Orgo Lab Midterm CHM2211L Exam Questions With
Complete Answers
Ch.2: What is the melting point? - ANSWER The temperature range over which the first
crystal of a solid begins to melt and the last crystal completes its melting
Ch.2: How would a small amount of impurity effect a compound's melting point? -
ANSWER It would lower the compound's melting point (lower temp needed to melt) and
the temperature range would broaden
Ch.2: Melting point is dependent on... - ANSWER the stacking arrangement of the
crystalline lattice
Ch.2: Impurities must be __________ in order to affect the melting point range - ANSWER
soluble
Ch.2: List 3 compounds that are NOT classified as impurities - ANSWER Sand / Glass /
Decolorized Carbon
Ch.2: How can we effectively use the melting point test to differentiate between 2
compounds? - ANSWER Grind both compounds into equal amounts to mix-take the
melting point of this mixture
Identical: MP range is sharp
,Different: MP range is low and broad
Ch.2: Can a substance melt at a higher temperature than its melting point? - ANSWER
No.
Ch.2: What is the accurate way to perform a melting point test? - ANSWER Sample is
finely powdered
Is firmly packed in capillary tube
Sample must be completely dry
Ch.3: An unknown compound was determined to be 74.97% C, 8.39% H, and no N.
Calculate the empirical formula. - ANSWER First: Find the amount of Oxygen:
74.97% C + 8.39% H = 83.36%
100% - 83.36% = 16.64% Oxygen
Second: Divide each percentage to it's Molecular Weight:
C: 74.97/12 = 6.04
H: 8.39/1 = 8.39
O: 16.64/16 = 1.04
Third: Divide each by Oxygen's product:
C: 6.04/1.04 = 6
, H: 8.39/1.04 = 8
O: 1.04/1.04 = 1
Empirical Formula: C6H8O
Ch.3: Can IR spectroscopy be used to distinguish 2-pentanone from 2-hexanone? Why or
why not? - ANSWER No, b/c IR is used to distinguish between functional groups and
both molecules contain the same functional group so they would look the same on the
IR spectroscopy
Ch.3: An unknown organic compound gives the following data: 54.53% C, 9.15% H, no N.
What is the empirical formula? If the molecular weight is determined to be 88g/mol,
what is the molecular formula for this compound? - ANSWER 1st: Find amount of
Oxygen:
54.53% + 9.15% = 63.68%
100% - 63.68% = 36.32%
2nd: Divide each by its molecular weight:
C: 54.54/12 = 4.54
H: 9.15/1 = 9.15
O: 36.32/16 = 2.27
3rd: Divide each by oxygen's product:
Complete Answers
Ch.2: What is the melting point? - ANSWER The temperature range over which the first
crystal of a solid begins to melt and the last crystal completes its melting
Ch.2: How would a small amount of impurity effect a compound's melting point? -
ANSWER It would lower the compound's melting point (lower temp needed to melt) and
the temperature range would broaden
Ch.2: Melting point is dependent on... - ANSWER the stacking arrangement of the
crystalline lattice
Ch.2: Impurities must be __________ in order to affect the melting point range - ANSWER
soluble
Ch.2: List 3 compounds that are NOT classified as impurities - ANSWER Sand / Glass /
Decolorized Carbon
Ch.2: How can we effectively use the melting point test to differentiate between 2
compounds? - ANSWER Grind both compounds into equal amounts to mix-take the
melting point of this mixture
Identical: MP range is sharp
,Different: MP range is low and broad
Ch.2: Can a substance melt at a higher temperature than its melting point? - ANSWER
No.
Ch.2: What is the accurate way to perform a melting point test? - ANSWER Sample is
finely powdered
Is firmly packed in capillary tube
Sample must be completely dry
Ch.3: An unknown compound was determined to be 74.97% C, 8.39% H, and no N.
Calculate the empirical formula. - ANSWER First: Find the amount of Oxygen:
74.97% C + 8.39% H = 83.36%
100% - 83.36% = 16.64% Oxygen
Second: Divide each percentage to it's Molecular Weight:
C: 74.97/12 = 6.04
H: 8.39/1 = 8.39
O: 16.64/16 = 1.04
Third: Divide each by Oxygen's product:
C: 6.04/1.04 = 6
, H: 8.39/1.04 = 8
O: 1.04/1.04 = 1
Empirical Formula: C6H8O
Ch.3: Can IR spectroscopy be used to distinguish 2-pentanone from 2-hexanone? Why or
why not? - ANSWER No, b/c IR is used to distinguish between functional groups and
both molecules contain the same functional group so they would look the same on the
IR spectroscopy
Ch.3: An unknown organic compound gives the following data: 54.53% C, 9.15% H, no N.
What is the empirical formula? If the molecular weight is determined to be 88g/mol,
what is the molecular formula for this compound? - ANSWER 1st: Find amount of
Oxygen:
54.53% + 9.15% = 63.68%
100% - 63.68% = 36.32%
2nd: Divide each by its molecular weight:
C: 54.54/12 = 4.54
H: 9.15/1 = 9.15
O: 36.32/16 = 2.27
3rd: Divide each by oxygen's product:
