Answer
The empirical formula is $P_{4}$$Se_{3}$
Work Step by Step
Step 1. First you need to calculate the mass of selenium.
131.6-45.2=86.4 mg selenium
Step 2. Now you need to convert selenium and phosphorus mass to moles. The easiest way to do this is to use a process known as dimensional analysis.
Formula= mass x $\frac{1g}{1000mg}$x$\frac{1 mol}{molar mass}$
Selenium - 86.4mgSe x$\frac{1g Se}{1000mg Se}$x$\frac{1 mole Se}{78.96g Se}$ = 0.001094225 mole Se
Phosphorus- 45.2 mgP $\frac{1 gP}{1000 mg P}$x$\frac{1 mole P}{30.97g P}$ = 0.001459477 mole P
Step 3. Now you are going to take both the calculated mole for selenium and phosphorus and divided it by the smallest more calculated.
Selenium= $\frac{0.001094225 mol Se}{0.001094225 mol}$= 1Se
Phosphorus= $\frac{0.001459477molP}{0.001094225Mol}$=1.33379972126391P
Step 4. Lastly you want to multiply the coefficients by 3 so that you can get a whole number for phosphorus.
3(1)=3 Se
3(1.33379972126291)=4 P
Empirical Formula=$P_{4}$$Se_{3}$
