Chapter 9.5, Problem 48SP

Interpretation Introduction

Interpretation: The lowest whole-number mass ratio of lead in the two compounds is to be stated.

Concept Introduction : According to the law of multiple proportions, the two elements form more compounds, than the ratios of the second element’s masses, which combined with the fixed first’s element mass will always be ratios of small whole numbers.

Answer to Problem 48SP

The lowest whole-number mass ratio of lead in the two compounds is 1:2.

Explanation of Solution

It is given that lead forms two compounds with oxygen. One contains 2.98 g of lead and 0.461 g of oxygen. The other contains 9.89 g of lead and 0.763 g of oxygen.

The ratio of lead and oxygen in both compounds is:

  $\begin{array}{l}\text{Compound}\text{1}=\frac{\text{2}\text{.98}\text{g}\text{Pb}}{\text{0}\text{.461}\text{g}\text{O}}\\ \text{Compound}\text{2}=\frac{9.89\text{g}\text{Pb}}{\text{0}\text{.763}\text{g}\text{O}}\end{array}$

The denominator of each compound is to be made 1. So,

  $\begin{array}{l}\text{Compound}\text{1}=\frac{\text{2}\text{.98}\text{g}\text{Pb}}{\text{0}\text{.461}\text{g}\text{O}}=\frac{\frac{\text{2}\text{.98}\text{g}\text{Pb}}{\text{0}\text{.461}\text{g}}}{\frac{\text{0}\text{.461}\text{g}\text{O}}{\text{0}\text{.461}\text{g}}}=\frac{6.464\text{g}\text{Pb}}{\text{1}\text{g}\text{O}}\\ \text{Compound}\text{2}=\frac{9.89\text{g}\text{Pb}}{\text{0}\text{.763}\text{g}\text{O}}=\frac{\frac{9.89\text{g}\text{Pb}}{\text{0}\text{.763}\text{g}}}{\frac{\text{0}\text{.763}\text{g}\text{O}}{\text{0}\text{.763}\text{g}}}=\frac{12.96\text{g}\text{Pb}}{\text{1}\text{g}\text{O}}\end{array}$

Now, compound the mass of each Pd separately.

  $\begin{array}{c}\text{Ratio}=\frac{6.464\text{g}\text{Pb}}{12.96\text{g}\text{Pb}}\\ =1:2\end{array}$

Therefore, the lowest whole-number mass ratio of lead in the two compounds is 1:2.