To find an empirical formula, convert the amounts of each element into a mole ratio, then express that ratio using the smallest whole numbers. To find a molecular formula, compare the empirical formula mass with the supplied relative molecular mass. Keeping those two jobs separate prevents most common errors.
These empirical and molecular formula worked examples use original theoretical composition datasets. They are calculation exercises, not claims about laboratory measurements or instructions for preparing compounds. Use the stated relative atomic masses throughout: H = 1, C = 12, N = 14, O = 16, Na = 23 and S = 32.
Understand what each formula tells you
An empirical formula gives the simplest whole-number ratio of atoms of the elements in a compound. A molecular formula gives the actual number of each type of atom in one molecule. Several molecular formulae can share an empirical formula, so composition alone does not necessarily identify a unique molecule.
For example, both C2H4 and C3H6 simplify to CH2. To choose between possible multiples, you need additional information such as relative molecular mass. OpenStax’s formula-determination reference explains this distinction and the mass-to-mole method.
An ionic compound does not consist of separate molecules in the same sense. You may calculate its simplest formula from composition, but should not automatically describe that as a molecular formula. The sodium sulfate practice task below is an example where this language matters.
The working table that keeps the logic visible
- Write the mass of each element, in the same units.
- Divide each mass by its corresponding molar mass to find amount in moles.
- Divide every amount by the smallest non-zero amount.
- If necessary, multiply the entire ratio to obtain small whole numbers.
- Write the formula and check its composition against the original information.
If percentages are given, imagine a 100 g sample: 18% becomes 18 g within that convenient sample. This does not claim that an actual 100 g sample was weighed. Choosing a different total mass would give the same ratio. If a compound contains only specified elements and one percentage is omitted, subtract the others from 100%; do not assume an unmentioned element without justification.
Worked example 1: use masses, not their direct ratio
A theoretical 9.00 g sample contains 3.60 g carbon, 0.60 g hydrogen and 4.80 g oxygen. Determine the empirical formula.
| Element | Mass (g) | Amount (mol) | Ratio after dividing by 0.30 |
|---|---|---|---|
| C | 3.60 | 3.60 ÷ 12 = 0.30 | 1 |
| H | 0.60 | 0.60 ÷ 1 = 0.60 | 2 |
| O | 4.80 | 4.80 ÷ 16 = 0.30 | 1 |
The mole ratio is C:H:O = 1:2:1, giving CH2O. Subscript 1 is omitted. The direct mass ratio, 6:1:8 after division by 0.60, is not the atom ratio: carbon, hydrogen and oxygen atoms have different masses.
Check the result using an empirical formula mass of 12 + 2(1) + 16 = 30. The carbon mass fraction is 12/30 = 40%, agreeing with 3.60/9.00 = 40%. Hydrogen gives 2/30 and oxygen 16/30, matching the remaining masses. This reverse check can reveal a copied number or a misplaced subscript before you proceed.
Worked example 2: preserve a fractional ratio
A theoretical nitrogen oxide contains 7.00 g nitrogen and 20.00 g oxygen. Determine its empirical formula.
Nitrogen amount = 7.00 ÷ 14 = 0.500 mol. Oxygen amount = 20.00 ÷ 16 = 1.250 mol. Dividing by 0.500 gives N:O = 1:2.5.
The value 2.5 is not close enough to 2 or 3 to round to either. Multiply both parts by 2: 1 × 2 : 2.5 × 2 = 2:5. The empirical formula is therefore N2O5.
Check the mass relationship: two nitrogen atoms contribute relative mass 28 and five oxygen atoms contribute 80. Their ratio 28:80 simplifies to 7:20, exactly matching the supplied masses. Writing NO3 would predict 14:48, a different composition. Writing N2O2.5 would multiply only one part of the ratio and also be wrong.
Useful fractional patterns include 0.5, requiring multiplication by 2; approximately one third or two thirds, often requiring 3; and quarters, often requiring 4. Apply the multiplier to every element. Treat these as ratio clues, not a licence to force badly inconsistent data into a convenient formula.
Worked example 3: percentages to a molecular formula
A molecular compound has composition 62.07% carbon, 10.34% hydrogen and 27.59% oxygen by mass. Its relative molecular mass is 116. Determine both formulae.
On a 100 g basis, divide 62.07 by 12, 10.34 by 1 and 27.59 by 16. The amounts are approximately 5.1725 mol C, 10.3400 mol H and 1.724375 mol O. Keep these digits while forming the ratio.
Divide all three by 1.724375. The ratio is approximately 2.9996:5.9964:1. Because the supplied percentages were rounded, this supports the whole-number ratio 3:6:1. The empirical formula is C3H6O, with empirical formula mass 3(12) + 6(1) + 16 = 58.
The molecular multiplier is 116 ÷ 58 = 2. Multiply every empirical subscript by 2 to obtain C6H12O2. Its relative molecular mass is 6(12) + 12(1) + 2(16) = 116, as required.
Finally, its carbon percentage is 72/116 × 100 = 62.0689…%, which rounds to 62.07%. Hydrogen gives 12/116 × 100 = 10.3448…%, and oxygen gives 32/116 × 100 = 27.5862…%. Both round to the stated values. The molecular formula passes both the mass and composition checks, though it does not specify the molecule’s structural arrangement.
Know when to stop and question the data
The molecular multiplier should be a positive whole number, allowing for the precision of supplied measurements. Suppose an exercise gives empirical formula CH2O and relative molecular mass 75. Its multiplier would be 75/30 = 2.5. Rounding that to 3 would change the implied molecular mass to 90. Instead, recheck the empirical formula, the supplied mass and whether the question was copied correctly.
Relative atomic, formula and molecular masses are dimensionless. Molar mass has units such as g mol−1. They can have the same numerical value in the usual school calculation convention, but writing “relative molecular mass = 116 g” confuses the quantities.
Three solved practice tasks
Practice 1: a three-element ionic formula
A theoretical composition contains 4.60 g sodium, 3.20 g sulfur and 6.40 g oxygen. Find its empirical formula.
Solution: Amounts are 4.60/23 = 0.200 mol Na, 3.20/32 = 0.100 mol S and 6.40/16 = 0.400 mol O. Divide by 0.100 to obtain 2:1:4. The formula is Na2SO4. Do not call this a molecular formula for individual sodium sulfate molecules.
Practice 2: scale an empirical formula
A molecular compound has empirical formula CH and relative molecular mass 78. Determine the molecular formula.
Solution: Empirical formula mass = 12 + 1 = 13. Multiplier = 78/13 = 6. The molecular formula is C6H6. Both subscripts are multiplied, including the unwritten 1.
Practice 3: avoid premature rounding
A composition calculation gives C:H:O amounts of 0.080, 0.120 and 0.080 mol. Find the simplest formula.
Solution: Divide by 0.080 to obtain 1:1.5:1. Multiply all parts by 2 to obtain 2:3:2, giving C2H3O2. Rounding 1.5 to 2 would wrongly produce CH2O.
From correct formulae to balanced ionic equations
A formula fixes a substance’s composition; coefficients count whole units of that substance. To write an ionic compound’s formula, make the total positive and negative charges equal. Al³⁺ and SO₄²⁻ need a 2:3 ratio: 2(+3) + 3(−2) = 0, giving Al₂(SO₄)₃. Brackets preserve the sulfate group. AlSO₄ would retain a net positive charge and is not the neutral salt formula.
When balancing a reaction, change coefficients, never a correct formula’s subscripts. State symbols mean solid (s), liquid (l), gas (g) and aqueous solution (aq). Aqueous does not mean that the pure substance is a liquid. These are paper equations and supplied observations, with no preparation or testing activity.
Worked example: audit an existing formula equation
The carbonate evidence lesson explains why its fictional record becomes cloudy and later clear. Use its first equation: Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l). The atom inventory on each side is Ca = 1, C = 1, H = 2 and O = 4. In particular, the bracket in Ca(OH)₂ supplies two O and two H atoms. Every displayed formula unit is neutral, so both total charges are zero.
The reported cloudiness is the observation. Formation of CaCO₃(s) is the interpretation supported by the named reactants and reaction model. The equation’s (s) explains the solid, but the atom check alone cannot establish that this reaction actually occurred. Keep the source lesson’s later hydrogencarbonate reaction separate; it explains a different stage.
Worked example: cancel spectators without losing charge
A new fictional aqueous record specifies calcium chloride and sodium carbonate, with concentrations sufficient for calcium carbonate precipitation. A white solid is reported. The formula equation is CaCl₂(aq) + Na₂CO₃(aq) → CaCO₃(s) + 2NaCl(aq). Ca = 1, Cl = 2, Na = 2, C = 1 and O = 3 on each side.
Represent the soluble strong electrolytes as their dissolved ions, but keep the solid together. The complete ionic equation is Ca²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + CO₃²⁻(aq) → CaCO₃(s) + 2Na⁺(aq) + 2Cl⁻(aq). Both totals are zero charge. Sodium and chloride ions appear unchanged on both sides. Cancelling equal amounts of those spectators gives:
Ca²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s). Its left charge is +2 − 2 = 0, matching the neutral solid. Its atom inventory is Ca₁C₁O₃ on both sides. Splitting CaCO₃(s) back into aqueous ions would erase the very precipitation being explained.
Do not split every aqueous formula automatically: weak acids, weak bases and molecular solutes need their appropriate representation. Nor is a half-equation a complete ionic reaction. The electrolysis lesson checks Cu²⁺ + 2e⁻ → Cu at a cathode; a complete cell account also needs a corresponding oxidation. Electrons cancel when matched half-equations are combined.
Independent equation checks
- Using Mg²⁺ and Cl⁻, write the neutral formula. Then balance Al(s) + O₂(g) → Al₂O₃(s) without changing any formula.
- For the supplied strong-acid/strong-alkali reaction HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l), write the net ionic equation and check atoms and charge.
Checked answers
1. MgCl₂: +2 + 2(−1) = 0. The balanced paper equation is 4Al(s) + 3O₂(g) → 2Al₂O₃(s), with four Al and six O atoms on each side. 2. Cancel Na⁺ and Cl⁻ to obtain H⁺(aq) + OH⁻(aq) → H₂O(l). Both sides have H = 2, O = 1 and total charge zero. H⁺ is conventional aqueous proton notation; the hydrated form is H₃O⁺(aq) + OH⁻(aq) → 2H₂O(l), which also balances.
Background: OpenStax, precipitation and ionic reactions. The audits and independent questions are original; the first carbonate reaction is reused with its existing explanation.
Frequently asked questions
Why divide by the smallest mole amount?
It normalises the ratio so one component becomes 1. Other components may still be fractional, requiring a common multiplier.
Can I use percentages without knowing sample mass?
Yes. Use a convenient 100 g basis. You need relative composition rather than a particular sample size.
Must the empirical and molecular formulae differ?
No. If the molecular multiplier is 1, they are identical. Do not assume the molecular formula must be larger.
Should I round every ratio to one decimal place?
No fixed rounding rule works for every dataset. Retain intermediate precision and inspect whether a small whole-number ratio is supported by the data’s stated precision.
Can different compounds share an empirical formula?
Yes. It gives a ratio, not a unique identity. Even a molecular formula can correspond to different structural arrangements.
What if the percentages do not total exactly 100?
Small differences may result from rounding. A substantial discrepancy needs explanation, such as missing composition data; do not silently adjust numbers to force a result.
How can I check the final answer quickly?
Recalculate the formula mass and at least one percentage. For a molecular formula, also verify that its relative molecular mass matches the supplied value.
Choose your next calculation carefully
Practise explaining why the mole ratio differs from the mass ratio before attempting another question. The existing KCSE Chemistry topic-tests page currently requires login. The examples here provide complete theoretical practice without promising access to any particular gated file.
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