Except for very small alkanes (hydrocarbons), the boiling point mm-i 20-3(1 degr

Question

Except for very small alkanes (hydrocarbons), the boiling point mm-i 20-3(1 degrees for each additional carbon atom in the molecule Assume that the normal boiling point of the fuel in the lighter is 10 degree C, why was it not necessary to extend the table further (i.e., why was it unlikely that your unknown contained more than eight carbon atoms per molecule)? A gaseous hydrocarbon collected over water at a temperature of 21 degree C and a barometric pressure of 753 torr occupied a volume of 48.1 mL. The hydrocarbon in this volume weighs 0.1133 g. Calculate the molecular mass of the hydrocarbon. How is the ideal gas law related to the molar mass of a gas?

Explanation / Answer

1. The table is not given

The boiling point for a compound with eight carbons is around 120 oC.

If the boiling point of the unknown is less than 120 oC, then it may not contain more than 8 carbons per molecule.

2. Given:

Temperature, T = 21 oC = 21 + 273.15 = 294.15 K

Pressure, P = 753 torr = 753/760 = 0.99 atm

V = 48.1 mL = 0.0481 L

From ideal gas law: PV = nRT

or, n = PV/RT

n = (0.99 atm)(0.048 L)/(0.0821 L.atm/K.mol)(294.15 K) = 0.00197 mol

no. of moles, n = mass of hydrocarbon/molar mass

or molar mass = mass of hydrocarbon/no. of moles of hydrocarbon = 0.1133 g /0.00197 mol = 57.5 g/mol

molecular mass of hydrocarbon = 57.5 g

3.

Ideal gas law : PV = nRT
where, n= m/M

Here, n = no. of moles, M = molecular mass and m = mass of the substance

Then, PV = (m/M)RT

or, PV = mRT/M

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