Q.Calculate the average atomic mass of hydrogen using the following data: Isotope $^1H$ — % Natural abundance: 99.985, Molar mass: 1
Isotope $^2H$ — % Natural abundance: 0.015, Molar mass: 2

The key idea is that the average atomic mass is the weighted mean of the isotopic masses, using their natural abundances as weights.

Step 1: Convert each percentage abundance into a decimal fraction.

• $^1H$: $99.985\% = 0.99985$
• $^2H$: $0.015\% = 0.00015$

Step 2: Multiply each isotopic mass by its fractional abundance and sum the products. …

Why a weighted average?

An element's atomic mass on the periodic table isn't the mass of a single atom — it's the average mass of all naturally occurring atoms of that element. Hydrogen exists as two stable isotopes: protium ($^1H$, mass ≈ 1 u) and deuterium ($^2H$, mass ≈ 2 u). Since protium is vastly more common (99.985% of all hydrogen atoms), the average should be very close to 1, but slightly higher because of the tiny fraction of heavier deuterium atoms.

The formula is straightforward:

The division by 100 converts percentage to a decimal fraction.

Step-by-step calculation

1. Identify the data

   • $^1H$: mass = 1 u, abundance = 99.985%
   • $^2H$: mass = 2 u, abundance = 0.015%

2. Multiply each isotope's mass by its percentage abundance

   • For $^1H$: $1 \times 99.985 = 99.985$
   • For $^2H$: $2 \times 0.015 = 0.030$

3. Add these products

4. Divide by 100 (because abundances are in percent) $\frac{100.015}{100} = 1.00015$ …

Concept: Weighted Average (Isotopic Abundance)

The atomic mass of an element is not a simple average of its isotopes' masses — it is a weighted average, where each isotope's mass is multiplied by its fractional abundance (its percentage divided by 100).

Method: Weighted Average Formula

Steps

1. Convert percentages to decimal fractions

   • $^1H$: $99.985\% = 0.99985$
   • $^2H$: $0.015\% = 0.00015$

2. Multiply each isotope's mass by its fractional abundance

   • $^1H$ contribution: $1 \times 0.99985 = 0.99985$
   • $^2H$ contribution: $2 \times 0.00015 = 0.00030$

3. Sum the contributions

4. Result Average atomic mass of hydrogen = $1.00015 \, \text{u}$

Why this works …

Common Mistakes & How to Avoid Them

Mistake 1: Using the wrong formula (averaging the masses directly)

The error:

Students often take a simple average:

This is wrong because it ignores the fact that $^1H$ is far more abundant.

Why it's wrong:

Average atomic mass is a weighted average, not an arithmetic mean. The weight is the natural abundance (as a fraction or percentage).

How to avoid:

Always use the weighted average formula:

For this problem:

Mistake 2: Forgetting to convert percentage to decimal

The error:

Using abundances as decimals (0.99985 and 0.00015) but then not dividing by 100 — or vice versa.

Why it's wrong:

If you use percentages directly, you must divide by 100. If you use decimals, you don't divide again. Mixing these gives a wrong result.

How to avoid:

Pick one method and stick to it:

• Method A (percentages):

• Method B (decimals):

Both give the same answer: 1.00015 u

Mistake 3: Rounding too early

The error:

Rounding 99.985 to 100 or 0.015 to 0 before calculating.

Why it's wrong:

This destroys the precision. The contribution from $^2H$ (0.00015 u) is small but real — rounding it to zero gives exactly 1 u, which is incorrect.

How to avoid:

Carry all digits through the calculation. Round only at the final step to the required significant figures.

Mistake 4: Confusing "molar mass" with "mass number"

The error:

Using mass numbers (1 and 2) directly without realising they are approximate values of the actual isotopic masses.

Why it's wrong:

For hydrogen, the mass of $^1H$ is actually 1.007825 u, not exactly 1 u. The problem gives "molar mass: 1" as a simplification, but in real exam problems, using exact masses changes the answer slightly.

How to avoid:

• If the problem gives exact masses (like 1.007825), use those. …