Chemistry questions from 1411?
If the outside of the flask is not dried after vaporizing the liquid, will the unknown's calculated molecular
weight is too high or too low? Explain?
2.
Suppose the atmospheric pressure is assumed to be 1 atmosphere instead of its actual value. How wills this
error affects the molecular weight of the unknown? Explain.
3.
If the vapor's volume is assumed to be 125 mL instead of the measured volume, what is the percent error of
the unknown's calculated molecular weight? Show your work.
4.
If the entire unknown does not vaporize in the 125 mL Erlenmeyer flask, will the reported molecular weight
be too high or low? Explain.
5. Solve the following problems.
a) How many moles of gas are contained in 890.0 mL at 21.0 °C and 750.0 mm Hg pressure?
b) What volume will 1.27 moles of helium gas occupy at STP?
c) At what temperature will 0.654 moles of neon gas occupy 12.30 liters at 1.95 atmospheres?
Answer:
You need to give a little more information for this question: I am assuming you are describing the experiment where you vaporize an unknown volatile material in a flask of known (or determined) volume which is covered by a piece of foil with a pinhole in it to let the extra vapor out. You have a known volume, pressure, and temperature, which combined with knowing the weight of the contained vapor you can get molecular weight.
Note: if you're doing the experiment a different way, mistakes in doing it may affect the answer differently, so these answers might not apply, please check carefully
PV = nRT
Solve for n;
n = PV/RT note: n = mass(g)/mW(g); so substituting:
mass (g)/mW = PV/RT, rearanging;
mass = [PV/RT] x mW >>> mass/[PV/RT] = mW
(mass is weight of flask & vapor - weight of empty flask)
If the flask is wet (and you are getting vapor mass by mass of flask with vapor - empty flask) it will apear as if the known volume of vapor was heavier than it actually was and thus the mW will appear larger than it should be.
2] if you assumed 1 atm and the actual air pressure was higher, the mw would appear to be larger (since the bottom term of the fraction would be smaller than it should). If you assumed 1 atm but actual pressure was lower, the calculated mW would appear to be smaller than it should be (because the bottom term of the fraction would be larger than it should be.
3] Since you didn't give the actual volume (or any of the experimental values) I can only tell you how to do it. Do the calculation using the real (determined) volume of the flask. Then, re-do the calculation with v = 125ml. Take the two values and calculate % error; m = measured vol; g = guessed vol.
[mW (m) - mW (g)]/mW (m) x 100%
(they want % error so, if it is negative, just get rid of the sign)
4] I presume they mean there is still liquid unknown in the bottom of the flask. The factor: weight of flask & vapor - empty flask will be bigger than it should; since this is the top number of the fraction the mW will appear larger than it should.
Part 5: Note: most mistakes happen by not having terms in the proper units! If you're using R as 0.08206 latm/mol K; make certian V is in liters, P is in atm, and T is in Kelvin!
A] n = PV/RT = 0.987atm x 0.890l/0.08206latm/molK x 294K
= 0.0364 mole
(note: as a check, you can estimate your answer here, as the conditions are not far from STP, and knowing 1 mole occupies 22.4 liters at STP: 0.89/22.4 = 0.397
B] At STP 1 mole is 22.4 liters 1.27mol x 22.4 l/mol = 28.4l
C] PV=nRT; solve for T; T = PV/nR
T = 1.95atm x 12.30l/0.654mol x 0.8206latm/molK = 447 K
I hope this is of help to you.
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