# Mole Mass Relationship in a Chemical Reaction

A balanced chemical equation provides quantitative information regarding the consumption of reactants and production of products. The numbers which precede the chemical symbols and which balance the equation (with the understanding that if no number appears, it is equal to unity) are called the stoichiometric coefficients (or numbers) and are proportional to the number of molecules or the amounts of the constituents that change during the reaction. For a general reaction,

n

_{1}A + n

_{2}B n

_{3}C+ n

_{4}D

n

_{1}, n

_{2, }n

_{ 3 }and n

_{4}are the stoichiometric coefficients for the species A, B, C and D, respectively. The changes in the number of molecules or the amounts of constituents during the course of reaction will be related to each other through the expression

or |
||

here

N stands for the change in number of molecules and Î” n for the change in amount of substance. The negative and positive signs in the above expressions represent, respectively, the decrease and increase in the number of molecules or the amount of the substances. Taking a specific example of the reaction

2H

_{2}(g) + O

_{2}(g) 2H

_{2}O(g)

we can write

In general, for the above reaction, we can state that

- 2x molecules of H
_{2}(g), combine with 1x molecules O_{2}(g) to give 2x molecules of H_{2}O(g), where x can have any integral value. For example, 2 molecules of H_{2}combine with 1 molecule of O_{2}to give 2 molecules of H_{2}O or 4 molecules of H_{2}combine with 2 molecules of O_{2}to give 4 molecules of H_{2}O, and so on. - In terms of molecular masses, 2x 2.016 m
_{au}of H_{2}combine with 1x32.00m_{au}of O_{2}to give 2x(2.016+16.00) m_{au}of H_{2}O, where x can have integral value and m_{au}stands for the atomic mass unit (=1.660310-27 k kg). - In terms of amount of substance, 2y mol of H
_{2}combine with 1y mol of O_{2}to give 2y mol of H_{2}O, where y can have any numerical value (need not be only an integer). For example, 2 mol of H_{2}combine with 1 mol of O_{2}to give 2 mol of H_{2}O or 3 mol of H_{2}combine with 1.5 mol of O_{2}to give 3 mol of H_{2}O, and so on.

# Chemical Arithmetic

In terms of molar masses, 2y2.016 g of H

_{2}(where 2.016 g mol

^{-1}is the molar mass of H

_{2}) combine with 1y32.00 g of O

_{2}(where 32.00 g mol

^{-1}is the molar mass of O

_{2}) to give 2y18.016 g of H

_{2}O (where 18.016 g mol

^{-1}is the molar mass of H

_{2}O). Here y can have any numerical value (need not be only integrals).

The above interpretations also hold good for the reactions taking place in solution. If it is assumed that the volume of the solution does not change during the course of a reaction, then the changes in the reactants and products can also be stated in terms of molar concentrations. By definition, molar concentration is equal to the amount of substance present per dm

^{3}of the solution. The molar concentration is represented by the symbol M and its unit mol dm

^{-3}is represented by M (roman type).

A given reaction may be started with any amount of reactants, but the consumption will take in accordance with the stoichiometric coefficients appearing in the balanced chemical reaction. For example, let the reaction

_{2}SO

_{4}+ 2NaOH Na

_{2}SO

_{4}+ 2H

_{2}O

_{2}SO

_{4}and 0.7 mol of NaOH. From the chemical reaction given above, we find that 1 mol of H

_{2}SO

_{4}will combine with 2 mol of NaOH. But we are provided with only 0.5 mol of H

_{2}SO

_{4}, which will react with 20.5 mol, i.e. 1 mol of NaOH. But we are provided with only 0.7 mol of NaOH. Therefore, we can conclude that only 0.35 mol out of 0.5 mol of H

_{2}SO

_{4}will react and the remaining 0.15 mol will remain unreacted. Accordingly, the amounts of H

_{2}SO

_{4}and water formed will be 0.35 mol and 20.35 mol, respectively. The reactant present in a lesser amount than the required one is known as limiting reactant because its amount limits the amount of products.

We now solve a few problems to illustrate the mole-mass relationship in a chemical reaction.

**Problem**

Potassium bromide contains 32.9 mass % of potassium. If 6.40 g of bromine is made to react with 3.60 g of potassium, calculate the mass of potassium, which combines with bromine to form potassium bromide.

**Solution**

From the given percentage of potassium in potassium bromide, we can proceed as follows:

67.1g of bromine will react with 32.9 g of potassium to give 100 g of potassium bromide. Hence, for the given mass of bromine the mass of potassium that can combine is