Relative atomic mass

Relative atomic mass, written Ar, is one of the first numbers you meet in SPM Chemistry, and it quietly underpins almost every calculation that follows. We teach it carefully in our online one-to-one lessons because a shaky grasp here costs marks later in the mole chapter, in titration and in every formula question.

When you use it

You reach for Ar whenever a question gives an element and expects a mass comparison, reading a value from the Periodic Table, or working one out from isotope data. It is also the building block of relative molecular mass and of molar mass, so the idea returns again and again.

The definition and the formula

The relative atomic mass of an element is the average mass of one atom of the element compared with one-twelfth of the mass of one carbon-12 atom. Because it compares two masses, Ar has no unit.

When an element has more than one isotope, you calculate a weighted average:

Ar = ( sum of [ relative abundance x isotopic mass ] ) / ( total relative abundance )

If the abundances are given as percentages, the total is 100.

Units

Ar has no unit. The isotopic masses are themselves relative (unit-less) masses, and the abundances are ratios or percentages.

Worked example 1 (easy), chlorine

Chlorine has two isotopes: chlorine-35 (75 %) and chlorine-37 (25 %).

• Step 1, multiply each isotopic mass by its abundance: (35 x 75) + (37 x 25) = 2625 + 925 = 3550.
• Step 2, divide by the total abundance: 3550 / 100 = 35.5.

Ar of chlorine = 35.5 (no unit).

Worked example 2 (medium), copper

Copper has two isotopes: copper-63 (69.2 %) and copper-65 (30.8 %).

• Step 1: (63 x 69.2) + (65 x 30.8) = 4359.6 + 2002 = 6361.6.
• Step 2: 6361.6 / 100 = 63.6.

Ar of copper = 63.6 (rounded to one decimal place). Notice the answer sits closer to 63 because copper-63 is the more abundant isotope.

Worked example 3 (SPM level), finding a missing abundance

Boron has two isotopes, boron-10 and boron-11, and its relative atomic mass is 10.8. Find the percentage abundance of each isotope.

• Step 1, let the abundance of boron-11 be x %; then boron-10 is (100 − x) %.
• Step 2, write the weighted average: [ 11x + 10(100 − x) ] / 100 = 10.8.
• Step 3, expand the bracket: 11x + 1000 − 10x = 1080.
• Step 4, collect and solve: x = 80.

So boron-11 is 80 % and boron-10 is 20 %. Always check that your two abundances add up to 100.

Common traps

• Writing a unit after Ar. It has none, it is a pure ratio, so answers such as “35.5 g” lose the mark.
• Confusing Ar with the nucleon (mass) number of one isotope. Ar is a weighted average and is often not a whole number.
• Forgetting to divide by the total abundance (100 when percentages are used).
• Rounding the isotopic products too early; keep the figures until the final line, then round.
• In the “find the missing abundance” type, forgetting that the two abundances must sum to 100 before you solve.

How we help

In our lessons, our teachers give you isotope-abundance questions in the exact phrasing that SPM Chemistry uses, including the harder “find the missing abundance” style of Worked example 3, so the algebra never surprises you in the examination hall. Lessons are one-to-one and taught in English, from RM50 per hour, with a paid one-hour trial so you can judge the method before you commit. Master Ar now and relative molecular mass, molar mass and the whole mole chapter fall into place far more easily.