Topics

The molecular ion and molar mass

Covers AS91388

What a mass spectrometer does

The sample is ionised, usually by knocking off one electron to form a positive ion.
The ions are accelerated and then deflected by a magnetic field. Lighter ions deflect more.
A detector records how many ions arrive at each mass-to-charge ratio (m/z).
Because almost every ion has a charge of +1, the m/z value is effectively the mass of the ion.

The molecular ion peak

The molecular ion, M⁺, is the whole molecule with one electron removed:

$\text{M} + e^- \rightarrow \text{M}^+ + 2e^-$

Its m/z value is the molar mass of the compound.
It is normally the peak at the highest m/z, apart from small isotope peaks just to its right.
Some molecules fragment so readily that the molecular ion peak is very small or absent. If the highest significant peak looks implausible as a molar mass, consider that possibility.

The base peak

The base peak is the tallest peak in the spectrum, assigned a relative intensity of 100%.
It is the most stable fragment ion, not necessarily the molecular ion.
Do not confuse the two: the base peak tells you about fragmentation, the molecular ion tells you the molar mass.

Reading off the molecular formula

Once you have the molar mass and know a functional group from the IR, you can usually fix the formula.

Work out how much mass the functional group accounts for, then see what is left for the hydrocarbon part.
Check the count with the general formulae:

Family	General formula	Example (M)
Alkane	$\text{C}_n\text{H}_{2n+2}$	butane, 58
Alkene	$\text{C}_n\text{H}_{2n}$	butene, 56
Alcohol	$\text{C}_n\text{H}_{2n+2}\text{O}$	propan-1-ol, 60
Aldehyde / ketone	$\text{C}_n\text{H}_{2n}\text{O}$	propanone, 58
Carboxylic acid / ester	$\text{C}_n\text{H}_{2n}\text{O}_2$	ethanoic acid, 60
Primary amine	$\text{C}_n\text{H}_{2n+3}\text{N}$	ethylamine, 45

Useful atomic masses to have memorised: H = 1, C = 12, N = 14, O = 16, Cl = 35, Br = 79.

Degrees of unsaturation

Compare the hydrogen count with the saturated alkane of the same carbon number, $\text{C}_n\text{H}_{2n+2}$ .
Every two hydrogens missing means one double bond or ring.
- C₃H₆O has 6 H where propane C₃H₈ has 8 — two missing, so one double bond. With an oxygen present, that is almost certainly a C=O.
- C₃H₈O has the full complement, so no double bonds — an alcohol or an ether.
This is a fast, powerful check. If your IR shows a C=O but the formula has no degrees of unsaturation, one of the two readings is wrong.

Worked ExampleDetermining a molecular formula from the molar mass

A compound gives a molecular ion peak at m/z 74. Its infrared spectrum shows a strong absorption at 1740 cm⁻¹ and no absorption above 3000 cm⁻¹. Determine a possible molecular formula and state the degrees of unsaturation.

Step 1 — Read the molar mass

The molecular ion is at m/z 74, so the molar mass is 74 g mol⁻¹.

Step 2 — Use the IR to identify the functional group

A strong absorption at 1740 cm⁻¹ is a C=O stretch, and its position in the 1735–1750 range points to an ester.

The absence of any absorption above 3000 cm⁻¹ is equally informative: there is no O–H (which would appear as a broad band at 2500–3550) and no N–H. This eliminates alcohols, carboxylic acids, amines and amides.

Step 3 — Account for the mass

An ester contains the group –COO–, which accounts for:

$12 + 16 + 16 = 44$

Subtracting from the molar mass leaves:

$74 - 44 = 30$

for the rest of the molecule — the two alkyl groups attached either side.

Thirty mass units of C and H is C₂H₆, since $2(12) + 6(1) = 30$ .

Step 4 — Assemble the formula

$\text{C}_2\text{H}_6 + \text{COO} = \text{C}_3\text{H}_6\text{O}_2$

Checking: $3(12) + 6(1) + 2(16) = 36 + 6 + 32 = 74$ . ✓

This matches the general formula for an ester, $\text{C}_n\text{H}_{2n}\text{O}_2$ , with $n = 3$ .

Step 5 — Degrees of unsaturation

The saturated alkane with three carbons is C₃H₈. Our compound has C₃H₆ in its hydrocarbon skeleton — two hydrogens fewer, so there is one degree of unsaturation.

That one degree is accounted for by the C=O already identified in the IR. The two readings agree, which is a check that neither is wrong.

Step 6 — Note what is still undetermined

C₃H₆O₂ as an ester has two possible structures: methyl ethanoate, CH₃COOCH₃, and ethyl methanoate, HCOOCH₂CH₃. Both have the same formula and the same functional group, so neither the mass spectrum nor the IR can separate them. ¹³C NMR is needed, and it distinguishes them cleanly.

Answer: C₃H₆O₂, an ester, with one degree of unsaturation accounted for by the C=O group. Two isomers remain possible.

The molecular ion and molar mass

Covers AS91388

What a mass spectrometer does

The sample is ionised, usually by knocking off one electron to form a positive ion.
The ions are accelerated and then deflected by a magnetic field. Lighter ions deflect more.
A detector records how many ions arrive at each mass-to-charge ratio (m/z).
Because almost every ion has a charge of +1, the m/z value is effectively the mass of the ion.

The molecular ion peak

The molecular ion, M⁺, is the whole molecule with one electron removed:

$\text{M} + e^- \rightarrow \text{M}^+ + 2e^-$

Its m/z value is the molar mass of the compound.
It is normally the peak at the highest m/z, apart from small isotope peaks just to its right.
Some molecules fragment so readily that the molecular ion peak is very small or absent. If the highest significant peak looks implausible as a molar mass, consider that possibility.

The base peak

The base peak is the tallest peak in the spectrum, assigned a relative intensity of 100%.
It is the most stable fragment ion, not necessarily the molecular ion.
Do not confuse the two: the base peak tells you about fragmentation, the molecular ion tells you the molar mass.

Reading off the molecular formula

Once you have the molar mass and know a functional group from the IR, you can usually fix the formula.

Work out how much mass the functional group accounts for, then see what is left for the hydrocarbon part.
Check the count with the general formulae:

Family	General formula	Example (M)
Alkane	$\text{C}_n\text{H}_{2n+2}$	butane, 58
Alkene	$\text{C}_n\text{H}_{2n}$	butene, 56
Alcohol	$\text{C}_n\text{H}_{2n+2}\text{O}$	propan-1-ol, 60
Aldehyde / ketone	$\text{C}_n\text{H}_{2n}\text{O}$	propanone, 58
Carboxylic acid / ester	$\text{C}_n\text{H}_{2n}\text{O}_2$	ethanoic acid, 60
Primary amine	$\text{C}_n\text{H}_{2n+3}\text{N}$	ethylamine, 45

Useful atomic masses to have memorised: H = 1, C = 12, N = 14, O = 16, Cl = 35, Br = 79.

Degrees of unsaturation

Compare the hydrogen count with the saturated alkane of the same carbon number, $\text{C}_n\text{H}_{2n+2}$ .
Every two hydrogens missing means one double bond or ring.
- C₃H₆O has 6 H where propane C₃H₈ has 8 — two missing, so one double bond. With an oxygen present, that is almost certainly a C=O.
- C₃H₈O has the full complement, so no double bonds — an alcohol or an ether.
This is a fast, powerful check. If your IR shows a C=O but the formula has no degrees of unsaturation, one of the two readings is wrong.

Worked ExampleDetermining a molecular formula from the molar mass

Step 1 — Read the molar mass

The molecular ion is at m/z 74, so the molar mass is 74 g mol⁻¹.

Step 2 — Use the IR to identify the functional group

A strong absorption at 1740 cm⁻¹ is a C=O stretch, and its position in the 1735–1750 range points to an ester.

Step 3 — Account for the mass

An ester contains the group –COO–, which accounts for:

$12 + 16 + 16 = 44$

Subtracting from the molar mass leaves:

$74 - 44 = 30$

for the rest of the molecule — the two alkyl groups attached either side.

Thirty mass units of C and H is C₂H₆, since $2(12) + 6(1) = 30$ .

Step 4 — Assemble the formula

$\text{C}_2\text{H}_6 + \text{COO} = \text{C}_3\text{H}_6\text{O}_2$

Checking: $3(12) + 6(1) + 2(16) = 36 + 6 + 32 = 74$ . ✓

This matches the general formula for an ester, $\text{C}_n\text{H}_{2n}\text{O}_2$ , with $n = 3$ .

Step 5 — Degrees of unsaturation

The saturated alkane with three carbons is C₃H₈. Our compound has C₃H₆ in its hydrocarbon skeleton — two hydrogens fewer, so there is one degree of unsaturation.

That one degree is accounted for by the C=O already identified in the IR. The two readings agree, which is a check that neither is wrong.

Step 6 — Note what is still undetermined

Answer: C₃H₆O₂, an ester, with one degree of unsaturation accounted for by the C=O group. Two isomers remain possible.