How do you calculate the% by mass of acetic acid present? The table below demonstrates to find one answer, but I have 3 samples. The instructions sheet is attached. Thanks

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How do you calculate the% by mass of acetic acid present? The table below demonstrates to find one answer, but I have 3 samples. The instructions sheet is attached. Thanks
a. Analysis of a Vinegar Solution
Vinegar is a dilute solution of acetic acid and can be effectively titrated with NaOH using the
phenolphthalein endpoint.
Clean and dry a small beaker, and obtain 25-30 mL of the unknown vinegar solution. Cover the
vinegar solution with a watch glass to prevent evaporation. Record the code number of the
sample. If the vinegar is a commercial product, record its brand name. Clean three Erlenmeyer
flasks, and label as samples 1, 2, and 3. Rinse the flasks with small portions of distilled water.
Using the rubber safety bulb to provide suction, rinse the 5-mL pipet with small portions of the
vinegar solution, and discard the rinsings.
Using the rubber safety bulb, pipet a 5-mL sample of the vinegar solution into each of the
Erlenmeyer flasks. Add approximately 100 mL of distilled water to each flask, as well as 2-3
drops of phenolphthalein indicator solution.
Refill the buret with the NaOH solution, and record the initial reading of the buret to the nearest
0.02 ml.. Titrate sample 1 of vinegar in the same manner as in the standardization until a single
drop of NaOH causes the pale pink color to appear and persist.
Record the final reading of the buret to the nearest 0.02 mL.
Repeat the titration for the other two vinegar samples.
From the volume of vinegar sample taken, and from the volume and average concentration of
NaOII titrant used, calculate the concentration of the vinegar solution in moles per liter.
Example: Suppose a 5.00 mL sample of vinegar requires 19.31 mL of 0.121 M/NaOH to reach
the phenolphthalein endpoint
mol NaOH = 0.01931 L NaOH sol.x
0.121 mol NaOH
1 1. NaOll sol.
= 0.002337 mol NaOH
At the endpoint, mol acetic acid mol NaOII, so
Molarity of acetic acid =
= 0.0167 M
Given that the molar mass of acetic acid is 60.0 g and that the density of the vinegar solution is
1.01 g/mL, calculate the percent by weight of acetic acid in the vinegar solution.
Example:
0.002337 mol acetic acid
0.00500 Lacetic acid used
0.467 mol HOẶC
1 I.
Using the data in the previous example, and abbreviating acetic acid as "HOAC,"
1 L
0.467 M HOAc =
60.0 g HOẶC
1 mol ||OAC10108
x100% 2.77%
Transcribed Image Text:a. Analysis of a Vinegar Solution Vinegar is a dilute solution of acetic acid and can be effectively titrated with NaOH using the phenolphthalein endpoint. Clean and dry a small beaker, and obtain 25-30 mL of the unknown vinegar solution. Cover the vinegar solution with a watch glass to prevent evaporation. Record the code number of the sample. If the vinegar is a commercial product, record its brand name. Clean three Erlenmeyer flasks, and label as samples 1, 2, and 3. Rinse the flasks with small portions of distilled water. Using the rubber safety bulb to provide suction, rinse the 5-mL pipet with small portions of the vinegar solution, and discard the rinsings. Using the rubber safety bulb, pipet a 5-mL sample of the vinegar solution into each of the Erlenmeyer flasks. Add approximately 100 mL of distilled water to each flask, as well as 2-3 drops of phenolphthalein indicator solution. Refill the buret with the NaOH solution, and record the initial reading of the buret to the nearest 0.02 ml.. Titrate sample 1 of vinegar in the same manner as in the standardization until a single drop of NaOH causes the pale pink color to appear and persist. Record the final reading of the buret to the nearest 0.02 mL. Repeat the titration for the other two vinegar samples. From the volume of vinegar sample taken, and from the volume and average concentration of NaOII titrant used, calculate the concentration of the vinegar solution in moles per liter. Example: Suppose a 5.00 mL sample of vinegar requires 19.31 mL of 0.121 M/NaOH to reach the phenolphthalein endpoint mol NaOH = 0.01931 L NaOH sol.x 0.121 mol NaOH 1 1. NaOll sol. = 0.002337 mol NaOH At the endpoint, mol acetic acid mol NaOII, so Molarity of acetic acid = = 0.0167 M Given that the molar mass of acetic acid is 60.0 g and that the density of the vinegar solution is 1.01 g/mL, calculate the percent by weight of acetic acid in the vinegar solution. Example: 0.002337 mol acetic acid 0.00500 Lacetic acid used 0.467 mol HOẶC 1 I. Using the data in the previous example, and abbreviating acetic acid as "HOAC," 1 L 0.467 M HOAc = 60.0 g HOẶC 1 mol ||OAC10108 x100% 2.77%
Quantity of
vinegar
taken
Initial
NaOH buret
reading
Final
NaOH buret
reading
Volume of
NaOH used
Molarity of
vinegar
Mean
molarity of
vinegar
•%. by mass
acetic acid
present
Titration of vinegar
Sample 2
5.0ml
Sample 1
5.0ml
0.0mL
18.0mL
18.0mL
0.406 M
0.0mL
39.5mL
39.5mL
0.892 M
0.546M
Sample 3
5.0ml
0.0mL
15.0mL
15.0mL
0.340 M
Transcribed Image Text:Quantity of vinegar taken Initial NaOH buret reading Final NaOH buret reading Volume of NaOH used Molarity of vinegar Mean molarity of vinegar •%. by mass acetic acid present Titration of vinegar Sample 2 5.0ml Sample 1 5.0ml 0.0mL 18.0mL 18.0mL 0.406 M 0.0mL 39.5mL 39.5mL 0.892 M 0.546M Sample 3 5.0ml 0.0mL 15.0mL 15.0mL 0.340 M
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