3.5.5 Determination of Relative Atomic Mass

This topic explains how to calculate the relative atomic mass of an element from the masses and natural abundances of its isotopes.

Definition

The relative atomic mass of an element can be calculated from the relative masses of its isotopes and their relative abundance.

Simple explanation: Relative atomic mass is the weighted average of the masses of all naturally occurring isotopes of an element, taking into account how much of each isotope is present.

Main Idea

Most elements exist as a mixture of isotopes. Since each isotope has a different mass, the relative atomic mass is not simply the mass of one isotope — it is a weighted average based on how abundant each isotope is.

General formula:

Relative atomic mass = (abundance of isotope 1 × mass of isotope 1) + (abundance of isotope 2 × mass of isotope 2) + …100

Simple explanation: Multiply each isotope's mass by its percentage abundance, add them all up, and divide by 100.

Core Concept

The Weighted Average Formula

If an element has isotopes with masses m₁, m₂, m₃, … and abundances a₁\%, a₂\%, a₃\%, …, then:

Relative atomic mass = a₁ m₁ + a₂ m₂ + a₃ m₃ + …100

Detailed Explanation

Example from the Book: Carbon

Natural abundance of isotopes of carbon:

¹²C = 98.8\%, ¹³C = 1.1\%, ¹⁴C = 0.009\%

Calculate relative atomic mass of carbon.

Solution:

The relative atomic mass is a weighted average of all the naturally occurring isotopes of an element, taking into consideration their natural abundance.

Use the general formula:

Relative atomic mass of C = (98.8 × 12) + (1.1 × 13) + (0.009 × 14)100

= 1185.6 + 14.3 + 0.126100

= 1200.026100

= 12.00 amu

Final Answer: The relative atomic mass of carbon is 12.00 amu.

Simple explanation: Even though carbon has three isotopes, its relative atomic mass is very close to 12 because carbon-12 is so abundant (98.8%).

Worked Example: Finding the Isotopic Mass of an Unknown Isotope

Problem (from the book):

An element has two isotopes A and B. The relative atomic mass of the element is 35.5 amu. Relative abundance of isotope A is 75.77% and its isotopic mass is 35. Find the isotopic mass of B if its relative abundance is 24.23%.

Solution:

Let the isotopic mass of B be x.

35.5 = (75.77 × 35) + (24.23 × x)100

Multiply both sides by 100:

3550 = 2651.95 + 24.23x

24.23x = 3550 - 2651.95 = 898.05

x = 898.0524.23 = 37.06 amu

Final Answer: The isotopic mass of B is approximately 37 amu.

Simple explanation: This is essentially the chlorine problem — two isotopes, one at 35 amu (75.77%) and one at ~37 amu (24.23%), giving a weighted average of 35.5 amu.

Intermediate Level

Why Divide by 100?

Because the abundances are given as percentages (out of 100). Dividing by 100 converts the weighted sum back to a single atomic mass value.

If abundances are given as fractions (e.g., 0.75 and 0.25), you do not divide by 100 — you just add:

Relative atomic mass = (0.75 × 35) + (0.25 × 37) = 35.5 amu

Why Relative Atomic Mass Is Often Not a Whole Number

• Elements exist as mixtures of isotopes.
• Each isotope has a different mass.
• The relative atomic mass is a weighted average.
• So it is often a **decimal** (e.g., Cl = 35.5, Cu = 63.5, C = 12.00).

Common Examples

Solving for an Unknown Isotope

If you know the relative atomic mass and all but one isotope's mass, you can solve for the unknown mass.

General approach:

• Write the weighted average equation.
• Substitute all known values.
• Solve for the unknown mass.

Example (from the book):

• Element with two isotopes A and B
• Relative atomic mass = 35.5 amu
• A: 75.77%, mass = 35 amu
• B: 24.23%, mass = x

35.5 = (75.77 × 35) + (24.23 × x)100

Solving gives x ≈ 37 amu.

Important Points

• - Relative atomic mass = weighted average of isotope masses.
• - Formula:
• Relative atomic mass = ∑ (abundance × isotopic mass)100
• - The more abundant isotope has a bigger effect on the average.
• - Relative atomic mass is often not a whole number (because of isotopes).
• - Example: Carbon = 12.00 amu; Chlorine = 35.5 amu.
• - You can solve for an unknown isotope mass if you know the relative atomic mass and other isotope data.

Remember

• - Relative atomic mass = weighted average of isotopes.
• - Formula: ∑ (abundance × mass)100.
• - Carbon: ¹²C (98.8%), ¹³C (1.1%), ¹⁴C (0.009%) → 12.00 amu.
• - Chlorine: ³⁵Cl (75%), ³⁷Cl (25%) → 35.5 amu.
• - The more abundant isotope has a bigger effect.
• - Relative atomic mass is often not a whole number.
• - You can solve for an unknown isotope mass using the formula.

Common Mistakes

• 1. Forgetting to divide by 100. The formula uses percentages, so you must divide by 100.
• 2. Thinking the most common isotope's mass is the relative atomic mass. It is a weighted average.
• 3. Forgetting to include all isotopes. All naturally occurring isotopes must be included.
• 4. Using mass numbers instead of isotopic masses. For most problems, mass numbers are close enough, but for precise work, use actual isotopic masses.
• 5. Confusing relative atomic mass with mass number. Mass number is a whole number (protons + neutrons); relative atomic mass is a weighted average (may be decimal).
• 6. Forgetting to multiply abundance by mass. Each isotope contributes based on both its mass and its abundance.

Quick Revision

• Relative atomic mass can be calculated from isotope masses and abundances.
• Formula: ∑ (abundance × isotopic mass)100.
• **Carbon example:** (98.8 × 12) + (1.1 × 13) + (0.009 × 14) ÷ 100 = 12.00 amu.
• **Chlorine example:** (75 × 35) + (25 × 37) ÷ 100 = 35.5 amu.
• The more abundant isotope has a bigger effect on the average.
• Relative atomic mass is often not a whole number because it is a weighted average of isotopes.
• You can solve for an unknown isotope mass if you know the relative atomic mass and other isotope data.

60-Second Revision

• Relative atomic mass = weighted average of isotopes.
• Formula: ∑ (abundance × mass)100.
• Carbon: 12.00 amu; Chlorine: 35.5 amu.
• The more abundant isotope has a bigger effect.
• Warning: always divide by 100 when using percentage abundances.