SME
Organic Chemistry I · Problem type
Exact mass and ppm error
Use this for “What is the exact mass of C₇H₈O, and how many ppm off is a peak measured at m/z 108.0573?”
Table
Atom
n
m
(u)
n × m
(u)
C
8
12
96
H
8
1.0078
8.063
N
0
14.0031
0
O
1
15.9949
15.99
M
120.05752
Measured mass Mₘₑₐₛ = 120.0572 u
Mass error error = −2.62 ppm
This is an example. Type your own values to start a new problem.
M (u)
=
12
×
nC
+
1.007825
×
nH
+
14.003074
×
nN
+
15.994915
×
nO
Mₘₑₐₛ
Measured mass
example · u
error
Mass error
worked out (not typed) · ppm
Formulas
M
=
12
×
nC
+
1.007825
×
nH
+
14.003074
×
nN
+
15.994915
×
nO
120.05752 = 12 × 8 + 1.007825 × 8 + 14.003074 × 0 + 15.994915 × 1
error
=
10
6
×
Mₘₑₐₛ −
M
M
−2.62
=
10
6
×
120.0572 − 120.05752
120.05752
Assumptions
•
Exact (monoisotopic) masses use the most common isotope of each element: ¹²C = 12.0000, ¹H = 1.007825, ¹⁴N = 14.003074 and ¹⁶O = 15.994915 u.
•
The ion carries one charge, so its m/z is its mass in u; the electron’s mass is left out.
•
A formula fits when the measured mass is within about 5 ppm of its exact mass.
Step-by-step
WE KNOW
nC = 8, nH = 8, nN = 0, nO = 1, Mₘₑₐₛ = 120.0572 u
FIND
exact mass (M), mass error (error)
Step 1 · Find exact mass (M)
Use M = 12 × nC + 1.007825 × nH + 14.003074 × nN + 15.994915 × nO
Add the mass of each atom: the count times the exact mass of its most common isotope.
M
=
12
×
8
+
1.007825
×
8
+
14.003074
×
0
+
15.994915
×
1
M
=
96
+
8.0626
+
15.9949
M
=
120.05752
u
Step 2 · Find mass error (error)
Use error = 10⁶ × (Mₘₑₐₛ − M) ÷ M
Divide the gap between the measured and exact masses by the exact mass, in parts per million.
error
=
10
6
×
120.0572 − 120.05752
120.05752
error
=
10
6
×
−0.00032
120.0575
error
=
1,000,000
×
−0.00032
120.0575
error
=
−
320
120.0575
error = −2.62 ppm → Within 5 ppm: the formula fits the peak.
Check
Put the numbers back in. Both sides should match.
120.05752 = 12 × 8 + 1.007825 × 8 + 14.003074 × 0 + 15.994915 × 1 ✓
−2.62
=
10
6
×
120.0572 − 120.05752
120.05752
✓
Topic check
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