Respuesta :
Answer:
[tex]C_9H_{12}[/tex]
Explanation:
[tex]Moles =Given\ mass \times {Molar\ mass}[/tex]
Mass of water obtained = 42.8 mg
1 mg = 10⁻³ g
Moles of [tex]H_2O[/tex] = 42.8/18 = 2.3778×10⁻³ moles
2 moles of hydrogen atoms are present in 1 mole of water. So,
Moles of H = 2 x 2.4 = 4.7556×10⁻³ moles
Molar mass of H atom = 1.008 g/mol
Mass of H in molecule = 4.7556×10⁻³ x 1.008 = 4.7936×10⁻³ g
Given that the cumene only contains hydrogen and carbon. So,
Mass of C in the sample = Total mass - Mass of H
Mass of the sample = 47.6 mg = 47.6×10⁻³ g
Mass of C in sample = 47.6×10⁻³ - 4.80×10⁻³ = 42.8064×10⁻³ g
Moles of C = 42.8064×10⁻³ / 12 = 3.5672×10⁻³ moles
Taking the simplest ratio for H and C as:
4.7556×10⁻³ : 3.5672×10⁻³ = 4 : 3
The empirical formula is = [tex]C_3H_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 = 3×12 + 4×1 = 40 g/mol
115 g/mol < Molar mass > 125 g/mol
So,
Molecular mass = n × Empirical mass
115 g/mol < 40 n > 125 g/mol
⇒ n ≅ 3
The formula of cumene = [tex]C_9H_{12}[/tex]
