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

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Answer 1 · 2016-05-29T19:39:53

$PCl_3$

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

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

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

$P:$ $(22.6*g)/(30.9737*g*mol^-1)$ $=$ $0.730*mol$ $P$

$Cl:$ $(77.4*g)/(35.45*g*mol^-1)$ $=$ $2.18*mol$ $Cl$

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

$PCl_3$ .

What is the empirical formula for a compound that is found to contain 10.15 mg PP and 34.85 mg ClCl?