Question

Naphthalene balls contain $93.71\%$ carbon and $6.29\%$ hydrogen. If its molar mass is$128{\text{ }}gmo{l^{ - 1}}$, calculate its molecular formula.

Answer 1

Molecular formulas give the exact number of atoms of each element present in the molecular compound whereas Empirical formula gives the simplest whole-number ratio of atoms in a compound.
If the empirical formula and molecular mass of the compound is known we can calculate the Molecular formulas of the compound.

Complete answer:
The molecular formula is the formula derived from molecules and is representative of the total number of individual atoms present in a molecule of a compound. The molecular formula and the empirical formula expressed as,
$Molecular{\text{ }}formula{\text{ }} = {\text{ }}n{\text{ }} \times {\text{ }}empirical{\text{ }}formula$
First we find the empirical formula.
Step 1: If percentages are given, assume that the total mass is $100{\text{ }}grams$
Then the mass of each element = the percent given.
Now, the Mass of carbon = $93.71$
Mass of hydrogen = $6.29$
Step 2: Convert the mass of each element to moles
Number of moles (carbon) = Mass of carbon / atomic weight of carbon = $\dfrac{{93.71}}{{12{\text{ }}}} = 7.8$
Number of moles (hydrogen) = Mass of hydrogen / atomic weight of hydrogen = ${\text{ }}\dfrac{{6.29}}{1} = 6.29$
Step 3: Divide each mole value by the smallest number of moles calculated.
Hence, most simple ratio for carbon $= {\text{ }}\dfrac{{7.8}}{{6.29{\text{ }}}} = 1.25\;$
Most simple ratio for hydrogen $= \dfrac{{6.29}}{{6.29}} = 1$
Step 4: Round to the nearest whole number.
(it is not a whole number we multiply the ratio by $4$ to get a whole number ratio.
Now, Lowest whole number ratio for carbon = $5$
Lowest whole number ratio for hydrogen = $4$
Empirical formula = ${C_5}{H_4}$
Step 5: Now, we can find the molecular formula by finding the mass of the empirical formula.
$\Rightarrow M_{C_5}{H_4} = \;\left( {5{\text{ }} \times 12} \right) + \left( {4 \times 1} \right) = 64\,g$
$\Rightarrow n = \dfrac{{{\text{ }}Molar{\text{ }}mass{\text{ }}}}{{empirical{\text{ }}formula}}$ $= \dfrac{{128\,g}}{{64\,g}} = {\text{ }}2$
Putting value of $n = 2$ in the empirical formula we get molecular formula as

Molecular formula of Naphthalene balls $= 2{\text{ }} \times {\text{ }}{C_5}{H_{4}} = {C_{10}}{H_{8}}$

Note:
Divide the molar mass of the compound by the empirical formula mass. The result should be a whole number or very close to a whole number. Sometimes, the empirical formula and molecular formula both can be the same.