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(D) Let the mass of the element be $100 \ g$ in both cases.In the first oxide,mass of oxygen $= 53.33 \ g$,so mass of element $= 100 - 53.33 = 46.67 \

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Multiple proportions

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(D) Let the mass of the element be $100 \ g$ in both cases.In the first oxide,mass of oxygen $= 53.33 \ g$,so mass of element $= 100 - 53.33 = 46.67 \ g$.In the second oxide,mass of oxygen $= 36.36 \ g$,so mass of element $= 100 - 36.36 = 63.64 \ g$.Calculate the mass of oxygen combined with $1 \ g$ of the element:For oxide $1$: $\frac{53.33}{46.67} \approx 1.143 \ g$.For oxide $2$: $\frac{36.36}{63.64} \approx 0.571 \ g$.The ratio of the masses of oxygen that combine with a fixed mass of the element is $1.143 : 0.571 \approx 2 : 1$.Since the ratio is a simple whole number,this illustrates the Law of Multiple Proportions.

An element forms two oxides containing respectively $53.33\%$ and $36.36\%$ of oxygen. These figures illustrate the law of

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