# What is the empirical formula for a compound that is found to contain 10.15 mg P and 34.85 mg Cl?

May 29, 2016

$P C {l}_{3}$

#### Explanation:

%P $=$ (10.15*mg)/((34.85+10.15)*mg)xx100% $=$ 22.6%

%Cl $=$ (34.85*mg)/((34.85+10.15)*mg)xx100% $=$ 77.4%

We use the percentages and assume $100$ $g$ of compound.

$P :$ $\frac{22.6 \cdot g}{30.9737 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.730 \cdot m o l$ $P$

$C l :$ $\frac{77.4 \cdot g}{35.45 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.18 \cdot m o l$ $C l$

We divide thru by the SMALLER molar quantity to give an empirical formula of:

$P C {l}_{3}$.