# Chemistry

The question is asking the molarity of vinegar, and I am not sure if i did it right.
NaOH + HCH3COO = CH3COONa + H2O

NaOH C= 0.70mol/L
V= 14.6mL --> 0.0146L
n, NaOH = CxV
= 0.70 (0.0146)
= 0.01022mol
1 mol NaOH = 0.01022mol NaOH
----------------- -------------------------
1 molCH3COONa x mol CH3COONa
x= 1(0.01022)
x = 0.01022mol
C= n = 0.01022mol
----- ------------- = 1.02mol/L
v 0.01L

The next step is to convert the molarity to percentage by volume and I am confused on how to do that.

1. 👍
2. 👎
3. 👁
1. You should have written the problem and you didn't. You supplied data and I have tried to reconstruct the problem. I don't know if I've done that right or not. Here is what I think the problem must have said.

Calculate the molarity of vinegar if 10 mL of vinegar were neutralized by 14.6 mL of 0.70 M NaOH. Convert to percent w/v.

Your calculation of M seems to be correct at 1.02 mols/L. The molar mass of vinegar (acetic acid or CH3COOH is 60 grams/mol).
1.02 mols/L x 60 grams/mol = about 60 grams CH3COOH/1000 mL solution.
% w/v = grams solute/100 mL. Your 60 g/1000 mL is 6 g/100 = about 6% w/v. You will need to redo this last part since I've estimated. I usually see vinegar on the store shelves as 4% or 5% so the about 6% sounds a little hight to me.

1. 👍
2. 👎
2. Sorry for not stating the question clearly, this was a lab report.
The lab was finding the concentration of vinegar. Maybe this will give you more of the idea what it is about.

This was the trial data we did: Trail 1 Trail 2 Trail 3 Trail 4 Trail 5
Concentration of base(NaOH) solution (M) 0.70 0.70 0.70 0.70 0.70
Initial reading (mL) 0mL 14.8 0 14.3 29.3
Final reading (mL) 14.8 29.5 14.3 29.3 43.7
Volume of soduim Hydroxide (mL) 14.8 14.7 14.3 15 14.4
Volume of acid (Vinegar) solution(mL) 10.0 10.0 10.0 10.0 10.0

The average volume of NaOH = 14.64 ...14.6

Balanced equation: NaOH + HCH3COO ----> CH3COONa + H2O

1) The next question was to calculate the molarity of the vinegar. I got 1.02mol/L
2) Convert the molarity above to percentage by volume (v/v%)?
Sorry, for the confusion. The question was not asking for the weight per volume.

1. 👍
2. 👎
3. Your 1.02 M looks ok to me. However, your last sentence in your first post said "The next step is to convert the molarity to percentage by volume and I am confused on how to do that." and I've shown you how to do that. 1.02 mols/L x 60 g/mol = about 60 g/1000 mL which is about 6 g/100 mL and that is the definition of percent by volume so your sample is about 6% w/v.

1. 👍
2. 👎
4. Oh, I was confused on where you got the 60g from. Did you find the molar mass of vinegar and multiply it by the number of moles to get around 60.07g .....60g. Also the question gave the density of the acetic acid 1.05g/mL, am I suppose to incorporate that with anything?

1. 👍
2. 👎
5. Yes, the 60 g is the molar mass of CH3COOH. If you want the percent weight/volume then you don't need the density. You would need that if you want percent weight/weight.

1. 👍
2. 👎
6. Sir, I think you are getting the percentage weight/volume and the volume/volume mixed up.
As far as I know shouldn't it be set up more like this..
v/v % = volume of acetic acid divided by volume of vinegar × 100
And the density is 1.05g/mL

1. 👍
2. 👎
7. Well, first you are right that I have confused v/v with w/v but when I read the problem I misconstrued the meaning because w/v is the usual question. I just didn't read the problem closely enough.

1. 👍
2. 👎
8. Also, I read the density of the SOLUTION as 1.05 g/mL and the problem may have meant the density of pure acetic acid is 1.05 g/mL.That wasn't clear in your post. I looked up the density of glacial acetic acid and it gave 1.05 g/mL; therefore, if you have about 0.6 g acetic acid in that 10 mL, the volume of the acetic acid is v = m/d = about 0.6/1.05 = ?, then
%v/v = (volume acetic acid/volume solution)*100 or (? mL/10 mL)*100 = xx

1. 👍
2. 👎
9. When I did the equation I got 0.0583 as the percentage and I am sure I did it wrong.
This is what I did...

First I calculated the molar mass of HCH3COO = 60.06g/mol
Then I found the moles by C X V = 1.02mol/l (0.01L) = 0.0102mol
mass of acetic acid--> n x M
= 0.0102 (60.06)
= 0.6126g
Volume of acetic mass= mass acid divided by density acetic aid
=0.6126 divided by 1.05
= 0.583mL---> 0.00583L
v/v% = 0.000583 divided by 0.01L x 100 = 0.0583

Can you please check my claculation? Also my teacher left this info to guide us.
1.05 mol acid
------------------- --> ----- -------- =L acid
1L L L L solid
1.05mol--->mass ---> volume

1. 👍
2. 👎
10. Il rewrite the last part in case it wasn't clear 1.05mol acid/1L --> ?/L ?/L ---> Lacid/Lsolid
1.05 mol --> mass--> volume

1. 👍
2. 👎
11. OK. First, you calculated the vinegar in your first post as 1.02 M and that is correct. The last post is confusing how you did that so let me start over. I will use 0.0146 L for the volume of NaOH, 0.01 L for the volume of vinegar used, 0.70 M as the molarity of the NaOH and 60.1 as the molar mass of CH3COOH.

CH3COOH + NaOH ==> CH3COONa + H2O

mols NaOH = c*v = 0.70 x 0.0146 = 0.0102 mols.
mols acid = mols NaOH (because the equation shows 1 mol NaOH = 1 mols acid).
Then M acid in the vinegar = mols/L = 0.0102/0.01 = 1.02 M which is your answer in your first post.

So you have 1.02 mols acid in 1L of 1.02 M acid and grams = mols x molar mass = 1.02 x 60.1 = 61.3 grams CH3COOH in 1L, The volume of CH3COOH then is volume= mass/density = 61.3 g/1.05 g/mL = 58.4 mL in that 1000 mLvinegar. The definition of % v/v is (volume solute/volume solution)*100 = (58.4 mL acid/1000 mLsolution)*100 =5.84% v/v
You may want to use 60.06 so your numbers will change a little. I think you converted mols in 1 L to mols in the 10 mL and that's ok but I find that the more often I change units the more mistakes I make so I choose to keep everything in 1 L and work in mols/L. I hope this helps. I think I see your error. I will copy the last part of your work, then bold what you should have done.

"First I calculated the molar mass of HCH3COO = 60.06g/mol
Then I found the moles by C X V = 1.02mol/l (0.01L) = 0.0102mol(<.b>this is 0.0102 mols in 10 mL sample).
mass of acetic acid--> n x M
= 0.0102 (60.06)
= 0.6126g
Volume of acetic mass= mass acid divided by density acetic aid
=0.6126 divided by 1.05
= 0.583mL---> 0.00583L
0.583 mL in the 10 mL sample or 5.83 mL in 100 which is 5.83% v/v or 58.3 mL in 1L (not 0.00583) but that is still 5.83 mL/100 mL which is 5.83%.
v/v% = 0.000583 divided by 0.01L x 100 = 0.0583"

1. 👍
2. 👎
12. Thank you so much that make so much, but one last question. I see that you are using CH3COOH for the acetic acid instead of HCH3COO. Which one do I use because I'm a little confused

1. 👍
2. 👎
13. Either is correct but I prefer the one I use because it shows the COOH group together and that is the active group on acetic acid. It is often written as HC2H3O2 and abbreviated as either HAc or HOAc. If you draw it out as a Lewis structure it will be shown as CH3COOH. Here is how it looks a skeleton form.https://en.wikipedia.org/wiki/Acetic_acid The main reason some like the HCH3COO form is because it shows that H (for the acid) out front and students don't get confused with all of the other stuff. If you write it as CH3COOH you must remember it is the H on the right hand side that is the acid hydrogen and not any of the H atoms on the CH3 group.

1. 👍
2. 👎
14. Oh okay, thanks again!

1. 👍
2. 👎
15. Hi again, I have a question. Why did you divide by 1000?

1. 👍
2. 👎

## Similar Questions

1. ### chemistry

A vinegar contains acetic acid, CH3COOH. Titration of 5.000g of vinegar with 0.100, NaOH requires 33.0 ml to reach equivalent point. What is the weight percentage of CH3COOH in vinegar?

2. ### Chemistry

Vinegar is a solution of acetic acid, CH3COOH, dissolved in water. A 4.69 g sample of vinegar was neutralized by 32.97 mL of 0.100 M NaOH. What is the percent by weight of acetic acid in the vinegar?

3. ### chemistry

If 3.15 mL of vinegar needs 42.5 mL of 0.115 M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.40 qt sample of this vinegar?

4. ### CHM 105

Vinegar contains ethanoic acid (CH3OOH).titration of 3.00g of vinegar with 0.12M NAOH Solution requires 23.0cm3 to reach the equivalent ppoint.what is the weight percentage of the ethanioc acid in vinegar?

1. ### chem

vinegar is 5.0% acetic acid, CH3COOH, by mass. vinegar has a density of 1.02 g/ml what is the madd of acetic acid in 50.0 ml of vinegar?

2. ### ENGINEERING

a vinegar contains acetic acid,CH3COOH.Titration of 5.00g of vinegar with 0.100M of NAOH requires 33.00cm to reach the equivalence point.a)what is the weight percentage of CH3COOH in vinegar?

3. ### Chemistry

Assuming the density of vinegar is 1.0 g/mL ' what is the molarity of vinegar? (use the percent by mass of acetic acid to vinegar which is 5%.) The molar mass of acetic acid is 60.05 g/mol?

4. ### chemistry

10 ml sample of vinegar an aqueous solution 0f acetic acid( HC2H3O2) is titrated with 0.5062 M and 16.58 ml is required to reach equivalence point what is the molarity of the acetic acid b. if the density of vinegar is 1.006 g/cm3

1. ### chemistry

12.5g mass of the vinegar sample and the density of vinegar is 1.05g/mL determine the volume of the vinegar sample in liters?

2. ### Chemistry

100.0 mL Vinegar (d=1.06g/mL) contains 4% acetic acid. a. How many moles of acid are in 100 mL of vinegar? b. How many mL of 0.11 M NaOH are required to neutralize the vinegar?

3. ### Chemistry

Commercial vinegar was titrated with NaOH solution to determine the content of acetic acid, HC2H3O2. For 20.0 milliliters of the vinegar 26.7 milliliters of 0.600-molar NaOH solution was required. What was the concentration of

4. ### Chemistry

Vinegar is a dilute solution of acetic acid. In the titration of 5.00 mL of vinegar, 39.75 mL of 0.137 M sodium hydroxide solution was required to neutralize the vinegar to a phenolphthalein end point. Calculate each of the