How this calculator works
A saturated acyclic hydrocarbon has two more hydrogens than twice its carbon count. Each ring or double bond removes one hydrogen pair; a triple bond removes two. Ordinary trivalent nitrogen adds one to the hydrogen limit, each monovalent halogen counts like hydrogen and divalent oxygen leaves the count unchanged. A negative or half-integer answer is rejected for this closed-shell model. A whole result does not prove that a stable structure exists, and ions, radicals and unusual valences require a different analysis.
Formula
DBE = (2C + 2 + N - H - X)/2. Each ring or double bond contributes 1; each triple bond contributes 2. Oxygen contributes 0.
Worked example
Benzene, C6H6, gives (12 + 2 - 6)/2 = 4: one ring plus three double bonds. C4H6Br2 gives (8 + 2 - 6 - 2)/2 = 1.
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Frequently asked questions
Does DBE locate the bonds?
No. It constrains the total number of rings and pi-bond equivalents, leaving many possible structures.
Why enter oxygen if it cancels?
It documents the complete supported formula and verifies that its atom count is a whole non-negative number.
Can I use this for a charged formula?
This equation assumes a neutral molecule with ordinary valences. Use a charge-aware structural analysis for ions or radicals.