Worked example 1
A compound is 40.00 % C, 6.71 % H and 53.29 % O by mass. Find its empirical formula.
Try it first: What mass of each element would a 100.00 g sample contain?
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Empirical formulaCourse level: High School / HonorsCourse level: AP Chemistry
Empirical and Molecular Formulas
What you'll be able to do: Turn mass or percent data into an empirical formula, then use a molar mass to find the molecular formula.
Best after: The Mole and Molar Mass, Percent Composition
Introduction
Percent composition goes from a formula to masses. Empirical formula work runs that process backwards: you start with laboratory masses and finish with the simplest whole-number ratio of atoms. This is how unknown compounds were identified long before modern instruments, and it is still how most introductory analysis questions are framed.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Convert element masses or mass percentages into moles of each element.
- Determine the empirical formula from a mole ratio.
- Clear fractional ratios into whole numbers correctly.
- Use a measured molar mass to convert an empirical formula into a molecular formula.
Empirical versus molecular
The empirical formula is the simplest whole-number atom ratio; the molecular formula is the actual number of atoms in one molecule. Glucose has the empirical formula CH₂O and the molecular formula C₆H₁₂O₆. Both describe the same substance, but only the molecular formula gives the true molecule.
Ionic formulas describe electrically neutral ratios of ions rather than discrete molecules. Preserve the identities of polyatomic ions: Na₂O₂ contains peroxide O₂²⁻ and must not be reduced to NaO as a chemical formula.
An ionic formula describes the compound’s ion stoichiometry; do not imply that every ionic substance has discrete formula-unit molecules.
The percent-to-mass shortcut
If you are given percentages rather than masses, assume a 100.00 g sample. Then each percentage is directly a mass in grams: 40.00 % C becomes 40.00 g C.
Nothing is lost, because the ratio of atoms does not depend on how much of the compound you have.
Write "assume 100.00 g" explicitly. Graders and your future self both need to see that step.
Divide by the smallest
After converting each mass to moles, divide every mole value by the smallest of them. That forces the least-abundant element to 1 and expresses the others relative to it. Small deviations such as 1.98 or 3.01 are experimental noise and round to 2 and 3.
Use a table with columns for element, measured mass, molar mass, moles, and normalized mole ratio. The smallest mole amount is a scaling choice, not necessarily the element with the smallest mass. For example, 12.0 g C and 2.02 g H are nearly 1.00 mol C and 2.00 mol H: carbon has more mass but fewer atoms. The ratio is CH₂, not C₆H.
Keep several digits until you inspect all ratios together. Measurement precision determines how close a ratio must be to a simple fraction; there is no universal rounding window that makes every dataset valid. If no small common multiplier gives plausible integers, check masses, missing elements, sample purity and molar masses before assigning a formula.
ratioX = nX / n_smallest
When the ratio is not whole
A result near 0.5 means multiply everything by 2; 0.33 or 0.67 means multiply by 3; 0.25 or 0.75 means multiply by 4. Multiply every element, not just the fractional one. A value like 2.5 that you round to 3 changes the compound entirely, so resist rounding anything between about 0.2 and 0.8 away from its fraction.
For a normalized ratio A:B:C = 1.00:1.33:2.00, multiply all entries by three to obtain 3:4:6. For 1.00:1.75, multiply both by four to obtain 4:7. A ratio of 1.77 is not close enough to 1.50 to justify replacing it by 1.50 from ordinary two-significant-figure data. Finally calculate the proposed formula's theoretical mass percentages and compare them with the original observations. This reverse calculation can expose a plausible-looking but unsupported integer ratio.
Never round 1.5 to 2. Multiply the whole ratio by 2 instead.
From empirical to molecular
Compute the empirical formula mass, then divide the measured molar mass by it. The result should be very close to a whole number n, and the molecular formula is the empirical formula with every subscript multiplied by n.
If n comes out as 2.6 rather than 3, either the empirical formula or the molar mass is wrong.
n = M(molecular) / M(empirical)
Combustion analysis (Challenge extension)
In combustion analysis a hydrocarbon is burned and the CO₂ and H₂O produced are weighed.
All carbon in the CO₂ came from the sample, and all hydrogen in the H₂O came from the sample, so mol C = mol CO₂ and mol H = 2 x mol H₂O.
Any oxygen in the compound is found by difference: mass O = sample mass minus mass C minus mass H. This material goes beyond the core requirement and is optional on a first pass.
Challenge material. Core learners can skip this section and still meet every objective above.
Recover oxygen by difference in combustion data
When a compound containing C, H and O burns completely, the collected CO₂ tells you the starting moles of carbon and the H₂O tells you twice the starting moles of hydrogen.
These product masses do not directly tell you the sample's oxygen, because combustion also adds oxygen from O₂. Calculate the masses of C and H originally present, subtract both from the initial sample mass, and convert the remaining oxygen mass to moles.
Only then divide all three mole amounts by the smallest. The subtraction is a mass balance on the original sample, not on the total products.
Key ideas
Definition
Empirical formula
The simplest whole-number ratio of atoms in a compound.
Definition
Molecular formula
The actual number of each kind of atom in one molecule; always a whole-number multiple of the empirical formula.
Rule
Mass to moles first
Atom ratios only appear after masses are converted to moles. Comparing grams directly gives a meaningless ratio.
Rule
Multiply, do not round, near-halves
Ratios ending in 0.5, 0.33 or 0.25 are real fractions. Scale the entire ratio up instead of rounding.
Assumption
The sample is pure
Empirical formula analysis assumes every gram of the sample belongs to one compound.
Equation
Mole ratio from mass
nX = mX / AX
- nX = moles of element X
- mX = mass of X in the sample, in g
- AX = atomic mass of X, in g/mol
- use when = converting each element's mass into moles at the start of the problem
- limits = use the atomic mass of the element, not the molar mass of the compound
Equation
Molecular formula multiplier
n = M(molecular) / M(empirical)
- n = whole-number multiplier applied to every subscript
- M(molecular) = measured molar mass of the compound, in g/mol
- M(empirical) = molar mass of the empirical formula unit, in g/mol
- use when = an experimental molar mass is supplied alongside composition data
- limits = n must be close to a whole number, otherwise something upstream is wrong
Worked examples
Worked example 2
The compound above has a measured molar mass of 180.2 g/mol. What is its molecular formula?
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Worked example 3
A 2.50 g compound contains 1.75 g Fe and 0.75 g O. Determine its empirical formula. Use Fe = 55.85 and O = 16.00 g/mol.
Try it first: Find the oxygen mass if needed, then convert both elements to moles before choosing a multiplier.
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Worked example 4
A 4.60 g compound containing only C, H and O yields 8.80 g CO₂ and 5.40 g H₂O on complete combustion. Find its empirical formula.
Try it first: Obtain C from CO₂, H from H₂O and O by subtracting their masses from 4.60 g.
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Common mistakes
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Building the atom ratio directly from grams. | Grams weigh atoms of different mass, so 1 g of H and 1 g of O are not equal numbers of atoms. | Convert every mass to moles first, then form the ratio. |
| Rounding a ratio of 1.5 up to 2. | 1.5 is a real 3:2 ratio; rounding it produces a different compound, for example FeO instead of Fe₂O₃. | Multiply every element in the ratio by 2, 3 or 4 to clear the fraction. |
| Multiplying only the fractional subscript when clearing a fraction. | Scaling one element changes the composition instead of rewriting the same ratio. | Apply the multiplier to every element, then confirm the ratio is unchanged. |
| Reporting the empirical formula when a molar mass was given. | A supplied molar mass is a signal that the molecular formula is wanted. | If the problem gives M, always divide by the empirical formula mass and check the multiplier. |
| Ignoring oxygen when it is not listed explicitly. | In many analyses the oxygen mass is only available by subtracting the other elements from the sample mass. | Add up the listed element masses; any shortfall against the sample mass is usually oxygen. |
Scroll horizontally if needed to read all columns.
After the lesson · Question bank
Independent practice
Work through new problems on your own. The lesson's quick checks and worked examples stay above if you need to review a step.
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What you should now know
Convert each element's mass to moles, divide every value by the smallest, and clear any fraction to whole numbers. That gives the empirical formula. Comparing the empirical formula mass with a measured molar mass gives the whole-number multiplier that produces the molecular formula.
- Empirical formula is the simplest whole-number atom ratio; molecular formula is the actual count.
- Assume a 100.00 g sample to turn percentages straight into grams.
- Convert to moles, divide by the smallest, then clear fractions by multiplying the whole ratio.
- Divide the measured molar mass by the empirical formula mass to get the molecular multiplier.
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Sources and further reading
- Chemistry 2e, Section 3.2: Determining Empirical and Molecular FormulasPaul Flowers, Klaus Theopold, Richard Langley, William R. Robinson · OpenStax, Rice University · Chapter 3.2View source
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Source rights and review status - IUPAC Commission on Isotopic Abundances and Atomic Weights: standard atomic weightsIUPAC (CIAAW)View source
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Source rights and review status - General Chemistry bookshelf (LibreTexts)LibreTexts, University of California, DavisView source
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