Answer
a.) 1.1 x $10^{22}$ atoms
b.) 3.9 x $10^{21}$ atoms
c.) 4.2 x $10^{21}$ atoms
Work Step by Step
a.) $\frac{1.0 g P_{4}O_{6}}{1}$$\times$$\frac{1 mol P_{4}O_{6}}{216.82 g P_{4}O_{6}}$$\times$$\frac{4 mol P^{3+}}{1 mol P_{4}O_{6}}$$\times$$\frac{6.022x10^{23}atoms}{1 mol P^{3+}}$
b.) $\frac{1.0 g Ca_{3}(PO_{4})_{2}}{1}$$\times$$\frac{1 mol Ca_{3}(PO_{4})_{2}}{310.1 g Ca_{3}(PO_{4})_{2}}$$\times$$\frac{2 mol P^{6+}}{1 mol Ca_{3}(PO_{4})_{2}}$$\times$$\frac{6.022x10^{23}atoms}{1 mol P^{6+}}$
c.) $\frac{1.0 g Na_{2}HPO_{4}}{1}$$\times$$\frac{1 mol Na_{2}HPO_{4}}{141.92 g Na_{2}HPO_{4}}$$\times$$\frac{1 mol P^{5+}}{1 mol Na_{2}HPO_{4}}$$\times$$\frac{6.022x10^{23}atoms}{1 mol P^{5+}}$
You must convert grams to moles. Then using mole to mole ratios within the compound, relate the compound to the phosphorus ion. Then take this to atoms using Avogadro's Number. ALWAYS pay attention to sig figs.