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Percent Composition Calculator

Chemistry

Calculate the percent composition by mass for each element in a compound. Enter up to 3 elements with atom counts to get elemental mass percentages and molar mass.

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Element 1 (% mass)

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Element 2 (% mass)
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Element 3 (% mass)
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Molar Mass
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Breakdown

How the total splits

Element 1 (% mass)
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Element 2 (% mass)
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Element 3 (% mass)
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This calculator computes your Element 1 (% mass), Element 2 (% mass), Element 3 (% mass), Molar Mass from the values you enter.

Inputs
Element 1Count 1Element 2Count 2Element 3 (optional)Count 3
Outputs
Element 1 (% mass)Element 2 (% mass)Element 3 (% mass)Molar Mass

What is a Percent Composition?

The Percent Composition Calculator determines the mass percentage of each element in a compound. Select up to 3 elements with atom counts and the calculator computes %element = (n ร— M_element / M_total) ร— 100 for each, plus the total molar mass. The default shows glucose (Cโ‚†Hโ‚โ‚‚Oโ‚†): 40% C, 6.7% H, 53.3% O.

Percent composition connects the chemical formula of a compound to the mass fractions of its elements โ€” the information you would measure in combustion analysis or elemental analysis. It is the starting point for determining empirical formulas from experimental data: convert percentages to mole ratios (divide by atomic mass), then find the simplest whole-number ratio. The Empirical Formula Calculator automates this reverse calculation.

For building the calculation from element atomic masses, the Molar Mass Calculator is the complementary tool โ€” it calculates total molar mass with the same element-select interface but highlights the molar mass rather than the elemental percentages. For compounds with more than three elements, combine these calculations manually using the atomic masses from the Atomic Mass Calculator.

How to use this Percent Composition calculator

  1. Select Element 1 and enter Count 1. For glucose: C ร— 6.
  2. Select Element 2 and enter Count 2. For glucose: H ร— 12.
  3. Select Element 3 and enter Count 3. For glucose: O ร— 6.
  4. Read Element 1 (% mass) as the primary result โ€” the mass percentage of the first element.
  5. Check that elements 1 + 2 + 3 percentages sum to 100% โ€” verify the calculation is complete.

Formula & Methodology

Percent composition by mass:

%element_i = (nแตข ร— Mแตข / M_compound) ร— 100 M_compound = nโ‚Mโ‚ + nโ‚‚Mโ‚‚ + nโ‚ƒMโ‚ƒ

Worked example โ€” Urea (CHโ‚„Nโ‚‚O):

Urea can be entered as Cร—1, Hร—4, Nร—2, Oร—1 โ€” but this calculator only handles 3 elements. Use the first three (C, H, N) to find their percentages, then compute O% by subtraction (100 โˆ’ %C โˆ’ %H โˆ’ %N):

M(urea) = 1ร—12.011 + 4ร—1.008 + 2ร—14.007 + 1ร—15.999 = 60.055 g/mol %C = 12.011/60.055 ร— 100 = 19.998% โ‰ˆ 20.00% %H = 4.032/60.055 ร— 100 = 6.714% %N = 28.014/60.055 ร— 100 = 46.646% %O = 100 โˆ’ 20.00 โˆ’ 6.71 โˆ’ 46.65 = 26.64%

Urea's 46.65% nitrogen content (the highest of common fertilisers) is why it is the most widely used nitrogenous fertiliser in India, applied to rice, wheat, sugarcane, and cotton. IFFCO (Indian Farmers Fertiliser Cooperative) is the world's largest producer and distributor of urea.

Frequently Asked Questions

Percent composition (percent by mass or mass percent) is the percentage of the total molar mass contributed by each element in a compound. For element i: %element_i = (nแตข ร— Mแตข / M_compound) ร— 100, where nแตข is the atom count, Mแตข is the atomic mass, and M_compound is the total molar mass. The percentages of all elements must sum to exactly 100%. Percent composition characterises the elemental makeup of a compound and is used to determine empirical formulas from combustion analysis.
%element = (mass of element in 1 mol of compound / molar mass of compound) ร— 100 = (n ร— M_element / M_compound) ร— 100. For water (Hโ‚‚O, M = 18.015): %H = (2 ร— 1.008 / 18.015) ร— 100 = 11.19%; %O = (15.999 / 18.015) ร— 100 = 88.81%. Sum = 100.00%. For glucose Cโ‚†Hโ‚โ‚‚Oโ‚† (M = 180.156): %C = 72.066/180.156 ร— 100 = 40.00%; %H = 12.096/180.156 ร— 100 = 6.71%; %O = 95.994/180.156 ร— 100 = 53.29%.
Select Element 1, enter its atom count, repeat for Element 2 and optionally Element 3. Default shows glucose (Cโ‚†Hโ‚โ‚‚Oโ‚†): Cร—6, Hร—12, Oร—6. The calculator returns the mass percentage of each element and the total molar mass. The pie chart visualises the elemental composition by mass fraction.
Percent composition โ†’ empirical formula: (1) Divide each element's percentage by its atomic mass to get mole ratios. (2) Divide all mole ratios by the smallest. (3) If ratios are not whole numbers, multiply by the smallest integer to get whole numbers. Example: 40% C, 6.71% H, 53.29% O: C = 40/12.011 = 3.33; H = 6.71/1.008 = 6.66; O = 53.29/15.999 = 3.33. Ratio = 1:2:1 โ†’ empirical formula CHโ‚‚O (formaldehyde/glucose share this empirical formula). The [Empirical Formula Calculator](/empirical-formula-calculator/) automates this process.
Water (Hโ‚‚O): H = 11.19%, O = 88.81%. Carbon dioxide (COโ‚‚): C = 27.29%, O = 72.71%. Ammonia (NHโ‚ƒ): N = 82.24%, H = 17.76%. Sodium chloride (NaCl): Na = 39.34%, Cl = 60.66%. Glucose (Cโ‚†Hโ‚โ‚‚Oโ‚†): C = 40.00%, H = 6.71%, O = 53.29%. Aspirin (Cโ‚‰Hโ‚ˆOโ‚„): C = 60.00%, H = 4.48%, O = 35.52%. Urea (CHโ‚„Nโ‚‚O): C = 20.00%, H = 6.71%, N = 46.65%, O = 26.64% (4 elements โ€” beyond this calculator's 3-element limit).
Urea (CHโ‚„Nโ‚‚O): 46.65% N โ€” highest nitrogen content of common fertilisers; India is the world's second-largest urea consumer. DAP (diammonium phosphate, (NHโ‚„)โ‚‚HPOโ‚„): 18% N, 46% Pโ‚‚Oโ‚… (phosphate). KCl (muriate of potash): 60% Kโ‚‚O equivalent (52.4% K). India's fertiliser subsidy system (managed by MoF and Ministry of Chemicals) is based on nutrient content โ€” knowing percent composition of N, P, K elements determines the subsidy rate per metric tonne. Companies like IFFCO and Coromandel specify fertilisers by elemental percentages.
Combustion analysis determines organic compound composition: a weighed sample is burned in excess Oโ‚‚; the COโ‚‚ and Hโ‚‚O produced are collected and weighed. %C = (mass COโ‚‚ / mass sample) ร— (12.011/44.009) ร— 100; %H = (mass Hโ‚‚O / mass sample) ร— (2.016/18.015) ร— 100; %O = 100 โˆ’ %C โˆ’ %H โˆ’ %N (by difference). This gives percent composition from which the empirical formula is derived. Combustion analysis is the standard method for characterising new organic compounds in chemistry research labs.
Yes โ€” compounds with the same empirical formula but different molecular formulas have the same percent composition. Formaldehyde (CHโ‚‚O) and glucose (Cโ‚†Hโ‚โ‚‚Oโ‚†) and acetic acid (Cโ‚‚Hโ‚„Oโ‚‚) all have the same empirical formula CHโ‚‚O and identical percent composition: 40.0% C, 6.7% H, 53.3% O. Distinguishing between them requires molecular mass determination (by mass spectrometry, osmometry, or vapour density). This is why percent composition alone cannot uniquely identify a compound.
Hematite (Feโ‚‚Oโ‚ƒ): M = 2ร—55.845 + 3ร—15.999 = 159.687 g/mol; %Fe = 111.69/159.687 ร— 100 = 69.94%; %O = 47.997/159.687 ร— 100 = 30.06%. Magnetite (Feโ‚ƒOโ‚„): M = 3ร—55.845 + 4ร—15.999 = 231.531 g/mol; %Fe = 167.535/231.531 ร— 100 = 72.36%. Magnetite has slightly higher iron content than hematite โ€” relevant for blast furnace calculations at SAIL plants in Bhilai, Rourkela, and Bokaro, which process Odisha and Chhattisgarh iron ores.
Percent composition โ†’ empirical formula โ†’ molecular formula (with additional molecular mass information). Empirical formula: the simplest whole-number ratio of atoms. Molecular formula: the actual number of atoms. Both have the same percent composition. Benzene (Cโ‚†Hโ‚†) and acetylene (Cโ‚‚Hโ‚‚) have the same empirical formula CH (92.26% C, 7.74% H). To get the molecular formula from the empirical formula: n = M_molecular / M_empirical; molecular formula = n ร— empirical formula. This is a standard NCERT Class 11 problem type.