Question

Calculate $\%$ composition of all elements in sodium carbonate $\left(N{a}_{2}C{O}_3\right)$.

Answer 1

Formula: Na2CO3
Na Molecular Mass: 22.98 grams
C Molecular Mass= 12.01 grams
O Molecular Mass= 16 grams
Now, the molecular mass of the compound: 2(22.98) + 12.01 + 3(16)= 105.97 grams

Now, when you do percent composition all you are doing is this: Amount of Element in Grams/ Total Compound Mass in Grams x 100%.

So, lets do that for each of these elements:

$Na:\frac{2\times 22.98}{105.97}\times 100\%=43.37\%$
$C:\frac{12.01}{105.97}\times 100\%=11.33\%$
$O:\frac{3\times 16}{105.97}\times 100\%=45.30\%$

Now, add up these percentages and make sure that they add up to exactly 100%.

43.37% + 11.33% + 45.30%= 100.00 %