Question

The molecular formula of acetylsalicylic acid (aspirin), one of the most commonly used pain relievers, is ${\mathbf{C}}_{\mathbf{9}}{\mathbf{H}}_{\mathbf{8}}{\mathbf{O}}_{\mathbf{4}}$

a. Calculate the molar mass of aspirin.

b. A typical aspirin tablet contains 500. mg${\mathbf{C}}_{\mathbf{9}}{\mathbf{H}}_{\mathbf{8}}{\mathbf{O}}_{\mathbf{4}}$. What amount (moles) of${\mathbf{C}}_{\mathbf{9}}{\mathbf{H}}_{\mathbf{8}}{\mathbf{O}}_{\mathbf{4}}$molecules and what number of molecules of acetylsalicylic acid are in a 500. mg tablet?

Short answer

a. The aspirin molar mass is $180.15{\mathrm{gmol}}^{-1}$.

b. The molecules of acetylsalicylic acid are$1.671\times {10}^{21}\mathrm{molecules}.$

Step-by-step solution

Molar mass and the number of molecules

The molar mass of a compound is the sum of the atomic masses of each constituent element in it.

The number of molecules present in the given amount of a compound depends on its sample mass and the molar mass of the compound, and Avogadro’s number.

$\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{molecules}\mathbf{=}{\mathbf{N}}_{\mathbf{A}}\mathbf{\times }\frac{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{sample}}{\mathbf{its}\mathbf{}\mathbf{molar}\mathbf{}\mathbf{mass}}$

The ${\mathbf{N}}_{\mathbf{A}}$ represents Avogadro’s number.

a. Calculation of aspirin molar mass

The molar mass is given as:

$$\begin{gathered} \mathrm{Molar}\mathrm{mass}=9\times \left(12.01\mathrm{g}/\mathrm{mol}\right)+8\times \left(1.008\mathrm{g}/\mathrm{mol}\right)+4\times \left(16.00\mathrm{g}/\mathrm{mol}\right) \\ \mathrm{Molar}\mathrm{mass}=\left(108.09+8.064+64\right)\mathrm{g}/\mathrm{mol} \\ \mathrm{Molar}\mathrm{mass}=180.15\mathrm{g}/\mathrm{mol} \end{gathered}$$

b. Calculation of molecules

The number of moles in 500.0 mg as:

$$\begin{gathered} {\mathrm{n}}_{\mathrm{molecules}}=\left(500.0\mathrm{mg}\times \frac{1\mathrm{g}}{{10}^3\mathrm{mg}}\right)\times \frac{1{\mathrm{molC}}_9{\mathrm{H}}_8{\mathrm{O}}_4}{180.15\mathrm{g}} \\ =2.775\times {10}^{-3}{\mathrm{molC}}_9{\mathrm{H}}_8{\mathrm{O}}_4 \end{gathered}$$

Hence, the number of molecules as:

$$\begin{gathered} {\mathrm{n}}_{\mathrm{molecules}}=\left(2.775\times {10}^{-3}{\mathrm{molC}}_9{\mathrm{H}}_8{\mathrm{O}}_4\right)\times \frac{6.022\times {10}^{23}\mathrm{molecules}}{1{\mathrm{molC}}_9{\mathrm{H}}_8{\mathrm{O}}_4} \\ =1.671\times {10}^{21}\mathrm{molecules} \end{gathered}$$