Question

A student started with 1.409 grams of copper (II) oxide and produced 3.500 g of copper (II) sulfate. What is the percent yield? The product is a hydrate. Use appropriate significant figures. Do not put a percent sign in the answer box. Please ignore extra zeros that will be automatically added to your answer by the system.

Useful information:

chemical equation: CuO(s) + H₂SO₄ (aq) ---> CuSO₄ (aq) +H₂O(l)

Formula weight of hydrated copper (II) sulfate = 249.677 g/mol

Formula weight of copper (II) oxide = 79.545 g/mol

Answer 1

Consider a reaction , CuO (s) + H₂SO₄ (aq) $\rightarrow$ Hyd. CuSO ₄ (aq) + H₂O (l)

From reaction, we can write 1 mol CuO $\equiv$ 1 mol Hyd. CuSO ₄

$\therefore$ 79.545 g CuO $\equiv$ 249.677 g Hyd. CuSO ₄

$\therefore$ 1.409 g CuO $\equiv$ 249.677 $\times$ 1.409 / 79.545 g Hyd. CuSO ₄

1.409 g CuO $\equiv$ 4.4225 g Hyd. CuSO ₄

Theoretical yield of Hyd. CuSO ₄ = 4.4225 g

Actual yield of Hyd. CuSO ₄ = 3.500 g

% yield of Hyd. CuSO ₄

We have, % yield = [ Actual yield of product / Theoretical yield of product ] 100

$\therefore$ % yield of Hyd. CuSO ₄ = [ 3.500 g / 4.4225 g ] 100

= 79.14 %

ANSWER : Percent yield = 79.14