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D. Quantities in Chemical Reactions · Quantities in Chemical Reactions

Empirical Formulas from Percentage Composition and Mass Data

D2.4 D3.3 D2.1 D2.2About 70 minutes
Learning goals
  • I can explain what an empirical formula is.
  • I can find an empirical formula from percentage composition.
  • I can find an empirical formula from masses measured in a lab.
  • I can deal with mole ratios such as 1 : 1.5 or 1 : 1.33 by multiplying to whole numbers.

Words to know

empirical formula
The simplest whole-number ratio of atoms (or moles of atoms) of each element in a compound, for example CH₂O.
molecular formula
The actual number of atoms of each element in one molecule, for example C₆H₁₂O₆. It is a whole-number multiple of the empirical formula.
mole ratio
The ratio of the amounts (in moles) of two substances or elements.
combustion analysis
Finding the composition of a compound that contains carbon and hydrogen by burning it and weighing the CO₂ and H₂O produced.

Working backwards

In the last two lessons you went forwards: from a formula to its percentage composition. Chemists who discover or make a new compound have to go backwards. A lab instrument tells them the percentage of each element; they must work out the formula. The first formula they can find is the empirical formula, the simplest whole-number ratio of atoms.

Why moles, not grams?

A formula counts atoms, and moles count atoms. Grams do not, because atoms of different elements have different masses. Water is 88.8% oxygen by mass, but only one atom in three is oxygen. So the key step is always to convert masses to moles, and then compare the moles.

1. Percentassume 100 g→ grams2. Grams÷ molar mass→ moles3. Moles÷ smallest→ ratio4. Ratio× to makewhole numbers"Percent to mass, mass to mole, divide by small, multiply 'til whole."
The four steps to an empirical formula, with a rhyme to help you remember them.
Worked example

Example 1: From percentage composition

A sugar found in some fruits is 40.0% C, 6.7% H and 53.3% O by mass. Find its empirical formula.

  1. Percent to mass: assume 100 g, so there are 40.0 g C, 6.7 g H and 53.3 g O.
  2. Mass to mole: C: 40.0 ÷ 12.01 = 3.331 mol. H: 6.7 ÷ 1.01 = 6.63 mol. O: 53.3 ÷ 16.00 = 3.331 mol.
  3. Divide by small (3.331): C: 1.00. H: 1.99. O: 1.00.
  4. 'Til whole: 1.99 rounds to 2 (it is within about 0.1 of a whole number, which is normal experimental rounding). The ratio is 1 : 2 : 1.

The empirical formula is CH₂O.

Worked example

Example 2: From lab masses

In a lab, 5.585 g of iron combines with 2.400 g of oxygen. Find the empirical formula of the iron oxide.

  1. Masses are already given, so skip step 1.
  2. Fe: 5.585 ÷ 55.85 = 0.1000 mol. O: 2.400 ÷ 16.00 = 0.1500 mol.
  3. Divide by the smaller (0.1000): Fe: 1.000. O: 1.500.
  4. 1.5 is not close to a whole number, so do not round it. Multiply both by 2: Fe: 2, O: 3.

The empirical formula is Fe₂O₃, iron(III) oxide.

121.25× 4 → 51.33× 3 → 41.50× 2 → 31.67× 3 → 51.75× 4 → 7If a ratio ends in one of these decimals, multiply ALL the ratios by the number shown.
Ratios that end in .25, .33, .5, .67 or .75 are real fractions, not rounding errors. Multiply every ratio by the same small number to make them whole.
Worked example

Example 3: A ratio ending in .5

A phosphorus oxide is 43.64% P and 56.36% O. Find its empirical formula.

  1. Assume 100 g: 43.64 g P and 56.36 g O.
  2. P: 43.64 ÷ 30.97 = 1.409 mol. O: 56.36 ÷ 16.00 = 3.5225 mol.
  3. Divide by 1.409: P: 1.000. O: 2.500.
  4. Multiply both by 2: P: 2, O: 5.

The empirical formula is P₂O₅.

Rounding rules

  • Keep at least three or four significant figures in the moles. Rounding too early can turn 1.33 into 1.3 or 1.4 and hide the real ratio.
  • Round a ratio to a whole number only if it is within about 0.1 (1.96 → 2, 3.04 → 3).
  • If a ratio is near 1.25, 1.33, 1.5, 1.67 or 1.75, multiply all ratios by 4, 3, 2, 3 or 4.
  • Write the elements in the order the question gives them (usually metal or carbon first).

Combustion analysis

Many compounds from living things contain carbon and hydrogen (and often oxygen). Chemists burn a weighed sample in excess oxygen. All the carbon ends up in CO₂ and all the hydrogen ends up in H₂O. The CO₂ and H₂O are trapped and weighed. Then:

  • mol C = mol CO₂ (one C per CO₂)
  • mol H = 2 × mol H₂O (two H per H₂O)
  • if the compound has oxygen too, mass O = mass of sample − mass C − mass H

The rest is the usual empirical-formula method. You will meet this in the Challenge questions.

Just for fun

Why are empirical formulas so modest?

They always reduce themselves to the simplest terms.

Hands-on activity

A coin "compound"

This models the empirical-formula method with coins. You need: a kitchen scale (grams), at least 20 nickels and 20 dimes, two bowls, and a partner.

  1. Weigh 10 nickels together and divide by 10 to get the mass of one nickel. Do the same for dimes. These are your "molar masses".
  2. Your partner secretly builds a "compound": a pile of coins in a fixed whole-number ratio, such as 2 nickels for every 3 dimes, repeated several times (for example 8 nickels and 12 dimes).
  3. Your partner sorts the pile and tells you only the mass of all the nickels and the mass of all the dimes.
  4. Divide each mass by the mass of one coin to get the number of each coin ("moles"). Divide by the smaller number. Multiply if needed to get whole numbers. Write the empirical formula, like N₂D₃.
  5. Swap roles and try a tricky ratio such as 3 : 4.

What to look for: you should get your partner's ratio back exactly, even though you never counted the coins. Notice that the empirical formula does not tell you how big the pile was (8 and 12 coins, or 2 and 3, give the same N₂D₃). That is exactly the difference between an empirical formula and a molecular formula, which is the next lesson.

5¢5¢10¢10¢10¢one "formula unit": 2 nickels + 3 dimesempirical formula N₂D₃Weigh all the nickels, then all the dimes.÷ mass of ONE coin = number of coins(like ÷ molar mass = moles)÷ smallest → 2 : 3
The coin model: masses divided by the mass of one coin give numbers of coins, just as masses divided by molar masses give moles of atoms.

Your turn

1

A gas is 75.0% carbon and 25.0% hydrogen. Find its empirical formula.

2

2.43 g of magnesium combines with 1.60 g of oxygen. Find the empirical formula.

3

An oxide of aluminium is 52.9% Al and 47.1% O. Find its empirical formula.

4

A compound is 27.3% C and 72.7% O. Find its empirical formula.

5

A hydrocarbon is 85.6% C and 14.4% H. Find its empirical formula.

6

An orange compound used in some older lab tests is 26.6% K, 35.4% Cr and 38.1% O. Find its empirical formula.

More practice: warm-up

1

What does an empirical formula show?

2

What is the empirical formula of glucose, C₆H₁₂O₆?

3

What is the empirical formula of hydrogen peroxide, H₂O₂?

4

What is the empirical formula of ethene, C₂H₄?

5

A mole ratio comes out as 1 : 1.5. What should you multiply by?

More practice: on your own

1

A white solid is 32.4% Na, 22.6% S and 45.0% O. Find its empirical formula.

2

A black iron oxide is 72.4% Fe and 27.6% O. Find its empirical formula.

3

Vitamin C is 40.92% C, 4.58% H and 54.50% O. Find its empirical formula.

4

Sam calculated mole ratios of C : H = 1 : 1.33 and wrote the empirical formula CH. What did Sam do wrong? What is the correct formula?

5

Explain why you must convert grams to moles before finding the ratio.

Challenge

1

Aspirin is 60.0% C, 4.48% H and 35.5% O. Find its empirical formula.

2

A 1.000 g sample of a hydrocarbon (only C and H) is burned completely. It produces 3.137 g of CO₂ and 1.284 g of H₂O. Find the empirical formula.

3

A 5.000 g sample of a zinc sulfate hydrate, ZnSO₄·xH₂O, loses 2.193 g of water when heated to constant mass. Find x. (M(ZnSO₄) = 161.45 g/mol)

Check your understanding

Quiz

1.

Which is an empirical formula?

2.

A compound has the mole ratio C : H = 1 : 2.5. What is its empirical formula?

3.

Why do we assume a 100 g sample when we start from percentages?

4.

2.00 g of calcium combines with 3.54 g of chlorine. What is the empirical formula?

5.

A mole ratio comes out as 1 : 1.33. What should you do?

6.

In combustion analysis, how do you find the moles of hydrogen in the sample?

For parents and teachers

Big idea: An empirical formula is the simplest whole-number ratio of atoms. To find it: percent → grams → moles → divide by the smallest → multiply to whole numbers. This lesson covers part of D2.4 (empirical and molecular formula problems) and D3.3 (the meaning of an empirical formula); the next lesson finishes them with molecular formulas.

How to help: Have your teen recite the rhyme and then do one problem aloud, step by step. The coin activity makes the method concrete: you divide by the mass of one coin just as chemists divide by molar mass.

Common difficulties:

  • Comparing grams instead of moles.
  • Rounding 1.5 or 1.33 to the nearest whole number instead of multiplying.
  • Rounding moles too early, which hides the true ratio.
  • Forgetting to multiply every ratio by the same number.
  • In combustion analysis, forgetting that each H₂O has two hydrogen atoms.