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Formula Mass and Percent Composition

How do you calculate percent composition?

BeginnerReactions & Stoichiometry

What is it?

Percent composition tells you what percentage of a compound’s mass comes from each element.

To work it out, you first need the formula mass: the sum of the atomic masses of all the atoms in the formula, in atomic mass units (u).

Key idea

Percent composition is about mass, not about the number of atoms. A light element can make up most of the atoms in a compound but only a small part of its mass.

Why does it matter?

  • Identifying substances. Chemists analyse an unknown compound to find the mass of each element it contains. Comparing those percentages with calculated values helps identify the compound, and it is the first step in working out an empirical formula.
  • Checking purity. If a sample’s measured composition doesn’t match the expected percentages, it may be impure.
  • Everyday labels. Products such as fertilisers state their contents as percentages by mass.

How does it work?

1. Find the formula mass

Add the atomic masses of every atom in the formula, using the subscripts to count atoms. For water, HX2O\ce{H2O}:

2(1.008)+16.00=18.016 u2(1.008) + 16.00 = 18.016\ \text{u}

A subscript after parentheses multiplies everything inside them. Calcium hydroxide, Ca(OH)X2\ce{Ca(OH)2}, contains one Ca, two O and two H:

40.08+2(16.00+1.008)=74.10 u40.08 + 2(16.00 + 1.008) = 74.10\ \text{u}

For molecules this is also called the molecular mass. Ionic compounds like NaCl\ce{NaCl} have no molecules, so we say formula mass (the mass of one formula unit).

2. Work out each element’s share

% element=n×Aformula mass×100%\%\ \text{element} = \frac{n \times A}{\text{formula mass}} \times 100\%

where nn is the number of atoms of that element in the formula and AA is its atomic mass.

The percentages of all the elements in a compound add up to 100% (apart from tiny rounding differences).

3. It doesn’t matter how much you have

A pure compound always has the same composition by mass. Every sample of pure water is 11.19% hydrogen by mass, whether it is a drop or a lake. So you can calculate percent composition from the formula alone, using one formula unit or one mole.

Think of it like this

A bag of mixed nuts might contain more peanuts than almonds, but if the almonds are much larger they could still make up most of the bag’s mass. Percent composition works the same way: it weighs each element’s share, it doesn’t count atoms.

More precisely

The atomic masses on the periodic table are weighted averages of the masses of each element’s naturally occurring isotopes. For some elements, IUPAC reports the standard atomic weight as a small range because isotope proportions vary slightly between natural sources. Textbook values such as H = 1.008 are conventional rounded values. They are accurate enough for these calculations.

Visualise it

Two bar charts for water. By number of atoms, hydrogen is 66.7 percent and oxygen 33.3 percent, because each molecule has two hydrogen atoms and one oxygen atom. By mass, hydrogen is only 11.19 percent and oxygen 88.81 percent, because an oxygen atom is much heavier than a hydrogen atom.
In water, two out of three atoms are hydrogen, but hydrogen is only about 11% of the mass.

Worked example

Worked example: Percent composition of glucose

Question: Calculate the percent composition by mass of glucose, CX6HX12OX6\ce{C6H12O6}. (C = 12.01, H = 1.008, O = 16.00)

  1. Mass of each element in one formula unit: C: 6 × 12.01 = 72.06 u; H: 12 × 1.008 = 12.096 u; O: 6 × 16.00 = 96.00 u
  2. Formula mass: 72.06 + 12.096 + 96.00 = 180.156 u
  3. % C = 72.06180.156×100%=40.00%\dfrac{72.06}{180.156} \times 100\% = 40.00\%
  4. % H = 12.096180.156×100%=6.71%\dfrac{12.096}{180.156} \times 100\% = 6.71\%
  5. % O = 96.00180.156×100%=53.29%\dfrac{96.00}{180.156} \times 100\% = 53.29\%

Answer: glucose is 40.00% C, 6.71% H and 53.29% O by mass. Check: 40.00 + 6.71 + 53.29 = 100.00%.

Worked example: Mass of an element in a sample

Question: What mass of iron is contained in 50.0 g of iron(III) oxide, FeX2OX3\ce{Fe2O3}? (Fe = 55.85, O = 16.00)

  1. Formula mass: 2(55.85) + 3(16.00) = 111.70 + 48.00 = 159.70 u
  2. Fraction of the mass that is iron: 111.70159.70=0.6994\dfrac{111.70}{159.70} = 0.6994 (that is, 69.94% Fe)
  3. Mass of iron: 0.6994×50.0 g=35.0 g0.6994 \times 50.0\ \text{g} = 35.0\ \text{g}

Answer: 50.0 g of FeX2OX3\ce{Fe2O3} contains 35.0 g of iron.

Common mistake

Common mistake: Ignoring subscripts and parentheses

In FeX2OX3\ce{Fe2O3} there are two iron atoms, so iron contributes 2 × 55.85 = 111.70 u, not 55.85 u. Leaving out the 2 gives half the correct mass of iron (17.5 g instead of 35.0 g in the example above).

In Ca(OH)X2\ce{Ca(OH)2}, the 2 applies to both O and H: the formula mass is 74.10 u, not 58.10 u.

Common mistake: Counting atoms instead of weighing them

Two-thirds of the atoms in water are hydrogen, but hydrogen is not 66.7% of water. Percent composition is by mass: hydrogen is 11.19% of water’s mass.

Common mistake: Rounding too early

If you round water’s formula mass to 18.02 u before dividing, you get 11.19% H and 88.79% O, which add up to only 99.98%. Keep extra digits (18.016 u) until the final step to get 11.19% and 88.81%.

Notation note

Different textbooks use different names for the same quantity:

  • Formula weight and molecular weight are traditional names; formula mass and molecular mass are more common in newer texts. All are given in u (also written amu or Da).
  • Some courses use relative formula mass, MrM_\text{r}. It has the same number but no units, because it is measured relative to 112\frac{1}{12} of the mass of a carbon-12 atom.
  • Percent composition is sometimes called mass percent or percentage by mass.

Remember this

Remember this

  • Formula mass = sum of the atomic masses of all atoms in the formula (watch subscripts and parentheses).
  • % element = (atoms of element × atomic mass) ÷ formula mass × 100%.
  • Percent composition is by mass, and it is the same for any sample of a pure compound.
  • Keep extra digits until the end; the percentages should add to 100%.

Test yourself

Check your understanding before moving on.

Flashcards

9 cards

  1. Question
    What does percent composition tell you?
    Answer

    The percentage of a compound's mass that comes from each element.

  2. Question
    How do you calculate the mass percent of an element in a compound?
    Answer

    (number of atoms of the element × atomic mass) ÷ formula mass × 100%

  3. Question
    What is a formula mass?
    Answer

    The sum of the atomic masses of all the atoms in a chemical formula, in u.

  4. Question
    What is the formula mass of Ca(OH)X2\ce{Ca(OH)2}? (Ca = 40.08, O = 16.00, H = 1.008)
    Answer

    40.08+2(16.00+1.008)=74.1040.08 + 2(16.00 + 1.008) = 74.10 u. The 2 multiplies both O and H.

  5. Question
    What is the mass percent of hydrogen in water, HX2O\ce{H2O}?
    Answer

    2(1.008)18.016×100%=11.19%\dfrac{2(1.008)}{18.016} \times 100\% = 11.19\%

  6. Question
    Two-thirds of the atoms in water are hydrogen. Why is hydrogen only 11.19% of its mass?
    Answer

    Hydrogen atoms are much lighter than oxygen atoms. Percent composition is by mass, not by number of atoms.

  7. Question
    Does the percent composition of a pure compound depend on the size of the sample?
    Answer

    No. A pure compound always has the same composition by mass, so a drop and a lake of pure water have the same percentages.

  8. Question
    What should the mass percents of all elements in a compound add up to?
    Answer

    100%, apart from small rounding differences. Keep extra digits until the end to avoid them.

  9. Question
    What is relative formula mass, MrM_\text{r}?
    Answer

    The formula mass expressed without units, relative to 112\frac{1}{12} of the mass of a carbon-12 atom. It has the same number as the formula mass in u.

Quiz

7 questions

  1. Question 1EasyWhat is the formula mass of calcium hydroxide, Ca(OH)X2\ce{Ca(OH)2}? (Ca = 40.08, O = 16.00, H = 1.008)
    Show answer

    Answer: 74.10 u

    The subscript 2 after the parentheses applies to both O and H: 40.08+2(16.00+1.008)=74.1040.08 + 2(16.00 + 1.008) = 74.10 u. The answer 58.10 u doubles only the H.

  2. Question 2EasyWhat is the mass percent of oxygen in water, HX2O\ce{H2O}?
    Show answer

    Answer: 88.81%

    16.0018.016×100%=88.81%\dfrac{16.00}{18.016} \times 100\% = 88.81\%. The answer 33.3% is the fraction of atoms that are oxygen (1 of 3), not the fraction of the mass.

  3. Question 3MediumWhat is the mass percent of carbon in carbon dioxide, COX2\ce{CO2}?
    Show answer

    Answer: 27.29%

    Formula mass = 12.01 + 2(16.00) = 44.01 u. % C = 12.0144.01×100%=27.29%\dfrac{12.01}{44.01} \times 100\% = 27.29\%. The answer 72.71% is the percentage of oxygen.

  4. Question 4MediumWhat is the mass percent of nitrogen in ammonia, NHX3\ce{NH3}?
    Show answer

    Answer: 82.25%

    Formula mass = 14.01 + 3(1.008) = 17.034 u. % N = 14.0117.034×100%=82.25%\dfrac{14.01}{17.034} \times 100\% = 82.25\%. The answer 25.0% counts atoms (1 of 4) instead of mass.

  5. Question 5HardWhat mass of iron is in 50.0 g of iron(III) oxide, FeX2OX3\ce{Fe2O3}? (Fe = 55.85, O = 16.00)
    Show answer

    Answer: 35.0 g

    Formula mass = 2(55.85) + 3(16.00) = 159.70 u. Mass of Fe = 111.70159.70×50.0 g=35.0\dfrac{111.70}{159.70} \times 50.0\ \text{g} = 35.0 g. The answer 17.5 g forgets that there are two Fe atoms in the formula.

  6. Question 6EasySample A is 1.00 g of pure water and sample B is 1.00 kg of pure water. How do their mass percents of hydrogen compare?
    Show answer

    Answer: They are the same

    A pure compound always has the same composition by mass. Both samples are 11.19% hydrogen; only the total mass is different.

  7. Question 7MediumWhat is the mass percent of calcium in calcium carbonate, CaCOX3\ce{CaCO3}? (Ca = 40.08, C = 12.01, O = 16.00)
    Show answer

    Answer: 40.04%

    Formula mass = 40.08 + 12.01 + 3(16.00) = 100.09 u. % Ca = 40.08100.09×100%=40.04%\dfrac{40.08}{100.09} \times 100\% = 40.04\%. The answer 20.0% counts atoms (1 of 5).

References

  1. Brown, T. L.; LeMay, H. E., Jr.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M.; Stoltzfus, M. W. Chemistry: The Central Science, 15th ed.; Pearson, 2022.

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