A 1-g mixture of cuprous oxide and cupric oxide was quantitatively reduced to 0.839 g of metallic copper . What was t he weight of cupric oxide in the original sample ? (Cu = 63.5, O = 16)

Explanation

Let the weight of `CuO be x g`. The weight of `Cu_(2) O` will be (1 - x) g . As the Cu atoms are conserve, applying POAC for Cu atoms,
moles of Cu in CuO + moles of Cu in `Cu_(2)O`
= moles of Cu in the product
`1 xx ` moles of `CuO + 2 xx "moles of " Cu_(2) O` = moles of Cu
`(x)/( 79 . 5) + 2 xx (1 - x)/( 143) = (0.839)/( 63.5) [{:(CuO = 79 . 5),(Cu_(2) O = 143):}] `
x = 0.55 g