Question

Two samples 'm' and 'n' of slaked lime, $Ca(OH{)}_2$, were obtained from two different reactions. The details about their composition are as follows:

$\underset{–}{Sample{}^{\prime }{m}^{\prime }}$

Total mass: 7 g

Mass of constituent oxygen: 2 g

Mass of constituent calcium: 5 g

$\underset{–}{Sample{}^{\prime }{n}^{\prime }}$

Total mass: 1.4 g

Mass of constituent oxygen: 0.4 g

Mass of constituent calcium: 1 g

Which law of chemical combination does this prove? Explain.

Answer 1

Slaked lime is $Ca(OH{)}_2$

$Sample{}^{\prime }{m}^{\prime }$	$Sample{}^{\prime }{n}^{\prime }$
Mass = 7 g	Mass = 1.4 g
Oxygen = 2 g	Oxygen = 0.4 g
Calcium = 5 g	Calcium = 1 g

Proportion of oxygen in sample 'm' $=\frac{2}{7}$

Proportion of oxygen in sample 'n' $=\frac{0.4}{1.4}$ $=\frac{2}{7}$

Proportion of calcium in sample 'm' $=\frac{5}{7}$

Proportion of calcium in sample 'n' $=\frac{1}{1.4}$ $=\frac{5}{7}$

Proportion of oxygen in both samples is same and that of calcium is also same. This proves the law of constant proportion, i.e. “In a chemical substance the elements are always present in definite proportions by mass”.