Atomic Mass Calculator
Calculate atomic mass from protons and neutrons, including mass defect, binding energy, and mole-to-atom conversions using nuclear physics formulas.
Atomic Mass Calculator
Calculator
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Formula: Atomic Mass = Sum of nucleon masses - Mass defect
Worked example โ 4.002603 amu (actual: 4.002602 amu)
Formula
Atomic Mass = Sum of nucleon masses - Mass defect
Atomic mass equals the sum of individual proton, neutron, and electron masses minus the mass defect (mass converted to nuclear binding energy via E=mc2). Mass defect = Total Binding Energy / 931.494 MeV per amu.
Worked Examples
Example 1: Helium-4 Atomic Mass
Problem:Calculate atomic mass of He-4 (2 protons, 2 neutrons, BE/nucleon = 7.074 MeV)
Solution:Sum = 2(1.007276) + 2(1.008665) + 2(0.000549) = 4.032980 amu Total BE = 7.074 * 4 = 28.296 MeV Mass defect = 28.296/931.494 = 0.030377 amu Atomic mass = 4.032980 - 0.030377 = 4.002603 amu
Result:4.002603 amu (actual: 4.002602 amu)
Example 2: Iron-56 Mass from 10 Grams
Problem:How many moles and atoms in 10g of Fe-56 (26p, 30n)?
Solution:Atomic mass ~ 55.935 amu = 55.935 g/mol Moles = 10/55.935 = 0.17878 mol Atoms = 0.17878 * 6.022e23 = 1.0765e23 atoms
Result:0.1788 mol | 1.077e23 atoms
Frequently Asked Questions
What is atomic mass and how is it different from mass number?
Atomic mass is the actual mass of an atom measured in atomic mass units (amu), where 1 amu is defined as 1/12 the mass of a carbon-12 atom. It includes the mass of protons, neutrons, and electrons, minus the mass defect due to nuclear binding energy. Mass number (A) is simply the count of protons plus neutrons and is always a whole number. Atomic mass is always slightly less than the sum of individual nucleon masses because some mass is converted to binding energy according to Einstein's E=mc2. For example, helium-4 has mass number 4 but atomic mass 4.002602 amu.
What is mass defect and binding energy?
Mass defect is the difference between the sum of individual proton and neutron masses and the actual measured mass of the nucleus. This missing mass has been converted to nuclear binding energy that holds the nucleus together, as described by Einstein's mass-energy equivalence E=mc2. The conversion factor is 931.494 MeV per amu. For example, a helium-4 nucleus has a mass defect of 0.03038 amu, corresponding to 28.3 MeV of binding energy. Higher binding energy per nucleon indicates greater nuclear stability, with iron-56 having the maximum at about 8.8 MeV per nucleon.
How is the semi-empirical mass formula used?
The semi-empirical mass formula (Bethe-Weizsacker formula) estimates nuclear binding energy using five terms: volume energy (proportional to A), surface energy (proportional to A^(2/3)), Coulomb energy (proportional to Z^2/A^(1/3)), asymmetry energy (proportional to (N-Z)^2/A), and pairing energy. Each term has an empirically fitted coefficient. This formula successfully predicts binding energies and atomic masses for most nuclides within 1% accuracy, and it explains why nuclei with certain proton-to-neutron ratios are more stable than others.
How do you convert between atomic mass units and grams?
One atomic mass unit equals 1.66054 x 10^-24 grams, which is the reciprocal of Avogadro number expressed in appropriate units. This means the atomic mass in amu is numerically equal to the molar mass in grams per mole. For example, carbon-12 has an atomic mass of exactly 12 amu, and one mole of carbon-12 weighs exactly 12 grams. To find the mass of a single atom in grams, multiply the atomic mass in amu by 1.66054 x 10^-24. To find the number of atoms in a given mass, divide the mass by the molar mass and multiply by Avogadro number.
Why does binding energy per nucleon peak near iron and nickel?
Two competing terms shape the curve. The attractive strong force saturates, so the volume term adds roughly the same binding for each added nucleon, while Coulomb repulsion between protons grows with the square of Z and reaches across the whole nucleus. Below the iron region the volume term dominates and adding nucleons increases binding per nucleon; above it the Coulomb penalty wins. The maximum sits near 8.8 MeV per nucleon around iron-56 and nickel-62, which is why fusion releases energy below that peak and fission above it.
What are nuclear magic numbers and why does the semi-empirical formula miss them?
The magic numbers 2, 8, 20, 28, 50, 82 and 126 mark closed shells of protons or neutrons, and nuclides at those counts are measurably more tightly bound than their neighbours. Lead-208 is doubly magic with 82 protons and 126 neutrons. The Bethe-Weizsacker formula treats the nucleus as a charged liquid drop with no shell structure, so it cannot reproduce these peaks and its residuals show a systematic pattern at the magic numbers rather than random scatter.
Why does the sum of parts include electron mass as well as protons and neutrons?
Because the quantity being computed is an atomic mass rather than a nuclear mass. Tabulated values in unified atomic mass units, including the 12 u definition of carbon-12, refer to the neutral atom with all Z of its electrons present. The calculator therefore adds Z times 0.000549 u alongside 1.007276 u per proton and 1.008665 u per neutron. Omitting the electrons would understate a heavy atom noticeably: the 92 electrons of uranium contribute about 0.05 u.
What is the packing fraction and how does it relate to the mass defect?
Packing fraction, introduced by Francis Aston, is the fractional deviation of an atomic mass from its whole-number mass number: divide the atomic mass minus A by A, usually quoted scaled by ten thousand. It is negative wherever binding is strong, so it dips to about minus 11 near iron-56 and turns positive for hydrogen-1 and for the very heaviest nuclides. It carries the same information as the mass defect but normalised per nucleon, which makes it a convenient way to compare stability across the chart of nuclides.
What do the Coulomb and asymmetry terms in the semi-empirical mass formula control?
The Coulomb term subtracts binding in proportion to Z times Z minus one divided by the cube root of A, representing electrostatic repulsion among protons. The asymmetry term subtracts a penalty proportional to N minus Z squared divided by A, reflecting the Pauli exclusion principle, which forces surplus neutrons into higher energy levels. Together they explain why light stable nuclides have N close to Z while heavy ones need a neutron excess to dilute proton repulsion, tracing the valley of stability.
When should I enter a measured binding energy per nucleon instead of letting the formula estimate it?
Supply the measured value whenever precision matters or the nuclide sits near a shell closure, since the liquid-drop estimate carries no shell terms and typically lands within about one percent rather than to six decimal places. Measured binding energies from nuclear data tables are the right input for helium-4, oxygen-16, calcium-48 or lead-208. The estimate is adequate for mid-mass nuclides away from magic numbers, or when you only want the order of magnitude of the mass defect.
References
Background & Theory
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Reviewed for accuracy by Manoj Kumar, Mathematics Educator ยท Editorial policy
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