Calculating Mass of Acid in Vinegar

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SUMMARY

The discussion focuses on calculating the mass of acetic acid (CH3COOH) in vinegar using a neutralization reaction with sodium hydroxide (NaOH). The concentration of NaOH is 0.20 mol/L, and the volume used is 79 mL, resulting in 0.0158 moles of NaOH, which corresponds to the same amount of acetic acid due to a 1:1 reaction ratio. The molar mass of acetic acid is confirmed to be 60 g/mol, leading to a calculated mass of 0.948 grams of acetic acid in the 25 mL of vinegar used.

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Hello this is my question. Ok I need to find the molar mass of acetic acid and then calculate the mass of acid in the volume of vinegar I used. Now the equation for the lab i did was
CH3COOH + NaOH produces NaCH3COO + H20

The information I have is that the concentration of the sodium hydroxide is 0.20 mol/L, the volume of sodium hydroxide is 79mL, the volume of acetic acid I used was 25mL, I calculated the number of moles use in the sodium hydroxide and it was 0.0158 mol and since everything is on a 1 to 1 basis the number of moles for the acetic acid is 0.0158mol. From that I calculated the number of moles of acetic acid and it is 0.632 mol/L. I calculated the molar mass of the acetic acid by taking the 0.0158 mol of CH3COOH x 60g (the weight of the acid) which got me an answer of 0.948 grams. Now what I do not understand is how to calculate the mass of acid in the volume of vinegar. I may have everything to answer the question here but i just don't know what to do.

Plz Help me!
 
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Originally posted by majinknight
Hello this is my question. Ok I need to find the molar mass of acetic acid and then calculate the mass of acid in the volume of vinegar I used. Now the equation for the lab i did was
CH3COOH + NaOH produces NaCH3COO + H20
Hi Majinknight! Welcome to the forums!

So to summarize: You neutralized 25 ml of CH3COOH with 79 ml of NaOH [0.20 mol/l]

You calculated the number of moles of CH3COOH in your solution: 79 * 0.20*10-3 = 0.0158. You do not have to calculate the concentration of the CH3COOH!

Next, you calculated molar mass of CH3COOH, you do not explain how you did that? I can do it out of my head: the molar mass of C=12, O=16, H=1. Combine that and CH3COOH = 60 grams per mole.

Without knowing it you already answered your own question: "Now what I do not understand is how to calculate the mass of acid in the volume of vinegar".

You multiplied the molar mass of CH3COOH with the number of moles in the volume and that is your answer: 0.0158 * 60 = 0.948 g.
 
Oh really the question said find the molar mass of acetic acid. Then calculate the mass of acid in the volume of vinegar. I thought that the molar mass of acetic acid was 0.948g, but the molar mass is just the 60grams. Ok Thank You so much.
 
Yes, the molar mass (mw) is the mass of a mole of a substance
 

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