Question

A compound has the forumula M3N Where M is a metal element and N is nitrogen. It contains 0.673 g of N per gram of M. Determine the relative atomic mass of of M and thus its identity.

Answer 1

Question#46777 – Chemistry – Other

Question:

A compound has the formula $M_3N$ Where M is a metal element and N is nitrogen. It contains 0.673 g of N per gram of M. Determine the relative atomic mass of M and thus its identity.

Answer:

The amount of moles of Nitrogen can be calculated:

$$n(N) = \frac{m(N)}{M(N)} = \frac{0.673\,g}{14\,\frac{g}{mol}} = 0.048\,mol$$

The amount of moles of metal is three times greater, as the compound contains three atoms of metal per one atom of Nitrogen.

$$n(M) = 3n(N) = 0.048 \times 3 = 0.144\,mol$$

Hence, the molar mass of metal can be calculated:

$$M(M) = \frac{m(M)}{n(M)} = \frac{1\,g}{0.144\,mol} = 6.94\,g/mol$$

This molar mass references to Lithium (Li). The unknown compound is lithium nitride, $\mathrm{Li}_3\mathrm{N}$.