Molar Mass, Moles and Converting a Mass to an Amount of Substance

To convert a mass of peptide into an amount of substance, divide the mass by the molar mass: n = m ÷ M. With the mass in milligrams and the molar mass in grams per mole, the answer comes out directly in millimoles. That is the whole calculation. Every error in practice comes from the two inputs: a mass that includes material which is not the molecule being counted, or a molar mass that belongs to a different entity from the one the mass describes.

This page covers the conversion and the choice of inputs. How the peptide fraction of a weighed solid is determined sits in the companion reference on net peptide content and gross weight; carrying a molar amount through a set of dilutions sits in the reference on stock solutions. The worked figures below use an invented decapeptide that corresponds to no product and exists only to make the arithmetic visible.

The quantities and their units

Amount of substance, symbol n, is a measure of the number of specified elementary entities in a sample. The entity must be stated: an atom, a molecule, an ion, or a defined group of particles 1. For a peptide, the entity is normally one molecule of the free base, but a salt, a counterion or a single residue type can equally be counted, and each gives a different number from the same sample. The SI unit is the mole, fixed since 2019 by defining the Avogadro constant as exactly 6.02214076 × 10²³ mol⁻¹ 3.

Molar mass, symbol M, is mass divided by amount of substance, in kg·mol⁻¹ in strict SI and g·mol⁻¹ in laboratory use. Relative molecular mass, Mr, is the same number without a unit: the ratio of the average mass of the entity to one twelfth of the mass of a carbon-12 atom. Molecular mass in daltons is the mass of one molecule. The three share a numerical value and differ in meaning, which is why the Green Book treats them as separate quantities with separate symbols 2.

The conversion and its unit shortcuts

Because M is in grams per mole, the prefix on the mass carries straight through to the amount. No powers of ten need be written out if the prefixes are kept consistent.

The last row is the one most often fumbled. A mass concentration in mg·mL⁻¹ is numerically equal to g·L⁻¹, so dividing it by M gives mol·L⁻¹ directly. A solution at 3.60 mg·mL⁻¹ of a species with M = 1,169.35 g·mol⁻¹ is 3.08 × 10⁻³ mol·L⁻¹, or 3.08 mmol·L⁻¹.

Calculating the molar mass of a sequence

The illustrative peptide is H-Ala-Gly-Ser-Lys-Leu-Glu-Tyr-Phe-Val-Arg-OH, written in one-letter code as AGSKLEYFVR: a linear decapeptide with a free N-terminal amine and a free C-terminal acid. It was chosen for arithmetic, not biology. It carries one tyrosine, so it absorbs at 280 nm, and three basic sites, so it forms a salt with three counterions.

There are two routes to M. The first sums the residue masses and adds one water for the free termini. The second writes the molecular formula and multiplies each element count by its standard atomic weight. The formula route is easier to audit. For AGSKLEYFVR the formula is C54H84N14O15. The standard atomic weights of hydrogen, carbon, nitrogen and oxygen are published by IUPAC as intervals that reflect natural variation in isotopic abundance; for calculation, the conventional single values below are used 4.

The residue route gives the same answer within the last digit, the difference coming from rounding in whichever residue table is used. If the two routes disagree by more than a few hundredths, one of them has a counting error, most often a missing water or a residue entered twice.

Terminal modifications, bridges and counterions

Modifications change the formula, so they change M. Apply them to the linear free-acid value before anything else.

Average and monoisotopic mass

The average mass uses atomic weights, which are abundance-weighted over the natural isotopes of each element. The monoisotopic mass uses only the lightest stable isotope of each. For AGSKLEYFVR the average mass is 1,169.35 and the monoisotopic mass is 1,168.62, a difference of 0.73, or 0.062%.

For converting a weighed mass to an amount, the average mass is the correct one, and the choice is not a matter of taste. A sample on a balance contains the natural isotope distribution, so the mass per mole of that sample is the abundance-weighted mass. The monoisotopic value describes one peak in a resolved mass spectrum. The gap is small for a decapeptide and grows with size: for a small protein it runs to several daltons. The variation built into the atomic-weight intervals is smaller still and is negligible against every other input to this calculation 4.

Salt, water and which molar mass goes with which mass

A synthetic peptide purified by reversed-phase chromatography is isolated as a salt, usually with trifluoroacetate, one equivalent on each protonated basic site 6. With three basic sites, the tris-trifluoroacetate of AGSKLEYFVR has M = 1,169.35 + 3 × 114.02 = 1,511.41 g·mol⁻¹. The free base is 77.4% of that mass before any water or residual salt is counted.

Two consistent pairings exist. Peptide mass, meaning gross mass multiplied by net peptide content, goes with the free-base molar mass. A mass that is known to be the pure, dry salt goes with the salt molar mass. The second pairing is rarely usable in practice, because a real lyophilised solid also holds water and residual salt that a stoichiometric salt mass does not include. Net content measured by amino acid analysis captures all of it, which is why it is the input that makes the free-base pairing work 5.

Row three is the tempting shortcut. It removes the counterion but leaves water and residual salt in the numerator, so it is closer than row two and still wrong by several per cent. Row four is the double correction: net content has already taken the counterion out of the mass, and the salt molar mass takes it out again. Each error is systematic. It will not average away over replicates, and it will pass unchanged into every concentration made from the result.

Worked conversion

Illustrative figures throughout. They describe no real preparation. Substitute the values from the certificate and the balance record in hand.

1. State the entity: AGSKLEYFVR, free base, C54H84N14O15.
2. Compute the average molar mass from the formula: 1,169.35 g·mol⁻¹. Record the atomic-weight values used.
3. Record the weighed or stated gross mass and its source: 10.0 mg, weighed in house.
4. Record net peptide content, its method and its source: 72.0%, amino acid analysis, batch certificate.
5. Compute peptide mass: 10.0 mg × 0.720 = 7.20 mg.
6. Convert: 7.20 mg ÷ 1,169.35 g·mol⁻¹ = 6.16 × 10⁻³ mmol = 6.16 µmol.
7. Check the magnitude with the Avogadro constant if useful: 6.16 × 10⁻⁶ mol × 6.022 × 10²³ mol⁻¹ ≈ 3.71 × 10¹⁸ molecules.
8. Dissolve to a verified volume and convert to concentration: 6.16 µmol ÷ 2.00 mL = 3.08 mmol·L⁻¹, equivalent to 3.60 mg·mL⁻¹ of peptide.
9. Round only at the end. Carry the full calculator value through each step and report to the precision the least precise input supports, which here is the three-figure content value.

Going the other way: amount to mass

The reverse calculation, m = n × M, is used when a molar target has been set and a mass must be weighed. The same pairing rule applies in reverse, with one extra step. Multiplying the target amount by the free-base M gives the peptide mass required; dividing that by the net content fraction gives the gross mass to weigh. For 5.00 µmol of AGSKLEYFVR at 72.0% content: 5.00 µmol × 1,169.35 g·mol⁻¹ = 5.85 mg of peptide, and 5.85 ÷ 0.720 = 8.12 mg of solid to weigh.

The balance limits what that result can mean. A mass near 8 mg on a balance with a realistic uncertainty of a few hundredths of a milligram carries well under one per cent of relative error, which is small beside the uncertainty of the content figure itself. A target of a few hundred micrograms is a different matter, and is the reason molar work is done from a stock rather than from individual weighings.

Recording

A molar result is only reproducible if its inputs are written beside it. The entity and molar mass basis are the entries most often missing from bench records, and they are the ones that cannot be recovered later 5.

• Sequence, termini and any modifications, with the molecular formula.
• Molar mass, stated as average, and as free base or as a named salt.
• Gross mass and whether it was weighed or taken from a fill statement.
• Net peptide content, its method, and whether it was measured, supplied or assumed.
• Counterion form, and any exchange performed after the certificate was issued.
• The derived peptide mass, amount and concentration, each with its unit.
• Date, operator and the certificate or record the inputs came from.

When any input changes — a new lot, a revised certificate, a counterion exchange — recompute from the recorded inputs rather than scaling an old result. The division is trivial. The record of what was divided by what is the part that keeps the number meaningful.