3. Calculate the molarity of acetic acid in the diluted vinegar sample using the calculated moles of HC₂H3O2 in the 15.00 mL diluted vinegar sample. Remember to find molarity the volume must be in liters. (Show work below) M of HC₂H3O₂ in diluted vinegar sample 4. Give the molarity of acetic acid in the vinegar before it was diluted. The original vinegar from the store is 10 times more concentrated than the diluted vinegar sample used in the titration. M of HC₂H3O₂ in undiluted or original vinegar CALCULATIONS FOR MASS PERCENT OF ACETIC ACID IN VINEGAR 5. Convert the moles of acetic acid (HC₂H3O₂) in the diluted vinegar sample (previously calculated in question 2) to grams of HC₂H₂O₂. Show your work. (You will need to use the molar mass of acetic acid for this calculation.) g of HC₂H₂O₂ 6. Using the equation below, calculate the mass percent of acetic acid in the diluted vinegar sample. The density of the vinegar is 1.00 g/mL so you can assume that the mass of vinegar in grams is essentially equal to the volume of vinegar in mL. Show your work below. Percent by mass=- -× 100 mass of acetic acid mass of vinegar sample % by mass of HC₂H3O₂ in diluted vinegar sample 7. Give the percent by mass of acetic acid in the vinegar before it was diluted. The original vinegar from the store is 10 times more concentrated than the diluted vinegar sample used in the titration. .% by mass of HC₂H3O₂ in undiluted or original vinegar

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15. Subtract the initial volume from the final volume to calculate the volume of NaOH(aq)
required to neutralize the acetic acid in each sample of vinegar.
16. Complete the calculations to find the molarity and percentage of acetic acid in vinegar.
DATA
Initial base buret reading
Final base buret reading
TITRATION DATA (REPORT YOUR VOLUMES TO 1/100TH ML)
TRIAL 1
TRIAL 2
TRIAL 3
о
13.5
13.49
2.
Convert the average volume of NaOH required to reach endpoint from mL to L
(Show calculations here.) 13.36
= 0.01336
1000
Molarity of NaOH (recorded from the standard solution bottle)
13.49
Volume of sodium hydroxide
n
(NaOH) required to reach
endpoint in mL
13.5mL 13.49mL 13.49mL 13.1 m
Volume of vinegar used
15.00 mL
15.00 mL
15.00 mL
15.00 mL
Average volume of NaOH required to reach endpoint in mL
(Show calculations here.) 13.49+13.49+13.1 = 13.36
3
EXTRA TRIAL
M=
13.1
13.36mL
CALCULATIONS FOR MOLARITY OF ACETIC ACID IN VINEGAR
1. Using the known molarity of NaOH(aq) and the average volume of NaOH(aq) in liters
required to reach the endpoint, calculate the moles of NaOH used in the titration. Show
your calculations using dimensional analysis or the following equation:
0.01336
0.100m
moles of NaOH
Liters of solution
0.001386
_moles of NaOH used to reach endpoint of titration
Since the acid/base mole ratio is 1:1 for this neutralization reaction, how many moles of
acetic acid (HC₂H₂O₂) are present in a 15.00 mL sample of diluted vinegar?
0.001886
moles of HC₂H₂O₂ in diluted vinegar sample
K-11
Transcribed Image Text:15. Subtract the initial volume from the final volume to calculate the volume of NaOH(aq) required to neutralize the acetic acid in each sample of vinegar. 16. Complete the calculations to find the molarity and percentage of acetic acid in vinegar. DATA Initial base buret reading Final base buret reading TITRATION DATA (REPORT YOUR VOLUMES TO 1/100TH ML) TRIAL 1 TRIAL 2 TRIAL 3 о 13.5 13.49 2. Convert the average volume of NaOH required to reach endpoint from mL to L (Show calculations here.) 13.36 = 0.01336 1000 Molarity of NaOH (recorded from the standard solution bottle) 13.49 Volume of sodium hydroxide n (NaOH) required to reach endpoint in mL 13.5mL 13.49mL 13.49mL 13.1 m Volume of vinegar used 15.00 mL 15.00 mL 15.00 mL 15.00 mL Average volume of NaOH required to reach endpoint in mL (Show calculations here.) 13.49+13.49+13.1 = 13.36 3 EXTRA TRIAL M= 13.1 13.36mL CALCULATIONS FOR MOLARITY OF ACETIC ACID IN VINEGAR 1. Using the known molarity of NaOH(aq) and the average volume of NaOH(aq) in liters required to reach the endpoint, calculate the moles of NaOH used in the titration. Show your calculations using dimensional analysis or the following equation: 0.01336 0.100m moles of NaOH Liters of solution 0.001386 _moles of NaOH used to reach endpoint of titration Since the acid/base mole ratio is 1:1 for this neutralization reaction, how many moles of acetic acid (HC₂H₂O₂) are present in a 15.00 mL sample of diluted vinegar? 0.001886 moles of HC₂H₂O₂ in diluted vinegar sample K-11
3. Calculate the molarity of acetic acid in the diluted vinegar sample using the calculated
moles of HC₂H₂O2 in the 15.00 mL diluted vinegar sample. Remember to find molarity
the volume must be in liters. (Show work below)
M of HC₂H3O₂ in diluted vinegar sample
4. Give the molarity of acetic acid in the vinegar before it was diluted. The original vinegar
from the store is 10 times more concentrated than the diluted vinegar sample used in
the titration.
M of HC₂H3O₂ in undiluted or original vinegar
CALCULATIONS FOR MASS PERCENT OF ACETIC ACID IN VINEGAR
5. Convert the moles of acetic acid (HC₂H₂O₂) in the diluted vinegar sample (previously
calculated in question 2) to grams of HC₂H3O₂. Show your work. (You will need to use
the molar mass of acetic acid for this calculation.)
g of HC₂H₂O2
6. Using the equation below, calculate the mass percent of acetic acid in the diluted
vinegar sample. The density of the vinegar is 1.00 g/mL so you can assume that the
mass of vinegar in grams is essentially equal to the volume of vinegar in mL. Show
your work below.
Percent by mass=
mass of acetic acid
mass of vinegar sample
% by mass of HC₂H3O₂ in diluted vinegar sample
7. Give the percent by mass of acetic acid in the vinegar before it was diluted. The original
vinegar from the store is 10 times more concentrated than the diluted vinegar sample
used in the titration.
by mass of HC₂H3O₂ in undiluted or original vinegar
x 100
Transcribed Image Text:3. Calculate the molarity of acetic acid in the diluted vinegar sample using the calculated moles of HC₂H₂O2 in the 15.00 mL diluted vinegar sample. Remember to find molarity the volume must be in liters. (Show work below) M of HC₂H3O₂ in diluted vinegar sample 4. Give the molarity of acetic acid in the vinegar before it was diluted. The original vinegar from the store is 10 times more concentrated than the diluted vinegar sample used in the titration. M of HC₂H3O₂ in undiluted or original vinegar CALCULATIONS FOR MASS PERCENT OF ACETIC ACID IN VINEGAR 5. Convert the moles of acetic acid (HC₂H₂O₂) in the diluted vinegar sample (previously calculated in question 2) to grams of HC₂H3O₂. Show your work. (You will need to use the molar mass of acetic acid for this calculation.) g of HC₂H₂O2 6. Using the equation below, calculate the mass percent of acetic acid in the diluted vinegar sample. The density of the vinegar is 1.00 g/mL so you can assume that the mass of vinegar in grams is essentially equal to the volume of vinegar in mL. Show your work below. Percent by mass= mass of acetic acid mass of vinegar sample % by mass of HC₂H3O₂ in diluted vinegar sample 7. Give the percent by mass of acetic acid in the vinegar before it was diluted. The original vinegar from the store is 10 times more concentrated than the diluted vinegar sample used in the titration. by mass of HC₂H3O₂ in undiluted or original vinegar x 100
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