2.2 Atomic Weight
2.2 Atomic Weight
How do we measure something as impossibly small as an atom? This chapter builds the idea of relative atomic mass, and shows how Dulong & Petit turned a simple lab measurement into an atomic weight estimate.
🔬 Why do we need atomic mass?
Atoms are extremely small, so their actual masses are extremely small — far too small to measure directly on a laboratory balance. Instead, chemists compare the mass of one atom against another, using a relative scale.
⚖️ Relative Atomic Mass
The lightest element, hydrogen, was originally taken as the standard used to express the masses of other elements.
Problem: the isotope of hydrogen used wasn’t specified, and pure hydrogen is expensive and difficult to handle.
🧪 The modern C-12 standard ›
IUPAC later adopted the Carbon-12 (¹²C) isotope as the standard.
“The atomic mass of an element is a number which indicates how many times one atom of the element is heavier than one-twelfth of an atom of ¹²C isotope.”
In simple words: it tells us how heavy an atom is compared with the carbon-12 standard. This value is not the actual mass of the atom — it’s a relative atomic mass, also called atomic weight. Since it’s a ratio of two masses, atomic mass technically has no unit — but the atomic mass unit (amu) is used for convenience.
Don’t confuse atomic mass with mass number!
- Mass number = total protons + neutrons in one specific atom (always a whole number)
- Relative atomic mass = average mass compared to C-12 (often a decimal, because of isotopes)
Make a personal study card listing the first 30 elements with their name, symbol, atomic number, and atomic mass. Review it daily until you know them from memory — this knowledge is used throughout all of chemistry!
🧫 First 30 Elements — Reference Table
Use this as your study checklist for the assignment above.
| # | Element | Symbol | Atomic weight | Valency |
|---|---|---|---|---|
| 1 | Hydrogen | H | 1 | 1 |
| 2 | Helium | He | 4 | 0 |
| 3 | Lithium | Li | 7 | 1 |
| 4 | Beryllium | Be | 9 | 2 |
| 5 | Boron | B | 11 | 3 |
| 6 | Carbon | C | 12 | 4 |
| 7 | Nitrogen | N | 14 | 3 |
| 8 | Oxygen | O | 16 | 2 |
| 9 | Fluorine | F | 19 | 1 |
| 10 | Neon | Ne | 20 | 0 |
| 11 | Sodium | Na | 23 | 1 |
| 12 | Magnesium | Mg | 24 | 2 |
| 13 | Aluminium | Al | 27 | 3 |
| 14 | Silicon | Si | 28 | 4 |
| 15 | Phosphorus | P | 31 | 3 |
| 16 | Sulphur | S | 32 | 2 |
| 17 | Chlorine | Cl | 35.5 | 1 |
| 18 | Argon | Ar | 40 | 0 |
| 19 | Potassium | K | 39 | 1 |
| 20 | Calcium | Ca | 40 | 2 |
| 21 | Scandium | Sc | 45 | 3 |
| 22 | Titanium | Ti | 48 | 2, 3 |
| 23 | Vanadium | V | 51 | 2, 3 |
| 24 | Chromium | Cr | 52 | 3, 4 |
| 25 | Manganese | Mn | 55 | 3 |
| 26 | Iron | Fe | 56 | 2, 3 |
| 27 | Cobalt | Co | 59 | 2, 3 |
| 28 | Nickel | Ni | 59 | 2, 3 |
| 29 | Copper | Cu | 63.5 | 1, 2 |
| 30 | Zinc | Zn | 65.5 | 2 |
🧮 Determination by Dulong & Petit’s Method
In 1819, Dulong and Petit studied the specific heat of solid elements and found a striking pattern: atomic weight × specific heat ≈ 6.4 for almost every solid element they tested.
📊 See the illustrating data table ›
Specific heat is the amount of heat required to raise the temperature of 1 gram of a substance by 1°C. The product of atomic weight and specific heat is called the atomic heat — and Dulong & Petit’s law says this is nearly always ≈ 6.4:
| Solid element | Atomic weight | Specific heat | Atomic heat (wt × sp. heat) |
|---|---|---|---|
| Sodium (Na) | 22.98 | 0.29 | 6.6 |
| Magnesium (Mg) | 24.30 | 0.25 | 6.0 |
| Iron (Fe) | 55.84 | 0.12 | 6.7 |
| Zinc (Zn) | 65.40 | 0.093 | 6.0 |
| Tin (Sn) | 118.71 | 0.054 | 6.4 |
| Silver (Ag) | 107.86 | 0.056 | 6.0 |
| Gold (Au) | 196.96 | 0.032 | 6.3 |
📐 Steps in Dulong and Petit’s method ›
- Find the equivalent weight of the solid element (by any suitable method).
- Find the specific heat of the element in the lab, then calculate the approximate atomic weight:
Approx. atomic weight = 6.4 ÷ specific heat
- Divide the approximate atomic weight by the equivalent weight to get the approximate valency:
Approx. valency = Approx. atomic weight ÷ Equivalent weight
- Round the approximate valency to the nearest whole number (valency is always a whole number). Use this as the element’s valency.
- Find the correct atomic weight:
Correct atomic weight = Equivalent weight × Valency
🚧 Limitations of Dulong & Petit’s Law
- The law holds especially for solid elements, including metals.
- The atomic weight of non-metals like beryllium, boron, carbon, silicon, and sulphur cannot be determined this way, because their specific heat varies with temperature.
- The law only applies to heavier elements — it cannot be applied to lighter elements with high melting points.
✍️ Worked Example
The specific heat of a metal is 0.259, and its equivalent weight is 12 g. Find its exact atomic weight.
Given: specific heat = 0.259, equivalent weight = 12
Approx. atomic weight = 6.4 ÷ 0.259 = 24.710
Approx. valency = 24.710 ÷ 12 = 2.059 → rounded to 2
Correct atomic weight = 12 × 2 = 24
✅ The atomic weight of the metal is 24 amu.
🎯 Test yourself
Try the quick quiz on atomic weight before moving to the next topic.
