CHEM 123 LAB FINAL ACTUAL EXAM/CHEM
123 LAB FINAL ACTUAL EXAM QUESTIONS AND
CORRECT DETAILED ANSWERS LATEST VERSION
(VERIFIED ANSWERS)ALREADY GRADED A+
ratio of ascorbic acid to KIO3 - ANSWERS-3:1
Exp 9 Calc - ANSWERS-get mass of KIO3,
change to moles (divide by 214g) and divide by the volume (usually 0.250L) to
get KIO3 Concentration
Multiply the concentration of KIO3 by the volume of KIO3 used in titration,
convert the KIO3 mols to ascorbic acid by multiplying by 3. Then, convert AA to
mass (g) by multiplying by 176.
To find out how much vitamin C is in 25mL - ANSWERS-Take the percentage of
vitamin C and divide by 100
multiply by 90mg (daily intake)
divide by 10 to get in 25mL increments.
eg. 150%
(150/100 * 90 )/10 = 13.5mg per 25mL sample
,100%
(100/100 * 90)/10 = 9mg per 25mL sample
To find out how much KIO3 is required after finding out how much vitamin C is in
25mL. - ANSWERS-eg. 0.009g AA per 25mL sample
divide by molar mass of AA and divide by 3 to get mols of KIO3,
then divide by (25/1000 L) and multiply by (0.250L) and then multiply by the
molar mass of KIO3 (214g), now you have your measured Kio3 required.
Note that wording can be tricky sometimes. Suppose that the amount of
ascorbic acid in the sample you are about to titrate is 96 mg. (this means 96mg
in 1 25mL sample, do not divide by 10 to get 9.6mg per 25mL sample, thats
incorrect ) - ANSWERS-
iodometric titration - ANSWERS-Iodometric titration is performed to determine
the concentration of substances in solution if those substances are capable of
being oxidized in an oxidation-reduction reaction.
In an iodometric titration, molecular iodine, I2, is used as an oxidizing agent
against a solute of unknown concentration. Iodide gains electrons. The endpoint
is reached when all of the organic
substrate has been oxidized.
Vitamin C as the organic reducing agent in an iodometric titration. Iodine is
reduced while vitamin C is oxidized. This is the reaction that you will be carrying
out in Experiment 9. Again, the endpoint is reached when all of the vitamin C
has been oxidized, making this reaction suitable for determining the
concentration of vitamin C in a solution.
, Why are Liquid Liquid extractions used? - ANSWERS-These are performed by
transferring a dissolved substance (solute) from one solvent into another. For
example, let's say that we are trying to isolate some organic compound as a dry
solid from an aqueous solution. We may first be tempted to simply boil off the
water by heating the solution to 100°C. However, the dissolved organic
compound may be thermally unstable and decompose at water's boiling point.
Another strategy is more appropriate: transfer the compound into a low-boiling,
organic solvent such as dichloromethane through a liquid-liquid extraction. Then
boil away the dichloromethane at a much lower temperature (~40°C), ensuring
that the organic compound does not decompose. This will leave us with a dry
and chemically intact solid.
Liquid liquid extraction does NOT work if - ANSWERS-the compound to be
purified from an aqueous solution decomposed at a low temperature and had
low solubility in dichloromethane.
why choose water over DCM? - ANSWERS-dcm is more dense than water and
can sink to the bottom and this will cause separation needed to separate the
crude caffeine substance from its original solvent. ALSO, crude caffeine is more
soluble in DCM than in water, so DCM must be used because the solute wants to
migrate to the substance (solvent) in which it is more soluble in. So the crude
caffeine migrates to the DCM because it has high solubility.
Liquid-liquid extractions work because solutes preferentially migrate to or
remain in the solvent in which they are most soluble. By requiring that the
solvents used be immiscible (e.g. water and dichloromethane), physical
separation of the solutions becomes easy: the less dense solution floats on top
of the more dense one without mixing. The separatory funnels used in
123 LAB FINAL ACTUAL EXAM QUESTIONS AND
CORRECT DETAILED ANSWERS LATEST VERSION
(VERIFIED ANSWERS)ALREADY GRADED A+
ratio of ascorbic acid to KIO3 - ANSWERS-3:1
Exp 9 Calc - ANSWERS-get mass of KIO3,
change to moles (divide by 214g) and divide by the volume (usually 0.250L) to
get KIO3 Concentration
Multiply the concentration of KIO3 by the volume of KIO3 used in titration,
convert the KIO3 mols to ascorbic acid by multiplying by 3. Then, convert AA to
mass (g) by multiplying by 176.
To find out how much vitamin C is in 25mL - ANSWERS-Take the percentage of
vitamin C and divide by 100
multiply by 90mg (daily intake)
divide by 10 to get in 25mL increments.
eg. 150%
(150/100 * 90 )/10 = 13.5mg per 25mL sample
,100%
(100/100 * 90)/10 = 9mg per 25mL sample
To find out how much KIO3 is required after finding out how much vitamin C is in
25mL. - ANSWERS-eg. 0.009g AA per 25mL sample
divide by molar mass of AA and divide by 3 to get mols of KIO3,
then divide by (25/1000 L) and multiply by (0.250L) and then multiply by the
molar mass of KIO3 (214g), now you have your measured Kio3 required.
Note that wording can be tricky sometimes. Suppose that the amount of
ascorbic acid in the sample you are about to titrate is 96 mg. (this means 96mg
in 1 25mL sample, do not divide by 10 to get 9.6mg per 25mL sample, thats
incorrect ) - ANSWERS-
iodometric titration - ANSWERS-Iodometric titration is performed to determine
the concentration of substances in solution if those substances are capable of
being oxidized in an oxidation-reduction reaction.
In an iodometric titration, molecular iodine, I2, is used as an oxidizing agent
against a solute of unknown concentration. Iodide gains electrons. The endpoint
is reached when all of the organic
substrate has been oxidized.
Vitamin C as the organic reducing agent in an iodometric titration. Iodine is
reduced while vitamin C is oxidized. This is the reaction that you will be carrying
out in Experiment 9. Again, the endpoint is reached when all of the vitamin C
has been oxidized, making this reaction suitable for determining the
concentration of vitamin C in a solution.
, Why are Liquid Liquid extractions used? - ANSWERS-These are performed by
transferring a dissolved substance (solute) from one solvent into another. For
example, let's say that we are trying to isolate some organic compound as a dry
solid from an aqueous solution. We may first be tempted to simply boil off the
water by heating the solution to 100°C. However, the dissolved organic
compound may be thermally unstable and decompose at water's boiling point.
Another strategy is more appropriate: transfer the compound into a low-boiling,
organic solvent such as dichloromethane through a liquid-liquid extraction. Then
boil away the dichloromethane at a much lower temperature (~40°C), ensuring
that the organic compound does not decompose. This will leave us with a dry
and chemically intact solid.
Liquid liquid extraction does NOT work if - ANSWERS-the compound to be
purified from an aqueous solution decomposed at a low temperature and had
low solubility in dichloromethane.
why choose water over DCM? - ANSWERS-dcm is more dense than water and
can sink to the bottom and this will cause separation needed to separate the
crude caffeine substance from its original solvent. ALSO, crude caffeine is more
soluble in DCM than in water, so DCM must be used because the solute wants to
migrate to the substance (solvent) in which it is more soluble in. So the crude
caffeine migrates to the DCM because it has high solubility.
Liquid-liquid extractions work because solutes preferentially migrate to or
remain in the solvent in which they are most soluble. By requiring that the
solvents used be immiscible (e.g. water and dichloromethane), physical
separation of the solutions becomes easy: the less dense solution floats on top
of the more dense one without mixing. The separatory funnels used in
