What is the degree of unsaturation?
Type how many carbon, hydrogen, nitrogen, halogen and oxygen atoms the molecular formula has. The degree of unsaturation calculator gives the number of rings plus π bonds, with the equivalent hydrocarbon and the saturated alkane it is compared with.
- Degree of unsaturation
- 2
C6H10 has a degree of unsaturation of 2.
- Molecular formula (X = halogen)
- C6H10
- Equivalent hydrocarbon
- C6H10
- Saturated hydrocarbon
- C6H14
Degree of unsaturation: 2. C6H10 has a degree of unsaturation of 2.
How to calculate
Computes the degree of unsaturation (the number of rings plus π bonds) of a compound from the counts of carbon, hydrogen, nitrogen, halogen and oxygen atoms in its molecular formula.
Example with the default inputs (Carbon atoms (C) 6, Hydrogen atoms (H) 10, Nitrogen atoms (N) 0, Halogen atoms (F, Cl, Br, I) 0, Oxygen atoms (O) 0): C6H10 has a degree of unsaturation of 2.
Method: Degree of unsaturation = (2C + 2 + N − H − X) ÷ 2, where X is the number of halogen atoms; oxygen is ignored.
- The compound is a neutral molecule: carbon forms 4 bonds, nitrogen 3, oxygen 2, and hydrogen and the halogens 1.
- Each ring or π bond takes 2 hydrogens away from the saturated formula CnH2n+2; a double bond counts 1 and a triple bond 2.
- A formula that gives a half or a negative value has no answer, and so does a formula with no atoms.
Worked examples
Each example is checked against the calculator on every build.
- Carbon atoms (C) 6, Hydrogen atoms (H) 10, Nitrogen atoms (N) 0, Halogen atoms (F, Cl, Br, I) 0, Oxygen atoms (O) 0 gives Degree of unsaturation 2, Molecular formula (X = halogen) C6H10, Equivalent hydrocarbon C6H10, Saturated hydrocarbon C6H14.Source: OpenStax, Organic Chemistry, §7.2 Calculating the Degree of Unsaturation (C6H10 has 2; C4H6Br2 is equivalent to C4H8, 1; C5H8O to C5H8, 2; C5H9N to C5H8, 2), https://openstax.org/books/organic-chemistry/pages/7-2-calculating-the-degree-of-unsaturation (retrieved 2026-10-18)
- Carbon atoms (C) 4, Hydrogen atoms (H) 6, Nitrogen atoms (N) 0, Halogen atoms (F, Cl, Br, I) 2, Oxygen atoms (O) 0 gives Degree of unsaturation 1, Molecular formula (X = halogen) C4H6X2, Equivalent hydrocarbon C4H8, Saturated hydrocarbon C4H10.Source: OpenStax, Organic Chemistry, §7.2 Calculating the Degree of Unsaturation (C6H10 has 2; C4H6Br2 is equivalent to C4H8, 1; C5H8O to C5H8, 2; C5H9N to C5H8, 2), https://openstax.org/books/organic-chemistry/pages/7-2-calculating-the-degree-of-unsaturation (retrieved 2026-10-18)
- Carbon atoms (C) 5, Hydrogen atoms (H) 8, Nitrogen atoms (N) 0, Halogen atoms (F, Cl, Br, I) 0, Oxygen atoms (O) 1 gives Degree of unsaturation 2, Molecular formula (X = halogen) C5H8O, Equivalent hydrocarbon C5H8.Source: OpenStax, Organic Chemistry, §7.2 Calculating the Degree of Unsaturation (C6H10 has 2; C4H6Br2 is equivalent to C4H8, 1; C5H8O to C5H8, 2; C5H9N to C5H8, 2), https://openstax.org/books/organic-chemistry/pages/7-2-calculating-the-degree-of-unsaturation (retrieved 2026-10-18)
- Carbon atoms (C) 5, Hydrogen atoms (H) 9, Nitrogen atoms (N) 1, Halogen atoms (F, Cl, Br, I) 0, Oxygen atoms (O) 0 gives Degree of unsaturation 2, Molecular formula (X = halogen) C5H9N, Equivalent hydrocarbon C5H8.Source: OpenStax, Organic Chemistry, §7.2 Calculating the Degree of Unsaturation (C6H10 has 2; C4H6Br2 is equivalent to C4H8, 1; C5H8O to C5H8, 2; C5H9N to C5H8, 2), https://openstax.org/books/organic-chemistry/pages/7-2-calculating-the-degree-of-unsaturation (retrieved 2026-10-18)
- Carbon atoms (C) 0, Hydrogen atoms (H) 4, Nitrogen atoms (N) 2, Halogen atoms (F, Cl, Br, I) 0, Oxygen atoms (O) 0 gives Degree of unsaturation 0, Molecular formula (X = halogen) H4N2, Equivalent hydrocarbon H2, Saturated hydrocarbon H2.Source: OpenStax, Organic Chemistry, §7.2 Calculating the Degree of Unsaturation (C6H10 has 2; C4H6Br2 is equivalent to C4H8, 1; C5H8O to C5H8, 2; C5H9N to C5H8, 2), https://openstax.org/books/organic-chemistry/pages/7-2-calculating-the-degree-of-unsaturation (retrieved 2026-10-18)
- Carbon atoms (C) 6, Hydrogen atoms (H) 6, Nitrogen atoms (N) 0, Halogen atoms (F, Cl, Br, I) 0, Oxygen atoms (O) 0 gives Degree of unsaturation 4, Saturated hydrocarbon C6H14.Source: OpenStax, Organic Chemistry, §7.2 Calculating the Degree of Unsaturation (C6H10 has 2; C4H6Br2 is equivalent to C4H8, 1; C5H8O to C5H8, 2; C5H9N to C5H8, 2), https://openstax.org/books/organic-chemistry/pages/7-2-calculating-the-degree-of-unsaturation (retrieved 2026-10-18)
How it works
A saturated open-chain alkane with n carbons has the formula CnH2n+2: it holds 2n + 2 hydrogens. Each ring or π bond (a double bond counts 1, a triple bond 2) removes two hydrogens. Other atoms are handled by their bonds:
- Halogens (F, Cl, Br, I) take the place of hydrogens: add them to the hydrogen count.
- Oxygen forms two bonds and changes nothing: ignore it.
- Nitrogen forms three bonds, so a nitrogen compound has one more hydrogen than the related hydrocarbon: subtract the nitrogens from the hydrogens.
The equivalent hydrogen count is H + X − N, and
Degree of unsaturation = (2C + 2 − (H + X − N)) ÷ 2 = (2C + 2 + N − H − X) ÷ 2
With no carbon, n = 0 and the saturated reference is H2, so the same formula covers compounds such as hydrazine, N2H4 (0), or oxygen, O2 (1, its double bond).
Outputs. The degree of unsaturation (a whole number); the formula typed, written in the order C, H, N, O, X with X for all halogens (C4H6X2, C5H8O, H4N2); the equivalent hydrocarbon, with C carbons and H + X − N hydrogens; and the saturated alkane, with C carbons and 2C + 2 hydrogens. In every formula an atom with a count of 0 is left out and a count of 1 is not written (CH4, C5 for C5H0, H2 with no carbon). The equivalent hydrocarbon is not shown when H + X − N is below 0, or when C and H + X − N are both 0.
Rules. Each count is a whole number from 0 to 10,000, and at least one count must be above 0. There is no answer when 2C + 2 + N − H − X is below 0 (more hydrogens and halogens than a saturated molecule holds) or is odd (the result would end in a half, which no neutral molecule has). Oxygen never changes the result.
Worked examples by hand
C6H10. (2 × 6 + 2 + 0 − 10 − 0) ÷ 2 = 4 ÷ 2 = 2. Hexane, C6H14, has two more pairs of hydrogens.
C4H6Br2. Add the bromines: H + X = 8, the formula of C4H8. (8 + 2 − 6 − 2) ÷ 2 = 1.
C5H8O. Ignore the oxygen: the equivalent hydrocarbon is C5H8. (10 + 2 − 8) ÷ 2 = 2.
C5H9N. Subtract the nitrogen: 9 − 1 = 8, so C5H8. (10 + 2 + 1 − 9) ÷ 2 = 2.
Benzene, C6H6. (12 + 2 − 6) ÷ 2 = 4: one ring and three double bonds.
Hydrazine, N2H4. No carbon, so the reference is H2. H + X − N = 4 − 2 = 2, the equivalent hydrocarbon H2. (0 + 2 + 2 − 4) ÷ 2 = 0: no rings or π bonds.
Other questions people ask
What is the degree of unsaturation?
It is the number of rings plus π bonds in a molecule. Each ring or double bond takes two hydrogens away from the saturated alkane formula CnH2n+2, and a triple bond takes four. It is also called the index of hydrogen deficiency (IHD) or double bond equivalents.
How do I calculate the degree of unsaturation?
Work out (2C + 2 + N − H − X) ÷ 2, where C, H, N and X are the numbers of carbon, hydrogen, nitrogen and halogen atoms. For C6H10: (12 + 2 − 10) ÷ 2 = 2.
How do halogens, oxygen and nitrogen change the count?
A halogen takes the place of a hydrogen, so add halogens to the hydrogens. Oxygen forms two bonds and does not change the count, so ignore it. Nitrogen forms three bonds, so subtract one hydrogen for each nitrogen. C4H6Br2 then counts as C4H8, with 1 degree of unsaturation.
What does a degree of unsaturation of 4 mean?
The molecule has four rings and π bonds in total, in any mix. Benzene, C6H6, has one ring and three double bonds. A value of 4 or more often points to a benzene ring.
Why does the calculator give no answer for some formulas?
If H + N + X is odd, the result would end in a half, and no neutral molecule has that formula. If there are more hydrogens and halogens than a saturated molecule can hold, the result would be negative. Both mean a count is wrong, or the formula is an ion or a radical.
Does the degree of unsaturation tell me the structure?
No. It counts rings and π bonds together but does not say which. C6H10 could be a cyclic alkene, a diene, or an alkyne. Spectra or reactions such as hydrogenation tell them apart.