in the description

Question

The identity of an unknown monoprotic organic acid is determined by titration. A 0.436g sample of the acid is titrated with 0.185 M NaOH. What is the molar mass of the compound if 13.1 mL of the NaOH solution is required to neutralize the sample?

molar mass:

The compound is 60.0% C, 4.48% H, and 35.5% O. What is the molecular formula of the organic acid?

formula:

Answer 1

Let the monoprotic acid be HA, then...

HA + NaOH ==> H2O + NaA mole ratio of NaOH to HA is 1:1

moles HA present = moles NaOH used = 13.1 ml x 1 L / 1000 ml x 0.185 mol / L = 2.42x10^-3 mols HA

molar mass = g / mole = 0.436 g / 2.42x10^-3 mols = 180 g / mole

If you, for some reason, wanted to use the % composition data that is provided, you could do that also.

Assume you have 100 g of the unknown. Then the % of each element is equal to grams of each, from which we will determine moles of each.

moles C = 60.0 g C x 1 mol C / 12 g = 5 moles C

moles H = 4.48 g H x 1 mol H / 1 g = 4.48 moles H

moles O = 35.5 g O x 1 mol O / 16 g = 2.22 moles O

Divide all by 2.22 in an attempt to get whole numbers of moles

moles C = 2.25

moles H = 2.02

moles O = 1.00

Multiply all by 4 to get whole numbers

moles C = 9

moles H = 8

moles O = 4

Empirical formula = C₉H₈O₄ (molar mass = 108 + 8 + 64 = 180 g / mole)

In this case, the empirical formula is the same as the molecular formula.