A 1.17 g sample of an alkane hydrocarbon gas occupies a volume of 674 mL at 28°C and 741 mmHg. Alkanes are known to have the general formula C nH 2n+2. What is the molecular formula of the gas in this sample? (R = 0.08206 L×atm/K×mol)

C3H8

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Mass of alkane = 1.17g

Volume (V) = 674 mL

Temperature (T) = 28°C

Pressure (P) = 741 mmHg.

Gas constant (R) = 0.08206 atm.L/Kmol

Step 2:

Conversion to appropriate unit.

For Volume:

1000mL = 1L

Therefore, 674mL = 674/1000 = 0.674L

For Temperature:

Temperature (Kelvin) = Temperature (celsius) + 273

Temperature (celsius) = 28°C

Temperature (Kelvin) = 28°C + 273 = 301K

For Pressure:

760mmHg = 1atm

Therefore, 741 mmHg = 741/760 = 0.975atm

Step 3:

Determination of the number of mole of the alkane..

The number of mole of the alkane can be obtained by using the ideal gas equation. This is illustrated below:

Volume (V) = 0.674L

Temperature (T) = 301K

Pressure (P) = 0.975atm

Gas constant (R) = 0.08206 atm.L/Kmol

Number of mole (n) =?

PV = nRT

n = PV /RT

n = (0.975 x 0.674)/(0.08206x301)

n = 0.0266 mole

Step 4:

Determination of the molar mass of the alkane.

Mass of alkane = 1.17g

Number of mole of the alkane = 0.0266mole

Molar Mass of the alkane =?

Number of mole = Mass/Molar Mass

Molar Mass = Mass/number of mole

Molar Mass of the alkane = 1.17/0.0266 = 44g/mol

Step 5:

Determination of the molecular formula of the alkane.

This is illustrated below:

The general formula for the alkane is CnH2n+2

To obtain the molecular formula for the alkane we shall assume n = 1, 2, 3 or more till we arrive at molar Mass of 44.

When n = 1

CnH2n+2 = CH4 = 12 + (4x1) = 16g/mol

When n = 2

CnH2n+2 = C2H6 = (12x2) + (6x1) = 30g/mol

When n = 3

CnH2n+2 = C3H8 = (12x3) + (8x1) = 44g/mol

We can see that when n is 3, the molar mass is 44g/mol.

Therefore, the molecular formula for the alkane is C3H8.

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