Question

A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molecular mass is 98.96. What are its emperical and molecular formula

Answer 1

Molecular weight of H is 1
Molecular weight of C is 12
Molecular weight of Cl is 35.5
mass composition for H = 4.07/1 = 4.07
mass composition for C = 24.27/12 = 2.02
mass composition for Cl = 71.65/35.5 = 2.02
for to get the empirical formula divide all mass compositions by the lowest one for this case divide by 2.02
number of mole of hydrogen in molecule = 4.07/2.02 = 2 (approximately)
number of mole of carbon in molecule = 2.02/2.02 = 1
number of mole of chlorine in molecule = 2.02/2.02 = 1
So empirical formula is $CH_{2}Cl$ 
molecular formula = empirical formula × n
and 
n = (molecular mass)/(empirical formula mass)
empirical formula mass = 12 + 2×1 + 35.5 = 49.5
n = 98.96/49.5 = 2 (approximately)
So molecular formula is $C_{2} H_{4}Cl_{2}$

Answer 2

Hello!

A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molecular mass is 98.96. What are its emperical and molecular formula ?

• We have the following data:

Hydrogen (H) ≈ 1 a.m.u (g/mol)

Carbon (C) ≈ 12 a.m.u (g/mol)

Chlorine (Cl) ≈ 35.5 a.m.u (g/mol)

• We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)

H: 4.07 % = 4.07 g

C: 24.27 % = 24.27 g

Cl: 71.65 % = 71.65 g

• The values ​​(in g) will be converted into quantity of substance (number of mols), dividing by molecular weight (g / mol) each of the values, we will see:

$H: \dfrac{4.07\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 4.07\:mol$

$C: \dfrac{24.27\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} = 2.0225\:mol$

$Cl: \dfrac{71.65\:\diagup\!\!\!\!\!g}{35.5\:\diagup\!\!\!\!\!g/mol} \approx 2\:mol$

• We realize that the values ​​found above, some are not integers, so we divide these values ​​by the smallest of them, so that the proportion does not change, let's see:

$H: \dfrac{4.07}{2}\to\:\:\boxed{H \approx 2}$

$C: \dfrac{2.0225}{2}\to\:\:\boxed{C \approx 1}$

$Cl: \dfrac{2}{2}\to\:\:\boxed{Cl = 1}$

• Thus, the minimum or Empirical Formula (E.F) found for the compound will be:

$\boxed{\boxed{C_1H_2Cl_1}}\:\:\:or\:\:\:\boxed{\boxed{CH_2Cl}}\Longleftarrow(Empirical\:Formula)\:\:\:\:\:\:\bf\blue{\checkmark}$

• We are going to find the Molar Mass (MM) of the Empirical Formula (EF), let's see:

if: CH2Cl

C = 1*(12 a.m.u) = 12 a.m.u

H = 2*(1 a.m.u) = 2 a.m.u

Cl = 1*(35.5 a.m.u) = 35.5 a.m.u

-------------------------------------

$MM(E.F) = 12 + 2 + 35.5\to \boxed{MM(E.F) = 49.5\:g/mol}$

• Knowing that the Molar Mass of the Molecular Formula is 98.96 (in g/mol) and that the Molar Mass of the Empirical Formula is 49.5 (in g/mol), then we will find the number of terms (n) for the molecular formula of the compound, let us see:

$n = \dfrac{MM_{M.F}}{MM_{E.F}}$

$n = \dfrac{98.96}{49.5}$

$\boxed{n \approx 2}$

• The Molecular Formula is the Empirical Formula times the number of terms (n), then, we have:

$M.F = (E.F)*n$

$M.F = (CH_2Cl)*2$

$\boxed{\boxed{M.F = C_2H_4Cl_2}}\Longleftarrow(molecular\:formula\:of\:the\:compound)\:\:\:\:\:\:\bf\green{\checkmark}$

Answer:  

Empirical Formula: CH2Cl

Molecular Formula: C2H4Cl2

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