Question 3.9: CALCULATING A YIELD IN GRAMS, GIVEN A PERCENT YIELD Diethyl ...

CALCULATING A YIELD IN GRAMS, GIVEN A PERCENT YIELD

Diethyl ether (C_{4}H_{10}O), the “ether” used medically as an anesthetic, is prepared commercially by treatment of ethyl alcohol (C_{2}H_{6}O) with an acid. How many grams of diethyl ether would you obtain from 40.0 g of ethyl alcohol if the percent yield of the reaction is 87%?

\underset{Ethyl  alcohol}{2  C_{2}H_{6}O(l)}  \underrightarrow{Acid}  \underset{Diethyl  ether}{C_{4}H_{10}O(l)}  +  H_{2}O(l)

STRATEGY
Treat this as a typical stoichiometry problem to find the amount of diethyl ether that can theoretically be formed from 40.0 g of ethyl alcohol, and then multiply the answer by 87% to find the amount actually formed.

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First, calculate the molar masses of the reactant and product:

Ethylalcohol, C_{2}H_{6}O: Molec. mass = (2 × 12.0 amu) + (6 × 1.0 amu) + 16.0 amu

= 46.0 amu

Molar mass of ethyl alcohol = 46.0 g/mol

Diethyl ether, C_{4}H_{10}O: Molec.mass = (4 × 12.0 amu) + (10 × 1.0 amu) + 16.0 amu

= 74.0 amu

Molar mass of diethyl ether = 74.0 g/mol

Next, find how many moles of ethyl alcohol are in 40.0 g by using molar mass as a conversion factor:

40.0  \cancel{g  ethyl  alcohol}  ×  \frac{1  mol  ethyl  alcohol}{46.0  \cancel{g  ethyl  alcohol}} = 0.870 mol ethyl alcohol

Because we started with 0.870 mol of ethyl alcohol, and because the balanced equation indicates that 2 mol of ethyl alcohol yield 1 mol of diethyl ether, we can theoretically obtain 0.435 mol of product:

0.870  \cancel{mol  ethyl  alcohol}  ×  \frac{1  mol  diethyl  ether}{2  \cancel{mol  ethyl  alcohol}} = 0.435 mol diethyl ether

We therefore need to find how many grams of diethyl ether are in 0.435 mol, using molar mass as the conversion factor:

0.435  \cancel{mol  diethyl  ether}  ×  \frac{74.0  g  diethyl  ether}{1  \cancel{mol  diethyl  ether}} = 32.2 g diethyl ether

Finally, we have to multiply the theoretical amount of product by the observed yield (87% = 0.87 ) to find how much diethyl ether is actually formed:

32.2 g diethyl ether × 0.87 = 28 g diethyl ether

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