Long but quick chemistry lab that I need confirmation on answers?
OK so I have this lab that went pretty quick and smooth and I have an online assessment that I needed to put all my figures in. I am all complete but am still having trouble with 1 problems that the assessment is saying is wrong and I am looking for someone to verify or explain what I am doing incorrect.
Here were the steps of the lab and my findings:
Determining the molecular weight of Acetone:
Step 1: Obtain a Dumas bulb from the Equipment menu.
Step 2: Record the volume of the bulb in liters (volume is marked on bulb). .056L
Step 3: Record the mass the empty bulb using an electronic balance. 50.g
Step 4: Place 20 ml of acetone into the bulb.
Step 5: Obtain a 600-ml beaker.
Step 6: Select Beaker and place about 500 ml of water into beaker or enough to cover neck of bulb.
Step 7: Combine bulb with beaker (using combine menu option).
Step 8: Obtain a hot plate from Equipment menu.
Step 9: Place combined Dumas bulb and beaker assembly onto hot plate.
Step 10: Turn on hot plate to maximum heating setting.
Step 11: Allow all acetone to boil away and wait 1-2 minutes before removing bulb from beaker.
Step 12: After removing bulb from beaker, allow Dumas bulb to cool until acetone vapors have condensed (2-3 minutes).
Step 13: Weigh bulb and condensed vapor using an electronic scale and record result. 50.1062g
Then the lab told me to use this info to answer the assessment:
(i) Volume of Dumas bulb (L): 56ml or .056L
(ii) Mass of empty Dumas bulb (g): 50g
(iii) Mass of Dumas bulb + condensed acetone (g):50.106
(iv) Mass of acetone gas required to fill bulb ((iii) - (ii)): .106
(v) Molecular weight of Acetone: Mass / n = Mass / (PV/RT) = (Mass *R *T) /(P * V) =.00183
Given:
Pressure: 1atm
Temperature: 373 Kelvin (boiling point of water)
Then Here was the question I needed to answer (which I am getting number 5 incorrect):
1)Weight of empty Dumas flask=50g
2)Weight of acetone=? 15.714
3)Weight of liquid vapor after condensing=.106
4)Moles of acetone vapor based on ideal gas law=.00183
5)Calculated molecular weight of acetone based on experiment. 58
6)%error from the theoretical molecular of acetone and the experimental value.=which to me seems like there is a 0% error
I think I would take the calculation on 5 to three significant places, since your fraction .106/.00183 indicates that.
I just cant figure out the theoretical yield???
There is no theoretical yield which may explain why you can calculate it. There is a theoretical molar mass for acetone. That is (CH3)2CO and you can calculate the molar mass easily. Its about 58 or so but you need to do it exactly.
To help you understand where the mistake might be, let's go through the calculations step by step:
Step 1: Volume of Dumas bulb (L): 0.056L
Step 2: Mass of empty Dumas bulb (g): 50g
Step 3: Mass of Dumas bulb + condensed acetone (g): 50.1062g
Step 4: Mass of acetone gas required to fill bulb ((iii) - (ii)): 0.1062g (You made a typo in your question, it should be 0.1062g instead of .106)
Step 5: Molecular weight of Acetone = Mass / n
Using the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature, we can rearrange the equation to solve for n:
n = (mass * R * T) / (P * V)
Given:
Pressure: 1 atm
Temperature: 373 Kelvin
R (gas constant): 0.0821 L·atm/(K·mol)
Plugging in the values:
n = (0.1062g * 0.0821 L·atm/(K·mol) * 373K) / (1 atm * 0.056L)
n = 0.0503 mol
Now, to find the molecular weight of acetone, we can rearrange the equation:
Molecular weight = Mass / n
Plugging in the values:
Molecular weight = 0.1062g / 0.0503 mol
Molecular weight = 2.11 g/mol
So, based on the given data and calculations, the calculated molecular weight of acetone should be 2.11 g/mol, not 58 g/mol. Therefore, it seems that your answer in question 5 is incorrect.
To find the percent error, you need to compare the calculated molecular weight (2.11 g/mol) with the theoretical molecular weight of acetone. However, you haven't provided the theoretical molecular weight of acetone in your question, so it's not possible to calculate the percent error accurately.