Atomic mass and the atomic mass unit
A single atom is far too light to weigh in grams (a hydrogen atom is about 1.67 × 10⁻²⁴ g). So we compare atoms with a standard: the carbon-12 atom, given a mass of exactly 12 u (unified atomic mass unit, older name amu).
1 u = one-twelfth of the mass of one carbon-12 atom = 1.66 × 10⁻²⁴ g.
On this scale H = 1.008 u, O = 16.00 u, Na = 23 u, Cl = 35.5 u.
Average atomic mass
Most elements are a mix of isotopes (atoms of the same element with different masses). The atomic mass in tables is a weighted average. Chlorine: 75% of ³⁵Cl and 25% of ³⁷Cl → (0.75 × 35) + (0.25 × 37) = 35.5 u. That is why 35.5 is not a whole number.
Molecular mass and formula mass
Molecular mass = sum of the atomic masses of all atoms in a molecule.
- H₂O = 2(1) + 16 = 18 u
- CO₂ = 12 + 2(16) = 44 u
- C₆H₁₂O₆ = 6(12) + 12(1) + 6(16) = 180 u
Ionic compounds like NaCl do not form separate molecules; they form a big 3D network of ions. For them we use formula mass: NaCl = 23 + 35.5 = 58.5 u.
The mole and Avogadro number
A mole is the SI unit for amount of substance. One mole contains exactly 6.022 140 76 × 10²³ particles (atoms, molecules, ions or electrons). This number is the Avogadro constant, NA.
It is just a counting word, like dozen (12) or gross (144), but huge. 6.022 × 10²³ grains of sand would cover India many metres deep.
Always say which particle: 1 mol of O₂ molecules has 2 mol of O atoms.
Molar mass
Molar mass (M) is the mass of one mole of a substance in grams. Its number is the same as the atomic, molecular or formula mass in u.
- Water: 18 u → M = 18 g mol⁻¹
- Carbon: 12 u → M = 12 g mol⁻¹
- NaCl: 58.5 u → M = 58.5 g mol⁻¹
The three key formulas
n = m ÷ M (moles = mass ÷ molar mass)
N = n × NA (particles = moles × Avogadro number)
For a gas at STP (273 K, 1 bar) one mole takes 22.7 L (22.4 L at 1 atm, used in older books).
Percentage composition
Percentage composition tells how much of each element is in 100 g of a compound. It is used to check purity.
Mass % of an element = (mass of that element in 1 mol ÷ molar mass) × 100
Water: H = (2 ÷ 18) × 100 = 11.11%, O = (16 ÷ 18) × 100 = 88.89%. The percentages always add to 100.
Ethanol C₂H₅OH (M = 46): C = 24/46 = 52.2%, H = 6/46 = 13.0%, O = 16/46 = 34.8%.
Empirical and molecular formula
The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula gives the actual number of atoms in one molecule.
| Compound | Molecular | Empirical |
|---|---|---|
| Glucose | C₆H₁₂O₆ | CH₂O |
| Benzene | C₆H₆ | CH |
| Water | H₂O | H₂O |
Steps: from % to formula
- Take 100 g, so each % becomes grams.
- Change grams to moles: divide by atomic mass.
- Divide every mole value by the smallest one.
- If you get values like 1.5 or 1.33, multiply all by 2 or 3 to make whole numbers. This is the empirical formula.
- Find n = molar mass ÷ empirical formula mass. Molecular formula = n × empirical formula.
Try it
In the 3D free play, pick water and set 18 g, then 36 g. Predict the moles before you look. Then weigh one teaspoon of sugar (about 4 g of C₁₂H₂₂O₁₁, M = 342) at home and work out how many molecules you hold: about 7 × 10²¹.
Key formulas and definitions
- 1 u = 1/12 × mass of one C-12 atom = 1.66 × 10⁻²⁴ g
- Average atomic mass = Σ(fraction × isotope mass)
- N_A = 6.022 × 10²³ mol⁻¹
- n = m / M; N = n × N_A
- Mass % = (mass of element in 1 mol / M) × 100
- n (multiple) = molar mass / empirical formula mass; molecular formula = n × empirical
Worked examples
1. Chlorine is 75.77% ³⁵Cl (34.97 u) and 24.23% ³⁷Cl (36.97 u). Find its average atomic mass.
Line 1: 0.7577 × 34.97 = 26.50. Line 2: 0.2423 × 36.97 = 8.96. Line 3: Sum = 35.46 u ≈ 35.5 u.
2. Find the molecular mass of glucose C₆H₁₂O₆ and the formula mass of CaCO₃ (Ca = 40).
Line 1: C₆H₁₂O₆ = 6 × 12 + 12 × 1 + 6 × 16 = 72 + 12 + 96 = 180 u. Line 2: CaCO₃ = 40 + 12 + 3 × 16 = 100 u.
3. How many moles and molecules are in 36 g of water?
Line 1: M(H₂O) = 18 g mol⁻¹. Line 2: n = 36 ÷ 18 = 2 mol. Line 3: N = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ molecules.
4. What is the mass of 3.011 × 10²³ atoms of carbon?
Line 1: n = N ÷ N_A = 3.011 × 10²³ ÷ 6.022 × 10²³ = 0.5 mol. Line 2: m = n × M = 0.5 × 12 = 6 g.
5. How many oxygen atoms are in 88 g of CO₂?
Line 1: n(CO₂) = 88 ÷ 44 = 2 mol. Line 2: Each CO₂ has 2 O atoms → n(O) = 4 mol. Line 3: O atoms = 4 × 6.022 × 10²³ = 2.409 × 10²⁴.
6. Find the percentage composition of ethanol, C₂H₅OH.
Line 1: M = 2(12) + 6(1) + 16 = 46 g mol⁻¹. Line 2: %C = 24/46 × 100 = 52.17%. Line 3: %H = 6/46 × 100 = 13.04%. Line 4: %O = 16/46 × 100 = 34.78%. Check: total = 100%.
7. A compound has 4.07% H, 24.27% C and 71.65% Cl. Its molar mass is 98.96 g. Find the empirical and molecular formulas.
Line 1: In 100 g: H 4.07 g, C 24.27 g, Cl 71.65 g. Line 2: Moles: H 4.07/1.008 = 4.04; C 24.27/12.01 = 2.021; Cl 71.65/35.45 = 2.021. Line 3: Divide by 2.021: H 2, C 1, Cl 1 → empirical CH₂Cl (mass 49.48). Line 4: n = 98.96 ÷ 49.48 = 2. Line 5: Molecular formula = C₂H₄Cl₂.
8. A hydrocarbon has 85.7% C and 14.3% H. Its molar mass is 42 g mol⁻¹. Find its formulas.
Line 1: C 85.7/12 = 7.14 mol; H 14.3/1 = 14.3 mol. Line 2: Divide by 7.14: C 1, H 2 → empirical CH₂ (mass 14). Line 3: n = 42 ÷ 14 = 3. Line 4: Molecular formula = C₃H₆.
Common mistakes
- Forgetting to say which particle. 1 mol of H₂ has 2 mol of H atoms.
- Using molecular mass for NaCl. Ionic compounds have formula mass, since there are no separate molecules.
- Rounding 1.5 to 2 when finding an empirical formula. Multiply every value by 2 instead (1.5 : 1 → 3 : 2).
- Mixing u and g. One molecule of water is 18 u; one mole of water is 18 g.