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enter image description here The overall yield for the synthesis of N-phenylethanamide from benzene was found to be $35.2\%$. Calculate the minimum volume of benzene, in $\mathrm{cm}^3$, required to make $10.0\, \mathrm{g}$ of N-phenylethanamide.

[Density of benzene = $0.879\, \mathrm{g\cdot cm}^{-3}$]

To work out the answer, I did the following.

  • Mr of N-phenyl... = $135\, \mathrm{g\cdot mol}^{-1}$
  • Mr of benzene = $78\, \mathrm{g\cdot mol}^{-1}$

Moles of N-phenyl... = $10/135 = 0.074074 = \mathrm{mol}$ of benzene due to 1:1 ratio

Mass of benzene = $0.074074 x 78 = 5.7778 \,\mathrm{g}$

Volume of benzene = $\frac{\mathrm{mass}}{\mathrm{density}} = \frac{5.7778}{0.879} = 6.573 \,\mathrm{cm}^3$ (This is the volume at 100% yield)

The next step is where I am unclear. I took the percentage yield equation as $(X/6.573)\times 100=35.2$ where $X$ is the volume of benzene to produce $35.2\%$ of N-phenylethanamide.

Answer = $2.31 \,\mathrm{cm}^3$.

However the mark scheme takes the volume of benzene at 100%, divides it by 35.2, then multiplies it by 100 to get $18.7 \,\mathrm{cm}^3$

What is the reason for dividing by 35.2 and multiplying by 100?

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  • $\begingroup$ Instead think of it as multiplying by 100, then dividing by 35.2. Multiplying by 100 would give the mass of benzene required if the yield was 1% (need 100 times more benzene for a 1% yield compared to a 100% yield), dividing by 35.2 then gives you the mass required with a yield of 35.2% $\endgroup$
    – Ariichu
    Commented Apr 16 at 9:08
  • $\begingroup$ It's not very clear to me why though... if volume : yield = 100% : 6.573 cm^3, then I should find for 35.2% : Volume. However, it still doesn't divide by 35.2. Also everything is in a 1:1 ratio why do I need 100 times more benene? $\endgroup$
    – Jane902
    Commented Apr 16 at 12:53

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If all the reactions were to give 100% yield then you would need $\pu{5,7778g}$ of benzene to get $\pu{10g}$ of N-phenylacetamide, but since only $\pu{35,2g}$ out of $\pu{100g}$ of benzene actually transform into N-phenylacetamide you have

$$\frac{\pu{5,7778 g of benzene transformed into product}}{\pu{X g of benzene introduced}}=\frac{35,2}{100}$$ we get $\pu{X = 16,414 g}$ of benzene (Notice how 35,2% of 16,414 is approximately 5,7778)

Because the density of benzene is $\pu{0,879\frac{g}{cm^3}}$ => $\pu{\frac{\pu{1 cm^3}}{\pu{0,879g}}}\times\pu{16,414g benzene\simeq \pu{18,7cm^3}}$

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