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A flammable gas made up of only carbon and hydrogen is found to effuse through a porous barrier in 1.50 min. Under the same conditions of temperature and pressure, it takes an equal volume of bromine vapor 4.73 min to effuse through the same barrier. Calculate the molar mass of the unknown gas, and suggest what this gas might be.

Answer :

Eduard22sly

Answer:

A. The molar mass of the unknown gas is 16g/mol

B. Compound is CH4 i.e methane

Explanation:

A. Step 1:

Representation:

Let t1 be time for unknown gas

Let t2 be the time for bromine vapor

Let M1 be molar mass of the unknown gas

Let M2 be the molar mass of bromine vapor

A. Step 2 :

Data obtained from the question.

Time for the unknown gas (t1) = 1.50 min

Time for Br2 (t2) = 4.73 min

Molar Mass of unknown gas (M1) =?

Molar Mass of Br2 (M2) = 80 x 2 = 160g/mol

A. Step 3:

Determination of the molar mass of the unknown gas.

Applying the equation:

t2/t1 = √(M2/M1)

The molar mass of can be obtained as follow:

t2/t1 = √(M2/M1)

4.73/1.5 = √(160/M1)

Take the square of both side

(4.73/1.5)^2 = 160/M1

9.94 = 160/M1

Cross multiply to express in linear form.

9.94 x M1 = 160

Divide both side by 9.94

M1 = 160/9.94

M1 = 16g/mol

Therefore, the molar mass of the unknown gas is 16g/mol

B. Identification of the gas.

The gas contains C and H only. From the calculations made above, the molar mass of the unknown gas is 16g/mol.

We know also that the molar mass of carbon is 12g/mol and that of Hydrogen is 1g/mol

Therefore,

C + H = 16

There would be only 1 atom of C in the compound since the molar mass of the compound is 16g/mol. With these understanding, let us determine the number of H atom in the compound. This is illustrated below :

C + H = 16

12 + H = 16

H = 16 - 12

H = 4

Divide by the molar mass of H i.e 1

H = 4/1 = 4

There are 4 atoms of H in the compound. Therefore, the compound is CH4 i.e methane

pstnonsonjoku

The molar mass of the unknown gas is 16 g/mol hence the unknown gas is methane.

We must note that the time taken for a gas to diffuse is directly proportional to the molar mass of the gas.

Hence;

t1/t2 = √M1/M2

Let t1 = time taken for the flammable gas to diffuse = 1.50 min

Let t2 = time taken for the bromine vapor to diffuse = 4.73 min

M1 = molar mass of the flammable gas = ?

M2 = molar mass of the bromine vapor = 160 g/mol

Substituting values;

1.50/4.73 = √M1/160

(1.50/4.73)^2 = M1/160

0.1006 =  M1/160

M1 = 0.1006 × 160

M1 = 16 g/mol

The unknown gas is methane.

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