Chapter 6, Problem 57A

Interpretation Introduction

Interpretation: The empirical formula of the two Mg compounds must be found out.

Concept Introduction :

Empirical formula of a molecule is the smallest unit which is repeated in the molecule.

Molecular formula of a compound is the representation of the molecule with the number of different atoms present in the molecule.

Molecular formula $\text{=empirical formula×n}$ where n is any positive integer like 1, 2, 3 and so on.

Answer to Problem 57A

The empirical formulae of the two Mg compounds are Mg₃N₂ and MgO.

Explanation of Solution

For the first compound mass of Mg is 0.9240 g and mass of N is $\text{(1}\text{.2791-0}\text{.9240) g = 0}\text{.3551 g}$

Moles of Mg $=\frac{0.9240}{24}=\mathrm{0.0385.}$

Moles of N $=\frac{0.3551}{14}=\mathrm{0.025.}$

Smallest mole ratio Mg: N $\begin{array}{l}=\frac{0.0385}{0.025}:\frac{0.025}{0.025}\\ =1.54:1=3.08:2\approx 3:2.\end{array}$

Thus the empirical formula of the compound is Mg₃N₂.

For the second compound % of Mg and O are 60.31 and 39.69 respectively.

Moles of Mg $=\frac{60.31}{24}=\mathrm{2.555.}$

Moles of O $=\frac{39.69}{16}=\mathrm{2.48.}$

Smallest mole ratio Mg: O $\begin{array}{l}=\frac{2.555}{2.48}:\frac{2.48}{2.48}\\ =1.03:1\approx 1:1.\end{array}$

Thus the empirical formula of the compound is MgO.

Conclusion

Empirical formulae of the two magnesium compounds are Mg₃N₂ and MgO.