Stoichiometry & gases · SL/HL
Empirical formula
Find a small whole-number ratio from elemental amounts.
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Number formats & units
Use decimal dots in datasets. Single values also accept decimal commas and scientific notation such as 1.2e-3 or 1.2 × 10⁻³.
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Result & working
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Interpretation & assumptions
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Formula & example
The relationship
What the variables mean
| Symbol | Meaning |
|---|---|
| amount | Mass or percentage assigned to each element. |
| atomic mass | Relative atomic mass used to obtain mole ratios. |
Worked example
- Element data
- C 40.00 12.011 H 6.71 1.008 O 53.29 15.999
- Molecular molar mass / g mol⁻¹ (optional)
- 180.16
Empirical formula: CH2O
Follow the example steps
- C: amount/Ar = 40 / 12.011 = 3.330281
- H: amount/Ar = 6.71 / 1.008 = 6.656746
- O: amount/Ar = 53.29 / 15.999 = 3.330833
- C: 3.330281 / 3.330281 = 1
- H: 6.656746 / 3.330281 = 1.998854
- O: 3.330833 / 3.330281 = 1.000166
- Multiply ratios by 1 → 1 : 2 : 1 → CH2O.
- Molecular multiplier ≈ 180.16 / 30.026 = 6.000133 ≈ 6; molecular formula C6H12O6.
Common mistakes
Multiply a fractional mole ratio before rounding.
Incorrect approach
1 : 1.5 → 1 : 2
Correct approach
1 : 1.5 → 2 : 3
More checks before you submit
- Do not round 1.50 to 2 before testing a common multiplier of 2.
- Percent composition can be treated as grams in an arbitrary 100 g sample.
Method notes
The integer ratio step requires chemical judgement. The calculator searches small multipliers for a defensible whole-number ratio and reports the residual.