Answer
a. $2.74*10^{21}\text{ molecules $P_4O_6$}$
b. $1.94*10^{21}\text{ molecules $Ca_3(PO_4)_2$}$
c. $4.24*10^{21}\text{ molecules $Na_2HPO_4$}$
Work Step by Step
a. $1.00 \text{ grams $P_4O_6$}*\frac{1\text{ mol $P_4O_6$}}{219.8914\text{ grams $P_4O_6$}}*\frac{6.0221*10^{23}\text{ molecules $P_4O_6$}}{1\text{ mol $P_4O_6$}}=2.74*10^{21}\text{ molecules $P_4O_6$}$
b. $1.00 \text{ grams $Ca_3(PO_4)_2$}*\frac{1\text{ mol $Ca_3(PO_4)_2$}}{310.1767\text{ grams $Ca_3(PO_4)_2$}}*\frac{6.0221*10^{23}\text{ molecules $Ca_3(PO_4)_2$}}{1\text{ mol $Ca_3(PO_4)_2$}}=1.94*10^{21}\text{ molecules $Ca_3(PO_4)_2$}$
c. $1.00 \text{ grams $Na_2HPO_4$}*\frac{1\text{ mol $Na_2HPO_4$}}{141.9588\text{ grams $Na_2HPO_4$}}*\frac{6.0221*10^{23}\text{ molecules $Na_2HPO_4$}}{1\text{ mol $Na_2HPO_4$}}=4.24*10^{21}\text{ molecules $Na_2HPO_4$}$