Respuesta :
Answer:
[tex]C_{10}H_{8}[/tex] - Molecular formula
[tex]C_{5}H_{4}[/tex] - 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 [tex]CO_2[/tex] = 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 = [tex]C_5H_4[/tex]
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 = [tex]C_{10}H_{8}[/tex]
