# Question 7fb69

Oct 10, 2015

$\text{5.42 g}$

#### Explanation:

The idea here is that you need to use the pecent composition of copper in copper(I) oxide, $\text{Cu"_2"O}$, and copper(II) oxide, $\text{CuO}$, to write a relationship between the mass of the mixture and the total mass of the copper in this mixture.

So, use the molar mass of copper and those of copper(I) oxide and copper(II) oxide, respectively, to find the percent compositions of copper in the two oxides

$\text{For Cu"_2"O: " ( 2 xx 63.55color(red)(cancel(color(black)("g/mol"))))/(143.1color(red)(cancel(color(black)("g/mol")))) xx 100 = "88.8% Cu}$

$\text{For CuO: " (63.55color(red)(cancel(color(black)("g/mol"))))/(79.55color(red)(cancel(color(black)("g/mol")))) xx 100 = "79.9% Cu}$

So, let's say that $x$ is the mass of $\text{Cu"_2"O}$ and $y$ is the mass of $\text{CuO}$ in the mixture. You know that

$x + y = \text{9.80 g}$

Now, use the percent compositions of copper to write a second equation using the mass of copper you know the mixture contains

$0.888 \cdot x + 0.799 \cdot y = \text{8.22 g}$

This system of equations can be solved by substitution

$x = 9.80 - y$

$0.888 \cdot \left(9.80 - y\right) + 0.799 \cdot y = 8.22$

$8.702 - 0.888 \cdot y + 0.799 \cdot y = 8.22$

$0.089 \cdot y = 0.482 \implies y = \text{5.4157 g}$

The mass of copper(II) oxide in the mixture is thus equal to - rounded to three sig figs

m_"CuO" = color(green)("5.42 g")#