Q156CP - Chemistry: Molecular Nature Of Matter And Change

Question

Aspirin (acetylsalicylic acid, C₉H₈O₄) is made by reacting salicylic acid (C₇H₆O₃) with acetic anhydride [(CH₃CO)₂O]:

${\mathit{C}}_{\mathbf{7}}{\mathit{H}}_{\mathbf{6}}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathit{S}\mathbf{\right)}\mathbf{+}{\mathbf{\left(}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{O}\mathbf{\right)}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{I}\mathbf{\right)}\mathbf{\to }{\mathit{C}}_{\mathbf{9}}{\mathit{H}}_{\mathbf{8}}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathit{S}\mathbf{\right)}\mathbf{+}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{O}\mathit{O}\mathit{H}\mathbf{\left(}\mathit{I}\mathbf{\right)}$

In one reaction, 3.077 g of salicylic acid and 5.50 mL of acetic anhydride react to form 3.281 g of aspirin. (a) Which is the limiting reactant (d of acetic anhydride = 1.080 g/mL)? (b) What is the percent yield of this reaction? (c) What is the percent atom economy of this reaction?

Step-by-Step Solution

Answer

1. The limiting reactant is Salicylic acid
2. The percent yield of this reaction is 81.74%.
3. The percent atom economy of this reaction is 75%.

Step 1: Finding the limiting reactant

On multiplying the given volume to its density,

$Mass of {\left(C{H}_{3}CO\right)}_{2}O=5.5mL{\left(C{H}_{3}CO\right)}_{2}O\times \frac{1.080g(CH}{}$₃CO)₂O1mL(CH₃CO)₂O Mass of (CH3CO)2O=5.94gMoles of aspirin=3.077gC7H6O3×1molC7H6O3138.12gC7H6O3×1molC9H8O41molC7H6O3Moles of aspirinfromC7H6O3=0.02228molC9H8O4Moles of aspirinfrom(CH3CO)2O=5.94g(CH3CO)2O×1mol(CH3CO)2O138.12g(CH3CO)2O×1molC9H8O41mol(CH3CO)2OMoles of aspirinfrom(CH3CO)2O=0.05818molC9H8O4

The limiting reactant is salicylic acid due to their less product amount compared to acetic anhydride.

Step 2: Finding the percent yield

To find the percent yield,

$$\begin{gathered} 0.0228mol{C}_9{H}_8{O}_4\times \frac{180.16g{C}_9{H}_8{O}_4}{1mol{C}_9{H}_8{O}_4}=4.014g{C}_9{H}_8{O}_4 \\ \%yield=\frac{actual yield}{theoretical yield}\times 100 \\ \%yield=\frac{3.281g}{4.014g}\times 100 \\ \%yield=81.74\% \end{gathered}$$

Step 3: Finding the percent atom economy

The percent atom economy can be derived as,

$$\begin{gathered} \%atom economy=\frac{Number of moles\times molar mass of desiredproduct}{sum of number of moles of products\times molar mass for all products}\times 100 \\ =\frac{{\left(1mol\times {\displaystyle \frac{180.16g}{1mol}}\right)}_{C_9{H}_8{O}_4}}{{\left(1mol\times \frac{180.16g}{1mol}\right)}_{C_9{H}_8{O}_4}+{\left(1mol\times \frac{180.16g}{1mol}\right)}_{C{H}_{3}COOH}}\times 100 \\ \%atomeconomy=75\% \end{gathered}$$