Q.2.27 P - Chemistry: Molecular Nature Of Matter And Change

Question

Show, with calculations, how the following data illustrate the law of multiple proportions:Compound 1: 77.6 mass % xenon and 22.4 mass % fluorineCompound 2: 63.3 mass % xenon and 36.7 mass % fluorine

Step-by-Step Solution

Answer

Ratio =$\frac{\mathbf{r}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\left(I\right)}{\mathbf{r}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\left(II\right)}\mathbf{=}\frac{\mathbf{3}\mathbf{.}\mathbf{46}}{\mathbf{1}\mathbf{.}\mathbf{72}}\mathbf{=}\mathbf{2}$

Here we can observe that the ratio differs by a factor of 2, therefore, this follows the law of multiple proportions.

Step1. Calculation of the mass ratio

To show that this follows the law of multiple proportions, we have to calculate the ratio of elements that involve forms of compounds.

For Compound I, the ratio is calculated as follows:

Ratio =$\frac{\mathbf{77}\mathbf{.}\mathbf{6}}{\mathbf{22}\mathbf{.}\mathbf{4}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{46}$

For Compound II, the Ratio is

Ratio = $\frac{\mathbf{63}\mathbf{.}\mathbf{3}}{\mathbf{36}\mathbf{.}\mathbf{7}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{72}$

Step2. Calculation of ratio of Compound(I) to Compound(II).

The calculated ratio of Compound I and Compound II is as follows:

Ratio =$\frac{\mathbf{r}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\left(I\right)}{\mathbf{r}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\left(II\right)}\mathbf{=}\frac{\mathbf{3}\mathbf{.}\mathbf{46}}{\mathbf{1}\mathbf{.}\mathbf{72}}\mathbf{=}\mathbf{2}$

Here we can observe that the ratio differs by a factor of 2, therefore, this follows the law of multiple proportions.