SME
Grade 10 Science · Problem type
Average atomic mass from isotopes
Use this for “Boron is 19.9% boron-10 (10.01 u) and 80.1% boron-11 (11.01 u). Find its atomic mass.”
Diagram
average = 10.01 × 19.9% + 11.01 × 80.1% = 1.992 + 8.819 = 10.811 u
The beam balances nearer ¹¹B, the more common isotope.
The grid rounds 19.9% to 20 of the 100 atoms.
This is an example. Type your own values to start a new problem.
A (u)
=
m₁ (u)
×
f₁
100
+
m₂ (u)
×
f₂
100
Formulas
f₂
=
100
−
f₁
80.1 = 100 − 19.9
A
=
m₁
×
f₁
100
+
m₂
×
f₂
100
10.8 = 10.01 × 19.9/100 + 11.01 × 80.1/100
Assumptions
•
The element has two isotopes; their percents add to 100%.
•
The atomic mass on the periodic table is the average over the atoms, weighted by how common each isotope is.
Step-by-step
WE KNOW
m₁ = 10.01 u, m₂ = 11.01 u, f₁ = 19.9%
FIND
abundance of the second isotope (f₂), average atomic mass (A)
Step 1 · Find abundance of the second isotope (f₂)
Use f₂ = 100 − f₁
The two isotopes make up all the atoms: 100%.
f₂
=
100
−
19.9
f₂
=
80.1%
Step 2 · Find average atomic mass (A)
Use A = m₁ × f₁/100 + m₂ × f₂/100
Each isotope counts as much as its share of the atoms.
A
=
10.01
×
19.9/100
+
11.01
×
80.1/100
A
=
199.199/100
+
881.901/100
A
=
1.992
+
8.819
A
=
10.8
u
Check
Put the numbers back in. Both sides should match.
80.1 = 100 − 19.9 ✓
10.8 ≈ 10.01 × 19.9/100 + 11.01 × 80.1/100 ✓
Topic check
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