A5.0 g sample of copper(ll) oxide containing an inert contaminant is analyzed and found to contain 3.0 g copper. using molar masses of 64 g/mol for copper and 16 g/mol for oxygen, which of the following is the percent purity of the copper(ii) oxide sample? elimination tool a 80% 75% c 60% 50%

The percent purity is 75%

Explanation:

To obtain the purity of the copper (II) oxide sample there is necessary to calculate the amount of copper in sample. And then, by stoichiometry, the real amount of CuO.

The amount of copper in sample is:

3,0g × (1 mol Cu / 64 g) = 0,047 mol of Copper ≡ 0,047 mol Copper oxide

The copper moles are the same for CuO.

0,047 mol CuO × ( 80g / 1 mol CuO) = 3,75 g CuO

The real amount of copper (ll) oxide in the sample is 3,75 g.

Thus, the percent purity is:

3,75g of CuO / 5,0g of Sample × 100 = 75%

I hope it helps!

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