Question

a gaseous hydrogen and carbon containing compound is decomposed and found to contain 79.85% C and 20.15% H by mass

A.) What is the empirical formula of the compound?
I already found the the empirical formula, its CH3

B.) at 43.0 °C and 0.984 atm, the gas occupies a volume of 1.8 L and has a mass of 2.04g. what is the molecular formula of the compound?

Answer 1

A)

we have mass of each elements as:

C: 79.85 g

H: 20.15 g

Divide by molar mass to get number of moles of each:

C: 79.85/12.01 = 6.649

H: 20.15/1.008 = 19.99

Divide by smallest to get simplest whole number ratio:

C: 6.649/6.649 = 1

H: 19.99/6.649 = 3

So empirical formula is:CH3

Answer: CH3

B)

Given:

P = 0.984 atm

V = 1.8 L

T = 43.0 oC

= (43.0+273) K

= 316 K

find number of moles using:

P * V = n*R*T

0.984 atm * 1.8 L = n * 0.08206 atm.L/mol.K * 316 K

n = 6.83*10^-2 mol

mass(solute)= 2.04 g

use:

number of mol = mass / molar mass

6.83*10^-2 mol = (2.04 g)/molar mass

molar mass = 29.87 g/mol

This is molar mass of molecule.

Molar mass of CH3,

MM = 1*MM(C) + 3*MM(H)

= 1*12.01 + 3*1.008

= 15.034 g/mol

Now we have:

Molar mass = 29.87 g/mol

Empirical formula mass = 15.034 g/mol

Multiplying factor = molar mass / empirical formula mass

= 29.87/15.034

= 2

So molecular formula is:C2H6

Answer: C2H6