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Mass Percent


What is the mass percents of carbon, hydrogen, and oxygen in pure ethanol C2H6O?

First we calculate the mass of one mole of C2H6O...
mass of 2 moles of C is2 X 12.011 g
mass of 6 moles of H is6 X 1.008 g
mass of 1 mole of O is15.999 g

Molecular weight of C2H6O is

46.069 g

Next we calculate the mass percents


Note that the mass percentages should add up to 100%.

Combustion Analysis

When chemists make new compounds one of the first things they often do is determine the mass % for the different elements in the compound. To analyze the mass percent of carbon and hydrogen chemist use a combustion device.


The sample is burned in the presence of excess oxygen which converts all the carbon to carbon dioxide and all the hydrogen to water. The CO2 and H2O produced are absorbed in two different stages and their masses determined by measuring the increase in weight of the absorbers.

Ascorbic acid (vitamin C) contains only C, H, and O. Combustion of 1.000 g of Ascorbic acid produced 40.9% C and 4.5% H. What is the empirical formula for Ascorbic Acid?

First we need to calculate the mass percent of Oxygen. Since the sample contains C, H, and O, then the remaining

100% - 40.9% - 4.5% = 54.6% is Oxygen

Now we need to express the composition in grams and determine the number of moles of each element:


Next we divide by the smallest number of moles to obtain the mole ratio which is also the atom ratio. In this case carbon has the smallest number of moles, so...

C:0.0340 moles/0.0340 moles=1
H:0.045 moles/0.0340 moles=1.32 ~ 1 1/3
O:0.0341 moles/0.0340 moles~1

Finally we calculate the smallest whole integer ratios by multiplying each number above by 3 to get

C: 3H:4O:3

thus we obtain the empirical formula C3H4O3

Remember the empirical formula has the smallest whole integer ratios, the molecular formula can be different, e.g., C6H8O6, or C9H12O9, or C12H16O12, ... are all possible molecular formulas.

Now let's look at a related question.

What is the molecular formula if the molecular weight of Ascorbic Acid was formed to be 176 g/mole?

In this case we need to find the multiplication factor between the molecular formula and the empirical formula:

factor = (molecular weight)/(empirical formula weight)

The empirical formula weight of Ascorbic Acid is

mass of 3 moles of C is 3 X 12.011 g
mass of 4 moles of H is4 X 1.008 g
mass of 3 moles of O is3 X 15.999 g

mass of 1 mole of CH4 is


Therefore the multiplicative factor is (176 g/mole)/(88.062 g/mole) ~ 2, and the molecular formula for Ascorbic Acid is C6H8O6

  • Empirical Formulas - Combustion Analysis of Compounds of C, H and O:
  • Empirical Formulas - Combustion Analysis of Compounds of C, H and N:
  • Empirical Formulas - Combustion Analysis of Three Element Compounds:
  • Empirical Formulas - Combustion Analysis of Complex Compounds:
  • Molecular Weight Determination:
  • Calculating the Percentage Composition from Formula:
  • Composition from Molecular Weight and Number of Atoms:

Homework from Chemisty, The Central Science, 10th Ed.

3.21, 3.23, 3.25, 3.29, 3.31, 3.33, 3.35, 3.37, 3.39, 3.41, 3.43, 3.45, 3.47, 3.49