Final answer:
The moles of CH3COOH in vinegar can be determined by subtracting the moles of NaOH used in a titration from the original moles of CH3COOH, using the 1:1 stoichiometric ratio of their reaction. For example, if 5.00 mmol of NaOH has been used and is now in excess, the vinegar sample originally contained 5.00 mmol CH3COOH.
Step-by-step explanation:
To determine the moles of CH3COOH (acetic acid) in a vinegar sample from the moles of NaOH, we need to use the stoichiometry of the reaction between NaOH and CH3COOH.
Given that the reaction between NaOH and CH3COOH is 1:1, this means that for every 1 mole of NaOH used in the reaction, 1 mole of CH3COOH has reacted. If, for example, you have 5.00 mmol (millimoles) of NaOH titrated with vinegar, and after titration, you've determined there is an excess of CH3COOH, you can calculate the original moles of CH3COOH in the vinegar sample by subtracting the moles of NaOH from the excess moles of CH3COOH, as shown below:
5.00 mmol CH3COOH - 1.00 mmol NaOH = 4.00 mmol CH3COOH
This calculation shows that the original vinegar sample contained 5.00 mmol of CH3COOH, assuming that all of the NaOH has reacted with the CH3COOH present in the sample.