1. Chemistry 11 Challenge Question
A mixture of Cu2O and CuO of mass g is reduced to copper metal with hydrogen. If the mass of pure copper isolated was g, determine the percent (by mass) of CuO in the original sample.
Let Ω equal the mass of Cu2O
(8.828 – Ω) g
Cu2O Ω g
CuO
Ω g Cu2O
x 1mol
x 2 mol Cu
x g
= Ω g Cu
143.0 g
1 mol Cu2O
1 mole
(8.828 – Ω) g CuO
x 1mol
x 1 mol Cu
x g
= Ω g Cu
79.5 g
1 mol CuO
1 mole
Ω g Ω g
= total grams Cu
Ω = total grams Cu

2. Grams Cu = Grams Cu Ω =
Ω =
Ω = mass Cu2O = g
– Ω = mass CuO = g
Do not round until the end!
% CuO = g
x 100 %
= %
8.828 g
3 sig figs due to the molar masses!

3. A container of nickel II sulphate has been accidentally contaminated with
nickel III sulphate. The total mass of both sulphates was g. Through a single replacement reaction with Zn, the nickel was extracted from both sulphates and was found to have a mass of g. What was the original masses of the nickel II sulphate and nickel III sulphate before they were mixed?
Let Ω equal the mass of NiSO4
NiSO4 Ω g
Ni2(SO4)3
(24.44 – Ω) g
Ω g NiSO4
x 1mol
x 1 mol Ni
x g
= Ω g Ni
154.8 g
1 mol NiSO4
1 mole
(24.44 – Ω) g Ni2(SO4)3
x 1mol
x 2 mol Ni
x g
= Ω g Ni
405.7 g
1 mol Ni2(SO4)3
1 mole
Ω g Ω g
= total grams Ni
Ω = total grams Ni

4. 2 4 . 4 4 - 9. 7 7 1 4 . 6 7 Round to 2nd decimal Grams Ni = Grams NiΩ =
A =
Ω = mass NiSO4 = 9.77 g
– Ω = mass Ni2(SO4)3 = g
Round to 2nd decimal
A has 3 sig figs - molar masses!
Ni2(SO4)3 has 4 sig figs!