In a gas effusion experiment conducted at constant temperature and pressure, a sample of s… — JAMB UTME

In a gas effusion experiment conducted at constant temperature and pressure, a sample of sulfur(IV) oxide ( $SO_2$ ) effuses through a micro-porous aperture at a rate of $0.625\text{ cm}^3\text{ s}^{-1}$ . A gaseous alkane, designated as hydrocarbon $X$ , effuses through the exact same aperture at a rate of $1.25\text{ cm}^3\text{ s}^{-1}$ . Given relative atomic masses ( $S = 32$ , $O = 16$ , $C = 12$ , $H = 1$ ), determine the molecular formula of hydrocarbon $X$ .

Answer: CH4 / CH_4 / methane / CH4 (methane)

Answer

The molecular formula of hydrocarbon X is CH4 (methane).

According to Graham's Law of Effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The ratio of the rate of effusion of hydrocarbon X to that of sulfur(IV) oxide (

SO_2

, molar mass

64\text{ g/mol}

) is

1.25 / 0.625 = 2

. Squaring this ratio gives

4 = 64 / M_X

, yielding a molar mass of

16\text{ g/mol}

for hydrocarbon X. For an alkane with general formula

C_n H_{2n+2}

, a molar mass of

16\text{ g/mol}

corresponds to

n = 1

, giving the molecular formula

CH_4

(methane).

Step-by-Step Solution

Calculate the molar mass of sulfur(IV) oxide (

SO_2

M(SO_2) = 32 + 2(16) = 64\text{ g/mol}

The molar mass of the reference gas is needed to apply Graham's law.

Apply Graham's Law of Diffusion/Effusion relating effusion rates to molar masses.

\frac{r_X}{r_{SO_2}} = \sqrt{\frac{M(SO_2)}{M_X}} \implies \frac{1.25}{0.625} = \sqrt{\frac{64}{M_X}}

Graham's law states that the rate of diffusion/effusion of a gas is inversely proportional to the square root of its molar mass.

Solve for the molar mass of hydrocarbon

X

(

M_X

2 = \sqrt{\frac{64}{M_X}} \implies 4 = \frac{64}{M_X} \implies M_X = 16\text{ g/mol}

Squaring both sides allows direct calculation of the unknown molar mass.

Determine the molecular formula of the alkane with molar mass

16\text{ g/mol}

Alkane general formula:

C_n H_{2n+2} \implies 12n + 1(2n+2) = 16 \implies 14n + 2 = 16 \implies 14n = 14 \implies n = 1

. Thus, formula is

CH_4

Comparing the calculated molar mass to the general molecular formula of alkanes identifies the specific hydrocarbon.

Key Concept

Graham's Law of Diffusion and Effusion combined with alkane stoichiometry

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