Relative Atomic and Isotopic Mass

Learning Objectives (OCR Spec 2.1.2)

• Define relative isotopic mass and relative atomic mass based on the ¹²C scale
• Define relative molecular mass (Mr) and relative formula mass
• Calculate relative atomic mass from isotopic abundances
• Understand why relative atomic masses are usually non-integer values
• Work backwards from Ar to find isotopic abundances
• Appreciate why ¹²C was chosen as the reference

Why Use Relative Masses?

Individual atoms are fantastically small — a single ¹²C atom has a mass of about 1.993 × 10⁻²³ g. Working with such numbers is cumbersome, so chemists use relative masses: the mass of an atom or molecule compared with a standard reference.

Since 1961, that standard has been carbon-12 (¹²C). One atom of ¹²C is defined as having a mass of exactly 12 atomic mass units (12 u). Every other atom is measured against this.

Historical Note

The earliest atomic weights, measured by Dalton in 1803, were based on hydrogen = 1. In 1898, Wilhelm Ostwald proposed using O = 16 — convenient because oxygen combines with so many elements. From 1929 to 1961 chemists and physicists actually used different oxygen scales: chemists used "natural oxygen" while physicists used the pure ¹⁶O isotope. This caused small but infuriating discrepancies in the literature. In 1961 the IUPAC and IUPAP agreed on ¹²C = 12 exactly, resolving the conflict. ¹²C was chosen because:

• It is the most abundant carbon isotope
• Carbon forms the basis of organic chemistry
• It gives atomic masses that are almost identical (within 1%) to the old oxygen scale
• Carbon atoms can be counted very precisely using mass spectrometry

Key Definitions

You must learn these OCR definitions precisely — they are frequently asked verbatim. Marks are routinely lost for imprecise wording.

Relative Isotopic Mass

The mass of an atom of an isotope compared with one-twelfth of the mass of an atom of carbon-12.

Relative isotopic masses are effectively whole numbers (within 0.1%) because the dominant contribution to atomic mass comes from whole nucleons. For example, relative isotopic mass of ³⁵Cl ≈ 35, and of ³⁷Cl ≈ 37.

In precise experimental work, exact relative isotopic masses differ very slightly from whole numbers because of the mass defect: when nucleons bind together in a nucleus, a tiny amount of mass is converted into binding energy (E = mc²). For example, the relative isotopic mass of ³⁵Cl is actually 34.9689 — very close to but not exactly 35. For A-Level calculations, the whole number is usually sufficient, but if OCR gives you a decimal value you must use it.

Relative Atomic Mass (Ar)

The weighted mean mass of an atom of an element compared with one-twelfth of the mass of an atom of carbon-12.

The word weighted is critical: you must account for natural abundance. Because most elements exist as a mixture of isotopes, Ar values are rarely whole numbers. For chlorine, Ar = 35.5 because of the natural mix of ³⁵Cl and ³⁷Cl.

Relative Molecular Mass (Mr)

The weighted mean mass of a molecule compared with one-twelfth of the mass of an atom of carbon-12.

For molecular compounds — e.g. Mr(H₂O) = 2(1.0) + 16.0 = 18.0.

Relative Formula Mass

Used for ionic compounds which do not exist as discrete molecules. The definition is analogous but uses "formula unit" instead of "molecule". For example, relative formula mass of NaCl = 23.0 + 35.5 = 58.5.

Note: In OCR exams the symbol Mr is used for both relative molecular and relative formula mass — no distinction is made at A-Level unless the question specifically asks.

A Key Subtlety — Dimensionless

All four of these quantities are ratios of masses — they are dimensionless and have no units. Candidates who write "Ar = 35.5 g" lose a mark. If you want to express mass per mole, use molar mass (e.g. 35.5 g mol⁻¹), not relative atomic mass.

Calculating Relative Atomic Mass from Abundances

The formula you must know:

$A_r = \frac{\sum (\text{isotopic mass} \times \text{abundance})}{\sum \text{abundances}}$Ar​=∑abundances∑(isotopic mass×abundance)​

If abundances are given as percentages, the denominator is 100. If given as relative abundances (e.g. peak heights on a mass spectrum), you divide by their total.

Worked Example 1: Chlorine

Naturally occurring chlorine contains 75.78% ³⁵Cl and 24.22% ³⁷Cl. Calculate Ar(Cl) to 1 d.p.

$A_r = \frac{(35 \times 75.78) + (37 \times 24.22)}{100}$Ar​=100(35×75.78)+(37×24.22)​

$A_r = \frac{2652.3 + 896.14}{100} = \frac{3548.44}{100} = 35.5$Ar​=1002652.3+896.14​=1003548.44​=35.5

Worked Example 2: Copper

Copper exists as 69.2% ⁶³Cu and 30.8% ⁶⁵Cu. Calculate Ar(Cu).

$A_r = \frac{(63 \times 69.2) + (65 \times 30.8)}{100} = \frac{4359.6 + 2002.0}{100} = 63.6$Ar​=100(63×69.2)+(65×30.8)​=1004359.6+2002.0​=63.6

Worked Example 3: Magnesium (Three Isotopes)

Magnesium has three stable isotopes: