Answer:
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!
