The principal component of mothballs is naphthalene, a compound with the molecular mass of about 130 amu containing only carbon and hydrogen. a 3.000 mg sample of naphthalene burns to give 10.3 mg of co2. determine its empirical and molecular formulas.

- Molecular formula

- Empirical formula

Explanation:

Mass of carbon dioxide obtained = 10.3 mg

1 mg = 10⁻³ g

Molar mass of carbon dioxide = 44.01 g/mol

Moles of = 10.3 × 10⁻³ g  /44.01 g/mol = 0.2340×10⁻³ moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 0.2340×10⁻³ moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 0.2340×10⁻³ x 12.0107 = 2.8105×10⁻³ g

Given that the naphthalene only contains hydrogen and carbon. So,

Mass of H in the sample = Total mass - Mass of C

Mass of the sample = 3.000 mg = 3.00×10⁻³ g

Mass of H in sample = 3.00×10⁻³ - 2.8105×10⁻³ = 0.1895×10⁻³ g

Molar mass of H = 1.0078 g/mol

Moles of H  = 0.1895×10⁻³  / 1.0078  = 0.1880×10⁻³ moles

Taking the simplest ratio for H and C as:

0.1880×10⁻³ : 0.2340×10⁻³   = 4 : 5

The empirical formula is =

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 5×12 + 4×1 = 64 g/mol

Molar mass = 130 g/mol

So,  

Molecular mass = n × Empirical mass

130 g/mol  = n ×  64 g/mol

⇒ n ≅ 2

The formula of naphthalene =