# Which one of the following contains 35% carbon by mass?

Which one of the following contains 35% carbon by mass?

• You need to find out the mass percentage of carbon in each of the compounds. Do this by using the following formula:

w% (C) = [m (C) . 100] : m(compound) [%]

We`ll do the calculations using a quantity of 1 mol. Here`s how the rest of the calculations go starting with fluoromethane:

A)

w% (C) = [m (C) . 100] : m(CH3F)

m(C) = M(C) = 12,01 g (the molar mass being 12,01 g/mol)

m(CH3F) = M(CH3F) = 12,01 + 3.1,01 + 18,99 = 34,03 g

w% (C) = [12,01 . 100] : 34,03

w% (C) = 35,29% or approximately 35%

B)

w%(C) = [m (C) . 100] : m(CH4)

m(C) = 12,01 g

m(CH4) = M(CH4) = 12,01 + 4. 1,01 = 16,05 g

w%(C) = [12,01 . 100] : 16,05

w%(C) = 74,82% or approximately 75%

C)

w%(C) = [m (C) . 100] : m(CO2)

m(C) = 12,01 g

m(CO2) = M(CO2) = 12,01 + 2.15,99 = 43,99 g

w%(C) = [12,01 . 100] : 43,99

w%(C) = 27,30% or approximately 27%

D)

You have 2 mols of carbon in 1 mol of acethylene, so you have double the mass of C in the formula:

w%(C) = [2m (C) . 100] : m(C2H2)

m(C) = 12,01 g

m(C2H2) = M(C2H2) = 2.12,01 + 2.1,01 = 26,04 g

w%(C) = [2. 12,01 . 100] : 26,04

w%(C) = [24,02 . 100] :26,04

w%(C) = 92,24% or approximately 92%

Judging by the caluclations above I`d say that compound A (CH3F) contains 35% carbon by mass.

Source(s): Stoichiometry course
• Molar mass CH3F = 12+3+19 = 34g/mol

12/34*100 = 35.2% yes – That is the one

Try the others:

Molar mass CH4 = 12+4=16 = 12/16*100 = 75% no

Molar mass CO2 = 12+32 =44 = 12/44*100 = 27.3% NO

Molar mass C2H2 = 24+2 = 26 = 24/26*100 = 92.3% no.