# Minimum volume of benzene with percentage yield

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?

• 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% Commented Apr 16 at 9:08
• 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? Commented Apr 16 at 12:53

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