Percent composition formula
Percent composition gives the mass percent of each element in a compound.
mass % of element = (number of atoms × atomic mass ÷ molar mass) × 100
The table shows the percent composition of glucose, C6H12O6. Its molar mass is 180.156 g/mol.
| Element | Atoms | Mass in formula (g/mol) | Mass percent |
|---|---|---|---|
| Carbon (C) | 6 | 6 × 12.011 = 72.066 | 40.00% |
| Hydrogen (H) | 12 | 12 × 1.008 = 12.096 | 6.71% |
| Oxygen (O) | 6 | 6 × 15.999 = 95.994 | 53.28% |
How to find the empirical formula from percent composition
- Assume a 100 g sample. Each percent becomes a mass in grams.
- Divide each mass by the atomic mass of the element to get moles.
- Divide every mole value by the smallest mole value.
- If a ratio ends near .5, .33, or .25, multiply all ratios by 2, 3, or 4.
- Round to whole numbers and use them as subscripts.
Percent composition and empirical formula problems
Problem 1
Find the percent composition of CO2.
- The molar mass of CO2 is 12.011 + 2 × 15.999 = 44.009 g/mol.
- Carbon: 12.011 ÷ 44.009 × 100 = 27.29%.
- Oxygen: 31.998 ÷ 44.009 × 100 = 72.71%.
Answer: 27.29% C and 72.71% O
Problem 2
A compound is 40.0% C, 6.7% H, and 53.3% O. Find the empirical formula.
- In 100 g: C = 40.0 ÷ 12.011 = 3.33 mol, H = 6.7 ÷ 1.008 = 6.65 mol, O = 53.3 ÷ 15.999 = 3.33 mol.
- Divide by 3.33: C = 1, H = 2.0, O = 1.
Answer: CH2O
Problem 3
A compound is 72.4% Fe and 27.6% O. Find the empirical formula.
- In 100 g: Fe = 72.4 ÷ 55.845 = 1.296 mol, O = 27.6 ÷ 15.999 = 1.725 mol.
- Divide by 1.296: Fe = 1, O = 1.33.
- Multiply both values by 3: Fe = 3, O = 4.
Answer: Fe3O4
Problem 4
2.43 g of magnesium burns to form 4.03 g of magnesium oxide. What is the empirical formula of magnesium oxide?
- Mass of oxygen = 4.03 g − 2.43 g = 1.60 g.
- Mg = 2.43 ÷ 24.305 = 0.100 mol. O = 1.60 ÷ 15.999 = 0.100 mol.
- The mole ratio is 1 : 1.
Answer: MgO
How to use this calculator
- Enter each element symbol with its mass percent or mass in grams.
- Separate the element-value pairs with commas.
- Calculate and check the mole ratios.
- Change the tolerance only when measurement uncertainty supports the change.
Formula and variables
ratioᵢ = (massᵢ/Aᵢ) / min(mass/A)
| Symbol | Meaning | Unit |
|---|---|---|
| massᵢ | element mass | g or % |
| Aᵢ | atomic mass | g/mol |
| ratioᵢ | whole-number ratio | dimensionless |
Worked example
A compound is 40.0% C, 6.7% H, and 53.3% O. In 100 g, the mole ratio is 3.33 : 6.65 : 3.33, which gives CH2O.
Interpretation
The accepted multiplier is the smallest one that moves every normalized ratio within the visible tolerance of an integer.
Common mistakes
- Use a consistent mass basis for all values.
- Do not hide noisy ratios with an excessively broad tolerance.
Frequently asked questions
What is percent composition?
Percent composition is the mass percent of each element in a compound. The percentages add up to 100%.
How do you find the empirical formula from percent composition?
Treat each percent as grams in a 100 g sample. Convert to moles, divide by the smallest value, and round to whole numbers.
What is the empirical formula of magnesium oxide?
The empirical formula of magnesium oxide is MgO. Magnesium and oxygen combine in a 1:1 mole ratio.
Do the percentages have to total exactly 100?
No. Small differences can result from rounding or measurement uncertainty.
Why does the calculator use a ratio multiplier?
A multiplier converts ratios such as 1.5 or 1.33 to defensible whole-number subscripts.
Can I use scientific notation?
Yes. Enter values such as 6.022e23 or 1.0e-6 in a numeric field.
How should I round the result?
Keep extra digits during the calculation. Round the final result to the precision required by your data.