Percentage Composition

When we know the molecular formula of a compound, we can calculate what percentage of the total mass is contributed by each element. This is called the mass percent or percentage composition.

Formula

Mass % of element=Mass of that element in 1 mol of compoundMolar mass of compound×100\text{Mass \% of element} = \frac{\text{Mass of that element in 1 mol of compound}}{\text{Molar mass of compound}} \times 100

Example: Percentage Composition of Water (H2O\text{H}_2\text{O})

Molar mass of H2O\text{H}_2\text{O} = 18 g/mol

Mass % of H=2×118×100=11.11%\text{Mass \% of H} = \frac{2 \times 1}{18} \times 100 = 11.11\%

Mass % of O=1618×100=88.89%\text{Mass \% of O} = \frac{16}{18} \times 100 = 88.89\%

Check: 11.11 + 88.89 = 100% ✓ (The sum should always be 100%.)

[Board Important] Percentage composition calculations are very common in Board exams. Always verify that your percentages add up to 100%.

Key Point: Mass percent tells you the fraction of total mass contributed by each element. It's calculated from the molecular formula and atomic masses.

Empirical Formula

The empirical formula of a compound represents the simplest whole number ratio of atoms of each element in the compound.

For example:

  • Glucose has the molecular formula C6H12O6\text{C}_6\text{H}_{12}\text{O}_6, but its empirical formula is CH2O\text{CH}_2\text{O} (ratio 1:2:1)
  • Benzene has the molecular formula C6H6\text{C}_6\text{H}_6, but its empirical formula is CH\text{CH} (ratio 1:1)
  • Water has both molecular and empirical formula H2O\text{H}_2\text{O} (already simplest ratio)

Steps to Determine Empirical Formula from Percentage Composition

Step 1: Write down the percentage composition of each element. If percentages don't add up to 100%, the remainder is usually oxygen.

Step 2: Convert each percentage to moles by dividing by the atomic mass of that element. Moles of element=Mass %Atomic mass\text{Moles of element} = \frac{\text{Mass \%}}{\text{Atomic mass}}

Step 3: Divide all mole values by the smallest mole value to get the simplest ratio.

Step 4: If the ratios are not whole numbers, multiply all ratios by a suitable integer to make them whole numbers.

Step 5: Write the empirical formula using these whole number ratios.

Key Point: The empirical formula gives the simplest whole-number ratio of atoms. It's determined from percentage composition → moles → simplest ratio.

Molecular Formula from Empirical Formula

The molecular formula gives the actual number of atoms of each element in one molecule of the compound. It is always a whole-number multiple of the empirical formula.

Molecular formula=n×Empirical formula\text{Molecular formula} = n \times \text{Empirical formula}

where: n=Molar mass of compoundEmpirical formula massn = \frac{\text{Molar mass of compound}}{\text{Empirical formula mass}}

Example

A compound has empirical formula CH2O\text{CH}_2\text{O} and molar mass = 180 g/mol.

  1. Empirical formula mass = 12 + 2(1) + 16 = 30 g/mol
  2. n=18030=6n = \frac{180}{30} = 6
  3. Molecular formula = 6×CH2O=C6H12O66 \times \text{CH}_2\text{O} = \text{C}_6\text{H}_{12}\text{O}_6 (glucose!)

When are they the same?

The empirical and molecular formulas are identical when n=1n = 1, i.e., the molar mass equals the empirical formula mass. Examples: H2O\text{H}_2\text{O}, CO2\text{CO}_2, NaCl.

[JEE Tip] For ionic compounds like NaCl, CaCl2\text{CaCl}_2, etc., only the empirical (or formula unit) is meaningful since discrete molecules don't exist. The molecular formula concept applies mainly to covalent compounds.

[NEET Important] The relationship n=Molar massEmpirical formula massn = \frac{\text{Molar mass}}{\text{Empirical formula mass}} is tested very frequently. Always find the empirical formula first, then scale up.

Key Point: Molecular formula = nn × Empirical formula, where nn = Molar mass ÷ Empirical formula mass. This nn is always a positive integer.

Worked Strategy: Complete Empirical & Molecular Formula Determination

Let's walk through a complete example — this is the type of problem you'll see in Board, JEE, and NEET exams.

Problem: A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen. Its molar mass is 180 g/mol. Find its empirical and molecular formula.

Step 1: Convert % to moles

Element % by mass Atomic mass Moles = % ÷ At. mass
C 40.0 12 40/12 = 3.33
H 6.7 1 6.7/1 = 6.7
O 53.3 16 53.3/16 = 3.33

Step 2: Divide by smallest

Smallest value = 3.33

Element Ratio
C 3.33/3.33 = 1
H 6.7/3.33 = 2.01 ≈ 2
O 3.33/3.33 = 1

Step 3: Empirical formula = CH2O\text{CH}_2\text{O}

Step 4: Find molecular formula

  • Empirical formula mass = 12 + 2 + 16 = 30
  • n=18030=6n = \frac{180}{30} = 6
  • Molecular formula = C6H12O6\text{C}_6\text{H}_{12}\text{O}_6

This is glucose!

[Board Important] This 4-step method (% → moles → ratio → empirical formula → molecular formula) is the standard approach for all such problems. Master it thoroughly.

Key Point: The complete method is: % composition → moles (÷ atomic mass) → simplest ratio (÷ smallest) → empirical formula → molecular formula (using molar mass).

Solved Examples

Example 1: Percentage Composition of Ethanol

Calculate the percentage composition of ethanol (C2H5OH\text{C}_2\text{H}_5\text{OH}).

Solution: Molecular formula: C2H6O\text{C}_2\text{H}_6\text{O}

Molar mass = 2(12)+6(1)+16=462(12) + 6(1) + 16 = 46 g/mol

Mass % of C=2446×100=52.17%\text{Mass \% of C} = \frac{24}{46} \times 100 = 52.17\%

Mass % of H=646×100=13.04%\text{Mass \% of H} = \frac{6}{46} \times 100 = 13.04\%

Mass % of O=1646×100=34.78%\text{Mass \% of O} = \frac{16}{46} \times 100 = 34.78\%

Check: 52.17 + 13.04 + 34.78 = 99.99 ≈ 100% ✓

Example 2: Percentage Composition of Calcium Carbonate

Find the mass percent of each element in CaCO3\text{CaCO}_3. (Atomic masses: Ca = 40, C = 12, O = 16)

Solution: Molar mass = 40 + 12 + 3(16) = 100 g/mol

Mass % of Ca=40100×100=40%\text{Mass \% of Ca} = \frac{40}{100} \times 100 = 40\%

Mass % of C=12100×100=12%\text{Mass \% of C} = \frac{12}{100} \times 100 = 12\%

Mass % of O=48100×100=48%\text{Mass \% of O} = \frac{48}{100} \times 100 = 48\%

Check: 40 + 12 + 48 = 100% ✓

Example 3: Empirical Formula from Percentage Composition

A compound has the following composition: C = 12.5%, H = 3.13%, O = 84.37%. Find the empirical formula.

Solution:

Element % At. mass Moles Ratio (÷ smallest)
C 12.5 12 1.042 1.042/1.042 = 1
H 3.13 1 3.13 3.13/1.042 = 3.0
O 84.37 16 5.273 5.273/1.042 = 5.06 ≈ 5

Empirical formula: CH3O5\text{CH}_3\text{O}_5

Wait — let's double check. This doesn't look like a typical compound. Let's recompute:

Actually, let's reconsider: O = 84.37%, moles = 84.37/16 = 5.273. Ratio = 5.273/1.042 ≈ 5.06.

Empirical formula: CH3O5\text{CH}_3\text{O}_5

Note: If this seems unusual, it could be a peroxide or similar compound. The empirical formula gives only the simplest ratio — the actual compound might have a different molecular formula.

Example 4: Empirical Formula — Classic NCERT Type

A compound contains 4.07% hydrogen, 24.27% carbon, and 71.65% chlorine. What is its empirical formula?

Solution:

Element % At. mass Moles Ratio (÷ smallest)
H 4.07 1 4.07 4.07/0.676 = 6.02 ≈ 6
C 24.27 12 2.02 2.02/0.676 = 2.99 ≈ 3
Cl 71.65 35.5 2.02 2.02/0.676 = 2.99 ≈ 3

Wait — smallest is 2.02, not 0.676.

Element Moles Ratio (÷ 2.02)
H 4.07 2.01 ≈ 2
C 2.02 1
Cl 2.02 1

Empirical formula: CH2Cl\text{CH}_2\text{Cl}

Verification: With molar mass = 98.96 g/mol, n=98.9649.48=2n = \frac{98.96}{49.48} = 2, so molecular formula = C2H4Cl2\text{C}_2\text{H}_4\text{Cl}_2 (1,2-dichloroethane).

Example 5: From Empirical to Molecular Formula

The empirical formula of a compound is BH3\text{BH}_3. Its molar mass is 27.66 g/mol. Find the molecular formula. (Atomic masses: B = 10.81, H = 1.008)

Solution:

  1. Empirical formula mass: 10.81+3(1.008)=10.81+3.024=13.83410.81 + 3(1.008) = 10.81 + 3.024 = 13.834 g/mol
  2. n=27.6613.834=2.0n = \frac{27.66}{13.834} = 2.0
  3. Molecular formula: 2×BH3=B2H62 \times \text{BH}_3 = \text{B}_2\text{H}_6 (diborane)

Final Answer: B2H6\text{B}_2\text{H}_6

Example 6: Empirical Formula with Non-Integer Ratios

A compound contains 69.94% iron and 30.06% oxygen. Find its empirical formula. (Atomic masses: Fe = 55.85, O = 16)

Solution:

Element % At. mass Moles Ratio (÷ smallest)
Fe 69.94 55.85 1.252 1.252/1.252 = 1
O 30.06 16 1.879 1.879/1.252 = 1.501

The ratio 1 : 1.5 is NOT a whole number ratio!

Fix: Multiply both by 2: 1×2:1.5×2=2:31 \times 2 : 1.5 \times 2 = 2 : 3

Empirical formula: Fe2O3\text{Fe}_2\text{O}_3 (iron(III) oxide / ferric oxide)

Takeaway: When you get ratios like x.5, multiply all by 2. For x.33, multiply by 3. For x.25, multiply by 4.

Example 7: Percentage Composition of a Hydrate

Calculate the mass percent of water of crystallisation in CuSO4⋅5H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O}.

Solution: Molar mass of CuSO4⋅5H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O}: =63.5+32+4(16)+5(18)=63.5+32+64+90=249.5 g/mol= 63.5 + 32 + 4(16) + 5(18) = 63.5 + 32 + 64 + 90 = 249.5 \text{ g/mol}

Mass of 5H2O5\text{H}_2\text{O} = 90 g

Mass % of water=90249.5×100=36.07%\text{Mass \% of water} = \frac{90}{249.5} \times 100 = 36.07\%

Final Answer: CuSO4⋅5H2O\text{CuSO}_4 \cdot 5\text{H}_2\text{O} contains 36.07% water by mass.

Example 8: Finding Formula from Combustion Data

0.48 g of an organic compound containing C, H, and O gave 0.88 g of CO2\text{CO}_2 and 0.36 g of H2O\text{H}_2\text{O} on complete combustion. Find the empirical formula.

Solution:

  1. Mass of C (from CO2\text{CO}_2): CO2\text{CO}_2 = 44 g/mol, C = 12 g/mol C=1244×0.88=0.24 g\text{C} = \frac{12}{44} \times 0.88 = 0.24 \text{ g}

  2. Mass of H (from H2O\text{H}_2\text{O}): H2O\text{H}_2\text{O} = 18 g/mol, H = 2 g/mol H=218×0.36=0.04 g\text{H} = \frac{2}{18} \times 0.36 = 0.04 \text{ g}

  3. Mass of O (by difference): O=0.48−0.24−0.04=0.20 g\text{O} = 0.48 - 0.24 - 0.04 = 0.20 \text{ g}

  4. Convert to moles:

Element Mass (g) At. mass Moles Ratio (÷ smallest)
C 0.24 12 0.02 0.02/0.0125 = 1.6
H 0.04 1 0.04 0.04/0.0125 = 3.2
O 0.20 16 0.0125 0.0125/0.0125 = 1
  1. Multiply by 5 to get whole numbers: C : H : O = 8 : 16 : 5

Empirical formula: C8H16O5\text{C}_8\text{H}_{16}\text{O}_5

Takeaway: Combustion analysis gives CO2\text{CO}_2 (for C) and H2O\text{H}_2\text{O} (for H). Oxygen is found by difference.

Example 9: Reverse Problem — Mass from Percentage

How many grams of calcium can be obtained from 500 g of calcium carbonate (CaCO3\text{CaCO}_3)?

Solution: Molar mass of CaCO3\text{CaCO}_3 = 100 g/mol

Mass % of Ca = 40100×100=40%\frac{40}{100} \times 100 = 40\%

Mass of Ca=40100×500=200 g\text{Mass of Ca} = \frac{40}{100} \times 500 = 200 \text{ g}

Final Answer: 500 g of CaCO3\text{CaCO}_3 contains 200 g of calcium.

Example 10: Molecular Formula Determination — Complete Problem

An oxide of nitrogen contains 30.4% nitrogen. Its molar mass is 92 g/mol. Find the empirical and molecular formula.

Solution:

  1. % Composition: N = 30.4%, O = 100 - 30.4 = 69.6%

  2. Moles:

Element % At. mass Moles Ratio
N 30.4 14 2.171 2.171/2.171 = 1
O 69.6 16 4.35 4.35/2.171 = 2.003 ≈ 2

  1. Empirical formula: NO2\text{NO}_2

  2. Empirical formula mass: 14 + 32 = 46 g/mol

  3. n=9246=2n = \frac{92}{46} = 2

  4. Molecular formula: N2O4\text{N}_2\text{O}_4 (dinitrogen tetroxide)

Takeaway: When only one element's percentage is given, the other is obtained by subtracting from 100%.

Example 11: Percentage of an Element in a Compound

Which has a higher percentage of nitrogen — NH3\text{NH}_3 or \text{(NH_4)}_2\text{SO}_4?

Solution:

In NH3\text{NH}_3: Molar mass = 14 + 3 = 17 g/mol % N=1417×100=82.35%\text{\% N} = \frac{14}{17} \times 100 = 82.35\%

In (NH4)2SO4(\text{NH}_4)_2\text{SO}_4: Molar mass = 2(14) + 8(1) + 32 + 4(16) = 28 + 8 + 32 + 64 = 132 g/mol % N=28132×100=21.21%\text{\% N} = \frac{28}{132} \times 100 = 21.21\%

Answer: NH3\text{NH}_3 has a much higher percentage of nitrogen (82.35% vs 21.21%).

Why this matters: This type of comparison is important in agriculture — fertilisers are rated by their nitrogen content. NH3\text{NH}_3 (anhydrous ammonia) is one of the most nitrogen-rich fertilisers.

Example 12: Empirical Formula with Equal Ratios

A compound has 50% sulphur and 50% oxygen by mass. Find its empirical formula.

Solution:

Element % At. mass Moles Ratio
S 50 32 1.5625 1.5625/1.5625 = 1
O 50 16 3.125 3.125/1.5625 = 2

Empirical formula: SO2\text{SO}_2 (sulphur dioxide)

Takeaway: Equal percentages by mass do NOT mean equal numbers of atoms — always convert to moles first!