Calculate the percentage composition (by mass) of the mixture?

A 1.500 g sample of a mixture containing only Cu2O and CuO was treated with hydrogen to produce 1.252g pure copper metal. Calculate the percentage composition (by mass) of the mixtureCalculate the percentage composition (by mass) of the mixture?
What's the matter, Jay. Does your teacher not like you any more. (Just kidding.) This was a slightly ugly problem. There are multiple steps, so stick with me. If there's something you don't follow, be sure to ask.



Givens:

mass of Cu2O + mass of CuO = 1.500 grams

mass of copper metal formed = 1.252 grams Cu

moles of copper metal formed = 0.0197 moles Cu



Define variables:

x moles Cu2O produces 2x moles Cu

y moles CuO produces y moles Cu

therefore 2x + y = 0.0197 mol Cu

and y = 0.0197 - 2x



Masses of each compound in mixture:

x moles Cu2O x 143.0g Cu2O / 1 mol Cu2O = 143x g Cu2O

y moles CuO x 79.5g CuO / 1 mol CuO = 79.5y g CuO



Total mass of mixture:

143x + 79.5y = 1.50 grams of the mixture



Substitute for y (recall y = 0.0197 - 2x):

143x + 79.5(0.0197 - 2x) = 1.50 g



Solve for moles:

143x +1.566 - 159x = 1.50 g

-16x = -0.066

x = 0.004125 moles Cu2O

y = 0.0197 - 2x = 0.01145 moles CuO



Convert moles to grams:

0.004125 mol Cu2O x 143.0 g Cu2O/1 mol Cu2O = 0.590 g Cu2O

0.01145 mole CuO x 79.5 g CuO/1 mol CuO = 0.910 g CuO



Now check the masses:

0.590 g Cu2O + 0.910 g CuO = 1.500 g of mixture



Compute percent composition:

0.590 g Cu2O / 1.500 g x 100 = 39.3% Cu2O

0.910 g CuO / 1.500 g x 100 = 60.7% CuO

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