Question

Find the mass of 3.27 x 10^23 molecules of H2SO4. Use 3 significant digits and put the units.

Answer 1

Answer:

Approximately $53.3\; \rm g$.

Explanation:

Lookup Avogadro's Number: $N_{\rm A} = 6.02\times 10^{23}\; \rm mol^{-1}$ (three significant figures.)

Lookup the relative atomic mass of $\rm H$, $\rm S$, and $\rm O$ on a modern periodic table:

• $\rm H$: $1.008$.
• $\rm S$: $32.06$.
• $\rm O$: $15.999$.

(For example, the relative atomic mass of $\rm H$ is $1.008$ means that the mass of one mole of $\rm H\!$ atoms would be approximately $1.008\!$ grams on average.)

The question counted the number of $\rm H_2SO_4$ molecules without using any unit. Avogadro's Number $N_{\rm A}$ helps convert the unit of that count to moles.

Each mole of $\rm H_2SO_4$ molecules includes exactly $(1\; {\rm mol} \times N_\text{A}) \approx 6.02\times 10^{23}$ of these $\rm H_2SO_4 \!$ molecules.

$3.27 \times 10^{23}$ $\rm H_2SO_4$ molecules would correspond to $\displaystyle n = \frac{N}{N_{\rm A}} \approx \frac{3.27 \times 10^{23}}{6.02 \times 10^{23}\; \rm mol^{-1}} \approx 0.541389\; \rm mol$ of such molecules.

(Keep more significant figures than required during intermediary steps.)

The formula mass of $\rm H_2SO_4$ gives the mass of each mole of $\rm H_2SO_4\!$ molecules. The value of the formula mass could be calculated using the relative atomic mass of each element:

$\begin{aligned}& M({\rm H_2SO_4}) \\ &= (2 \times 1.008 + 32.06 + 4 \times 15.999)\; \rm g \cdot mol^{-1} \\ &= 98.702\; \rm g \cdot mol^{-1}\end{aligned}$.

Calculate the mass of approximately $0.541389\; \rm mol$ of $\rm H_2SO_4$:

$\begin{aligned}m &= n \cdot M \\ &\approx 0.541389\; \rm mol \times 98.702\; \rm g \cdot mol^{-1}\\ &\approx 53.3\; \rm g\end{aligned}$.

(Rounded to three significant figures.)