Question

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

Answer 1

In order to this question, to show the various results as respect to this question, we will first calculate the weight of oxygen in each oxide and then again we will calculate the weight of oxygen which would combine with $1g$ of metal in each oxide. And finally we will compare the weights of oxygen which combine with the same weight of metal in the two oxides.

Complete answer:
The weight of oxygen combined with the fixed weight of the metal in the two separate oxides should bear a simple ratio to one another to check the law of multiple proportions in each case. To demonstrate this, we will proceed in the following order:
Step-1: Calculate the weight of oxygen in each oxide.
Here, we are given:
Weight of each oxide $= 2.0g$
Weight of water produced in case I $= 0.2517g$
Weight of water produced in case II $= 0.4526g$
$18g\,of\,{H_2}O = 16g\,of\,oxygen$
i.e. $18g\,of\,water\,con\operatorname{ta} ins\,oxygen = 16g\,$
$\therefore 0.2517g\,of\,water\,contains\,oxygen = \dfrac{{16}}{{18}} \times 0.2517 = 0.2237g$
And, $0.4526g\,of\,water\,contains\,oxygen = \dfrac{{16}}{{18}} \times 0.4526g = 0.4023g$
Step-2: Calculate the weight of oxygen which would combine with $1g$ of metal in each oxide.
in case I, Weight of metal oxide $= 2g$ and weight of oxygen $= 0.2237g$
$\therefore Weight\,of\,metal = 2 - 0.2237 = 1.7763g$
$\therefore Weight\,of\,oxygen\,wich\,combines\,with\,1.7763g\,of\,metal = 0.2237g$
$\therefore Weight\,of\,oxygen\,wich\,combines\,with\,1g\,of\,metal = 0.2237g/1.7763g = 0.1259g$
In case II, Weight of metal oxide $= 2g$ and weight of oxygen $0.4023g$
$\therefore Weight\,of\,metal = 2 - 0.4023 = 1.5977g$
$\therefore Weight\,of\,oxygen\,wich\,combines\,with\,1.7763g\,of\,metal = 0.4023g$
$\therefore Weight\,of\,oxygen\,wich\,combines\,with\,1g\,of\,metal = 0.4023g/1.5977g = 0.2518g$
Step-3: Compare the weights of oxygen which combine with the same weight of metal in the two oxides.
The weights of oxygen which combine with $1g$ of metal in the two oxides are respectively $0.1259g\,and\,0.2518g$ . These weights are in the ratio of $0.1259:0.2518\,or\,1:2$ .
Since, this is a simple ratio, so the above results establish the law of multiple proportions.

Note:
The oxygen atom has an atomic weight of 16. The oxygen atom has a valency of 2. As a result, the equivalent weight of oxygen is calculated as $\left( {atomic{\text{ }}weight} \right)/Valency$ . As a result, the oxygen atom's equivalent weight is '8'. The weight of the material that reacts with or produces $1g{\text{ }}{H_2}$ is called its equivalent weight.