Question

Question:Consider 100.0-g samples of two different compounds consisting only of carbon and oxygen.One compound contains 27.2 g of carbon, and the other has 42.9 g of carbon. How can thesedata support the law of multiple proportions if 42.9 is not a multiple of 27.2? Show that these datasupport the law of multiple proportions.

Short answer

Answer

Given data supports the multiple proportions law other than 27.2 but not a multiple of 42.9.

Step-by-step solution

Law of multiple proportions

The formation of combination of two elements by mass of one element in specific mass with other element in fixed ratio in all components called law of multiple proportions.

Mass of oxygen in compound I and II

The compound’s mass is 100 g. Carbon’s mass in compound 1 is 27.2 g and it can be denoted as,

$$\begin{gathered} \text{mass} \text{of} \text{oxygen} \text{in} \text{compound} \text{II = mass} \text{of} \text{compound - mass} \text{of} \text{carbon} \\ \text{= 100g - 27.2g} \\ \text{= 72.8g} \end{gathered}$$

The carbon’s mass in compound II is 42.9 g. The oxygen’s mass can be denoted as,

$$\begin{gathered} \text{mass} \text{of} \text{oxygen} \text{in} \text{compound} \text{II}=\text{mass} \text{of} \text{compound - mass} \text{of} \text{carbon}\backslash \\ =100g-42.9g \\ =57.1g \end{gathered}$$

ratio of Mass of carbon to oxygen in compound I and II

The ratio of mass of carbon to oxygen in both compounds can be expressed as,

$$\begin{gathered} \text{Ratio} \text{of} \text{mass} \text{of} \text{C} \text{to} \text{mass} \text{of} \text{O} \text{in} \text{compound} \text{I}=\frac{\text{mass} \text{of} \text{C}}{\text{mass} \text{of} \text{O}} \\ =\frac{27.2 g}{72.8 g} \\ =\text{0.374} \text{g} \text{C/g} \text{O} \end{gathered}$$

$$\begin{gathered} \text{Ratio} \text{of} \text{mass} \text{of} \text{C} \text{to} \text{mass} \text{of} \text{O} \text{in} \text{compound} \text{II}=\frac{\text{mass} \text{of} \text{C}}{\text{mass} \text{of} \text{O}}\backslash \\ =\frac{\text{42.9} \text{g}}{57.1 \text{g}} \\ =\text{0.751} \text{g} \text{C/g} \text{O} \end{gathered}$$

ratio of Mass of carbon to 1 g of oxygen in compound I and II

It can be denoted as,

$$\begin{gathered} \frac{\text{mass} \text{of} \text{C/g} \text{O} \text{in} \text{compound} \text{II}}{\text{Mass} \text{of} \text{C/g} \text{O} \text{in} \text{compound} \text{I}}\text{=}\frac{\text{0.751}}{\text{0.375}} \\ \text{=}\frac{\text{2}}{\text{1}} \end{gathered}$$

Conclusion

The ratio can be found as 2:1. Thus, given data supports the multiple proportions law other than 27.2 but not a multiple of 42.9.