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"#.
