[UG]

Hi everyone,
I am having a bit of trouble doing this question:

A 1.50 gram sample of a mixture containing only Cu₂O and CuO was treated with hydrogen to produce 1.252 grams of pure copper metal. Calculate the percentage composition (by mass) of the mixture.

Okay, so firstly, I wrote out two balanced equations:

Cu₂O + H₂  2Cu + H₂O
CuO + H₂ Cu + H₂O

Then I calculated the amount of copper, 1.252/63.54 = 0.0197 mole

From here is where I get a bit confused, I know that m(Cu₂O) + m(CuO) = 1.50 g and I also know the molar masses of the two compounds, 143.1 g mol^-1 and 79.5 g mol^-1
So I think I should get something like:
n(Cu₂O) x M(Cu₂O) + n(CuO) + M(CuO) = 1.50

But the trouble is, I have no idea how to rearrange the 0.0197 mole and 'n' in order to substitute it into the equation 
Can someone lend me a hand?
Many thanks 

[Borek]

n(Cu₂O) x M(Cu₂O) + n(CuO) + M(CuO) = 1.50

Typo, should be

n(Cu₂O) x M(Cu₂O) + n(CuO) x M(CuO) = 1.50

Prepare for a facepalm

n(Cu₂O) + n(CuO) = 0.0197

Two equations, two unknowns.

[UG]

n(Cu₂O) + n(CuO) = 0.0197

Does it?
But copper (I) oxide produces two moles of Cu atoms 

Cu₂O + H₂ 2Cu + H₂O

Or am I just getting confused...

[Borek]

Facepalm :/

2 n(Cu2O) + n(CuO) = 0.0197

There is nothing like to make an idiot out of oneself first thing in the morning, it helps keep proportions for the rest of the day.

The general idea - of getting two equations in two unknowns - remains correct.

[UG]

Cool, thanks Borek