Question

A gaseous hydrocarbon R of mass 7.0g occupies a volume of 2.24dm^3 at s.t.p.If the percentage composition by mass of hydrogen is 14.3,determine its:
1.Empirical formula
2.Molecular formula
[H=1.00,C=12.00,Molar volume of gas at s.t.p=22.4 dm^3]

Answer 1

Answer on question #61606, Chemistry, Inorganic Chemistry

A gaseous hydrocarbon R of mass 7.0g occupies a volume of 2.24dm^3 at s.t.p. If the percentage composition by mass of hydrogen is 14.3, determine its

1. Empirical formula

2. Molecular formula

[H=1.00, C=12.00, Molar volume of gas at s.t.p=22.4 dm^3]

Solution:

1 mole of gas R occupies a volume of 22.4 dm^3 at s.t.p,

If 2.24 dm^3 of R weighs 7.0 g at s.t.p, then 22.4 dm^3 of R will weigh y g

$$y = 7 \times 22.4 / 2.24 = 70g$$

Since R is a hydrocarbon, therefore, it contains only carbon and hydrogen. If the percentage composition by mass of H is 14.3%, it implies that the balance will be C is 100 - 14.3 = 85.7%

(1) Empirical Formula

Elements C and H% by mass 85.7% and 14.3%

Atomic mass C= 12.0, H= 1.0

Mole ratio

$$85.7 / 12.0 = 14.3 / 1.0 = 7.14 \times 14.3$$

by least ratio

$$7.14 / 7.14 = 1$$ $$14.3 / 7.14 = 2$$

Empirical formula = CH₂

(2) Since molecular formula is a multiple of the empirical formula, then:

(Empirical formula) n = Molecular mass/formula.

Therefore,

$$(CH_2)n = 70 \times (12 + 2)n = 70$$ $$14n = 70$$ $$n = 70 / 14 = 5$$

Substituting the value of n into the equation above gives the molecular formula as: $(CH_2)_5 = C_5H_{10}$

Answer: 1) Empirical formula = CH₂ 2) molecular formula = C₅H₁₀