You have two distinct gaseous compounds made from element X and element Y. The mass percents are as follows: Compound I: \(30.43 \% \mathrm{X}, 69.57 \% \mathrm{Y}\) Compound \(\mathrm{II} : 63.64 \% \mathrm{X}, 36.36 \% \mathrm{Y}\) In their natural standard states, element X and element Y exist as gases. (Monatomic? Diatomic? Triatomic? That is for you to determine.) When you react “gas X” with “gas Y” to make the products, you get the following data (all at the same pressure and temperature): 1 volume "gas \(\mathrm{X}^{\prime \prime}+2\) volumes "gas $\mathrm{Y}^{\prime \prime} \longrightarrow$ 2 volumes compound I 2 volumes \(^{4}\) gas \(\mathrm{X}^{\prime \prime}+1\) volume "gas \(\mathrm{Y}^{\prime \prime} \longrightarrow\) 2 volumes compound II Assume the simplest possible formulas for reactants and products in the chemical equations above. Then, determine the relative atomic masses of element X and element Y.

Step by step solution

01

Study the compound formation by analyzing the data

Look at the data given: For Compound I: 1 volume gas X'' + 2 volumes gas Y'' → 2 volumes compound I For Compound II: 2 volumes gas X'' + 1 volume gas Y'' → 2 volumes compound II Assume that X'' and Y'' represent moles of X and Y, respectively, in the simplest possible formulas for reactants.

02

Set up a proportion based on mass percent

Let's use the mass percent information to set up a proportion. For Compound I, we have 30.43% X and 69.57% Y. Let the mass of one mole of element X be Mx and the mass of one mole of element Y be My. We can then set up a proportion: \( \frac{M_\text{x}}{M_\text{x} + 2M_\text{y}} = \frac{30.43}{100} \) For Compound II, we have 63.64% X and 36.36% Y. We can set up another proportion: \( \frac{2M_\text{x}}{2M_\text{x} + M_\text{y}} = \frac{63.64}{100} \)

03

Solve the equations to determine atomic masses of X and Y

We have a system of two equations for Mx and My: \( \begin{cases}\frac{M_\text{x}}{M_\text{x} + 2M_\text{y}} = \frac{30.43}{100}\\ \\ \frac{2M_\text{x}}{2M_\text{x} + M_\text{y}} = \frac{63.64}{100}\end{cases} \) Now, multiply the first equation by \(2M_\text{x} + 2M_\text{y}\) and the second equation by \((M_\text{x} + 2M_\text{y})(2M_\text{x} + M_\text{y})\) to remove the denominators: \( \begin{cases}100M_\text{x}(2M_\text{x}+2M_\text{y})=30.43M_\text{x}(M_\text{x}+2M_\text{y})\\ \\ 100M_\text{x}(M_\text{x}+2M_\text{y})(2M_\text{x}+M_\text{y})=63.64(M_\text{x}+2M_\text{y})(2M_\text{x}+M_\text{y})\end{cases} \) Solve this system of two equations with two variables (Mx and My) to determine the relative atomic masses of elements X and Y.

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