A 0.04380 g sample of gas occupies 10.0 mL at 290.5 K and 1.10 atm. Upon further analysis, the compound is found to be 25.305% C and 74.695% Cl. What is the molecular formula of the compound?

THE MOLECULAR FORMULA FOR THE COMPOUND IS C2 Cl2

Explanation:

mass = 0.04380 g

Volume = 10 mL = 10 / 1000 L = 0.010 L

Pressure = 1.10 atm

R = 0.082 L atm mol^-1 K^-1

Temperature = 290.5 K

Carbon = 25.305 %

Chlorine = 74.695 %

Atomic mass of carbon = 12

Atomic mass of chlorine = 35.5

First calculate the empirical formula by following these steps:

1. Write the percentage composition of the elements involved and divide by its atomic mass

Carbon = 25.305 / 12 = 2.10875

Chlorine = 74.695 /35.5 = 2.1040

2. Divide each by the smaller value

Carbon = 2.10875 / 2.1040 = 1.002

Chlorine = 2.1040 / 2.1040 = 1

3. Round up to the whole number

The empirical formula is C Cl

Next is to calculate the molar mass of the compound using ideal gas equation

PV = mRT/ MM

Mm = mRT / PV

Mm = 0.04380 * 0.082 * 290.5 / 1.10 * 0.010

Mm = 1.0433 / 0.011

Mm = 94.845 g/mol

Mm = 94.85 g/mol

Now that we know the molar mass, we can go on to calculate the molecular formula:

(C Cl) n = Molar mass

( 12 + 35.5) n = 94.85

(47.5)n = 94.85

n = 94.85 / 47.5

n = 1.9968

n ~ 2

The molecular formula can then be written as C2Cl2.