Question

An oxide of copper contains 2.118 g of copper and 0.2666 g of oxygen. Another oxide of copper contains the same mass of copper but 0.5332 g of copper. What are the empirical formulas of the oxides?

Answer 1

The empirical formulas are `"Cu"_2"O"` and `"CuO"`.

Oxide 1

`color(white)(mmmmml)"Cu" +color(white)(m) "O" → "Oxide 1"`
`"Mass/g":color(white)(l) 2.118color(white)(ll) 0.2666`

Our job is to calculate the ratio of the moles of each element.

`"Moles of Cu" = 2.118 color(red)(cancel(color(black)("g Cul"))) × "1 mol Cu"/(63.55 color(red)(cancel(color(black)("g Cu")))) = "0.033 28 mol Cu"`

`"Moles of O" = 0.2666 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "0.016 66 mol O"`

To get the molar ratio, we divide each number of moles by the smaller number (`"0.01 666"`).

From here on, I like to summarize the calculations in a table.

`"Element"color(white)(X) "Mass/g"color(white)(X) "Moles"color(white)(XXl) "Ratio"color(white)(mll)"Integers"`
`stackrel(—————————————————-——)(color(white)(l)"Cu" color(white)(XXXX)2.118 color(white)(Xl)"0.033 28" color(white)(Xll)1.997color(white)(Xmm)2`
`color(white)(ll)"O" color(white)(XXXXl)0.2666 color(white)(ll)"0.016 66" color(white)(Xll)1 color(white)(XXXmlll)1`

The empirical formula of Oxide 1 is `"Cu"_2"O"`.

Oxide 2

`color(white)(mmmmml)"Cu" +color(white)(m) "O" → "Oxide 2"`
`"Mass/g":color(white)(l) 2.118color(white)(ll) 0.5332`

Working as before, we get the table.

`"Element"color(white)(X) "Mass/g"color(white)(X) "Moles"color(white)(XXl) "Ratio"color(white)(mll)"Integers"`
`stackrel(—————————————————-——)(color(white)(l)"Cu" color(white)(XXXX)2.118 color(white)(Xl)"0.033 28" color(white)(Xll)1color(white)(Xmmmm)2`
`color(white)(l)"O" color(white)(XXXXl)0.5332 color(white)(m)"0.033 32" color(white)(Xll)1.002 color(white)(XXX)1`

The ratio comes out as `"Cu:O"= 1:1`.

The empirical formula of Oxide 2 is `"CuO"`.