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

SOMEONE ASKED ๐Ÿ‘‡

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

HERE THE ANSWERS ๐Ÿ‘‡

  • 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.

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