Question

An organic compound containing 4.07% hydrogen 24.27% carbon and 71.65% chlorine. The molar mass of the organic compound is 98.96 g.What are the empirical and molecular formulas of the above compound?

Answer 1

Hello!

An organic compound containing 4.07% hydrogen 24.27% carbon and 71.65% chlorine. The molar mass of the organic compound is 98.96 g.What are the empirical and molecular formulas of the above compound?

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:

$\boxed{C_1H_2Cl_1\:\:\:or\:\:\:CH_2Cl}\Longleftarrow(Empirical\:Formula)\end{array}}\qquad\checkmark$

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:

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

$M.F = (CH_2Cl)*2$

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

Answer:

C2H4Cl2

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I Hope this helps, greetings ... Dexteright02! =)

Answer 2

Answer: empirical formula is CH2Cl and molecular formula is C2H4Cl2

Explanation:- given mass of H is 4.07%, Cl is 71.65% and C is 24.27%

Element. Mass. At.mass. at.ratio

H. 4.07. 1. 4.07/1=4.07

Cl. 71.65. 35.5. 71.65/35.5=2.01

C. 24.27. 12. 24.27/12=2.022

Now their simplest ratio

H= 4.07/2.01=2

Cl= 2.01/2.01=1

C=2.022/2.01=1

a) Hence the empirical formula is CH2Cl.

b) we know molar mass given is 98.96

So molar mass=2× vapour density

=> 98.96=2× V.D.

=> V.D= 98.96/2=49.5

So n= molar mass/empirical formula mass we know,

n= 98.96/49.5=2

Hence the molecular formula is C2H4Cl2.

Hope this helps you so

Hope It will help you so.......