When 12 g of methanol (CH3OH) was treated with excess oxidizing agent (MnO4-), 14 g of formic acid (HCOOH) was obtained. Using the following chemical equation, calculate the percent yield. (The reaction is much more complex than this; please ignore the fact that the charges do not balance.) 3CH3OH + 4MnO4- > 3HCOOH + 4MnO2
(a) 100%
(b) 92%
(c) 82%
(d) 70%
(e) 55%

(c) 82%

Explanation:

Using the chemical reaction:

3CH₃OH + 4MnO₄⁻ → 3HCOOH + 4MnO₂ + 4e⁻

3 moles of methanol produce 3 moles of formic acid (Reaction 1:1)

To find the percent yield we need to calculate theoretical yield of the reaction when 12g of methanol (Molar mass: 32.04g/mol) are reacting, thus:

12g ₓ (1mol / 32.04g) = 0.3745 moles of methanol are in 12g.

As 1 mole of methanol produce 1 mole of formic acid, assuming a 100% of yield in the reaction (Theoretical yield) moles produced of formic acid are 0.3745 moles.

Using molar mass of formic acid (46.03g/mol), mass of the acid in 0.3745moles are:

0.3745 moles ₓ (46.03g/mol) = 17.2g of formic acid are theoretical yield

Percent yield is calculated as:

(actual yield / Theoretical yield) × 100

As actual yield of the reaction is 14g. Theoretical yield is:

(14g / 17.2g) × 100 = 81.2% ≈ 82%

Right answer is:

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