Q. how to calculate m/z value in mass spectrometry

Definition. The mass-to-charge ratio, written as m/z, is the mass of an ion divided by its charge. In formulas we use the absolute value of the charge because instruments report a positive m/z value. The basic relation is

\[ \mathrm{m/z} = \frac{m_{\text{ion}}}{\lvert z \rvert} \]

Step 1 — Determine the neutral (molecular) mass. Obtain the neutral (monoisotopic) molecular mass of the analyte. Call this mass \(M\). Use high-precision monoisotopic atomic masses when working with exact-mass calculations.

Step 2 — Account for adducts and losses. When an ion forms, the ion mass equals the neutral mass plus the masses of any added species (adducts) minus the masses of any lost species (neutral losses). Write

\[ m_{\text{ion}} = M + \sum_i n_i\,m_{i,\mathrm{add}} – \sum_j l_j\,m_{j,\mathrm{loss}} \]

Here \(n_i\) is the number of each adduct type added and \(l_j\) is the number of each neutral species lost. Common adduct masses (monoisotopic values) you will use include the proton \(m_{\mathrm{H^+}} = 1.007276466879\ \mathrm{u}\) and the sodium ion \(m_{\mathrm{Na^+}} = 22.98976928\ \mathrm{u}\).

Step 3 — Determine the charge state. Determine the integer charge \(z\) of the ion (for example \(z=+1\) for [M+H]+, \(z=+2\) for [M+2H]2+, \(z=-1\) for [M-H]-). When reporting m/z, use the absolute value \(\lvert z\rvert\) in the denominator.

Step 4 — Combine into the m/z formula. Substitute \(m_{\text{ion}}\) and \(\lvert z\rvert\) into the m/z formula:

\[ \mathrm{m/z} = \frac{M + \sum_i n_i\,m_{i,\mathrm{add}} – \sum_j l_j\,m_{j,\mathrm{loss}}}{\lvert z\rvert} \]

Worked example 1 — singly protonated ion [M+H]+. Let the neutral monoisotopic mass be \(M = 300.123400\ \mathrm{u}\). For [M+H]+ we add one proton and the charge is \(z=+1\). Compute the ion mass and m/z:

\[ m_{\text{ion}} = 300.123400 + 1.007276466879 = 301.130676466879\ \mathrm{u} \]

\[ \mathrm{m/z} = \frac{301.130676466879}{1} = 301.130676466879 \]

Round as appropriate for your instrument, for example to six decimals: \(301.130676\ \mathrm{Th}\). (Thomson, Th, is the unit often used for m/z, equal to u per elementary charge.)

Worked example 2 — doubly protonated ion [M+2H]2+. Using the same \(M = 300.123400\ \mathrm{u}\), add two protons and take \(\lvert z\rvert = 2\):

\[ m_{\text{ion}} = 300.123400 + 2\times 1.007276466879 = 302.137952933758\ \mathrm{u} \]

\[ \mathrm{m/z} = \frac{302.137952933758}{2} = 151.068976466879 \]

Rounded to six decimals: \(151.068976\ \mathrm{Th}\).

Worked example 3 — sodium adduct [M+Na]+. With \(M = 300.123400\ \mathrm{u}\) and one sodium adduct \(m_{\mathrm{Na^+}} = 22.98976928\ \mathrm{u}\):

\[ m_{\text{ion}} = 300.123400 + 22.98976928 = 323.11316928\ \mathrm{u} \]

\[ \mathrm{m/z} = \frac{323.11316928}{1} = 323.11316928 \]

Worked example 4 — deprotonated negative ion [M-H]-. For the same \(M\) and one proton loss, with charge \(z=-1\) (use absolute value 1):

\[ m_{\text{ion}} = 300.123400 – 1.007276466879 = 299.116123533121\ \mathrm{u} \]

\[ \mathrm{m/z} = \frac{299.116123533121}{1} = 299.116123533121 \]

Practical notes.

– Use monoisotopic masses and the exact masses of adducts for high-resolution instruments.

– For multiply charged species, the m/z is lower and peaks appear at roughly \( (M+\text{adducts})/\lvert z\rvert \).

– Instruments report m/z to a finite precision. Round the computed value to the same number of decimal places or ppm precision used in your analysis.

– If you need sub-ppm accuracy, include small corrections such as the electron mass when appropriate; for most routine calculations using the monoisotopic proton mass above is sufficient.