Empirical and Molecular Formulas

General Chemistry · Reaction Stoichiometry

Dr. Ray’s Learning Center · covers: Empirical and Molecular Formulas

20–25 min
8 figures
3 worked examples
4 practice sets

Before this lesson:

Percent Composition by Mass

When chemists make or isolate a new compound, they do not see its formula; they measure it. An analysis gives the mass of each element in a sample, and from those masses comes the simplest whole-number ratio of atoms, the empirical formula. One more measurement, the molar mass, turns it into the molecular formula.

Every problem in this lesson follows the path in Figure 1: masses to moles, moles to the smallest whole-number ratio, and, when the molar mass is known, the whole-number multiple.

startMassesgrams of each element (from % or combustion)
$\times \dfrac{1}{\text{atomic weight}}$ grams to moles
Molesof each element
$\times \dfrac{1}{\text{smallest}}$ then whole numbers
answerEmpirical formulasimplest whole-number ratio
$\times k$ k from the molar mass
answerMolecular formulaactual atoms per molecule

Subscripts are mole ratios, never mass ratios.Multiply by 2, 3 or 4 when a ratio is not close to whole.

Figure 1. The formula roadmap: masses to moles, moles to the simplest ratio, then the whole-number multiple k.

Learning Objectives

After this lesson, you should be able to:

Derive an empirical formula from mass-percent or combustion-analysis data.
Find the molecular formula from the empirical formula and the molar mass.

Key Ideas

1. Empirical and Molecular Formulas

The molecular formula gives the actual number of each kind of atom in one molecule. The empirical formula gives only the simplest whole-number ratio. Glucose, $\mathrm{C_6\!H_{12}\!O_6}$ , has the empirical formula $\mathrm{CH_2\!O}$ : every subscript divided by 6 (Figure 2).

\text{molecular formula} = (\text{empirical formula})_k

k = 1, 2, 3\ \text{or more}

Figure 2. Dividing every subscript by the same whole number k gives the empirical formula, for example

\mathrm{C_6\!H_{12}\!O_6} = (\mathrm{CH_2\!O})_6

. Water is already in its simplest ratio (k = 1).

Key idea: different compounds can share one empirical formula. Glucose and acetic acid are both CH₂O; only the molar mass tells them apart.

2. Subscripts Are Mole Ratios

A formula counts atoms, so its subscripts are ratios of moles, not of grams. To use mass-percent data, assume a 100 g sample so each percent becomes a mass in g, convert each mass to moles with the atomic weight, and divide every amount by the smallest one.

Formaldehyde is 40.00% C, 6.714% H and 53.28% O by mass. In 100 g:

n_{\mathrm{C}} = \dfrac{40.00\,\text{g}}{12.011\,\text{g/mol}} = 3.330\,\text{mol}

n_{\mathrm{H}} = \dfrac{6.714\,\text{g}}{1.008\,\text{g/mol}} = 6.661\,\text{mol}

n_{\mathrm{O}} = \dfrac{53.28\,\text{g}}{15.999\,\text{g/mol}} = 3.330\,\text{mol}

Dividing by the smallest amount, 3.330 mol, gives 1.000 : 2.000 : 1.000, so the empirical formula is $\mathrm{CH_2\!O}$ . The masses alone look nothing like this ratio (Figure 3).

Figure 3. The masses hide the formula; the moles show it:

3.330 : 6.661 : 3.330 = 1 : 2 : 1

If a ratio is not within about 0.05 of a whole number, do not round it. Multiply every ratio by the smallest whole number that clears it: 2 for a ratio ending near .5, 3 for .33 or .67, 4 for .25 or .75.

Key idea: grams to moles first, then divide by the smallest amount. Round only ratios that are already within about 0.05 of a whole number.

3. Combustion Analysis

A compound of carbon and hydrogen, possibly with oxygen, can be burned in excess oxygen. Every C atom ends up in $\mathrm{CO_2}$ and every H atom in $\mathrm{H_2\!O}$ , and the two products are trapped and weighed (Figure 4).

$n_{\mathrm{C}} = \frac{m_{\mathrm{CO_2}}}{44.01\,\text{g/mol}}$ : one C in each $\mathrm{CO_2}$
$n_{\mathrm{H}} = 2 \times \frac{m_{\mathrm{H_2\!O}}}{18.02\,\text{g/mol}}$ : two H in each $\mathrm{H_2\!O}$
$m_{\mathrm{O}} = m_{\text{sample}}-m_{\mathrm{C}}-m_{\mathrm{H}}$ : oxygen by difference, since the O in the products also comes from the oxygen gas

Figure 4. Combustion analysis: weigh the trapped products, then

n_{\mathrm{C}} = \frac{m_{\mathrm{CO_2}}}{44.01}

n_{\mathrm{H}} = 2 \times \frac{m_{\mathrm{H_2\!O}}}{18.02}

and

m_{\mathrm{O}} = m_{\text{sample}}-m_{\mathrm{C}}-m_{\mathrm{H}}

Key idea: C comes from the CO₂, H from the H₂O (times 2), and O from whatever mass is left.

4. From Empirical to Molecular Formula

The molar mass of a compound is a whole-number multiple $k$ of its empirical formula mass. Divide to find $k$ , then multiply every subscript by it (Figure 5):

k = \dfrac{M_{\text{compound}}}{M_{\text{empirical}}}

For $\mathrm{CH_2\!O}$ (30.03 g/mol), acetic acid has $M = 60.05\,\text{g/mol}$ and $k = 2.000 \approx 2$ , so it is $\mathrm{C_2\!H_4\!O_2}$ ; glucose has $M = 180.16\,\text{g/mol}$ and $k = 5.999 \approx 6$ .

Figure 5. Both compounds sit on a multiple of the CH₂O mass:

k = \frac{60.05}{30.03} = 2.000\text{,}\ \ k = \frac{180.16}{30.03} = 5.999

Worked Example 1: Empirical Formula from Mass Percents

Problem: A red-brown iron oxide contains 69.94% iron and 30.06% oxygen by mass. Find its empirical formula. Use Fe = 55.845 and O = 15.999 g/mol.

Step 1. Assume a 100 g sample, so each percent becomes a mass in g.

$m_{\mathrm{Fe}} = 69.94\,\text{g}$
$m_{\mathrm{O}} = 30.06\,\text{g}$

Step 2. Convert each mass to moles.

n_{\mathrm{Fe}} = \dfrac{69.94\,\text{g}}{55.845\,\text{g/mol}} = 1.252\,\text{mol}

n_{\mathrm{O}} = \dfrac{30.06\,\text{g}}{15.999\,\text{g/mol}} = 1.879\,\text{mol}

Step 3. Divide each amount by the smallest (1.252 mol).

\mathrm{Fe}: \dfrac{1.252}{1.252} = 1.000

\mathrm{O}: \dfrac{1.879}{1.252} = 1.501

Step 4. Multiply every ratio by 2 to make them whole numbers. 1.501 is close to 1.5, too far from a whole number to round.

\mathrm{Fe}: 1.000 \times 2 = 2.000

\mathrm{O}: 1.501 \times 2 = 3.002

Step 5. Write the whole numbers as subscripts. 2 Fe for every 3 O: $\mathrm{Fe_2\!O_3}$ .

Final Answer: Fe₂O₃

Masses $69.94\,\text{g}\ \mathrm{Fe}\text{,}\ 30.06\,\text{g}\ \mathrm{O}$ in a 100 g sample
Step 2 $\times \dfrac{1}{\text{atomic weight}}$ grams to moles
Moles $1.252 : 1.879$ Fe : O
Step 3 $\times \dfrac{1}{\text{smallest}}$ by 1.252 mol
Ratio $1.000 : 1.501$ 1.501 is near 1.5: do not round
Step 4 $\times 2$ clear the half
Formula $\mathrm{Fe_2\!O_3}$ 2.000 : 3.002

Figure 6. Worked Example 1 on the roadmap (Figure 1), including the multiply-by-2 step.

This is iron(III) oxide, the compound whose percent composition you calculated in the previous lesson: the method simply runs that calculation backwards (Figure 6).

Your turn: empirical formula from mass percents

Same five steps as Worked Example 1. Type the formula with plain numbers, for example Fe2O3.

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Worked Example 2: Combustion Analysis

Problem: A 0.2500 g sample of a liquid that contains only carbon, hydrogen and oxygen is burned completely. It produces 0.4777 g of $\mathrm{CO_2}$ and 0.2934 g of $\mathrm{H_2\!O}$ . Find the empirical formula. Use $\mathrm{CO_2}$ = 44.01 g/mol, $\mathrm{H_2\!O}$ = 18.02 g/mol, C = 12.011, H = 1.008 and O = 15.999 g/mol.

Step 1. Find the moles of C and H from the products.

n_{\mathrm{C}} = \dfrac{0.4777\,\text{g}}{44.01\,\text{g/mol}} = 0.01085\,\text{mol}

n_{\mathrm{H}} = 2 \times \dfrac{0.2934\,\text{g}}{18.02\,\text{g/mol}} = 0.03256\,\text{mol}

Step 2. Find the mass of O by difference, then its moles.

$m_{\mathrm{C}} = 0.01085\,\text{mol} \times 12.011\,\text{g/mol} = 0.1303\,\text{g}$
$m_{\mathrm{H}} = 0.03256\,\text{mol} \times 1.008\,\text{g/mol} = 0.03282\,\text{g}$

\begin{aligned}m_{\mathrm{O}} &= 0.2500\,\text{g}-0.1303\,\text{g}-0.03282\,\text{g} \\ &= 0.08688\,\text{g}\end{aligned}

n_{\mathrm{O}} = \dfrac{0.08688\,\text{g}}{15.999\,\text{g/mol}} = 0.005430\,\text{mol}

Step 3. Divide each amount by the smallest (0.005430 mol).

\mathrm{C}: 1.998

\mathrm{H}: 5.996

\mathrm{O}: 1.000

Step 4. Round and write the formula. All three ratios are within 0.01 of a whole number: 2 : 6 : 1.

Final Answer: C₂H₆O

Products $0.4777\,\text{g}\ \mathrm{CO_2}\text{,}\ 0.2934\,\text{g}\ \mathrm{H_2\!O}$ from a 0.2500 g sample
Step 1 $\times \dfrac{1}{\text{M}}$ C from CO₂, 2 H per H₂O
Moles of C and H $0.01085\ \mathrm{C}\text{,}\ 0.03256\ \mathrm{H}$
Step 2mass of O by difference: $0.2500-0.1303-0.03282 = 0.08688$
Moles of O $0.08688\,\text{g} \to 0.005430\,\text{mol}\ \mathrm{O}$
Step 3 $\times \dfrac{1}{\text{smallest}}$ C : H : O
Formula $\mathrm{C_2\!H_6\!O}$ $1.998 : 5.996 : 1.000$

Figure 7. Worked Example 2 on the roadmap (Figure 1): moles from the products, oxygen by difference, then the ratio.

C₂H₆O is the formula of ethanol. The three element masses add back up to the 0.2500 g sample (Figure 7).

Your turn: combustion analysis

Same method as Worked Example 2. If the compound has no oxygen, skip the difference step.

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Worked Example 3: Molecular Formula from the Molar Mass

Problem: Vitamin C (ascorbic acid) has the empirical formula $\mathrm{C_3\!H_4\!O_3}$ and a molar mass of 176.1 g/mol. What is its molecular formula?

Step 1. Find the empirical formula mass.

\begin{aligned}M_{\mathrm{C_3\!H_4\!O_3}} &= 3 \times 12.011 + 4 \times 1.008 + 3 \times 15.999 \\ &= 36.033 + 4.032 + 47.997 \\ &= 88.062\,\text{g/mol} \approx 88.06\,\text{g/mol}\end{aligned}

Step 2. Divide the molar mass by the empirical formula mass.

k = \dfrac{176.1\,\text{g/mol}}{88.06\,\text{g/mol}} = 2.000 \approx 2

Step 3. Multiply every subscript by 2.

(\mathrm{C_3\!H_4\!O_3})_2 = \mathrm{C_6\!H_8\!O_6}

Final Answer: C₆H₈O₆

Empirical formula mass $\mathrm{C_3\!H_4\!O_3}:\ 88.06\,\text{g/mol}$ Step 1
Step 2 $k = \frac{176.1\,\text{g/mol}}{88.06\,\text{g/mol}}$ molar mass over empirical formula mass
Multiple $k = 2.000 \approx 2$ a whole number
Step 3 $\times 2$ every subscript
Molecular formula $\mathrm{C_6\!H_8\!O_6}$

Figure 8. Worked Example 3 finishes the roadmap (Figure 1): find k, then multiply the subscripts.

Exam tip: k must come out very close to a whole number. If it does not, recheck the empirical formula mass before going on.

Your turn: molecular formula

Same three steps as Worked Example 3: empirical formula mass, divide, multiply the subscripts.

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Common Mistakes

Common mistake: Using mass ratios as subscripts. For formaldehyde, dividing the percents by the smallest percent gives about 6 : 1 : 8 (5.958 and 7.936 for C and O), not 1 : 2 : 1. Convert every mass to moles first.

Common mistake: Rounding a ratio too early. In Worked Example 1, rounding 1.501 to 2 gives FeO₂. A ratio near 1.5 means multiply everything by 2.

Common mistake: Not reducing the formula. Fe₄O₆ has the right ratio but is not the simplest one. An empirical formula has no common factor in its subscripts.

Common mistake: Forgetting the 2 H in each water. In Worked Example 2, using $\frac{0.2934\,\text{g}}{18.02\,\text{g/mol}} = 0.01628\,\text{mol}$ for H gives a ratio of 2.998 and the wrong formula C₂H₃O.

Common mistake: Leaving out the oxygen. If the compound contains oxygen, the C and H masses do not add up to the sample mass. Ignoring the rest in Worked Example 2 gives C : H = 1 : 3.001, the formula CH₃.

Common mistake: Stopping at the empirical formula. When a molar mass is given, the question asks for the molecular formula. For vitamin C, C₃H₄O₃ is only the empirical formula.

Key Takeaway

Subscripts count atoms, so they come from mole ratios: turn masses (from percents or from combustion products) into moles, divide by the smallest amount, and clear any fractions with a small whole number. The molar mass then fixes how many empirical units make up one molecule.

Cheat sheet

Empirical and molecular formulas on one card

startMassesgrams of each element (from % or combustion)
$\times \dfrac{1}{\text{atomic weight}}$ grams to moles
Molesof each element
$\times \dfrac{1}{\text{smallest}}$ then whole numbers
answerEmpirical formulasimplest whole-number ratio
$\times k$ k from the molar mass
answerMolecular formulaactual atoms per molecule

n_{\mathrm{C}} = \frac{m_{\mathrm{CO_2}}}{44.01}

n_{\mathrm{H}} = 2 \times \frac{m_{\mathrm{H_2\!O}}}{18.02}

m_{\mathrm{O}} = m_{\text{sample}}-m_{\mathrm{C}}-m_{\mathrm{H}}

k = \frac{M_{\text{compound}}}{M_{\text{empirical}}}

Masses: percents become grams in a 100 g sample; or use the combustion products (O by difference).
Moles: divide each mass by its atomic weight.
Ratio: divide by the smallest amount; multiply by 2, 3 or 4 if a ratio is not within about 0.05 of a whole number.
Molecular formula: $k = \frac{M_{\text{compound}}}{M_{\text{empirical}}}$ ; multiply every subscript by k.

Practice: randomized problems

Each problem below uses fresh numbers every time: work it out, check your answer, then open the worked solution to compare your steps.

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Your results

This summary collects every practice problem you try on this page and suggests what to review next.

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Figures: original diagrams by Dr. Ray’s Learning Center, licensed CC BY 4.0. Atomic weights: IUPAC 2021 abridged standard atomic weights. The practice problems are generated and checked by an audited engine; see the lesson audit log.