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Empirical and Molecular Formulas

How do you find an empirical formula?

IntermediateReactions & Stoichiometry

What is it?

An empirical formula shows the simplest whole-number ratio of the atoms in a compound.

A molecular formula shows the actual number of each kind of atom in one molecule.

CompoundMolecular formulaEmpirical formula
GlucoseCX6HX12OX6\ce{C6H12O6}CHX2O\ce{CH2O}
Hydrogen peroxideHX2OX2\ce{H2O2}HO\ce{HO}
BenzeneCX6HX6\ce{C6H6}CH\ce{CH}
WaterHX2O\ce{H2O}HX2O\ce{H2O} (already the simplest ratio)

Key idea

The molecular formula is always a whole-number multiple of the empirical formula. Glucose, CX6HX12OX6\ce{C6H12O6}, is 6 × CHX2O\ce{CH2O}.

Why does it matter?

When chemists analyse an unknown compound, the experiment does not hand them a formula. It tells them how much of each element is present, usually as a percentage by mass.

Working out the empirical formula turns those measurements into a ratio of atoms. Add a measured molar mass and you get the molecular formula. This is one of the first steps in identifying a new or unknown substance.

How does it work?

From percent composition to empirical formula

  1. Assume a 100 g sample. Each percentage then becomes a mass in grams (40.00% → 40.00 g).
  2. Convert grams to moles by dividing by each element’s atomic mass.
  3. Divide every mole value by the smallest one. This gives the mole ratio.
  4. Make whole numbers. Values very close to whole numbers (such as 1.99 or 2.01) can be rounded. If a ratio ends in about .5, multiply every value by 2; about .33 or .67, multiply by 3; about .25 or .75, multiply by 4.
  5. Write the formula using these whole numbers as subscripts.

From empirical formula to molecular formula

You need the compound’s molar mass, which is measured separately. Then:

n=molar massempirical formula massn = \frac{\text{molar mass}}{\text{empirical formula mass}}

Multiply every subscript in the empirical formula by nn (which should be a whole number).

Think of it like this

A recipe that uses 2 cups of flour for every 1 cup of sugar gives the ratio, 2 : 1. That’s like the empirical formula. A baker who uses 6 cups of flour and 3 cups of sugar has the same ratio but a bigger batch. That’s like the molecular formula. To know the batch size you need one more piece of information: for a compound, its molar mass.

More precisely

Experimental percentages carry measurement error, so mole ratios rarely come out as exact whole numbers. A value such as 1.98 is safely rounded to 2, but 1.50 is not “nearly 2”: it signals a ratio of 3 : 2. For compounds containing carbon and hydrogen, the percentages often come from combustion analysis, in which a sample is burned and the masses of COX2\ce{CO2} and HX2O\ce{H2O} formed are measured.

Visualise it

Flow chart. Percent by mass: assume 100 grams so percentages become grams. Grams of each element: divide by atomic mass. Moles of each element: divide by the smallest number of moles. Mole ratio: round to whole numbers, multiplying by 2 if a ratio ends in .5 or by 3 if it ends in .33. This gives the empirical formula, the simplest whole-number ratio. Multiply by n, where n equals the molar mass divided by the empirical formula mass, to get the molecular formula, the actual numbers of atoms.
From percentages to a molecular formula. Every route goes through moles.

Worked example

Worked example: Empirical and molecular formula of a sugar

Question: A compound is 40.00% C, 6.71% H and 53.29% O by mass. Its molar mass is 180.16 g/mol. Find its empirical and molecular formulas. (C = 12.01, H = 1.008, O = 16.00)

  1. In 100 g: 40.00 g C, 6.71 g H, 53.29 g O.
  2. Moles: C: 40.0012.01=3.331\dfrac{40.00}{12.01} = 3.331; H: 6.711.008=6.657\dfrac{6.71}{1.008} = 6.657; O: 53.2916.00=3.331\dfrac{53.29}{16.00} = 3.331
  3. Divide by the smallest (3.331): C = 1.000, H = 1.999, O = 1.000
  4. Whole numbers: C : H : O = 1 : 2 : 1, so the empirical formula is CHX2O\ce{CH2O}.
  5. Empirical formula mass: 12.01 + 2(1.008) + 16.00 = 30.03 g/mol
  6. n=180.1630.03=6.00n = \dfrac{180.16}{30.03} = 6.00, so multiply every subscript by 6.

Answer: empirical formula CHX2O\ce{CH2O}; molecular formula CX6HX12OX6\ce{C6H12O6} (glucose).

Worked example: When the ratio isn't a whole number

Question: An oxide of iron is 69.94% Fe and 30.06% O by mass. Find its empirical formula. (Fe = 55.85, O = 16.00)

  1. Moles: Fe: 69.9455.85=1.252\dfrac{69.94}{55.85} = 1.252; O: 30.0616.00=1.879\dfrac{30.06}{16.00} = 1.879
  2. Divide by the smallest: Fe = 1.000, O = 1.500
  3. 1.500 ends in .5, so multiply both by 2: Fe = 2, O = 3

Answer: the empirical formula is FeX2OX3\ce{Fe2O3}.

Common mistake

Common mistake: Rounding 1.5 up to 2

A ratio of 1 : 1.5 is not “about 1 : 2”. Rounding it would give FeO₂, which is wrong. Multiply every value by 2 instead to get 2 : 3. Only round values that are within a few hundredths of a whole number.

Common mistake: Using the percentages as the ratio

The percentages are a ratio of masses, not of atoms. In the sugar example, 40.00 : 6.71 : 53.29 says nothing directly about the number of atoms. Always convert to moles first.

Common mistake: Dividing by the wrong number

Divide by the smallest number of moles, not the largest. Dividing by the smallest makes the smallest value exactly 1 and the others 1 or more.

Notation note

  • The empirical formula is also called the simplest formula.
  • Formulas of ionic compounds such as NaCl\ce{NaCl} and FeX2OX3\ce{Fe2O3} are empirical formulas, because ionic compounds are not made of separate molecules.
  • A molecular formula shows how many atoms there are, but not how they are connected. That is shown by a structural formula.

Remember this

Remember this

  • Empirical formula = simplest whole-number ratio; molecular formula = actual numbers of atoms.
  • Percent → grams (assume 100 g) → moles → divide by the smallest → whole numbers.
  • A ratio ending in .5 means multiply by 2; don’t round it.
  • Molecular formula = empirical formula × (molar mass ÷ empirical formula mass).

Test yourself

Check your understanding before moving on.

Flashcards

9 cards

  1. Question
    What is an empirical formula?
    Answer

    The simplest whole-number ratio of the atoms in a compound.

  2. Question
    What is a molecular formula?
    Answer

    The actual number of each kind of atom in one molecule of a compound.

  3. Question
    What is the empirical formula of glucose, CX6HX12OX6\ce{C6H12O6}?
    Answer

    CHX2O\ce{CH2O} (divide every subscript by 6)

  4. Question
    What is the empirical formula of hydrogen peroxide, HX2OX2\ce{H2O2}?
    Answer

    HO\ce{HO}

  5. Question
    What are the steps from percent composition to an empirical formula?
    Answer

    Assume 100 g (% → g), divide by atomic masses (g → mol), divide by the smallest number of moles, then make whole numbers.

  6. Question
    Your mole ratio comes out as 1 : 1.50. What should you do?
    Answer

    Multiply both by 2 to get 2 : 3. Never round 1.5 to 2.

  7. Question
    How do you get the molecular formula from the empirical formula?
    Answer

    Find nn = molar mass ÷ empirical formula mass, then multiply every subscript by nn.

  8. Question
    A compound has empirical formula CHX2\ce{CH2} and molar mass 42.08 g/mol. What is its molecular formula?
    Answer

    n=42.08÷14.03=3n = 42.08 \div 14.03 = 3, so CX3HX6\ce{C3H6}.

  9. Question
    Why can't you use the percentages directly as the atom ratio?
    Answer

    Percentages compare masses. Atoms of different elements have different masses, so you must convert to moles first.

Quiz

7 questions

  1. Question 1EasyWhat is the empirical formula of glucose, CX6HX12OX6\ce{C6H12O6}?
    Show answer

    Answer: CHX2O\ce{CH2O}

    Divide every subscript by the largest common factor, 6: CX6HX12OX6\ce{C6H12O6} → CHX2O\ce{CH2O}. CX2HX4OX2\ce{C2H4O2} has the right ratio but is not the simplest one.

  2. Question 2EasyWhich of these formulas is already an empirical formula?
    Show answer

    Answer: COX2\ce{CO2}

    In COX2\ce{CO2} the ratio 1 : 2 cannot be simplified. The others can: CX2HX6\ce{C2H6} → CHX3\ce{CH3}, NX2OX4\ce{N2O4} → NOX2\ce{NO2}, CX4HX10\ce{C4H10} → CX2HX5\ce{C2H5}.

  3. Question 3MediumAfter dividing by the smallest number of moles, you get Fe : O = 1 : 1.50. What is the empirical formula?
    Show answer

    Answer: FeX2OX3\ce{Fe2O3}

    A ratio ending in .5 must be multiplied by 2: 1 : 1.50 becomes 2 : 3, so FeX2OX3\ce{Fe2O3}. Rounding 1.5 up to 2 would give the wrong formula, FeOX2\ce{FeO2}.

  4. Question 4MediumA compound is 74.87% C and 25.13% H by mass. What is its empirical formula?
    Show answer

    Answer: CHX4\ce{CH4}

    Moles in 100 g: C = 74.87 ÷ 12.01 = 6.234; H = 25.13 ÷ 1.008 = 24.93. Divide by 6.234: C = 1.00, H = 4.00, so CHX4\ce{CH4}. The answer CX3H\ce{C3H} comes from using the percentages directly instead of moles.

  5. Question 5MediumA compound has empirical formula CHX2\ce{CH2} and molar mass 42.08 g/mol. What is its molecular formula?
    Show answer

    Answer: CX3HX6\ce{C3H6}

    Empirical formula mass = 12.01 + 2(1.008) = 14.03 g/mol. n=42.08÷14.03=3.00n = 42.08 \div 14.03 = 3.00, so the molecular formula is CX3HX6\ce{C3H6}.

  6. Question 6MediumBenzene has empirical formula CH\ce{CH} and molar mass 78.11 g/mol. What is its molecular formula?
    Show answer

    Answer: CX6HX6\ce{C6H6}

    Empirical formula mass = 12.01 + 1.008 = 13.02 g/mol. n=78.11÷13.02=6.00n = 78.11 \div 13.02 = 6.00, so CX6HX6\ce{C6H6}.

  7. Question 7EasyWhy do you divide every mole value by the smallest one?
    Show answer

    Answer: To make the smallest value 1 and find the ratio between elements

    Dividing by the smallest value scales the numbers so the smallest becomes exactly 1. The others then show how many atoms of each element there are for every one atom of that element.

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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