Question

1 g of a compound contains 0.433 g of sodium, 0.453 g oxygen by mass, and the rest is carbon. What is the molecular formula of the compound if its molecular weight is 106?

[1 mark]

• A. $N{a}_{2}C{O}_3$
• B. $N{a}_2{C}_2{O}_3$
• C. $N{a}_{2}C{O}_4$
• D. $N{a}_4{C}_4{O}_6$

Answer 1

The correct option is A $N{a}_{2}C{O}_3$

The empirical formula of the compound can be calculated as follows:

Element	Mass in grams	Atomic mass	Relative number of atoms	Simplest ratio
Na	$0.433g$	23	$\frac{43.3}{23}=1.88$	$\frac{1.88}{0.95}=2$
O	$0.453g$	16	$\frac{45.3}{16}=2.83$	$\frac{2.83}{0.95}=3$
C	$0.114g$	12	$\frac{11.4}{12}=0.95$	$\frac{0.95}{0.95}=1$

Hence, the empirical formula of the compound is $N{a}_{2}C{O}_3$.

Therefore, empirical formula mass
= (2 × 23) + (12 × 1) + (3 × 16)
= 46 +12 + 48
= 106

Now,
$n=\frac{Molecularmass}{Empiricalformulamass}=\frac{106}{106}=1$

Empirical formula is $N{a}_{2}C{O}_3$, n=1, hence the molecular formula is $N{a}_{2}C{O}_3$.