Question

a chemist obtains a sample of sodium sulfate, $\ce{na_{2}so_{4}}$, with a mass of 71.0 g. how many formula units of sodium sulfate are in the sample?\
$\dfrac{71.0\\,\text{g}\ce{na_{2}so_{4}}\\,\bigg|\\,1\\,\text{mol}\ce{na_{2}so_{4}}\\,\bigg|\\,6.02\times10^{23}\\,\text{f.un.}\ce{na_{2}so_{4}}}{\\,\bigg|\\,142.05\\,\text{g}\ce{na_{2}so_{4}}\\,\bigg|\\,1\\,\text{mol}\ce{na_{2}so_{4}}}$\
$?\times10^{?}$ f.un. $\ce{na_{2}so_{4}}$\
enter the coefficient in the green blank and the exponent in the yellow blank. report your answer to the appropriate number of significant figures.

Explanation:

Step1: Multiply the given mass by the molar mass conversion factor.

We have \( 71.0\ \text{g}\ \text{Na}_2\text{SO}_4 \times \frac{1\ \text{mol}\ \text{Na}_2\text{SO}_4}{142.05\ \text{g}\ \text{Na}_2\text{SO}_4} \). Calculating this gives \( \frac{71.0}{142.05}\ \text{mol} \approx 0.5\ \text{mol} \) (but we'll keep more precision for now).

Step2: Multiply by Avogadro's number.

Now multiply by \( \frac{6.02 \times 10^{23}\ \text{f.un.}\ \text{Na}_2\text{SO}_4}{1\ \text{mol}\ \text{Na}_2\text{SO}_4} \). So we have \( 71.0 \times \frac{6.02 \times 10^{23}}{142.05} \). First, calculate \( \frac{71.0 \times 6.02 \times 10^{23}}{142.05} \). Notice that \( 71.0 \) is approximately half of \( 142.05 \), so \( \frac{71.0}{142.05} \approx 0.5 \). Then \( 0.5 \times 6.02 \times 10^{23} = 3.01 \times 10^{23} \).

Answer:

The coefficient (green blank) is \( 3.01 \) and the exponent (yellow blank) is \( 23 \). So the answer is \( 3.01 \times 10^{23} \) formula units.