We can do this by dividing the mass of each compound by its molar mass.
For $\mathrm{CaCO}_{3}$, we have:
\[
\frac{350 \, \mathrm{mg}}{100.09 \, \mathrm{g/mol}} = 3.50 \times 10^{-3} \, \mathrm{mol}
\]
For $\mathrm{SrCO}_{3}$, we have:
\[
\frac{350 \,
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