Answer to Question #110916 in Chemistry for amrith

Question

Calculate the empirical formula and molecular formula of an organic compound containing 26.67% of carbon and 2.22% of Hydrogen with the rest being oxygen. The Mr of the compound is 90

Answer 1

w(C) = 26.67%;

w(H) = 2.22%;

w(O) = 100% - w(C) - w(H) = 71.11%

Solution:

Suppose that 100 grams of organic compound is given, then:

m(C) = 26.67 g;

m(H) = 2.22 g;

m(O) = 71.11 g.

Now, we must convert these masses to moles and find their ratios.

Moles of carbon = n(C) = m(C) / M(C) = 26.67 / 12 = 2.2225 moles.

Moles of hydrogen = n(H) = m(H) / M(H) = 2.22 / 1 = 2.2200 moles.

Moles of oxygen = n(O) = m(O) / M(O) = 71.11 / 16 = 4.4444 moles.

n(C) : n(H) : n(O) = 2.2225 : 2.2200 : 4.4444 = 1 : 1 : 2

The empirical formula of an organic compound is CHO₂.

M(CHO₂) = Ar(C) + Ar(H) +2*Ar(O) = 12 + 1 + 2*16 = 45 (g/mol)/

The empirical formula mass is 45 g/mol respectively.

n = Molecular mass / empirical mass;

n = 90 / 45 = 2.

Therefore, the molecular formula of the organic compound is 2×CHO_2​ = C_2​H₂O₄.

The molecular formula of an organic compound is C₂​H₂O₄.

Answer:

The empirical formula of an organic compound is CHO₂.

The molecular formula of an organic compound is C₂​H₂O₄.

​