藉由摩爾質量 \(M\)，物質的質量 \(m\) 與它的摩爾數 \(n\) 可進行換算：

\[n = \frac{m}{M}\]

\(2.00 \text{ mol}\) 乙二酸的質量是多少？

題解：

\[\begin{align} \because \; M(\ce{C2H2O4}) & = 90.0 \text{ g mol}^{−1} \\ \therefore\; m(\ce{C2H2O4}) & = n(\ce{C2H2O4}) \times M(\ce{C2H2O4})\\ & = 2.00 \text{ mol} \times 90.0 \text{ g mol}^{−1} \\ & = 180 \text{ g} \end{align}\]

\(0.250 \text{ mol}\) 五水合硫酸銅(II)是多少克？

題解：

\[\begin{align} \because \; M(\ce{CuSO4.5H2O}) & = 249.6 \text{ g mol}^{−1} \\ \therefore\; m(\ce{CuSO4.5H2O}) & = n(\ce{CuSO4.5H2O}) \times M(\ce{CuSO4.5H2O})\\ & = 0.250 \text{ mol} \times 249.6 \text{ g mol}^{−1} \\ & = 62.4 \text{ g} \end{align}\]

\(112.0 \text{ g}\) 鐵中含有多少摩爾鐵原子？

題解：

\[\begin{align} \because \; M(\ce{Fe}) & = 55.8 \text{ g mol}^{−1} \\ \therefore\; n(\ce{Fe}) & = \frac{m(\ce{Fe})}{M(\ce{Fe})}\\ & = \frac{112.0 \text{ g}}{55.8 \text{ g mol}^{−1} }\\ & = 2.01 \text{ mol} \end{align}\]

\(56.0 \text{ g}\) 氧化鐵(III)的摩爾數是多少？

題解：

\[\begin{align} \because \; M(\ce{Fe2O3}) & = 159.6 \text{ g mol}^{−1} \\ \therefore\; n(\ce{Fe2O3}) & = \frac{m(\ce{Fe2O3})}{M(\ce{Fe2O3})}\\ & = \frac{56.0 \text{ g}}{159.6 \text{ g mol}^{−1} }\\ & = 0.351 \text{ mol} \end{align}\]