Question

Calculate the number of atoms of each type in 5.3g of $N{a}_{2}C{O}_3$ ?

Answer 1

Na= 23g/mol
C= 12 g/mol
O = 16g/mol
Molecular weight of the compound = 106.
So $\frac{\left(5.3gramsofN{a}_{2}C{O}_3\right)}{\left(106gramsofN{a}_{2}C{O}_3/mol\right)}=0.05moles$.
So we have 0.05 moles of $N{a}_{2}C{O}_3$ and using the Avogadro's number we have:
$6.022\times {10}^{23}$ (Number of molecules per mol) $\times 0.05moles=3.011\times {10}^{22}$ molecules of $N{a}_{2}C{O}_3$
We have 2 atoms of Na, 1 atom of C and 3 atoms of O per molecule of $N{a}_{2}C{O}_3$.
And so we have the answer
Number of Na atoms = $6.022\times {10}^{22}$ atoms
Number of C atoms = $3.011\times {10}^{22}$ atoms
Number of O atoms = $9.033\times {10}^{22}$ atoms