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The molarity of vinegar depends on its acidity. A vinegar that contains 5 g of acetic acid per 100 mL is about 0.833 mol/L (0.833 M) in acetic acid, and a vinegar at the 4 g per 100 mL floor used in US federal guidance is about 0.666 M. This guide gives the lookup table, the formula, two worked examples and a short spreadsheet and code snippet so you can compute the molarity of acetic acid in vinegar for any labeled strength.
What is the molarity of vinegar?
Molarity is moles of solute per litre of solution. In vinegar the solute that matters is acetic acid (CH3COOH, molecular weight 60.05 g/mol according to PubChem’s computed value). There is no single “average molarity of acetic acid in vinegar”, because the figure moves with the acidity printed on the bottle. What a primary source supports is a floor: FDA Compliance Policy Guide Sec. 525.825 says each listed variety of vinegar should contain 4 grams of acetic acid per 100 mL at 20 °C, and that diluted vinegar should in no case be less than 4 percent (FDA page content current as of 08/24/2018; the CPG was last revised in March 1995). Applying the formula below to that floor gives a minimum of roughly 0.666 M acetic acid.
The same FDA guidance notes that natural vinegars as they come from the generators normally contain in excess of 4 g of acetic acid per 100 mL, and that vinegar diluted with water is labeled with the diluted percent acid strength. The practical step for a bench calculation is therefore to read the percent acidity on the container you actually hold and convert it.
Vinegar molarity table by percent acidity
The table assumes the percentage is weight per volume (g of acetic acid per 100 mL of vinegar), the unit used in the FDA wording above. Values use a molecular weight of 60.05 g/mol and are rounded to three significant figures.
| Acidity (g acetic acid per 100 mL) | Concentration (g/L) | Molarity (mol/L) |
|---|---|---|
| 4 | 40 | 0.666 |
| 5 | 50 | 0.833 |
| 6 | 60 | 0.999 |
| 7 | 70 | 1.17 |
| 8 | 80 | 1.33 |
Source of the inputs: 4 g per 100 mL is the FDA figure quoted above; the other rows are the same calculation at higher strengths, not values taken from a regulation. The molecular weight is from PubChem (Acetic acid, CID 176). Each molarity is the g/L figure divided by 60.05.
The formula for molarity of acetic acid in vinegar
Because 100 mL is one tenth of a litre, multiply the g/100 mL figure by 10 to get g/L, then divide by the molecular weight:
M = (% w/v × 10) / 60.05
- % w/v is grams of acetic acid per 100 mL of vinegar.
- × 10 converts g/100 mL to g/L.
- 60.05 is the molecular weight of acetic acid in g/mol.
- The result is in mol/L, written M.
The w/v versus w/w trap
A percentage can mean grams per 100 mL (w/v) or grams per 100 g of solution (w/w). The formula above is only correct for w/v. If a label or specification gives w/w, convert through the density of your vinegar: M = (% w/w × density in g/mL × 10) / 60.05. Use a measured density for the lot in front of you rather than assuming one, since vinegar is not pure acetic acid and its density is not the density of the acid. This guide does not supply a vinegar density for that reason.
Worked examples
Example 1: molarity of a 5 percent vinegar
A vinegar is labeled 5 percent acidity, read as 5 g of acetic acid per 100 mL.
- Convert to g/L: 5 × 10 = 50 g/L.
- Divide by the molecular weight: 50 / 60.05 = 0.8326 mol/L.
- Round to three significant figures: 0.833 M.
Example 2: diluting that vinegar to a working solution
You need 100.0 mL of a more dilute acetic acid solution made from the 0.833 M vinegar by taking 10.0 mL and bringing it to 100.0 mL with water. Use C1V1 = C2V2:
- C2 = (0.8326 M × 10.0 mL) / 100.0 mL = 0.08326 M.
- Round to three significant figures: 0.0833 M.
For more on preparing and diluting solutions in general, see the molarity and solution calculations guide.
Compute vinegar molarity in Excel, R and Python
Put the percent acidity (g per 100 mL) in cell A2 and use:
=(A2*10)/60.05In R or Python, with pct as the percent acidity:
# R
molarity = pct * 10 / 60.05
# Python
molarity = pct * 10 / 60.05Round the result to match the precision of the percentage you started from; a label that says 5 percent supports about two or three significant figures, not more.
Safety note and the safety data sheet
The arithmetic in this guide also applies to laboratory acetic acid reagents, which can be far more concentrated. Before you handle any acetic acid reagent, read the supplier’s safety data sheet (SDS) for that exact product; the SDS lists the hazards, protective equipment and first-aid information for it. PubChem points to NIOSH, OSHA and the ILO-WHO International Chemical Safety Cards as hazard sources for acetic acid. See how to read a safety data sheet and OSHA SDS requirements for employers. For a related acid-base calculation, see titration with a weak acid and a strong base.
Frequently asked questions
What is the average molarity of acetic acid in vinegar?
There is no fixed average. A 5 g per 100 mL vinegar is about 0.833 M, and the 4 g per 100 mL figure in FDA guidance corresponds to about 0.666 M. Use the table or formula with the acidity on your label.
How do I find the molarity of vinegar from its percent acidity?
Multiply the g per 100 mL by 10 to get g/L, then divide by 60.05 g/mol. If the percentage is w/w, include the measured density as shown above.
Is vinegar molarity the same as the molarity of acetic acid in vinegar?
The acidity figure on a vinegar label is expressed as acetic acid, so the molarity of vinegar in this context means the molarity of its acetic acid.
Sources
- U.S. Food and Drug Administration, CPG Sec. 525.825 Vinegar; Definitions; Adulteration with Vinegar Eels, https://www.fda.gov/regulatory-information/search-fda-guidance-documents/cpg-sec-525825-vinegar-definitions-adulteration-vinegar-eels (checked 2026-10-29).
- PubChem, Acetic acid (CID 176), National Library of Medicine, https://pubchem.ncbi.nlm.nih.gov/compound/176 (checked 2026-10-29).
