What was the alcohol percent by weight in your sample with the most gas production (glucose; which gave 30 mL of CO2)? Remember that the temperature was 40 degrees C and assume the atmospheric pressure was 101.3 kPa.
To determine the alcohol percent by weight in the sample, we need to know the amount of alcohol produced. Since you mentioned that the sample produced 30 mL of CO2, we can use the concept of stoichiometry to find the amount of alcohol, assuming glucose is being completely converted to alcohol.
First, let's start by understanding the reaction for the production of alcohol from glucose:
C6H12O6 → 2C2H5OH + 2CO2
From this balanced equation, we can see that for every molecule of glucose (C6H12O6) consumed, two molecules of ethanol (C2H5OH) and two molecules of carbon dioxide (CO2) are produced.
Now, to determine the moles of alcohol produced, we need to convert the volume of CO2 produced to moles. To do this, we will use the ideal gas law:
PV = nRT
Where:
P = pressure (101.3 kPa)
V = volume (30 mL = 0.03 L)
n = moles of CO2
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (40 + 273.15 = 313.15 K)
Rearranging the equation to solve for moles of CO2 (n):
n = (PV) / (RT)
Substituting the given values:
n = (101.3 kPa * 0.03 L) / (0.0821 L·atm/(mol·K) * 313.15 K)
Now, we know that for every two moles of ethanol produced, there is one mole of glucose consumed. Therefore, to find the moles of alcohol produced, we can divide the moles of CO2 by 2:
moles of alcohol = n / 2
Next, to find the weight of alcohol, we need to know the molecular weight of ethanol, which is 46.07 g/mol.
Weight of alcohol = moles of alcohol * molecular weight of ethanol
Finally, to find the alcohol percent by weight, we divide the weight of alcohol by the weight of the entire sample and multiply by 100:
Alcohol percent by weight = (Weight of alcohol / Weight of sample) * 100
To calculate this value, you will need to know the weight of the sample being used.
So, with the information provided, we can determine the moles of alcohol produced, but we need the weight of the sample to calculate the alcohol percent by weight accurately.