# The following is the percentage composition of a compound. $Na = 16.08 \%, C = 4.19 \%, O = 16.78 \%$ and $H_2O = 62.95 \%$ . Its molecular formula is

$(a)\;Na_2CO_3 \\(b)\;Na_2CO_3.H_2O \\(c)\;Na_2CO_3.10H_2O\\(d)\;Na_2CO_3.5H_2O$

Toolbox:
• Step 1. If you have masses go onto step 2. If you have % composition of the elements. Assume the mass to be 100g, so the % becomes grams.
• Step 2.. Determine the moles of each element.
• Step 3.. Determine the mole ratio by dividing each elements no. of moles by the smallest value from step 2.
• Step 4. Double, triple, … to get an integer if they are not all whole numbers.
$\therefore$ Empirical Formula = $Na_2CO_3.10H_2O$
Hence (C) is the correct answer.
edited Mar 25, 2014