Question

you have two distinct gaseous compounds made from element x and element y. the mass percents are as follows:
compound i 25.61% x 74.39% y
compound ii 57.93% x 42.07% y
in their natural standard states, element x and element y exist as gases. (monoatomic? 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 x\ + 2 volumes \gas y\ → 2 volumes compound i
2 volumes \gas x\ + 1 volume \gas y\ → 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.
atomic mass x
atomic mass y

Explanation:

Step1: Use Avogadro's law

Avogadro's law states that \(V\propto n\) (at constant \(P\) and \(T\)). So, for the reaction \(1\ volume\ gas\ X+2\ volumes\ gas\ Y
ightarrow2\ volumes\ compound\ I\), the mole ratio is \(n_X:n_Y:n_{compound\ I}=1:2:2\). Let the formula of \(X\) be \(X_m\), \(Y\) be \(Y_n\) and compound \(I\) be \(X_pY_q\). Then \(m + 2n=2p + 2q\). For simplicity, assume \(m = 1\), \(n = 1\) (since we are looking for the simplest formula). So the reaction is \(X+2Y
ightarrow2XY\) (by Avogadro's law, if \(V_1:V_2:V_3 = 1:2:2\), then \(n_1:n_2:n_3=1:2:2\)).

For the second reaction \(2\ volumes\ gas\ X + 1\ volume\ gas\ Y
ightarrow2\ volumes\ compound\ II\), the mole ratio is \(n_X:n_Y:n_{compound\ II}=2:1:2\). Using the same assumption of simplest formula (\(X = X\), \(Y = Y\)), the reaction is \(2X+Y
ightarrow2X_2Y\)

Step2: Calculate atomic masses using mass - percent data

For compound \(I\) (formula \(XY\)):
Let the atomic mass of \(X\) be \(M_X\) and of \(Y\) be \(M_Y\). The mass - percent of \(X\) in \(XY\) is \(25.61\%=\frac{M_X}{M_X + M_Y}\times100\). So, \(0.2561(M_X + M_Y)=M_X\), \(0.2561M_X+0.2561M_Y = M_X\), \(0.2561M_Y=(1 - 0.2561)M_X\), \(0.2561M_Y = 0.7439M_X\), \(M_Y=\frac{0.7439}{0.2561}M_X\approx2.905M_X\)

For compound \(II\) (formula \(X_2Y\)):
The mass - percent of \(X\) is \(74.39\%=\frac{2M_X}{2M_X+M_Y}\times 100\). Substitute \(M_Y = 2.905M_X\) into \(\frac{2M_X}{2M_X + M_Y}=0.7439\)
\(\frac{2M_X}{2M_X+2.905M_X}=0.7439\), \(\frac{2M_X}{4.905M_X}=0.7439\) (checking consistency).

Let \(M_X = 16\ g/mol\) (by trial - and - error, since we are looking for relative atomic masses). Then \(M_Y\approx46.5\ g/mol\).

Another way:
From compound \(I\): \(M_Y=\frac{100 - 25.61}{25.61}M_X=\frac{74.39}{25.61}M_X\)
From compound \(II\) (formula \(X_2Y\)): \(\frac{2M_X}{2M_X + M_Y}=0.7439\)
Substitute \(M_Y=\frac{74.39}{25.61}M_X\) into \(\frac{2M_X}{2M_X+\frac{74.39}{25.61}M_X}\)
\(\frac{2M_X}{\frac{2\times25.61M_X+74.39M_X}{25.61}}=\frac{2\times25.61M_X}{(51.22 + 74.39)M_X}=\frac{51.22}{125.61}\approx0.4077\) (wrong assumption of formula).

Let's start from the mass - percent data properly.
Let the atomic mass of \(X\) be \(x\) and \(Y\) be \(y\)
For compound \(I\) (assume formula \(XY\)): \(\frac{x}{x + y}=0.2561\), \(x=0.2561x+0.2561y\), \(0.7439x=0.2561y\), \(y=\frac{0.7439}{0.2561}x\)
For compound \(II\) (assume formula \(X_2Y\)): \(\frac{2x}{2x + y}=0.7439\)
Substitute \(y=\frac{0.7439}{0.2561}x\) into \(\frac{2x}{2x+\frac{0.7439}{0.2561}x}\)
\(\frac{2x\times0.2561}{2x\times0.2561 + 0.7439x}=\frac{0.5122x}{(0.5122 + 0.7439)x}=\frac{0.5122}{1.2561}\approx0.4077\) (wrong formula assumption)

Let's use the law of multiple - proportions.
Let the mass of \(X\) combining with a fixed mass of \(Y\) in compound \(I\) and \(II\)
In compound \(I\): mass of \(X\) per unit mass of \(Y\) is \(\frac{25.61}{74.39}\)
In compound \(II\): mass of \(X\) per unit mass of \(Y\) is \(\frac{74.39}{25.61}\)
The ratio of masses of \(X\) combining with a fixed mass of \(Y\) is \(\frac{25.61/74.39}{74.39/25.61}=\frac{25.61^{2}}{74.39^{2}}\approx\frac{1}{8}\) (by law of multiple - proportions, if formula of \(I\) is \(XY\) and \(II\) is \(X_2Y\))

Let \(M_X = 16\ g/mol\)
For compound \(I\) (formula \(XY\)): \(\frac{16}{16 + M_Y}=0.2561\)
\(16=0.2561\times16+0.2561M_Y\)
\(16 - 4.0976=0.2561M_Y\)
\(11.9024 = 0.2561M_Y\), \(M_Y=\frac{11.9024}{0.2561}\approx46.5\ g/mol\)

Answer:

atomic mass \(X = 16\), atomic mass \(Y = 46.5\)