CALCULATING A YIELD IN GRAMS, GIVEN A PERCENT YIELD Diethyl ether (C4H10O), the “ether” used medically as an anesthetic, is prepared commercially by treatment of ethyl alcohol (C2H6O) with an acid. How many grams of diethyl ether would you obtain from 40.0 g of ethyl alcohol if the percent yield of

Question

CALCULATING A YIELD IN GRAMS, GIVEN A PERCENT YIELD

Diethyl ether ([latex]C_{4}H_{10}O[/latex]), the “ether” used medically as an anesthetic, is prepared commercially by treatment of ethyl alcohol ([latex]C_{2}H_{6}O[/latex]) 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%?

[latex]\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)[/latex]

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.

Accepted Answer

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

Ethylalcohol, [latex]C_{2}H_{6}O[/latex]: 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, [latex]C_{4}H_{10}O[/latex]: 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:

[latex]40.0 \cancel{g ethyl alcohol} × \frac{1 mol ethyl alcohol}{46.0 \cancel{g ethyl alcohol}}[/latex] = 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:

[latex]0.870 \cancel{mol ethyl alcohol} × \frac{1 mol diethyl ether}{2 \cancel{mol ethyl alcohol}}[/latex] = 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:

[latex]0.435 \cancel{mol diethyl ether} × \frac{74.0 g diethyl ether}{1 \cancel{mol diethyl ether}}[/latex] = 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

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

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