volliquid

Study Problems

Experiment 2:

Formula Weight of a Volatile Liquid

> 1. The TA wrote on the board that the barometric pressure is 1.000±.001 atm today. When you measured the temperature of the water bath, the thermometer showed 100±1 degrees C. Your flask holds 260±.5mL of water, and you calculated that the weight of the condensed liquid was .275±.002g. What is the molecular weight of the compound? (don't forget to calculate the error!) R = .08206 L atm K-1 mol-1

> Solution:

> P = 1.000±.001 atm

> T = 100±1 degrees C = 373±1 K

> V = .2600±.0005 L

> m = .275 ±.002 g

> R = ideal gas constant = .08206 L atm K-1 mol-1

> Now use the ideal gas law. (Note: n is the number of moles of the liquid.)

> PV = nRT

> (1.000atm)(.260 L) = (.08206 L atm mol-1 K-1)(373K)

> MW = 32.4 g/mol

> Now rewrite the equation to facilitate your error calculation.

> MW =

> The relative error in MW is the sum of the relative errors in the measurements that define it. (We need to add relative errors when a calculation involves division or multiplication; when we're finished, we convert to absolute error.) R is the ideal gas constant, so we'll treat it as such. (Of course, it was experimentally determined too.) Constants can enter the error calculation as the last step (see below).

> rel err (MW) = rel err (wt) + rel err (T) + rel err (P) + rel err (V)

> rel err (MW) =

> rel err (MW) = .00727 + .002681 + .001 + .00192 = .012

> Multiply the relative error by our answer, the MW (which includes our constant R), to get back an absolute error.

> abs err = .4g

> Therefore, MW = 32.4±.4g

> 2. 2.000 grams of a volatile liquid A, of molecular weight 60.0 g/mol, were evaporated at 100 degrees C and 760 mm Hg in a device similar to the one you used last week. After condensation, .490g of liquid A were left in the flask.

> a. What is the volume of the flask? (R=.08206 L atm K-1 mol-1. Assume ideal behavior.)

> Solution:

> First, let's calculate how many moles with which we're dealing.

> = .008167 mol

> Now invoke the ideal gas law.

> PV = nRT

> (1 atm)V = (.008167mol)(.08206 L atm mol-1 K-1)(373.15K)

> Remember: The ideal gas law always deals with Kelvin. The rest of our units need to match the version of R which we're using. (Reference tables can provide us with at least 7 values of R expressed in different units, but with equivalent values.) For example, in this case, we've chosen .08206 L atm mol-1 K-1 , so we must convert any measurements given in mL to L.

> V = .25L = 250 mL

> Therefore, the volume of the flask is approximately 300 mL when our answer is rounded to the appropriate number of significant figures.

> b. 2.000g of a volatile liquid B, of unknown molecular weight, were evaporated in the same flask under the same conditions. .638g of liquid were left inside the flask after condensation. What is the molecular weight of B?

> Solution:

> Again, use the ideal gas law:

> PV = nRT

> where n (moles of the liquid) =

> (1 atm)(.250 L) = (.08206 L atm mol-1 K-1)(373.15 K)

> MW = 78.1 g/mol or 80 g/mol (again, rounding to the appropriate number of significant figures)

> 3. Suppose that the ideal gas law were PV2 = nR / T2. What would be the units of R if P is in atm, V in liters, and T in Kelvin?

> Solution:

> First we rewrite this new ideal gas law in terms of R.

> R =

> Now we plug in the appropriate units for each of these terms.

> R =

> 4. A balloon filled with 5.000±.001g of helium gas is immersed in a tank that contains a solvent Q at 283±1 K. The volume of the solvent displaced by the balloon is 3.00±.01L.

> a. Find the pressure of the gas. Assume ideal gas behavior. Don't forget to calculate the absolute error! (Atomic mass of He = 4.00 g/mol, R = .08206 L atm mol-1 K-1)

> Solution:

> Again we rely on the ideal gas law.

> PV = nRT

> Rewrite.

> P =

> P =

> P = 9.67 L

> Now we have to generate an error associated with this calculation. We'll take the AW of He and R to be a constants (obviously they were determined experimentally too once, but now they're used as standards until more accurate values are obtained). If we examine the above equation, we see that we're dealing with only multiplications and divisions; therefore we need to convert the all of our errors from absolute form (which is what we were given) to relative. We then add together all the relative errors as explained above briefly and in the section on uncertainty analysis in the general chemistry manual. Finally, although we needed to convert to relative errors in order to determine the error associated with the multiplication/division calculations, we must convert back to an expression of absolute error when we report our answer.

> rel err (P) = rel err (wt) + rel err (T) + rel err (V)

> rel err (P) = = .0002 + .003534 + .00333 = .007064

> As above, we multiply the relative error by our answer, here P, to get the absolute error in P.

> abs err (P) = .07 atm

> Therefore, P = 9.67 ±.07 atm

> b. If the density of solvent Q is 1.2 g/mL, find the weight of the displaced solvent.

> Solution:

> We were given the weight of the gas in the balloon. Now we need to figure out the weight of the solvent surrounding it.

> d= m / V

> where d = density, m = mass, and V = volume

> 1.2 g/mL =

> m = 3600g

> Two important points of which to take notice. First, we must remember to convert from L to mL here so our units cancel and we're left with an answer in grams alone. Second, we recorded the volume as 3.00x103 to emphasize the number of significant figures that our volume measurement had: three. Had we expressed our volume measurement as 3000 mL, we would have lost this information and probably would not have recorded the final answer with an appropriate number of significant figures.

> 5. In the laboratory a student filled a 250 mL container with an unknown gas until a pressure of 760 mm Hg was obtained. She then found that the sample of gas weighed .164 g. Calculate the molecular weight of the gas if the temperature in the laboratory was 25C.

> Solution:

> Once again, we utilize the ideal gas equation. Also, we recall that 760 mm Hg is 1.0 atm.

> PV = nRT

> (1.0 atm)(.250 L) = (.08206 L atm mol-1 K-1)(298.15 K)

> MW = 16 g / mol