is implicit because carbon has a valence of four. However, a more explicit method is to write H₂C=CH₂ or less commonly H₂C::CH₂. The two lines (or two pairs of dots) indicate that a double bondconnects the atoms on either side of them.

A triple bond may be expressed with three lines or pairs of dots, and if there may be ambiguity, a single line or pair of dots may be used to indicate a single bond.

Molecules with multiple functional groups that are the same may be expressed by enclosing the repeated group in round brackets. For example isobutane may be written (CH₃)₃CH. This condensed structural formula implies a different connectivity from other molecules that can be formed using the same atoms in the same proportions (isomers). The formula (CH₃)₃CH implies a central carbon atom attached to one hydrogen atom and three CH₃ groups. The same number of atoms of each element (10 hydrogens and 4 carbons, or C₄H₁₀) may be used to make a straight chain molecule, butane: CH₃CH₂CH₂CH₃.

The alkene called but-2-ene has two isomers which the chemical formula CH₃CH=CHCH₃ does not identify. The relative position of the two methyl groups must be indicated by additional notation denoting whether the methyl groups are on the same side of the double bond (cis or Z) or on the opposite sides from each other (trans or E). Such extra symbols violate the rules for chemical formulas, and begin to enter the territory of more complex naming systems.

For polymers in condensed chemical formulas, parentheses are placed around the repeating unit. For example, a hydrocarbon molecule that is described as CH₃(CH₂)₅₀CH₃, is a molecule with fifty repeating units. If the number of repeating units is unknown or variable, the letter n may be used to indicate this formula: CH₃(CH₂)_nCH₃.

For ions, the charge on a particular atom may be denoted with a right-hand superscript. For example Na⁺, or Cu²⁺. The total charge on a charged molecule or a polyatomic ion may also be shown in this way. For example: H₃O⁺ or SO₄²⁻.

For more complex ions, brackets are often used to enclose the ionic formula, as in [B₁₂H₁₂]²⁻, which is found in compounds such as Cs₂[B₁₂H₁₂]. Parentheses can be nested inside brackets to indicate a repeating unit, as in [Co(NH₃)₆]³⁺. Here (NH₃)₆ indicates that the ion contains six NH₃ groups, and encloses the entire formula of the ion with charge +3.

Although isotopes are more relevant to nuclear chemistry or stable isotope chemistry than to conventional chemistry, different isotopes may be indicated with a prefixed superscript in a chemical formula. For example, the phosphate ion containing radioactive phosphorus-32 is ³²PO₄^3-. Also a study involving stable isotope ratios might include the molecule ¹⁸O¹⁶O.

A left-hand subscript is sometimes used redundantly to indicate the atomic number. For example, ₈O₂ for dioxygen, and ¹⁶₈O₂ for the most abundant isotopic species of dioxygen. This is convenient when writing equations for nuclear reactions, in order to show the balance of charge more clearly.

The @ symbol (at sign) indicates an atom or molecule trapped inside a cage but not chemically bound to it. For example, a buckminsterfullerene (C₆₀) with an atom (M) would simply be represented as MC₆₀ regardless of whether M was inside the fullerene without chemical bonding or outside, bound to one of the carbon atoms. Using the @ symbol, this would be denoted M@C₆₀ if M was inside the carboetwork. A non-fullerene example is [As@Ni₁₂As₂₀]^3-, an ion in which one As atom is trapped in a cage formed by the other 32 atoms.

This notation was proposed in 1991 with the discovery of fullerene cages (endohedral fullerenes), which can trap atoms such as La to form, for example, La@C₆₀ or La@C₈₂. The choice of the symbol has been explained by the authors as being concise, readily printed and transmitted electronically (the at sign is included in ASCII, which most modern character encoding schemes are based on), and the visual aspects suggesting the structure of an endohedral fullerene.

Chemical formulas most often use integers for each element. However, there is a class of compounds, called non-stoichiometric compounds, that cannot be represented by small integers. Such a formula might be written using decimal fractions, as in Fe_0.95O, or it might include a variable part represented by a letter, as in Fe_1–xO, where x is normally much less than 1.

A chemical formula used for a series of compounds that differ from each other by a constant unit is called general formula. Such a series is called the homologous series, while its members are called homologs.

For example alcohols may be represented by: C_nH_{(2n + 1)}OH (n ≥ 1)

The Hill system is a system of writing chemical formulas such that the number of carbon atoms in a molecule is indicated first, the number of hydrogen atoms next, and then the number of all other chemical elements subsequently, in alphabetical order. When the formula contains no carbon, all the elements, including hydrogen, are listed alphabetically. This deterministic system enables straightforward sorting and searching of compounds.

Empirical formula: Just the elements

The empirical formula indicates the different types of elements in a molecule and the lowest whole-number ratio of each kind of atom in the molecule. For example, suppose that you have a compound with the empirical formula:

C₂H₆O

Three different kinds of atoms are in the compound, C, H, and O, and they’re in the lowest whole-number ratio of 2 C to 6 H to 1 O. So the actual formula (called the molecular formula or true formula) may be any of the following, or another multiple of 2:6:1.

C₄H₁₂O₂, C₄H₁₈O₃, C₈H₂₄O

Molecular or true formula: Inside the numbers

The molecular formula, or true formula, tells you the kinds of atoms in the compound and the actual number of each atom.

You may determine, for example, that the following empirical formula is actually the molecular formula, too, meaning that there are actually two carbon atoms, six hydrogen atoms, and one oxygen atom in the compound:

C₂H₆O

For ionic compounds, this formula is enough to fully identify the compound, but it’s not enough to identify covalent compounds. To write a formula that stands for the exact compound you have in mind, you often must write the structural formula instead of the molecular formula.

Structural formula:

The structural formula shows the elements in the compound, the exact number of each atom in the compound, and the bonding pattern for the compound. The electron-dot formula and Lewis formula are examples of structural formulas.

Look at the Lewis formulas presented in the following figure.

Both compounds in the figure have two carbon atoms, six hydrogen atoms, and one oxygen atom. The difference is in the way the atoms are bonded, or what’s bonded to what. These are two entirely different compounds with two entirely different sets of properties:

The formula on the left represents dimethyl ether. This compound is used in some refrigeration units and is highly flammable.

The formula on the right represents ethyl alcohol, the drinking variety of alcohol.

Simply knowing the molecular formula isn’t enough to distinguish between the two compounds.

Compounds that have the same molecular formula but different structures are called isomers of each other. To identify the exactcovalent compound, you need its structural formula.

Example 1

What is the formula weight of sufuric acid H₂SO₄?

Solution:
The formula also indicates a mass as the sum of masses calculate this way
2*1.008 + 32.0 + 4*16.0 = 98.0
where 1.008, 32.0 and 16.0 are the atomic weights of H, S, and O respectively.

Discussion: If the formula is a molecular formula, the mass associated with it is called molecular mass or molecular weight. As an exercise, work out the following problem.

What is the molecular weight of caffeine, C₈H₁₀N₄O₂?

The diagram shown here is a model of the caffeine molecule.

With the aid of a table of atomic weights, a formula indirectly represents the formula weight. If the formula is a molecular formula, it indirectly represents the molecular weight. For simplicity, we may call these weights molar masses, which can be formula weights or molecular weights.

A chemical formula not only represents what a substance is made of, it provides a great deal of information about the substance. Do you know that chemical formulas are used all over the world, regardless of the language? Chinese, Russian, Japanese, African, and South Americans use the same notations we do. Thus, H₂S is recognized as a smelly gas all over the world. Chemical formula is an international or universal language.

Weight percentage and mole percentage

A chemical formula not only gives the formula weight, it accurately represents the percentages of elements in a compound. On the other hand, if you know the percentage of a compound, you may figure out its formula. Percentage based on weights is called weight percentage, and percentage based on the numbers of atoms or moles is called mole percentage.

Example 2

What are the weight and mole percentages of S in sufuric acid?

Solution:
From example 1, we know that there are 32.0 g or S in 98.0 g or sulfuric acid. Thus the weight percentage is

Weight percentage = 32/98 = 32.7%

From the formula, there is one S atom among 7 atoms in H₂SO₄

Mole percentage = 1/7 = 14.3%

Discussion:
You have learned what weight and mole percentages are and how to evaluate them in this example. As an exercise, work out the the following problem:

What are the weight and mole percentages of C, H, N, and O for caffeine, C₈H₁₀N₄O₂?

Determination of Chemical Formulas

How would you find the chemical formula of a substance? If you know the substance, its formula and other information is usually listed in a handbook. Handbooks such as the CRC Handbook of Chemistry and Physicscontain information on millions of substances.

If you are a researcher and you made a new compound that no one has ever made it before, then you need to determine its empirical or molecular formula. For an organic compound, you burn it completely to convert all carbon (C) to CO₂, and all hydrogen (H) to H₂O.

C_xH_2y =(burned in O₂)=> x CO₂ + y H₂O

Thus, from the weight of CO₂ and H₂O produced by burning a definite amount of the substance, you can figure out the percent of C and H in the compound.

Nitrogen is determined by converting it to NH₃. The amount of NH₃ can be determined by titration, and the percentage can also be determined.

Percentage of O is usually obtained by subtracting all percentages of C, H, and N, if the compound does not contain any other element.

 

Example

Native thallium (atomic number 81, atomi c mass 204.383) consist of two isotopes:

Thallium-203

Thallium-205

 

Average atomic mass of thallium is:

Hydrogen isotopes have special symbols and names:

Chemical properties of isotopes of one element are equal. Isotopes having an identical mass numbers, but different nucleus charge are called isobars.

Questions

1. When a match burns, its mass decreases. Is this a violation of the Law of Conservation of Mass? Explain.

2. An 8.4g sample of sodium bicarbonate is added to a solution of acetic acid weighing 20.0g. The two substances react, releasing carbon dioxide to the atmosphere. After reaction, the contents of the reaction vessel weigh 24.0g. What is the mass of the carbon dioxide given off during the reaction?

3. Is the Law of Conservation of Mass obeyed in the following experiment: 10.00g of zinc dust is mixed with 2.00g of powdered sulfur and the mixture is heated carefully. The result is 6.08g of white zinc sulfide and 5.92g of unreacted zinc. Also, identify the limiting reagent.

4. Is the Law of Conservation of Mass obeyed in the following experiment: 10.00g of calcium carbonate is dissolved in 100.0 mL of hydrochloric acid solution which has density of 1.148 g/mL. The products are 120.40g of solution and 2.22 L of carbon dioxide gas, which has a density of 1.9769 g/L.

5. When heated directly, Phosphorus reacts completely with Selenium as follows: 

Phosphorus + Selenium Compound #1

(1.00g)                   (3.82g)       (4.82g)

When heated under carbon dioxide and nitrogen, Phosphorus reacts completely with Selenium as follows:

Phosphorus + Selenium Compound #2 

(1.00g)         (6.37g)        (7.37g) 

Determine the elemental composition, percentage by mass, of Compound #1 and Compound #2.

6. For each experiment, determine the elemental composition, percentage by mass, of the compound Sodium Chloride.

Expt. 1

1.00g of Sodium metal was allowed to completely react with 10.00g of Chlorine gas. The Sodium was completely consumed and 2.54g of sodium chloride was produced.

Expt. 2

In a second experiment, 1.00g of Chlorine was allowed to react with 10.00g of Sodium. The Chlorine was consumed completely and 1.65g of sodium chloride was produced.   Are the results consistent with the Law of Definite Proportions?

7. Elemental Magnesium reacts with elemental Oxygen to form magnesium oxide. Magnesium oxide has a composition of 60.32% Mg and 39.68% O. How many grams of Oxygen are required to react completely with 5.00g of Magnesium? 

Magnesium + Oxygen Magnesium oxide 

8. Hydrogen forms two different Oxides; water and hydrogen peroxide. The relative amounts of hydrogen and oxygen for these two compounds are:

  Water Hydrogen Peroxide

 mass Hydrogen    1.00g                             1.00g 

mass Oxygen         7.94g                             15.87g.

Form the following ratio: (mass O in Hyd. Per. / mass H) / (mass O in Water / mass H).

Is the result consistent with the Law of Multiple Proportions? (Challenge question: If you know the chemical formula of water is H2O, what is the chemical formula of Hydrogen Peroxide?)

 9. Mercury forms two different Oxides; Mercurous Oxide and Mercuric Oxide. The relative amounts of mercury and oxygen in these compounds are:  

Mercurous Oxide          Mercuric Oxide 

mass Mercury                 1.00g                                      1.00g 

mass Oxygen                  0.0395g                         0.0799g 

Form the following ratio:  (mass O in Mercuric Ox. / mass Hg) / (mass O in Mercurous Ox. / mass Hg)

Is the result consistent with the Law of Multiple Proportions?

10. Gold forms two different compounds with chlorine:  

Aurous chloride                Auric chloride 

84.76% Au                       64.94% Au 

15.24% Cl                        35.06% Cl 

 a) Calculate the mass of Chlorine required to react with 1.00g of Gold to form each of these compounds. 

b) Form the following ratio:  

(mass Cl in Auric Chl. / mass Au) / (mass Cl in Aurous Chl. / mass Au)

Is the result consistent with the Law of Multiple Proportions?

11. Use the data of Problem #5 to form the following ratio:  (mass Se in Cmpd. #2 / mass P) / (mass Se in Cmpd. #1 / mass P)

Is the result consistent with the Law of Multiple Proportions?

 

References:

1. The abstract of the lecture.

2. intranet.tdmu.edu.ua/auth.php

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Prepared by PhD Falfushynska H.