Determine the empirical formulas of the compounds with the following compositions by mass: 24.5% Na, 14.9% Si, and 60.6% F

Each percentage represents the number of grams of an element per 100.0 grams of that compound. Convert these masses to moles using molar masses.

$$\begin{align*} \begin{array}{c|c} 24.5\;\cancel{\mathrm{g\;Na}}& 1\;\mathrm{mol\;Na}\\ \hline & 22.99\;\cancel{\mathrm{g\;Na}} \end{array} &=1.07\;\mathrm{mol\;Na} \end{align*}$$

$$\begin{align*} \begin{array}{c|c} 14.9\;\cancel{\mathrm{g\;Si}}& 1\;\mathrm{mol\;Si}\\ \hline & 28.09\;\cancel{\mathrm{g\;Si}} \end{array} &=0.530\;\mathrm{mol\;Si} \end{align*}$$

$$\begin{align*} \begin{array}{c|c} 60.6\;\cancel{\mathrm{g\;F}}& 1\;\mathrm{mol\;F}\\ \hline & 19.00\;\cancel{\mathrm{ g\;F}} \end{array} &=3.19\;\mathrm{mol\;F} \end{align*}$$

To find the empirical formula of the compound that has 24.5% Na, 14.9% Si, and 60.6% F. We must:

1. Assume there is 100.0 g of compound to get the number of grams per element. Therefore:$\\$

Na = 24.5 g Si = 14.9 g F= 60.6 g$\\$

(b)

The % by mass of Nа is 24.5%, Si is 14.9 %, and of F is 60.6 %.

Let us assume that the mass of the compound is 100 grams, hence

The mass of Na is $m_{Na} = 24.5 \mathrm{g}$

The mass of Si is $m_{Si} = 14.9 \mathrm{g}$

The mass of F is $m_{F} = 60.6 \mathrm{g}$

Let us find the empirical formula of a given compound.

First, we have to find the number of moles of Na, Si, and F.

$$\begin{align*} n_{Na} &= \frac { m_{Na} } { M_{Na} } = \frac { 24.5 \mathrm{g} } { 22.990 \mathrm{g/mol} } = 1.066 \mathrm{mol}\\ n_{Si} &= \frac { m_{Si} } { M_{Si} } = \frac { 14.9 \mathrm{g} } { 28.086 \mathrm{g/mol} } = 0.531 \mathrm{mol}\\ n_{F} &= \frac { m_{F} } { M_{F} } = \frac { 60.6 \mathrm{g} } { 19 \mathrm{g/mol} } = 3.189 \mathrm{mol} \end{align*}$$

To find the empirical formula, we have to divide each number of moles by the smallest one.

$$\begin{align*} \text{Number of Na atoms } &= \frac { 1.066 \mathrm{mol} } { 0.531 \mathrm{mol} } = 2\\ \text{Number of Si atoms } &= \frac { 0.531 \mathrm{mol} } { 0.531 \mathrm{mol} } = 1\\ \text{Number of F atoms } &= \frac { 3.189 \mathrm{mol} } { 0.531 \mathrm{mol} } = 6\\ \end{align*}$$

Therefore, the empirical formula is

$${\color{#4257b2}\mathrm{Na_2SiF_6}}$$