Question

An element X has a tribromide with the empirical formula XBr3 and a trichloride with the empirical formula XCl3. The tribromide is converted to the trichloride according to the equation

XBr3 + Cl2→ XCl3 + Br2


If the complete conversion of 1.334 g of XBr3 results in the formation of 0.722 g of XCl3, what is the atomic mass of the element X?

Answer 1

An element X has a tribromide with the empirical formula XBr3 and a trichloride with the empirical formula XCl3. The tribromide is converted to the trichloride according to the equation

$$\mathrm{XBr}_3 + \mathrm{Cl}_2 \rightarrow \mathrm{XCl}_3 + \mathrm{Br}_2$$

If the complete conversion of $1.334\,\mathrm{g}$ of $\mathrm{XBr}_3$ results in the formation of $0.722\,\mathrm{g}$ of $\mathrm{XCl}_3$, what is the atomic mass of the element X?

Solution:

We use the equivalents law:

$$\frac{m(\mathrm{XBr}_3)}{m(\mathrm{XCl}_3)} = \frac{E(\mathrm{XBr}_3)}{E(\mathrm{XCl}_3)}$$

The equivalent mass of $\mathrm{XBr}_3$ is: $E(\mathrm{XBr}_3) = E(\mathrm{X}) + E(\mathrm{Br}) = E(\mathrm{X}) + 79.904$, and the equivalent mass of $\mathrm{XBr}_3$ is: $E(\mathrm{XCl}_3) = E(\mathrm{X}) + E(\mathrm{Cl}) = E(\mathrm{X}) + 35.453$.

So, $\frac{1.334}{0.722} = \frac{E(X) + 79.904}{E(X) + 35.453}$, from this $1.334 \cdot E(X) + 47.294 = 0.722 \cdot E(X) + 57.691$,

$$E(X) = \frac{10.397}{0.612} = 16.989\,\mathrm{g/mol}$$

Because, the element X is three valence, the atomic mass of this element is: $\mathrm{Ar(X)} = \mathrm{E(X)} \cdot 3 = 16.989 \cdot 3 = 50.967\,\mathrm{amu}$

Answer:

The atomic mass of the element X is 50.967 amu.