Question

what is the molecular formula of the compound which has the following percentage composition.carbon 80%, hydrogen 20%.the vapour density is 15

Answer 1

Hello!

What is the molecular formula of the compound which has the following percentage composition.carbon 80%, hydrogen 20%.the vapour density is 15

data:

Carbon (C) ≈ 12 a.m.u (g/mol)  

Hydrogen (H) ≈ 1 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: 80% = 80 g

H: 20% = 20 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:

$C: \dfrac{80\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 6,66\:mol$

$H: \dfrac{20\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 20\:mol$

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

$C: \dfrac{6,66}{6,66}\to\:\:\boxed{C = 1}$

$H: \dfrac{20}{6,66}\to\:\:\boxed{H \approx 3}$

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

$\boxed{C_1H_3\:\:or\:\:CH_3}\Longleftarrow(Empirical\:Formula)\end{array}}\qquad\checkmark$

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

if: CH3

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

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

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

$MW(E.F) = 12 + 3 \to \boxed{MW\:(E.F) = 15\:a.m.u}$

Knowing that it is given in vapor density of 15, 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 * 15

$\boxed{MW\:(M.F) = 30\:a.m.u}$

Knowing that the Molecular Weight of the Molecular Formula is 30 units of atomic mass and that the Molecular Weight of the Empirical Formula is 15 units of atomic mass, then we will find the number of terms (n) for the molecular formula of the compound, let us see:

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

$n = \dfrac{30}{15}$

$\boxed{n = 2}$

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

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

$M.F = (CH_3)*2$

$\boxed{\boxed{M.F = C_2H_6}}\Longleftarrow(molecular\:formula - ethane)\end{array}}\qquad\checkmark$

Answer:

Ethane - C2H6

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