Answer
$Pb_{3}(PO_{4})_{2}$
Pb = 76.6%
P = 7.63%
O = 15.77%
Work Step by Step
We can tell from the name that the compound is made of lead(II) and phosphate ions. Since phosphate ($PO_{4}^{-3}$) has a $-3$ charge and lead(II) ($Pb^{+2}$) has a $+2$ charge then we need 3 lead(II) and 2 phosphate to balance the charge. Therefore the formula is $Pb_{3}(PO_{4})_{2}$. Adding up the molar masses of the constituent elements we get $811.543\frac{g}{mol}$. So to find the percent composition we just multiply the molar mass of each element (found on the periodic table) by how many times it occurs, divide by the molar mass of the whole molecule and multiply by 100%. Here is $O$ as an example:
$$\frac{8(15.999\frac{g}{mol})}{811.543\frac{g}{mol}}\times100\%= 15.77\%$$
