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On Line Activity: Biofuels – Analysis
Submit your Group Responses from This Point On
CalculationsCalculate the volume of ethanol collected in all four distillations. Follow these steps:
A. Convert the measured density of the distillate from each fraction to % ethanol by volume
(concentration). Use the conversion chart in https://handymath.com/cgibin/ethanolwater3.cgi?submit=Entry
B. Calculate the volume of ethanol collected in each fraction. Multiply the distillate volume
in each fraction by the concentration of ethanol in each fraction.
C. What is the total volume of ethanol collected? Sum the volume of ethanol collected in
each fraction.
D. The total volume of ethanol collected = the % recovery of ethanol because you started
with 100 mL of fermentation broth (x/100 *100) = x%
Data for distillation of fermented apple juice
Total volume of fermentation broth: 256.32 mL
Volume for distillation: 100.00 mL
Volume of pot residue: 61.3 mL
Time for distillation: 35 min
Sugar content per serving size = 43 g glucose per 296 mL of juice
Fraction Temp (° C) Distillate
volume (mL)
Density of
Distillate (g/mL)
Concentration (% EtOH
by Volume)
Volume of EtOH (mL)
1 76 1.99 0.806
2 79-92 3.80 0.891
3 92-94 3.96 0.902
4 94-95 3.51 0.972
5 95-96 3.74 0.974
6 96-97 3.81 0.990
7 97-98 3.11 0.990
8 98 2.00 0.997
Total Volume of Ethanol =
% Recovery of Ethanol =
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Data for distillation of fermented grape juice
Total volume of fermentation broth: 249.88 mL
Volume for distillation: 100.00 mL
Volume of pot residue: 72.0 mL
Time for distillation: 40 min
Sugar content per serving size = 40 g glucose per 240 mL of juice
Fraction Temp (° C) Distillate
volume (mL)
Density
(g/mL)
Concentration (% EtOH
by Volume)
Volume of EtOH (mL)
1 78 1.32 0.809
2 78 1.70 0.852
3 86 1.32 0.870
4 87 1.36 0.872
5 88 1.29 0.890
6 90 0.71 0.890
7 90 1.80 0.905
8 91 1.20 0.920
9 92 1.31 0.921
10 92 1.44 0.960
11 94 2.74 0.974
12 94 1.74 0.980
13 95 1.73 0.980
14 95 1.70 0.990
15 96 1.47 0.990
16 97 1.045 1.000
Total Volume of Ethanol =
% Recovery of Ethanol =
Data for distillation of fermented potato starch
Total volume of fermentation broth: 219.52 mL
Volume for distillation: 100.00 mL
Volume of pot residue: 73.2 mL
Time for distillation: 40 min
Sugar content per serving size = potato starch is pure glucose in starch form. 50 g glucose in 50 g of starch
Fraction Temp (° C) Distillate
volume (mL)
Density
(g/mL)
Concentration (% EtOH
by Volume)
Volume of EtOH (mL)
1 78 1.98 0.809
2 78-80 2.15 0.852
3 80-87 2.30 0.872
4 87 2.13 0.895
5 87 1.85 0.922
6 87-89 1.59 0.948
7 89 3.20 0.957
8 89-93 1.87 0.987
9 95 2.97 0.987
10 95 2.56 1.000
Total Volume of Ethanol =
% Recovery of Ethanol =
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Data for distillation of fermented corn mash
Total volume of fermentation broth: 209.36 mL
Volume for distillation: 100.00 mL
Volume of pot residue: 68.5 mL
Time for distillation: 34 min
Sugar content per serving size = 16 g glucose per 90 g corn
Fraction Temp (° C) Distillate
volume (mL)
Density
(g/mL)
Concentration (% EtOH
by Volume)
Volume of EtOH (mL)
1 79-92 1.345 0.867
2 92 1.232 0.910
3 92 1.204 0.940
4 92 1.469 0.950
5 92 1.368 0.955
6 92 1.236 0.960
7 92-94 1.295 0.980
8 94 1.430 0.982
9 94-95 1.310 0.989
10 95 1.109 0.990
11 95 1.424 0.991
12 95 1.369 0.992
13 95 1.498 0.996
14 95 1.435 0.997
15 95 1.192 0.997
16 95 1.045 0.998
17 95 1.287 0.998
18 95 1.414 0.999
19 95 1.411 1.000
Total Volume of Ethanol =
% Recovery of Ethanol =
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Calculate the volume of ethanol produced per gram of starting material (corn, starch, or juice) (mL/g).
The amounts of starting material are as follows: 100.0 g of corn, 50.0 grams of potato starch, 100.0 mL
of juice. The density of Welch’s apple juice is 1.05 g/mL and Welch’s grape juice is 1.06 g/mL.
Example Calculation: There was 220 mL of total fermentation broth (100 mL was taken for distillation)
collected and the group calculated a percent recovery of 6.1% ethanol. Take the volume of ethanol
that would have been produced had all of the fermentation broth been distilled and divide it by the
grams of starting material (mass of corn, potato starch, juice):
220 �� � 6.1% = 13.42 �� ��ℎ����
13.42 �� ��ℎ����
100 � ���� = 0.13 ��/�
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Add the volume of distillate collected in all fractions and the remaining residue in the round bottom
flask. How does this compare to the original volume? If any was lost, calculate the percent volume lost
in the distillation process.
Example Calculation: Let’s say the total volume collected in the fractions was 7.1 mL and the total volume
remaining in the distillation flask was 90.5 mL (volume distilled was 100 mL). The percent loss can be calculated
by:
% ���� = (100 − 97.6)
100 � 100 = 2.4%
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Calculate the theoretical yield (in mL) of ethanol that can be produced in each fermentation. You
will need to consider how much of the carbohydrate source was used, how much sugar/carbohydrate is
contained in each source, and the chemical equation for the fermentation of sugar/glucose into ethanol.
Assume that all available more complex carbohydrates, such as starch, were converted into glucose via
the acid hydrolysis pre-fermentation step. After calculating the theoretical yield for all four
carbohydrate sources, calculate the percent yield of each.
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Calculate the energy supplied by the heating mantle to the distilling flask and compare to the energy
released when the ethanol is combusted. Which is greater? Is this procedure commercially viable?
Assume the power of the heating mantle is 130 Watts and Power=Energy/time (watts = J/s). How long
did it take you to complete the distillation (in seconds)? Assume that ethanol generates 27.3 kJ/g of
material.
Carbohydrate
Source
Time to
Complete
Distillation
Apple Juice 35 min
Grape Juice 40 min
Potato Starch 40 min
Corn 34 min
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Consider a hectare of land; calculate the amount of ethanol that could be obtained per year with
each crop: apples or grapes, potatoes, corn and grass. Note: 1 hectare = 2.47 acre
Grass (Example): Assume an average quarterly harvest of 3.5 tons of grass per acre (yields will be
greater in the summer and less in the fall. Assume 2.0 g of grass yields approximately 0.65 mL of
ethanol or 0.33 mL/g.
������ ℎ������ �� ����� �� �� ��� ℎ������ �� ����:
3.5 ���� �����
1 ���� • �������
�
4 �������
1 ����
�
2.47 ����
1 ℎ������
�
907 ��
1 ��� = 31364 �� �����
ℎ������ • ����
����������� ����� �� ��ℎ���� �� ������ ��� ℎ������ �� ���� ��� ����:
31,364 �� ����� �
1000 �
1 �� �
0.65 �� ��ℎ����
2.0 � �����
�
1 �
1000 �� = 10193 � ��ℎ����
ℎ������ • ����
Corn: A typical corn crop is 158 bushels per acre, about 45 pounds of corn per bushel. Use the amount
of ethanol (in mL/g) that was produced from corn in our experiment.
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Apples: A typical orchard can produce 450 bushels per acre, 42 pounds per bushel of apples. Apples
are approximately 90% juice. Use the amount of ethanol (in mL/g) that was produced from apple juice
in our experiment.
Grapes: A typical vineyard can produce 4 tons of grapes per acre. Grapes are approximately 90% juice.
Use the amount of ethanol (in mL/g) that was produced from grape juice in our experiment.
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Potatoes: Assume 300 bushels per acre with 60 pounds of potatoes per bushel. Potato starch
accounts for approximately 10% of the mass (the rest is mostly water). Use the amount of ethanol (in
mL/g) that was produced from potato starch in our experiment.
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Calculate the heat of combustion in kJ/L for 1.0 L of ethanol and 1.0 L of octane. The density of
ethanol is 0.789 g/mL and the density of octane is 0.703 g/mL. What does your answer imply about
the fuel mileage for ethanol versus octane (the primary hydrocarbon in gasoline).
DHf (kJ/mol)
C2H5OH (l) -277.7
C8H18 (l) -250.0
O2 (g) 0
CO2 (g) -393.5
H2O (l) -285.8
∆� ���������� = (∆� ��������) − (∆� ���������)
C8H18 (l) + 25/2 O2 (g) 8CO2 (g) + 9H2O (l)
C2H5OH (l) + 3O2 (g) 2CO2 (g) + 3H2O (l)
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Short Essay Question: Compare and contrast all five glucose sources. Consider the relative amounts
of ethanol produced per gram of material, and the amount of material and resources needed to
produce each crop.

Did the student’s data provided to you ever measure a density of any fraction that corresponded to
a percent ethanol of higher than 95.6% (azeotrope)? What do you think can explain this data?
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