Question

[Chemistry]

A sample of mixture containing only CuO & Cu2O has mass of 1.512g. When this sample reacted with excess hydrogen gas (H2), 1.275g of copper metal (CU(s)) formed.

What is the mass percent of CuO & Cu2O in the original mixture?

I know the equation can be set up as this:

CuO + Cu2O + H2 -----> Cu(s) ......

MM CuO = 143.10g CuO / mol CuO
MM Cu2O = 79.55g Cu2O / mol Cu2O
MM Cu(s) = 63.55g Cu / mol Cu

I let "x" represent the mol of CuO
I let "y" represent the mol of Cu2O

The problem I am having is cancelling out the units in the first equation. This is how I thought it would be:

Answer 1

[1]

x mol CuO + y mol Cu2O = 1.512 g (CuO + Cu2O)
x mol CuO (75.99g CuO / mol CuO) + y mol Cu2O (143.10g Cu2O / mol Cu2O) = 1.512g (CuO + Cu2O)

75.99x g CuO + 143.10y gCu2O = 1.512 g(CuO + Cu2O)

I wasn't able to cancel out the units here. How would you do it?


But for the second equation,

[2] I know that all of the hydrogen gas reacted with the two copper compounds and still remained, so that means all the copper transferred from the two compounds to that one metallic solid form.


molCu (s) = 1.275g Cu(s)*(1 mol Cu(s) / 63.55g Cu(s) )
molCu (s) = 0.02mol Cu(s)


x mol Cu + 2y mol Cu = 0.02 mol Cu
Here I was able to cancel out the units:

mol Cu (x + 2y) = (0.02) mol Cu
x + 2y = 0.02

But the real problem is the first equation. How would I go about with canceling the units in the first equation?

because:

x mol CuO * (1 mol Cu /1 mol CuO)

y mol Cu2O * (2 mol Cu / 1 mol Cu2O)