Empirical formula calculator (and molecular formula)
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Result
| Element | Mass (g) | Moles | Ratio | Whole number |
|---|---|---|---|---|
| C | 40 | 3.3303 | 1 | 1 |
| H | 6.7 | 6.6468 | 1.996 | 2 |
| O | 53.3 | 3.3315 | 1 | 1 |
Find the empirical formula of a compound from its percent composition by mass or from the masses of each element in a sample. The calculator converts the masses to moles, divides by the smallest amount, recognises ratios such as 1.5 or 1.33 that must be multiplied to whole numbers, and gives the empirical formula. Add the molar mass to get the molecular formula as well.
Step-by-step solution
Assume a 100 g sample
Convert each mass to moles
Divide by the smallest amount
Round to whole numbers
Empirical formula
Molecular formula
Key formulas
- Moles
- n = m/M
- Mole ratio
- use the coefficients of the balanced equation
How to find an empirical formula
- If the composition is given in percent, assume a 100 g sample, so each percentage becomes a mass in grams.
- Convert each mass to moles by dividing by the atomic mass of the element.
- Divide every amount of moles by the smallest one to get the mole ratio.
- If a ratio is close to a whole number, round it. If it ends in about .5, .33, .25 or .2, multiply all ratios by 2, 3, 4 or 5 first.
- Write the empirical formula with these whole numbers as subscripts. For the molecular formula, divide the molar mass by the empirical formula mass and multiply every subscript by the result.
Questions and answers
What is the difference between an empirical and a molecular formula?
The empirical formula gives the simplest whole-number ratio of the atoms; the molecular formula gives the actual number of atoms in a molecule. Glucose has the molecular formula C6H12O6 and the empirical formula CH2O.
Why can’t I just round 1.5 to 2?
A ratio of 1.5 is not a rounding error but a ratio of 3 to 2. Rounding would give the wrong compound. Multiply all ratios by 2 instead: 1 : 1.5 becomes 2 : 3, as in Fe2O3.
What if the percentages do not add up to 100%?
Often one element, usually oxygen, was not measured directly. Its percentage is 100% minus the sum of the others. Small deviations of a few tenths of a percent are just rounding in the data.
How accurate does the data need to be?
For small subscripts, mass percentages to one decimal place are usually enough. For compounds with many atoms, such as large organic molecules, rounding can hide the right ratio, and more precise data or the molar mass is needed.
Practise it yourself
Practise Stoichiometry
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