Question

Arachidic acid is an organic compound composed of 76.86% C, 12.90% H, and the rest oxygen. If 0.297 mol of arachidic acid has a mass of 92.8 g, what are the empirical and molecular formulas of arachidic acid?

Answer 1

Mass percent of C , H and O in Arachidic acid

C = 76.86 % , H = 12.90 % , O = 100 – ( 76.86 + 12.90 ) %

Solution.

Lets assume total mass of this is acid to be 100 g then the percent of each element becomes mass in g.

So

Mass of Carbon = 76.86 g, H = 12.90 g , O = 10.24 g

Calculation of moles of each.

Moles = Mass in g / molar mass

n C = 76.86 g / 12.011 g per mol = 6.399 mol

n H = 12.90 g / 1.0079 g per mol = 12.79889 mol

n O = 10.24 g / 15.999 g per mol = 0.6400 mol

Calculation of simple mole ratio.

This is done by dividing all of the moles by smallest moles.

Smallest is the moles of Oxygen so lets divide each by the moles of Oxygen.

Number of Carbon = 6.399 / 0.64 = 9.99 = 10

Number of H = 12.079889 / 0.64 = 19.99 = 20

Number of O = 0.64 / 0.64 = 1

So the empirical formula would be

C10 H20 O

Calculation of molecules formula

Molecular formula = n x ( empirical formula )

Here n is the ratio of the molecular formula weight to the empirical formula weight

We have empirical formula and we can get weight of it from the formula.

Empirical formula weight =156.267 g /mol

Molecular formula weight can be calculated by given data in the problem .

We have weight of the acid for 0.297 mol and we can get the mass for 1 mol.

Mass of acid for 1 mol = 1 mol x 92.8 g / 0.297 mol = 312.4579 g per mol

Calculation of n

n = molecular formulate weight / empirical formula weight.

= 312.4579 / 156.267 = 1.99 = 2

Therefor molecular formula becomes

Molecular formula = 2 x ( C10H20O)

C₂₀H₄₀O₂