# 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?

Feb 26, 2016

The empirical formulas are $\text{Cu"_2"O}$ and $\text{CuO}$.

Oxide 1

$\textcolor{w h i t e}{m m m m m l} \text{Cu" +color(white)(m) "O" → "Oxide 1}$
$\text{Mass/g} : \textcolor{w h i t e}{l} 2.118 \textcolor{w h i t e}{l l} 0.2666$

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

$\text{Moles of Cu" = 2.118 color(red)(cancel(color(black)("g Cul"))) × "1 mol Cu"/(63.55 color(red)(cancel(color(black)("g Cu")))) = "0.033 28 mol Cu}$

$\text{Moles of O" = 0.2666 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "0.016 66 mol O}$

To get the molar ratio, we divide each number of moles by the smaller number ($\text{0.01 666}$).

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

$\text{Element"color(white)(X) "Mass/g"color(white)(X) "Moles"color(white)(XXl) "Ratio"color(white)(mll)"Integers}$
stackrel(—————————————————-——)(color(white)(l)"Cu" color(white)(XXXX)2.118 color(white)(Xl)"0.033 28" color(white)(Xll)1.997color(white)(Xmm)2
$\textcolor{w h i t e}{l l} \text{O" color(white)(XXXXl)0.2666 color(white)(ll)"0.016 66} \textcolor{w h i t e}{X l l} 1 \textcolor{w h i t e}{X X X m l l l} 1$

The empirical formula of Oxide 1 is $\text{Cu"_2"O}$.

Oxide 2

$\textcolor{w h i t e}{m m m m m l} \text{Cu" +color(white)(m) "O" → "Oxide 2}$
$\text{Mass/g} : \textcolor{w h i t e}{l} 2.118 \textcolor{w h i t e}{l l} 0.5332$

Working as before, we get the table.

$\text{Element"color(white)(X) "Mass/g"color(white)(X) "Moles"color(white)(XXl) "Ratio"color(white)(mll)"Integers}$
stackrel(—————————————————-——)(color(white)(l)"Cu" color(white)(XXXX)2.118 color(white)(Xl)"0.033 28" color(white)(Xll)1color(white)(Xmmmm)2
$\textcolor{w h i t e}{l} \text{O" color(white)(XXXXl)0.5332 color(white)(m)"0.033 32} \textcolor{w h i t e}{X l l} 1.002 \textcolor{w h i t e}{X X X} 1$

The ratio comes out as $\text{Cu:O} = 1 : 1$.

The empirical formula of Oxide 2 is $\text{CuO}$.