# Question #7fb69

##### 1 Answer

#### Explanation:

The idea here is that you need to use the pecent composition of copper in *copper(I) oxide*, *copper(II) oxide*,

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

#"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"#

#"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 + y = "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 * x + 0.799 * y = "8.22 g"#

This system of equations can be solved by substitution

#x = 9.80 - y#

#0.888 * (9.80 - y) + 0.799 * y = 8.22#

#8.702 - 0.888 * y + 0.799 * y = 8.22#

#0.089 * y = 0.482 implies y = "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")#