Using Mass Spectrometry to Determine Molecular Formula Revision Notes | A Level AQA Chemistry

1. Why Mass Spectrometry Can Reveal Molecular Formula

You already know that combustion analysis or percentage composition gives you an empirical formula — the simplest ratio of atoms. But that ratio alone cannot tell you whether a molecule is $C_2H_4$, $C_4H_8$, or $C_6H_{12}$. You need the actual molecular mass to decide.

Mass spectrometry solves this problem directly. When a sample is ionised and accelerated through the instrument, the heaviest detectable ion is the intact molecule minus one electron — the molecular ion, written $M^+$. The m/z value of that peak is numerically equal to the molecular mass of the compound (in g mol$^{-1}$), and that single number is enough to pin down the molecular formula.

At A-Level, AQA requires you to use precise atomic masses alongside the precise molecular mass from the spectrum. This is the key distinction from GCSE-level work: precise masses let you discriminate between molecules that share the same nominal (integer) mass but differ in their exact composition.

2. The Molecular Ion Peak and How to Read It

In a low-resolution mass spectrum, the molecular ion peak gives you the nominal (integer) molecular mass. In a high-resolution instrument, the m/z value is reported to several decimal places — this is the precise molecular mass that AQA expects you to use when determining molecular formula.

Here is why precision matters: $CO$ and $N_2$ both have a nominal mass of 28, yet their precise masses differ ($CO = 27.9949$; $N_2 = 28.0062$). A high-resolution spectrum resolves them instantly. At A-Level, you will typically be given the precise molecular mass and a set of precise atomic masses, and your task is to find the formula that fits.

3. Precise Atomic Masses and Why They Matter

The AQA specification requires you to use precise atomic masses — not the rounded values you used at GCSE. The table below shows the precise masses you will typically be given or need to apply.

When you are given a precise molecular mass from a high-resolution spectrum, you calculate the mass that each candidate formula would produce using these precise values, then match it to the measured value. The formula whose calculated precise mass agrees with the spectrum (typically to 4 decimal places) is the molecular formula.

For most AQA exam questions, you will work with organic molecules containing C, H, O, and sometimes N. The strategy is always the same: start from the empirical formula, find the integer multiplier using the molecular mass, and confirm with precise masses if required.

4. The Two-Stage Method: Empirical Formula to Molecular Formula

The standard AQA route to molecular formula combines two pieces of information: the empirical formula (from combustion analysis or percentage composition) and the molecular mass (from the $M^+$ peak in the mass spectrum). Here is how the method works in two clear stages.

Stage 1 — find the empirical formula mass by adding up the precise atomic masses of the atoms in the empirical formula. Stage 2 — divide the precise molecular mass by the empirical formula mass to get the integer multiplier $n$, then multiply each subscript in the empirical formula by $n$.

Worked example: Determining Molecular Formula from Empirical Formula and Mass Spectrum

A compound has an empirical formula of $CH_2O$ and gives a molecular ion peak at m/z = 60.0211 in a high-resolution mass spectrum. Use precise atomic masses ($^{12}C = 12.0000$, $^1H = 1.00794$, $^{16}O = 15.9949$) to determine the molecular formula.

1. $M(CH_2O) = 12.0000 + (2 \times 1.00794) + 15.9949$\n$= 12.0000 + 2.01588 + 15.9949$\n$= 30.0108$ — Add the precise atomic masses for each atom in the empirical formula: one C, two H, one O.
2. $n = \frac{\text{precise molecular mass}}{\text{empirical formula mass}} = \frac{60.0211}{30.0108} \approx 2.00$ — Divide the molecular mass from the M⁺ peak by the empirical formula mass. The result should be a whole number (or very close to one). Here n = 2.
3. Molecular formula $= (CH_2O)_2 = C_2H_4O_2$ — Multiply every subscript in the empirical formula by n = 2.
4. $M(C_2H_4O_2) = (2 \times 12.0000) + (4 \times 1.00794) + (2 \times 15.9949)$\n$= 24.0000 + 4.03176 + 31.9898$\n$= 60.0216$ — The calculated precise mass (60.0216) agrees with the measured value (60.0211) to within acceptable rounding. The molecular formula is confirmed.

Answer: Molecular formula: $C_2H_4O_2$ (for example, this is the formula of glycolaldehyde or ethanoic acid — further structural evidence would be needed to distinguish them).

Notice that the verification step in Stage 4 is not optional in high-precision questions — it is the confirmation that your formula is correct. If the calculated precise mass does not match the spectrum value, you have either the wrong empirical formula or the wrong multiplier.

5. Using Precise Masses to Distinguish Between Candidate Formulae

This is where high-resolution mass spectrometry becomes genuinely powerful. Two different molecular formulae can share the same nominal (integer) mass but have different precise masses. The spectrum tells you which one is present.

Consider the nominal mass of 44. Both $C_2H_4O$ (ethanal, $M = 44.0262$) and $CO_2$ ($M = 43.9898$) have a nominal mass near 44, yet their precise masses differ by about 0.036 u — easily resolved by a modern high-resolution instrument. Matching the measured precise mass to the calculated precise mass for each candidate formula is the definitive test.

Worked example: Distinguishing Between Two Candidate Formulae Using Precise Masses

A high-resolution mass spectrum gives a molecular ion peak at m/z = 58.0419. Two candidate formulae are proposed: $C_3H_6O$ and $C_2H_2O_2$. Use precise atomic masses ($^{12}C = 12.0000$, $^1H = 1.00794$, $^{16}O = 15.9949$) to identify the correct molecular formula.

1. $M(C_3H_6O) = (3 \times 12.0000) + (6 \times 1.00794) + (1 \times 15.9949)$ $= 36.0000 + 6.04764 + 15.9949$ $= 58.0425$ — Three carbons, six hydrogens, one oxygen. Add each contribution carefully.
2. $M(C_2H_2O_2) = (2 \times 12.0000) + (2 \times 1.00794) + (2 \times 15.9949)$ $= 24.0000 + 2.01588 + 31.9898$ $= 58.0057$ — Two carbons, two hydrogens, two oxygens. Note how replacing CH₄ with O shifts the precise mass significantly.
3. Measured: $58.0419$ $C_3H_6O$: $58.0425$ — difference $= 0.0006$ $C_2H_2O_2$: $58.0057$ — difference $= 0.0362$ — The formula whose calculated precise mass is closest to the measured value is the correct one.
4. $C_3H_6O$ matches the spectrum (difference < 0.001 u, within instrument precision). — $C_2H_2O_2$ is clearly ruled out by the large discrepancy of 0.036 u. The molecular formula is $C_3H_6O$.

Answer: The molecular formula is $C_3H_6O$ (for example, propanal or acetone — further spectroscopic evidence such as IR or NMR would be needed to determine the structure).

6. Worked Exam-Style Calculation with Percentage Composition

AQA questions often combine percentage composition data with a molecular ion peak, requiring you to derive the empirical formula first and then use the mass spectrum to find the molecular formula. The sequence is: percentage composition → empirical formula → molecular formula using $M^+$ peak.

Worked example: Full Route: Percentage Composition + Mass Spectrum to Molecular Formula

A compound contains 40.00% C, 6.71% H, and 53.29% O by mass. Its mass spectrum shows a molecular ion peak at m/z = 90. Determine the molecular formula. (Use $A_r$: C = 12.01, H = 1.008, O = 16.00.)

1. $n(C) = \frac{40.00}{12.01} = 3.331$ $n(H) = \frac{6.71}{1.008} = 6.657$ $n(O) = \frac{53.29}{16.00} = 3.331$ — Divide each percentage by the relative atomic mass of that element. This gives the molar ratio of atoms.
2. $C: \frac{3.331}{3.331} = 1.000$ $H: \frac{6.657}{3.331} = 1.998 \approx 2$ $O: \frac{3.331}{3.331} = 1.000$ — Dividing by the smallest value (3.331) gives the simplest whole-number ratio. Round values within 0.05 of a whole number.
3. Empirical formula $= CH_2O$ — The ratio C: H: O = 1: 2: 1 gives the empirical formula $CH_2O$.
4. $M(CH_2O) = 12.01 + (2 \times 1.008) + 16.00$ $= 12.01 + 2.016 + 16.00$ $= 30.026$ — Use the same $A_r$ values as in Step 1 for consistency.
5. $n = \frac{90}{30.026} \approx 2.998 \approx 3$ — The molecular mass from the M⁺ peak is 90. Dividing by the empirical formula mass gives n = 3 (rounded to the nearest integer).
6. Molecular formula $= (CH_2O)_3 = C_3H_6O_3$ — Multiply each subscript in the empirical formula by n = 3. Verify: $M(C_3H_6O_3) = (3 \times 12.01) + (6 \times 1.008) + (3 \times 16.00) = 36.03 + 6.048 + 48.00 = 90.078 \approx 90$. This matches the M⁺ peak.

Answer: Molecular formula: $C_3H_6O_3$

7. Exam Technique and Common Mistakes

Mass spectrometry questions on AQA papers follow predictable patterns. The marks are distributed across identifying the molecular ion peak, calculating the empirical formula mass, finding n, and stating the molecular formula. Losing marks here is almost always due to one of a small set of avoidable errors.

Before the next quiz, connect the data to the structure: the molecular ion gives the total Mr, while the fragment peaks help you test whether a proposed molecular formula is realistic.

A mass spectrum is strongest when you move in sequence: identify the molecular ion peak, estimate the formula, then use fragment peaks to rule options in or out.