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1991 2)
Answer the following questions about a pure compound that contains only carbon, hydrogen, and oxygen.

: The molecular formula of a hydrocarbon is to be determined by analyzing its combustion products and
investigating its colligative properties.

(a) The hydrocarbon burns completely, producing 7.2 grams of water and 7.2 liters of CO2 at standard

(b) Calculate the mass in grams of O2 required for the complete combustion of the sample of the
hydrocarbon described in (a).

(c) The hydrocarbon dissolves readily in CHCl3. The freezing point of a solution prepared by mixing 100.
grams of CHCl3 and 0.600 gram of the hydrocarbon is -64.0 °C. The molal freezing-point depression
constant of CHCl3 is 4.68 °C / molal and its normal freezing point is -63.5 °C. Calculate the molecular
weight of the hydrocarbon.

(d) What is the molecular formula of the hydrocarbon?

1992 : 8 (d) use appropriate chemical principles to explain the why NaI(s) is very soluble in water
whereas I2(s) has a solubility of only 0.03 gram per 100 grams of water

1994 8a:, use appropriate chemical principles to explain the why sodium chloride may be spread on an
icy sidewalk in order to melt the ice; equimolar amounts of calcium chloride are even more effective.

1994 Discuss the following phenomena in terms of the chemical and physical properties of the substances
involved and general principles of chemical and physical

(a) As the system shown on the right approaches equilibrium,
what change occurs to the volume of water in beaker A ? What
happens to the concentration of the sugar solution in beaker B ?
Explain why these changes occur.

(b) A bell jar connected to a vacuum pump is shown on the right.
As the air pressure under the bell jar decreases, what behavior of
water in the beaker will be observed? Explain why this occurs.

(d) A water solution of I2 is shaken with an equal volume of a nonpolar solvent such as TTE
(trichlorotrifluoroethane). Describe the appearance of this system after shaking. (A diagram may be
helpful.) Account for this observation.

1996 4) Concentrated sulfuric acid (18.4-molar H2SO4) has a density of 1.84 grams per milliliter. After
dilution with water to 5.20-molar, the solution has a density of 1.38 grams per milliliter and can be used
as an electrolyte in lead storage batteries for automobiles.
(a) Calculate the volume of concentrated acid required to prepare 1.00 liter of 5.20-molar H2SO4.

(b) Determine the mass percent of H2SO4 in the original concentrated solution.

(c) Calculate the volume of 5.20-molar H2SO4 that can be completely neutralized with 10.5 grams of
sodium bicarbonate NaHCO3.

(d) What is the molality of the 5.20-molar H2SO4 ?

1998) An unknown compound contains only the three elements C,H, and O. A pure sample of the
compound is analyzed and found to be 65.60 percent C and 9.44 percent H by mass.

(a) Determine the empirical formula of the compound.

(b) A solution of 1.570 grams of the compound in 16.08 grams of camphor is observed to freeze
at a temperature 15.2 Celsius degrees below the normal freezing point of pure camphor.
Determine the molar mass and apparent molecular formula of the compound. (The molal
freezing-point depression constant, Kf, for camphor is 40.0 kg-K-mol¯1.)

(c) When 1.570 grams of the compound is vaporized at 300 °C and 1.00 atmosphere, the gas
occupies a volume of 577 milliliters. What is the molar mass of the compound based on this

(d) Briefly describe what occurs in solution that accounts for the difference between the results
obtained in parts (b) and (c).

1999) 7
Answer the following questions, which refer to the 100 mL samples of aqueous solutions at
                                                              2  ,0.10 M C 2 H 5 OH, and
0.10 M CH 3 COOH

(a) Which solution has the lowest electrical conductivity? Explain.
(b) Which solution has the lowest freezing point? Explain.
(c) Above which solution is the pressure of water vapor greatest? Explain.
(d) Which solution has the highest pH? Explain.

2005 form a)

(a) A 0.7549 g sample of the compound burns in O2(g) to produce 1.9061 g of CO2(g) and
0.3370 g of H2O(g).
(i) Calculate the individual masses of C, H, and O in the 0.7549 g sample.
(ii) Determine the empirical formula for the compound.

(b) A 0.5246 g sample of the compound was dissolved in 10.0012 g of lauric acid, and it was determined
Assume that the compound does not dissociate in lauric acid.
(i) Calculate the molality of the compound dissolved in the lauric acid.
(ii) Calculate the molar mass of the compound from the information provided.

(c) Without doing any calculations, explain how to determine the molecular formula of the compound
based on the answers to parts (a)(ii) and (b)(ii).
(d) Further tests indicate that a 0.10 M aqueous solution of the compound has a pH of 2.6 . Identify the
organic functional group that accounts for this pH.

2003 a)

a) When table salt (NaCl) and sugar (C12H22O11) are dissolved in water, it is observed that
(i) both solutions have higher boiling points than pure water, and
(ii) the boiling point of 0.10 M NaCl(aq) is higher than that of 0.10 M C12H22O11(aq).

(b) Water droplets form on the outside of a beaker containing an ice bath.


(d) Ammonia, NH3 , is very soluble in water, whereas phosphine, PH3 , is only moderately soluble in

2004 b)
3. Hydrogen peroxide decomposes according to the equation above.
(a) An aqueous solution of H2O2 that is 6.00 percent H2O2 by mass has a density of 1.03 g mL–1.
Calculateeach of the following.
(i) The original number of moles of H2O2 in a 125 mL sample of the 6.00 percent H2O2 solution
(ii) The number of moles of O2(g) that are produced when all of the H2O2 in the 125 mL sample
1991 answer)
a) three points
7.2 g H2O ÷ 18.0 g/mol = 0.40 mol H2O
0.40 mol H2O x (2 mol H / 1 mol H2O) = 0.80 mol H
7.2 L CO2 ÷ 22.4 L/mol = 0.32 mol CO2
0.32 mol CO2 x (1 mol C / 1 mol CO2) = 0.32 mol C
n = PV ÷ RT = [(1 atm) (7.2 L)] ÷ [(0.0821 L atm mol¯1 K1) (273 K)] = 0.32 mol CO2
0.80 mol H ÷ 0.32 = 2.5
0.32 mol C ÷ 0.32 = 1
2.5 x 2 = 5 mol H
1 x 2 = 2 mol C
empirical formula = C2H5

b) two points
mol O2 for combustion = mol CO2 + 1/2 mol H2O = 0.32 + 0.20 = 0.52 mol O2
0.52 mol O2 x 32 g/mol = 17 g O2
alternate approach for mol O2 from balanced equation
C2H5 + 13/4 O2 ---> 2 CO2 + 5/2 H2O
other ratio examples:
1, 6.5 ---> 4, 5
0.25, 1.625 ---> 1, 1.25
mol O2 = 0.40 mol H2O x (13/4 mol O2 / 5/2 mol H2O) = 0.52 mol O2
Note: starting moles of C2H5 = 0.16 mol C2H5

c) three points
MM stands for molar mass.
         f (g/MM)) / kg of solvent
0.5 °:C = ((4.68 °:C kg mol¯:1) x (0.60 g / MM)) / 0.1 kg
MM = (4.68 x 0.60) / (0.5 x 0.1) = 56 or 6 x 101
an alternate solution for (c)
molality = 0.5 °:C / (4.68 0.5 °:C/m) = 0.107 m
mol solute =( 0.107 mol / kg solvent) x 0.100 kg solvent = 0.0107 mol
MM = 0.60 g / 0.0107 mol = 56 or 6 x 101

d) one point
(56 g/mol of cmpd) / (29 g/mol of empirical formula) = 1.9 empirical formula per mol
6 x 101 / 29 = 2.1
empirical formula times 2 equals molecular formula = C4H10

1994) two points

Volume decreases in beaker A; the concentration decreases in beaker B (either obs. earns one point
provided the other is not wrong.)

The vapor pressure of pure H2O is greater than the vapor pressure of H2O in solution.


The rate of evaporation of H2O molecules from pure H2O is greater than that from the sugar solution,
while the condensation rates are the same.
b) two points

The water will begin to boil (or evaporate).

The external pressure on the water will become equal to the vapor pressure of the water, causing it to boil.


The drop in external pressure causes the boiling point to drop to the temperature of the water.

d) two points

Two layers will form, one of which is colored. Iodine is nonpolar and will dissolve in TTE. Water is polar
and will not dissolve in TTE.

Placment of I2 must be correctly indicated for second point.

1996: 4)

(a) one point

V1M1 = V2M2

(1.00L) (5.20 mol/L) = (x) (18.4 mol/L)

x = 5.2 mol / (18.4 mol/L) = 0.283 L (or 283 mL)

(b) two points

mass 1 liter of concentrated H2SO4 = 1 L x (1.84 g/mL) x (1,000 ml/L) = 1,840 g H2SO4

18.4 mol H2SO4 x 98.1 g/mol = 1,805 g H2SO4

mass percent H2SO4 = (1,805 g / 1,840 g) x 100 = 98.1%

(c) three points

Stoichiometric ratio of NaHCO3 to H2SO4 = 2:1

10.5 g NaHCO3 x (1 mol NaHCO3 / 84.0 g NaHCO3) = 0.125 mol NaHCO3

Since 1 mol H2SO4 reacts with 2 mol NaHCO3, 0.125 mol NaHCO3 reacts with 0.0625 mol H2SO4

0.0625 mol H2SO4 = V x M = (V) (5.20 M)
V = 0.0625 mol / (5.20 mol/L) = 0.0120 L (or 12.0 mL)

(d) three points

molality = moles solute / 1,000 g solvent = moles solute / 1 kg solvent

mass of 1 L of 5.20 M H2SO4 = 1 L x (1,000 mL / 1 L) x (1.38 g 1 mL) = 1,380 g

mass of H2SO4 in 1 L = (5.20 mol/L) (98.1 g.mol) = 510 g

mass of H2O in 1 L = 1,380 - 510 = 870 g

molality = (5.20 mol H2SO4 / 870 g) x (1,000 g / 1 kg) = 5.98 m

Note: no credit earned for 5.20 mol / 1.38 kg = 5.77 m


(a) Assume a 100 gram sample ( not necessary for credit ):

65.60g C x (1 mol C / 12.01 g C) = 5.462 mol C

9.44g H x (1 mol H / 1.0079 g H) = 9.366 mol H

mass O = [100 - (65.60 + 9.44)] = 24.96 g O

24.96 g O x (1 mol O / 15.9994 g O) = 1.560 mol O

C5.462H9.366O1.560 ---> C3.5H6.0O1.0 ---> C7H12O2

        One point earned for determining moles of C and moles of H
        One point earned for determining moles of O
        One point earned for correct empirical formula

                   f   = 15.2 °C /40.0 K kg mol¯1 = 0.380 mol / kg

0.01608 kg x (0.380 mo / 1 kg) = 0.00611 mol

molar mass = 1.570 g/ 0.00611 mol = 257 g / mol

        One point earned for determination of molarity
        One point earned for conversion of molarity to molar mass


                                          f   = 0.00611 mol (one point)

molar mass = 1.570 g / 0.00611 mol = 257 g / mol (one point)


molar mass = (mass x Kf
empirical mass of C7H12O2 = 7(12) + 12(1) + 2(16) = 128 g/mol

128 g/mol = 1/2 molar mass ---> molecular formula = 2x ( empirical formula) -----> molecular formula =
C14H24O4 (one point)

        One point earned if molecular formula is wrong but is consistent with empirical formula
        and molar mass
        No penalty for simply ignoring the van't Hoff factor
        Only one point earned for part (b) if response indicates that
        and molar mass = 13.6 g / mol

(c) n = (pV) / (RT) = [(1 atm) (0.577 L)] / [(0.0821 L atm mol°1 K°1) (573 K)] = 0.0123 mol (one point)

molar mass = mass of sample / moles in sample = 1.570 g / 0.0123 mol = 128 g/mol (one point)

        Only one point can be earned for part (c) if wrong value for R is used and/or T is not
        converted from C to K

(d) The compound must form a dimer in solution, because the molar mass in solution is twice that it is in
the gas phase,


the compound must dissociate in the gas phase ( A (g) --> 2B (g)) because the molar mass in the gas
phase is half that it is in solution.

One point earned for a reference to either or both the ideas of dimerization and dissociation,br> -No point
earned for a " non - ideal behavior " argument


        8 points:

            a. 1 point

                C2H5OH (Flask #3)

                1 point

                Ethanol, a nonelectrolyte, does not break up or dissociate in solution.

                    o     One point earned for identifying C2H5OH
                    o     One point earned for the correct explanation.
                    o     Explanation point earned for a description of a nonelectrolyte (e.g.,
                          something that does not break up or does not dissociate.)
                    o     No point earned for describing C2H5OH as organic, or as the compound
                          that contains the most hydrogens.

            b. 1 point
              MgCl2 (Flask #2)

              1 point

              The freezing-point depression is proportional to the concentration of solute
              particles. All solutes are at the same concentration, but the van't Hoff factor (i) is
              largest for MgCl2.

                  o     One point earned for identifying MgCl2 .
                  o     One point earned for the correct explanation.

           c. 1 point

              C2H5OH (Flask #3)

              1 point

              The lowering of vapor pressure of water is directly proportional to the
              concentration of solute particles in solution. C2H5OH is the only nonelectrolyte,
              so it will have the fewest solute particles in solution.

                  o     One point earned for identifying C2H5OH .
                  o     One point earned for the correct explanation.

           d. 1 point

              NaF (Flask #1)

              1 point

              The F – ion, generated upon dissolution of NaF, is a weak base. It is the only
              solution with pH > 7.

                  o     One point earned for identifying NaF.
                  o     One point earned for the correct explanation.


A) Mass of C =.5202g                  Mass of H= 0.03771 g H Mass of oxygen = 0.1970 g O
B) The empirical formula is C 7 H 6 O 2 .
C) i) 0.431 molal ii) 122 g mol –1
d) Since an aqueous solution of the compound is acidic, the compound must
be an organic acid. The functional group in an organic acid is the
carboxyl group – COOH.

2004) 0.227 mol H 2 O 2 , 0.114 mol O (g)

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