Question

A compound in a copper ore has the following percentage composition by mass:
55.6% copper, 16.4% iron, 28.0% sulfur.
Calculate the empirical formula of the compound.
Relative atomic masses (Ar): S = 32; Fe = 56; Cu = 63.5
You must show all of your working.
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please show all working, thanks x

Answer 1

1) since we are given percentages, we can assume we have 100 grams of the molecule.

55.6 % Cu ----> 55.6 grams Cu
16.4 % Fe------> 16.4 grams Fe
28.0% S--------> 28.0 grams S

2) convert each gram to moles using the molar masses given

$55.6 g Cu \frac{1 mol}{63.5 g} = 0.876 mol Cu$
$16.4 g Fe \frac{1 mol}{56.0 g} = 0.293 mol Fe$
$28.0 gS \frac{1 mol}{32.0} = 0.875 mol S$

3) we divide the smallest value of moles (0.293) to each one.

Cu --> 0.876 / 0.293= 3
Fe---> 0.293 / 0.293= 1
S-----> 0.875 / 0.293= 3

4) let's write the empirical formula

Cu₃FeS₃