Question

Barium and nitrogen form two binary compounds containing 90.745% and 93.634% barium, respectively. Determine the empirical formulas of these compounds.

Short answer

The empirical formulas of the compounds are BaN and${\mathbf{\text{Ba}}}_{\mathbf{3}}{\mathbf{N}}_{\mathbf{2}}$

Step-by-step solution

Empirical formula

There are two types of important formulas, and one among them is the empirical formula. The empirical formula provides the simplest ratio of the count of various atoms present.

Determining the empirical formula of the compounds formed from barium and nitrogen when barium is 90.745%.

In thefirst composition, the percentage of barium is 90.745%.

If the complete compound is of 100.0 g, then;

Mass of Ba = 90.745 g

$\begin{array}{rcl}& & \text{Mass of nitrogen = 100.0g - 90.745g}\\ & & \text{= 9.255g}\end{array}$

The number of moles of each element can be found as;

$\text{moles =}\frac{\text{mass}}{\text{mol.mass}}$

The number of moles of barium;

$$\begin{gathered} \text{Moles of Ba =}\frac{\text{90.745 g}}{\text{137.33 g}} \\ \text{= 0.6608 mol} \end{gathered}$$

The number of moles of nitrogen;

$$\begin{gathered} \text{Moles of N =}\frac{\text{9.255 g}}{\text{14.01g}} \\ \text{= 0.6606 mol} \end{gathered}$$

As, the ratio of barium and nitrogen becomes 1:1, the formula of the compound is

BaN.

Determining the empirical formula of the compounds formed from barium and nitrogen when barium is 93.634%.

In the second composition, the percentage of barium is 93.634%.

If the complete compound is of 100.0 g, then;

The mass of barium is 93.634gms

$$\begin{gathered} \text{Mass of nitrogen = 100.0g - 93.634g} \\ \text{= 6.366g} \end{gathered}$$

The number of moles of each element can be found as;

$\text{moles =}\frac{\text{mass}}{\text{mol.mass}}$

The number of moles of barium;

$$\begin{gathered} \text{Number of moles =}\frac{\text{93.634 g}}{{\text{137g mol}}^{\text{- 1}}} \\ \text{= 0.68 mol} \end{gathered}$$

The number of moles of nitrogen;

$$\begin{gathered} \text{Number of moles =}\frac{\text{6.366 g}}{{\text{14 g mol}}^{\text{- 1}}} \\ \text{= 0.45 mol} \end{gathered}$$

The simplest ratio of Ba and N is 1.5:1 or the simplest whole number ratio is 3:2.

Hence, the empirical formula of the compound is${\mathbf{\text{Ba}}}_{\mathbf{\text{3}}}{\mathbf{\text{N}}}_{\mathbf{\text{2}}}$