Just need help with the last part of my lab. Thank you! Lab notes: Flask 1: Gluc
ID: 94597 • Letter: J
Question
Just need help with the last part of my lab. Thank you!
Lab notes:
Flask 1: Glucose
Temp: 24.7 C
Initial gas volume: 0.1 mL
Final gas volume (after 5 min): 19.3 mL
Flask 2: Fructose
Temp: 21.7 C
Initial gas volume: 0.0 mL
Final gas volume (after 5 min) : 6.5 mL
Flask 3: Maltose
Temp: 25.3 C
Initial gas volume: 0.1 mL
Final gas volume (after 5 min): 20.5 mL
Flask 4: Maltotriose
Temp: 21.9 C
Initial gas volume: 0.0 mL
Final gas volume (after 5 min): 4.6 mL
Lab:
For each of the sugars fermented by yeast, fill in the chart below to determine CO2 production?
For each of the sugars fermented by yeast, fill in the chart below to determine the mg of sugar consumed per minute during fermentation.For column one use
to calculate the moles of CO2 produced
Use
to calculate the moles of sugar consumed
Use
to calculate the mg of sugar fermented per minute
Based on your results, which sugars should be provided to yeast grown commercially considering that maximizing efficiency is the goal of the industry?
Results Table Sugar Initial Gas Volumet=0 minutes (mL) Final Gas Volume
t=5 minutes (mL) Volume of CO2 produced
Final - Initial (mL) Glucose 0.1 19.3 19.2 Fructose 0.0 6.5 6.5 Maltose 0.1 20.5 20.4 Maltotriose 0.0 4.6 4.6
Explanation / Answer
Determine the mg of sugar consumed per minute during fermentation per minute during fermentation
we use this formula for calculate the volume of CO2 produced
n = (P × V) ÷ (R × T)
where- n=Number of moles
P=pressure=101.3kpa
T=temperature=273.15k
R=gas constant=8.3145/mole
Glucose-
n = (P × V) ÷ (R × T)
19.2 * 101.3 / 273.15 * 8.314 =1944.96/2270.9691 = 0.8564
Fructose-
6.5 * 101.3 / 273.15 * 8.314 = 658.45 / 2270.9691 = 0.3018
Maltose-
20.4 * 101.3 / 273.15 * 8.314 = 2066.52 / 2270.9691 = 0.9099
Maltotriose-
4.6 * 101.3 / 273.15 * 8.314 = 465.98 / 2270.9691 = 0.2051
2- To calculate the moles of sugar consumed
moles of sugar consumed = moles of CO2 produced ÷ (2 × number of simple sugars in that sugar)
Glucose- number of simple sugars in glucose-6 -because chemical formula of glucose-C6H12O6
0.8564 / 2*6 = 0.7136
Fructose- number of simple sugars in glucose-6 -because chemical formula of fructose will be the same
0.3018 / 2 * 6 = 0.0251
Maltose-number of simple sugars in maltose -12 because chemical formula C12H22O11
0.9099 / 2 * 12 = 0.379
Maltotriose--number of simple sugars in maltotriose -18 because chemical formula C18H32O16
0.2051 /2*18 = 0.0057
3-to calculate the mg of sugar fermented per minute
mg of sugar per minute = (moles sugar) × (MW g/mole) × (1000 mg/g) ÷ (5 minutes)
the volume of CO2 gas produced during each fermentation test with yeast,
Molecular mass of CO2 = 44.01 g/mol
Glucose-
0.7136 * 44.01 * 1000 / 5 = 6281.10
Fructose-
Glucose should be provided to yeast grown commercially considering that maximizing efficiency is the goal of the industry.
sugar Moles of CO2 produced moles of sugar consumed Mg of sugar /min Glucose 0.8564 0.7136 6281.10 Fructose 0.3018 0.0251 220.93 Maltose 0.9099 0.379 3335.958 Maltotriose 0.2051 0.0057 50.17Related Questions
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