Question

Hydrogen peroxide and water contains 5.93% and 11.2% of hydrogen respectively. Show that the data illustrates law of multiple proportions.

Answer 1

firstly find %age of oxygen in h2o2 and find %age in h20 of oxygen

Answer 2

Answer: Oxygen is present in the ratio of 2 : 1 in both the compounds.

Explanation: Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number.

We are given the percentages of hydrogen in both the compounds which are 5.93% in $H_2O_2$ and 11.2% in $H_2O$

Mass of hydrogen peroxide = 34 g/mol

Mass of water = 18 g/mol

Mass of hydrogen in $H_2O_2=\frac{5.93}{100}\times 34=2.016g/mol$

Mass of hydrogen in $H_2O=\frac{11.2}{100}\times 18=2.016g/mol$

As, the mass of hydrogen is fixed in both the compounds, so mass ratio of oxygen must be present in whole number ratio. Now, to calculate the mass ratio, we first need to find the mass of oxygen in both the compounds and for that we first find the percentages of oxygen in both the compounds.

Percentage of oxygen in $H_2O_2=100-5.93=94.07\%$

Mass of oxygen in $H_2O_2=\frac{94.07}{100}\times 34=31.98g/mol$

Percentage of oxygen in $H_2O=100-11.2=88.8\%$

Mass of oxygen in $H_2O=\frac{88.8}{100}\times 18=15.98g/mol$

Mass ratio of oxygen in both the compounds = 31.98 : 15.98 = 2 : 1

As, the mass ratio is present in whole number ratio. So, these two compounds follow Law of Multiple Proportions.