Calculate molecular weight and the percent composition

Calculate the mass (in gr) of $Pb_3(PO_4)_3$ that contains $0.3gr$ of $Pb$. Then calculate the mass of $P$ contained in the same quantity of $Pb_3(PO_4)_3$. What is the percentage of oxygen?

SOLUTION:
Remember the atomic weight of the single atoms:
$\left\{\begin{matrix} Pb=207.2\\ P=31\\ O=16\\ \end{matrix}\right.$

The molecular weight of $Pb_3(PO_4)_3$ is:
$MW=207.2\cdot 3+\left ( 31+16\cdot 4 \right )\cdot 2=811.6gr/mol$
So one mole of $Pb_3(PO_4)_3$ has a mass of $811.6gr$.

The percentage of $Pb$ is:
$\%Pb=\frac{3\cdot 207.2gr/mol}{\displaystyle 811.6gr/mol}\cdot 100=76.59\%$
So, in $100gr$ of $Pb_3(PO_4)_3$ there are $76.59gr$ of $Pb$; $X$ grams of $Pb_3(PO_4)_3$ contains $0.3gr$ of $Pb$. Then:
$100gr:76.59gr=X:0.3gr\Rightarrow X=\frac{100gr\cdot 0.3gr}{76.59gr}=0.392gr=392mg$

The percentage of $P$ is:
$\%Pb=\frac{2\cdot 31gr/mol}{811.6gr/mol}\cdot 100=7.64\%$
So, in $100gr$ of $Pb_3(PO_4)_3$ there are $7.64gr$ of $P$; $0.392gr$ of $Pb_3(PO_4)_3$ contains $X$ grams of $P$. Then:
$100gr:7.64gr=0.392gr:X\Rightarrow X=\frac{7.64gr\cdot 0.392gr}{100gr}=0.0299gr=29.9mg$

The percentage of oxygen is:
$\%O=\frac{4\cdot 2\cdot 16gr/mol}{811.6gr/mol}\cdot 100=15.7\%$

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PROBLEM 2:
Calculate the mass of barium sulfate that contains 1.2gr of sulfur. Then compute the percentage of oxygen.