= Ω g Cu x 63.5 g 1 mole x 2 mol Cu 1 mol Cu 2 O x 1mol g Ω g Cu 2 O = Ω g Cux 63.5 g 1 mole x 1 mol Cu 1 mol CuO x 1mol 79.5 g (8.828 – Ω) g CuO Ω g Ω g= total grams Cu Chemistry 11 Challenge Question A mixture of Cu 2 O 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 Cu 2 O Cu 2 O Ω gCuO Ω = total grams Cu (8.828 – Ω) g

Grams Cu=Grams Cu 3 sig figs due to the molar masses! – Ω = mass CuO = g Ω = mass Cu 2 O =1.823 g Ω = Ω = % CuO = g x 100 % g = 79.3 % Do not round until the end!

= Ω g Ni x 58.7 g 1 mole x 1 mol Ni 1 mol NiSO 4 x 1mol g Ω g NiSO 4 = Ω g Nix 58.7 g 1 mole x 2 mol Ni 1 mol Ni 2 (SO 4 ) 3 x 1mol g (24.44 – Ω) g Ni 2 (SO 4 ) Ω g Ω g= total grams Ni Let Ω equal the mass of NiSO 4 NiSO 4 Ω g Ni 2 (SO 4 ) Ω = total grams Ni (24.44 – Ω) g 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?

Grams Ni=Grams Ni A has 3 sig figs - molar masses! Ω = mass NiSO 4 =9.77 g A = Ω = – Ω = mass Ni 2 (SO 4 ) 3 = g Ni 2 (SO 4 ) 3 has 4 sig figs! Round to 2 nd decimal