How do I determine the empirical formula of succinic acid?

2 Answers

• need to know that succinic acid has Carbon, Hydrogen and Oxygen in it and you need to be able to match the % of your given numbers with those elements that it is composed of.

  Carbon happens to be the 40.68%

  Hydrogen is the 5.12 %

  Oxygen is the 54.19 %

  Say you had 100 grams and turn those percentages straight to grams and multiply those by the molar mass conversion factor for each element, such that you are dividing by the grams in the molar mass question. For instance 1mol O/15.994g not 15.994g/1mol O. It works out that grams cancel and we get moles and so in doing so we are actually dividing by the molar mass.

  40.68g x 1mol C/12.011 g = 3.39 mol C

  5.12 g x 1 mol H/1.0079 g = 5.08 mol H

  54.19 g x 1 mol O/15.994 = 3.39 mol O

  Now divide each one of thes by the smallest answer to calculate the lowest whole number ratio of moles in the empirical formula: 3.39/3.39 = 1 mol C, 5.08/3.39 = 1.5 mol of H and 3.39/3.39 = 1 mol O

  It is still not in lowest whole number ratio, so now double everything to get 1.5 a whole number.

  C2H3O2

• You have to tell us what these %’s represent. C4H6O4 is the accepted formula.

  I will assume that you’ve written them in the order that I have found them.

  Assume there 100 grams of sample.

  C = 40.68 / 12 = 3.39 moles.

  H = 5.12 / 1 = 5.12

  O = 54.19/16 = 3.39

  C_3.39 H_5.12 O_3.39 Divide through by 3.39

  C_1 H_1.5 O_1 Multiply 2 to get rid of the the fraction.

  C_2 H_3 O_2

  This is the empirical formula which is what the actual formula predicts.