SME
Organic Chemistry I
Mass spectrometry and structure determination
Use this for “The molecular ion at m/z 128 is 42% high and the M + 1 peak is 4.2%. About how many carbons are there?”
Chart
This is an example. Type your own values to start a new problem.
nC
=
IM+1 (%)
IM (%)
×
100
1.1
r
Ratio of the M + 1 peak to M
calculated · %
Formulas
r
=
100
×
IM+1
IM
6.7
=
100
×
6.7
100
nC
=
r
1.1
6.09
=
6.7
1.1
Assumptions
•
About 1.1% of carbon atoms are ¹³C, so each carbon adds 1.1% of the M peak to the M + 1 peak.
•
The small M + 1 shares from ²H, ¹⁵N and ¹⁷O are left out; nitrogen-rich compounds read a little high.
•
The M peak is the molecular ion, and both heights are read on the same scale, as a percent of the tallest peak.
Step-by-step
WE KNOW
IM = 100%, IM+1 = 6.7%
FIND
ratio of the M + 1 peak to M (r), carbons in the molecule (nC)
Step 1 · Find ratio of the M + 1 peak to M (r)
Use r = 100 × (IM+1) ÷ IM
Write the M + 1 height as a percent of the M height.
r
=
100
×
6.7
100
r
=
670
100
r
=
6.7%
Step 2 · Find carbons in the molecule (nC)
Use nC = r ÷ 1.1
Each carbon adds 1.1%, so divide the ratio by 1.1.
nC
=
6.7
1.1
nC = 6.09 → Round to 6 carbons: ¹³C gives each carbon a 1.1% chance.
Check
Put the numbers back in. Both sides should match.
6.7
=
100
×
6.7
100
✓
6.09
=
6.7
1.1
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.