# Writing Mole Ratio for Chemical Species in Balanced Chemical Equations

LAWS OF CHEMICAL COMBINATION

By studying the results of quantitative measurements of many reactions it was observed that whenever substances react, they do so according to certain laws. These laws are called the laws of chemical combination. These laws formed the basis of Dalton’s atomic theory of matter. Before we discuss this theory it would be appropriate to briefly study these laws.

LAW OF CONSERVATION OF MASS

This law was stated by the French chemist Antoine Laurent Lavoisier (1774).

This law states that

Antoine Lavoisier

(1743 – 1794)

During any physical or chemical change, the total mass of the products remains equal to the total mass of the reactants.

Lavoisier showed that when mercuric oxiae was heated it produced free mercury and oxygen. The sum of masses of mercury and oxygen was found to be equal to the mass of mercuric oxide

SOLVED EXAMPLES

Example 12.8. 10.0 g ofCaC03 on heating gave 4.4 g of C02 and 5.6 g of CaO. Show that these observations are in agreement with law of conservation of mass.

Solution.    Mass of the reactants = 10.0 g

Mass of the products= 4.4 + 5.6 = 10.0 g

Since mass of the reactants is equal to the mass of the products, therefore, these observations are in agreement with law of conservation of mass.

Example 12.9 When 4.2 g of NaHCO 3 is added to a solution of CH3COOH weighing 10 g, it is observed that 2.2 g of C02 is released into atmosphere. The residue is found to weigh 12.0 g. Show that these observations are in agreement with the law of conservation of mass.

Solution. According to law of conservation of mass the mass of reactants and products must be equal.

Thus,

mass of reactants  = 4.2 + 10    = 14.2 g

and mass of products       =  2.2 + 12.0 = 14.2 g

Hence, the law of conservation of mass is proved.

LAW OF CONSTANT COMPOSITION OR DEFINITE PROPORTIONS

This law deals with the composition of chemical compounds. It was discovered

by the French chemist, Joseph Proust (1799). This law states that:

A chemical compound always contains same elements combined together in same . proportion by mass.

It implies from this law that in a chemical compound the elements are present infixed and not arbitrary ratio by mass. For example, pure water obtained from different sources such as, river, well, spring, sea, etc., always contains hydrogen and oxygen combined together in

the ratio 1 : 8 by mass. Similarly. carbon dioxide can be obtained by different methods such as:

(a) burning of carbon,

(b) heating limestone, or

(c) the action of dilute hydrochloric acid on marble pieces.

It can be shown experimentally that different samples of carbon dioxide contain carbon and oxygen in the ratio of 3 : 8 by mass.

SOLVED EXAMPLE

example 12.10. 1.375 g of cupric oxide was reduced by heating in a current of hydrogen and the weight of copper that remained was 1.098g.lnanotherexperiment, 1.179 g of copper was dissolved in the nitric acid and the resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed was 1.476 g. Show that these results illustrate the law of constant composition.

Solution. First experiment:

Copper oxide= 1.375 g

Copper left= 1.098 g

. . Oxygen present= 1.375 – 1.098 = 0.277 g

Hence % of oxygen m. CuO = 0.277 x 100 / 1.375 = 20.14.

Second experiment:

Copper taken = 1.179 g

Copper oxide formed= 1.476 g

Oxygen present= 1.476- 1.179 g = 0.297 g

Hence % of oxygen m. CuO = 0.297 x 100 / 1.476 = 20.12.

Percentage of oxygen is the same both the above cases, so the law of constant composition is illustrated.

LAW OF MULTIPLE PROPORTIONS

This law was discovered by John Dalton (1803). This law states that:

When two elements combine with each other to form two or more than two

compounds, the masses of one of the elements which combine with fixed mass of the other, bear a simple whole number ratio to one John Dalton another.

For example, carbon and oxygen combine with each other ·to form carbon n:onoxide (CO) and carbon dioxide (CO2).

In carbon monoxide (CO2):    12 parts by mass of carbon combine with 16 parts by mass of oxygen.

In carbon dioxide (CO2):        12 parts by mass of carbon combine with 32 parts by mass of oxygen.

Ratio of the masses of the oxygen which combine with fixed mass of carbon (12 parts) in these compounds is 16: 32 or 1 : 2, which is a simple whole number ratio.

Similarly, copper and oxygen combine to form two oxides, the red cuprous oxide (C~O) and the black cupric oxide (CuO).

In red oxide·( Cu2O ):             16 parts by mass of oxygen combine with 63.5 x 2

parts by mass of copper.

In black oxide ( CuO ):           16 parts by mass of oxygen combine with 63.5 parts by mass of copper.

Ratio of masses of copper that combine with fixed mass of oxygen (16 parts) in these oxides is 63.5 x 2: 63.5 or 2: l , which is a simple whole number ratio.

SOLVED EXAMPLES

Example 12.11. Hydrogen and oxygen are known  to form two compounds. The hydrogen content in one of these is 5.93% while in the other it is 11.2%. Show that this data illustrates the law of multiple proportions.

Solution. In the first compound

Hydrogen= 5.93%

Oxygen=  (100 – 5.93)% = 94.07%.

In the second compound

Hydrogen = 11.2%

Oxygen = (100- 11.2)% = 88.8%.

In the first compound the number of parts by mass of oxygen that com me With one part by mass of hydrogen = 94.07 / 5.93 = 15.86 parts.

In the second compound the number of parts by mass of oxygen that combine with one part by mass of hydrogen = 88.8 / 11.2 = 7.9 parts.

The ratio of masses of oxygen that combine with fixed mass (1 part) by mass of hydrogen is 15.86: 7.9 or 2: 1.

Since this ratio is a simple whole number ratio, hence the given data illustrates the Jaw of multiple proportions.

Example 12.12.Carbon and oxygen are known to form two compounds. The carbon content in one of these is 42.9% while in the other it is 27.3%. Show that this data is in agreement with law

of multiple proportions.

solution.     In compound I             c = 42.9%; 0 = 57.1%

In compound II           c = 27.3%; 0 = 72.7%

For a fixed mass of carbon (or parts)= 1 the ratio of oxygen in ‘he two compounds would be

Oxygen in I : II = 57.1 / 42.9 : 72.7 / 27.3 = 1.33 : 2.66 = 1:2

Since it is a simple whole number ratio, it illustrates law of multiple proportions.