Answer
There are $1.60\times10^{23}$ O atoms in 12.0g asprin
There are $1.86\times10^{23}$ O atoms in 12.0g Ca$_3$(PO$_4$)$_2$
Work Step by Step
Calculate the molar mass of each compound, and use this to determine how many moles there are in each sample. Then use Avogadro's number to determine how many formula units (molecules) there are of that compound. Finally, multiply by the number of Oxygen atoms in one formula unit of that compound.
Apsrin, C$_9$H$_8$O$_4$: Molar mass is $12.0107 \times 9 + 1.0079 \times 8 + 15.9994 \times 4=180.1571$ g/mol
$12.0g \times \frac{1 mol}{180.1571g} \times \frac{6.022\times10^{23} molecules}{1 mol}\times\frac{4\space O\space atoms}{1 molecule}=1.60\times10^{23}$ O atoms
Ca$_3$(PO$_4$)$_2$: Expanding the parentheses there are 2 P atoms and 8 O atoms per formula unit. Molar mass is $40.078\times3+30.9738\times2+15.9994\times8=310.177$ g/mol
$12.0g\times\frac{1mol}{310.177g}\times\frac{6.022\times10^{23}units}{1mol}\times\frac{8\space O\space atoms}{1 unit}=1.86\times10^{23}$ O atoms.
Round each to 3 significant figures, the lowest number of significant figures in any non-exact number.
