Atomic and Molecular Masses: Formulas & Numericals
How chemists weigh the unweighable — the carbon-12 scale, chlorine’s famous 35.5, molecular versus formula mass, and the algebra of isotope abundances, with a live lab to run the numbers yourself.
What are Atomic and Molecular Masses? — Complete Theory
A single hydrogen atom weighs about 0.000 000 000 000 000 000 000 001 67 g — a number no balance displays and no exam expects you to carry. Chemistry’s fix is elegant: stop weighing atoms in grams, start comparing them to a standard atom. Since 1961 that standard is carbon-12. One ¹²C atom is assigned exactly 12 units, everything else is measured against it, and the unit is the unified mass unit (u): 1 u = 1/12 the mass of one ¹²C atom = 1.6605 × 10⁻²⁴ g. (History in one line: hydrogen’s scale → oxygen’s scale → ¹²C, chosen because it gave the cleanest, most reproducible mass-spectrometry reference.)
So atomic mass is a ratio, not a weight. Saying “oxygen’s atomic mass is 16 u” means one oxygen atom is sixteen times heavier than 1/12 of a carbon-12 atom — and, usefully, sixteen times heavier than a hydrogen atom. Because it is a ratio, the atomic-mass table carries no grams anywhere; grams enter only when you convert through 1 u or through the mole.
x̄ = Σ fᵢ·mᵢ
Molecular massM = Σ nᵢ·aᵢ
One entitym = M / NA
But why do atomic masses land on awkward decimals — chlorine at 35.5, never a whole number? Because natural samples are isotope mixtures. Chlorine occurs as Cl-35 (≈75%) and Cl-37 (≈25%), and every bottle of chlorine on Earth carries the same blend. The atomic mass is therefore a weighted average: x̄ = (0.75 × 35) + (0.25 × 37) = 35.5 u. Neon mixes three isotopes (20, 21, 22) and averages 20.18 u. The fraction fᵢ is the abundance as a fraction — the sum of all fᵢ must be 1, and that check is free marks.
| Element | Symbol | Atomic mass (u) | Exam note |
|---|---|---|---|
| Hydrogen | H | 1 | True value 1.008 — papers use 1 |
| Carbon | C | 12 | The scale’s anchor (¹²C = 12 exactly) |
| Nitrogen | N | 14 | — |
| Oxygen | O | 16 | — |
| Sodium | Na | 23 | — |
| Sulfur | S | 32 | — |
| Chlorine | Cl | 35.5 | Isotope average — 75% Cl-35 : 25% Cl-37 |
| Potassium | K | 39 | — |
| Calcium | Ca | 40 | — |
| Iron | Fe | 56 | — |
| Copper | Cu | 63.5 | Isotope average (63 : 65 mix) |
| Silver | Ag | 108 | — |
Step up from atoms to compounds and the same arithmetic adds instead of averages. The molecular mass of a covalent molecule is the sum of atomic masses over its atoms: water = 2(1) + 16 = 18 u, glucose = 6(12) + 12(1) + 6(16) = 180 u, ammonia = 14 + 3(1) = 17 u. For ionic compounds there are no molecules — a crystal of NaCl is a lattice of ions, so we speak of formula mass: one formula unit NaCl = 23 + 35.5 = 58.5 u, CaCO3 = 40 + 12 + 3(16) = 100 u, MgCl2 = 24 + 2(35.5) = 95 u.
| Species | Type | Working | Result |
|---|---|---|---|
| H2O | Molecular | 2(1) + 16 | 18 u |
| NH3 | Molecular | 14 + 3(1) | 17 u |
| CO2 | Molecular | 12 + 2(16) | 44 u |
| C6H12O6 | Molecular | 6(12) + 12(1) + 6(16) | 180 u |
| NaCl | Formula | 23 + 35.5 | 58.5 u |
| CaCO3 | Formula | 40 + 12 + 3(16) | 100 u |
| MgCl2 | Formula | 24 + 2(35.5) | 95 u |
Now the bridge that makes this whole section pay off: the molar mass of a substance in g/mol is numerically equal to its atomic or molecular mass in u — 18 u ↔ 18 g/mol. That single equality is what makes 1 u = 1/NA g = 1.6605 × 10⁻²⁴ g, and it lets you weigh out a countable army of molecules on a kitchen-scale instrument. Need one entity instead of a mole? Divide: mass of one atom or molecule = M ÷ NA grams — one water molecule is 18 ÷ (6.022 × 10²³) ≈ 2.99 × 10⁻²³ g. (The full machinery of that bridge lives in § 1.8; using these masses in recipes is § 1.10.)
Isotope algebra runs in both directions, and JEE loves the reverse: given the average and both isotope masses, recover the abundances. Let the lighter isotope’s fraction be x: x·m₁ + (1 − x)·m₂ = x̄, so x = (m₂ − x̄)/(m₂ − m₁). Chlorine: x = (37 − 35.5)/(37 − 35) = 0.75 → 75% Cl-35, 25% Cl-37. One line, full marks — the lab below runs this live.
Visualising the u Scale
Ek standard, saare masses — the ¹²C pie defines the unit, the ruler places the elements, and the lab below averages isotopes like NTA asks.
Isotope Abundance Lab
Reverse solve — the exam version
Average diya hai, abundance nahi — wahi NTA format. Algebra: x = (m₂ − x̄)/(m₂ − m₁) × 100, jahan x lighter isotope ka % hai.
Abundances follow exam convention (Cl 75 : 25, Ne 90.92 : 0.26 : 8.82 …). Sliders hold Σ = 100% automatically — the last isotope takes the remainder, exactly like the algebra.
Solved Examples (Step-by-Step)
Given → formula → substitute → verify. Write these same steps on your rough sheet — that is what earns full method marks.
Average atomic mass of chlorine
Chlorine occurs as Cl-35 (75%) and Cl-37 (25%). Using isotope masses 35 u and 37 u, calculate the atomic mass of chlorine. Compare with the periodic-table value.
- Givenm₁ = 35 u (f₁ = 0.75), m₂ = 37 u (f₂ = 0.25); f₁ + f₂ = 1 ✓
- Formula
x̄ = f₁m₁ + f₂m₂ - Substitute
x̄ = 0.75 × 35 + 0.25 × 37 = 26.25 + 9.25 - Verify
x̄ = 35.5 u— matches the table’s 35.45 (real masses 34.97/36.97 account for the gap).
x̄ = 35.5 u — an average no single atom carries
From average back to abundance
Boron has two isotopes, B-10 (10 u) and B-11 (11 u), with an average atomic mass of 10.8 u. Find the percentage abundance of each isotope.
- Letx = fraction of B-10 → (1 − x) is B-11.
- Equation
10x + 11(1 − x) = 10.8 - Solve
11 − x = 10.8 → x = 0.2— or one shot:x = (11 − 10.8)/(11 − 10) = 0.2 - Verify
0.2 × 10 + 0.8 × 11 = 2 + 8.8 = 10.8 ✓
B-10 = 20% · B-11 = 80%
Molecular mass → mass of one glucose molecule
Calculate the molecular mass of glucose, C6H12O6, and the mass of a single glucose molecule in grams. (C = 12, H = 1, O = 16, NA = 6.022 × 10²³)
- Sum
M = 6(12) + 12(1) + 6(16) = 72 + 12 + 96 - Result (u)
M = 180 u— the mass of one molecule on the relative scale. - ConvertOne entity in grams:
m = M/NA = 180 ÷ (6.022 × 10²³) - Verify
m ≈ 2.99 × 10⁻²² g— power of ten near 10⁻²², magnitude check passes ✓
M = 180 u · one molecule ≈ 2.99 × 10⁻²² g
Key Formulas & Takeaways
Eight lines that solve this topic
Anchors: ¹²C = 12 u exact · 1 u = 1.6605 × 10⁻²⁴ g · NA = 6.022 × 10²³ mol⁻¹ · Cl avg = 35.5 u (75 : 25) · Ne avg = 20.18 u · H = 1, O = 16, C = 12 (exam convention)
- Atomic mass is a ratio, not a weight — ¹²C defines the scale; grams arrive only through 1 u or NA.
- Decimal atomic masses are isotope averages — 35.5 u belongs to the mixture; weighted mean, then reverse-solve when the average is given.
- Molecular vs formula mass is graded terminology — covalent molecules vs ionic lattices; the number can match, the word decides.
- One bridge line connects to § 1.8 — M u ↔ M g/mol ↔ M/NA g per entity; every numerical here flows through it.
FAQs
What is atomic mass and what is its unit?
Atomic mass is the mass of an atom measured relative to one-twelfth of a carbon-12 atom; its unit is the unified mass unit (u), where 1 u = 1.6605 × 10⁻²⁴ g. On this scale carbon-12 is exactly 12 u, hydrogen is about 1 u and chlorine averages 35.5 u.
Why is the atomic mass of chlorine 35.5 u?
Natural chlorine is a mixture of two isotopes — Cl-35 (about 75%) and Cl-37 (about 25%). The atomic mass is the weighted average: 0.75 × 35 + 0.25 × 37 = 35.5 u. No single chlorine atom actually weighs 35.5 u.
What is the difference between molecular mass and formula mass?
Molecular mass is the sum of atomic masses in one covalent molecule, e.g. H₂O = 18 u. Formula mass is used for ionic compounds like NaCl, which have no molecules — it is the sum for one formula unit, e.g. NaCl = 58.5 u. Both are numerically equal to the molar mass in g/mol.
How do you calculate the mass of one atom or one molecule?
Divide the molar mass by Avogadro’s constant: mass of one entity = M ÷ N_A grams. One oxygen atom: 16 ÷ (6.022 × 10²³) ≈ 2.66 × 10⁻²³ g; one water molecule: 18 ÷ (6.022 × 10²³) ≈ 2.99 × 10⁻²³ g.
Practice Questions
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QCC Notes — Class 11 Chemistry
Strictly NCERT-aligned notes for JEE Main & NEET, prepared by QCC Notes (Padho Likho JEE). Content follows the latest NCERT edition and current NTA exam pattern.