Question

Determine the mass of 5.2 x 10 power of 21 molecules of propanol C3H7OH(l), on grams.

Answer 1

$n(\text{C}_3\text{H}_7\text{OH}) = N(\text{C}_3\text{H}_7\text{OH}) / N_A\\ \phantom{n(\text{C}_3\text{H}_7\text{OH})} = 5.2 \times 10^{21} / (6.02 \times 10^{23})\\ \phantom{n(\text{C}_3\text{H}_7\text{OH})} = 8.6 \times 10^{-3} \; \text{mol}$

where $N_A = 6.02 \times 10^{23}\; \text{mol}^{-1}$ the Avogadro's constant that relates the number of particles to their number, in the unit moles $\text{mol}$.

The molar mass of propanol- mass per mole propanol- can be directly deduced from its molecular formula with reference to a modern periodic table.

$M(\text{C}_3\text{H}_7\text{OH}) = \underbrace{3 \times 12.01}_{\text{carbon}} + \underbrace{8 \times 1.008}_{\text{hydrogen}} + \underbrace{1\times 16.00}_{\text{oxygen}} = 60.09 \; \text{g} \cdot \text{mol}^{-1}$

$8.6 \times 10^{-3} \; \text{mol}$ of propanol molecules would thus have a mass of $8.6 \times 10^{-3} \; \text{mol} \times 60.09 \; \text{g} \cdot \text{mol}^{-1} = 0.52 \; \text{g}$