Question

2 oxides of a certain metal was separately heated in a current of hydrogen until constant weights were obtained. The water produced in each case was carefully collected and weighed. 2 grams of each oxide gave respectively 0.2517 grams and 0.4526 grams of water. Show that these results establish the law of multiple proportion.

Answer 1

Step 1: Given data :

Weight of the each oxide = $2\mathrm{g}$
Weight of water produced in case 1 = $0.2517g$
Weight of water produced in case 2 =$0.4526g$.

Step 2: Calculating the weight of oxygen in each oxide

The law of multiple proportion in all the given cases can be verified by combining the weight of oxygen with the weight of the metal in two different oxides
$18g$of water is equivalent to $16g$of oxygen
Thus $18g$ of water contains $16g$of oxygen
Therefore, $0.2517g$of water contains $16/18\times 0.2517=0.2237g$
And, $0.4526g$ of water contains $\left(16\right)/\left(18\right)\times 0.4526=0.4023g$

Step 3: Calculating the weight of oxygen when combined with $1g$ of metal in each oxide

Now two cases arise :
Case 1
Weight of metal oxide = $2g$
Weight of oxygen =$0.2237g$

Thus Weight of metal,

$$\begin{gathered} 2-0.2237 \\ =1.7763g \end{gathered}$$
Weight of oxygen combined with $1.7763g$ of metal = $0.2237g$
and, Weight of oxygen combined with $1g$ of metal = $(0.2237)/(1.7763)=0.1259g$
Case 2
Now, Weight of metal oxide =$2g$
Weight of oxygen =$0.4023g$
Weight of metal = $2-0.4023=1.5977g$
Weight of oxygen combined with 1g of metal =

$(0.4023)/(1.5977)=0.2518g$
Step 4: Comparing the weights of oxygen
The ratio of$0.1259:0.2582$ is the same as $1:2$

This ratio explains the law of multiple proportions.