Question

An organic substance containing carbon, hydrogen and oxygen gave the percentage composition as: c=40.68%, h=5.085%, o=54.228%. The vapour density of the compund is 59. Calculate the molecular formula of the compound

Answer 1

Hello!

An organic substance containing carbon, hydrogen and oxygen gave the percentage composition as: c=40.68%, h=5.085%, o=54.228%. The vapour density of the compund is 59. Calculate the molecular formula of the compound

data:

Carbon (C) ≈ 12 a.m.u (g/mol)  

Hydrogen (H) ≈ 1 a.m.u (g/mol)

Oxygen (O) ≈ 16 a.m.u (g/mol)

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

C: 40.68 % = 40.68 g

H: 5.085 % = 5.085 g

O: 54.228 % = 54.228 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:

$C: \dfrac{40.68\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} = 3.39\:mol$

$H: \dfrac{5.085\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 5.085\:mol$

$O: \dfrac{54.228\:\diagup\!\!\!\!\!g}{16\:\diagup\!\!\!\!\!g/mol} = 3.38925\:mol$

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

$C: \dfrac{3.39}{3.38925}\to\:\:\boxed{C \approx 1}$

$H: \dfrac{5.085}{3.38925}\to\:\:\boxed{H \approx 1.5}$

$O: \dfrac{3.38925}{3.38925}\to\:\:\boxed{O = 1}$

convert number of atomic radio into whole number  

2*(1 : 1.5 : 1) =

= 2 : 3 : 2  ← whole number of atomic radio

C = 2

H = 3

O = 2

Thus, the minimum or empirical formula found for the compound will be:

$\boxed{C_2H_3O_2}\Longleftarrow(Empirical\:Formula)\end{array}}\qquad\checkmark$

We are going to find the Molecular Weight (MW) of the Empirical Formula (EF), let's see:

if: C2H3O2

C = 2*(12 a.m.u) = 24 a.m.u

H = 3*(1 a.m.u) = 3 a.m.u

O = 2*(16 a.m.u) = 32 a.m.u

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

$MW(E.F) = 24 + 3 + 32\to \boxed{MW\:(E.F) = 59\:a.m.u}$

Knowing that it is given in vapor density of 59, we will find the Molecular Weight of the Molecular Formula "MW (M.F)", see:

Molecular Weight (M.F) = 2 * Vapour density

MW (M.F) = 2 * 59

$\boxed{MW\:(M.F) = 118\:a.m.u}$

Knowing that the Molecular Weight of the Molecular Formula is 118 units of atomic mass and that the Molecular Weight of the Empirical Formula is 59 units of atomic mass, then we will find the number of terms (n) for the molecular formula of the compound, let us see:

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

$n = \dfrac{118}{59}$

$\boxed{n = 2}$

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

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

$M.F = (C_2H_3O_2)*2$

$\boxed{\boxed{M.F = C_4H_6O_4}}\Longleftarrow(molecular\:formula\:of\:the\:compound)\end{array}}\qquad\checkmark$

Answer:

C4H6O4

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