Problems #1-10

Return to the boiling point elevation discussion

**Problem #1:** What is the molecular mass of an organic compound if 16.00 g of the compound is dissolved in 225.0 g of carbon tetrachloride raises the boiling point to 85.36 °C?

**Solution:**

1) We need to look up the ebullioscopic constant of CCl_{4} as well as its boiling point:

ebullioscopic constant: 5.03 °C kg mol¯^{1}

boiling point: 76.72 °CIn addition:

85.36 − 76.72 = 8.64 °C <--- that's the Δt

2) We now utilize this formula:

Δt = i K_{b}m8.64 °C = (1) (5.03 °C kg mol¯

^{1}) (x / 0.2250 kg)8.64 °C = (22.36 °C mol¯

^{1}) (x)x = 0.3864 mol

3) Calculate the molar mass:

16.00 g / 0.3864 mol = 41.4 g/mol

**Problem #2:** A solution of 10.0 g of a nonvolatile, nondissociating compound dissolved in 0.200 kg of benzene boils at 81.2 °C. Calculate the molecular weight of the compound.

**Solution:**

1) The equation to use is this:

Δt = i K_{b}m

2) Look up necessary values:

boiling point of pure benzene: 80.1 °C

ebullioscopic constant for benzene: 2.53 °C/m

3) Insert values into equation:

1.1 °C = (1) (2.53 °C/m) (x / 0.200 kg)x = 0.0869565 mol

Remember that molality is moles solute / kilograms solvent. The kg will cancel as will the °C, leaving moles in the numerator.

We know the van 't Hoff factor is 1 from the problem stating that the compound id nondissociating. By the way, nonvolatile tells us that the solute is not present in the vapor above the solution. This tells us that all the compound remains in solution and our calculation for 'x' gives us the true amount of moles.

4) Calculate molecular weight:

10.0 g / 0.0869565 mol = 115 g/mol

**Problem #3:** Seawater is about 3.5% (by weight) dissolved solids, almost all of which is NaCl. Calculate the normal boiling point of seawater.

**Solution**

Some comments prior to starting:

a) 3.5% means 3.5 grams of solids per 100 grams total of solution. So that means 3.5 grams of solids are dissolved in 96.5 grams of water.

c) since we know most of the solids are NaCl, we can't go too wrong is we assume NaCl is 100%. Some error is introduced, but not too much.

1) Calculate moles of NaCl:

3.5 g / 58.5 g mol¯^{1}= 0.0598 mol

2) Calculate molality of NaCl:

m = 0.0598 mol / 0.0965 kg = 0.612 m

3) use boiling point elevation constant:

ΔT = i K_{b}mx = (2) (0.52 °C m¯

^{1}) (0.612 m) = 0.64 °CSo, the water boils at 100.64 °C

**Problem #4:** What is the new boiling point of a solution prepared by adding 96.0 g of sodium acetate to 383 mL of water? The boiling point constant for water is 0.52 °C/m.

**Solution:**

1) Determine the molality of the NaCl solution:

(96.0 g / 58.443 g/mol) / 0.383 kg = 4.28884 m

2) We utilize this formula:

Δt = i K_{b}mx = (2) (0.52 °C kg mol¯

^{1}) (4.28884 mol/kg)x = 4.46 °C

The new boiling point is 104.46 °C.

Note the use of the theoretical van 't Hoff factor of 2 for NaCl, as opposed to the experimentally determined value of 1.8, one which takes ion-pairing into account.

**Problem #5:** What is the boiling point elevation when 147 g of lactic acid (C_{6}H_{10}O_{5}) is dissolved in 647 g of cyclohexane (C_{6}H_{12})? The boiling point constant for cyclohexane is 2.79 °C/m.

**Solution:**

1) Determine the molality of the lactic acid solution:

(147 g / 162.14 g/mol) / 0.647 kg = 1.40127 m

2) We utilize this formula:

Δt = i K_{b}mx = (1) (2.79 °C kg mol¯

^{1}) (1.40127 mol/kg)x = 3.91 °C

**Problem #6:** How many grams of fucose (C_{6}H_{12}O_{5}) must be dissolved in 419 g of benzene to raise the boiling point by 3.8 °C? The boiling point constant for benzene is 2.67 °C/m.

**Solution:**

1) Determine moles of fucose:

Δt = i K_{b}m3.8 °C = (1) (2.67 °C kg mol¯

^{1}) (x/0.419 kg)3.8 °C = (6.372315 °C mol¯

^{1}) (x)x = 0.59633 mol

2) Determine grams of fuctose:

(0.59633 mol) (164.1565 g/mol) = 97.9 g (to three sig figs)

Note that the sugar involved is fucose, not fructose.

**Problem #7:** What is the boiling point of a solution prepared by adding 29.3 g of methol (C_{10}H_{20}O) to 0.0590 kg of chloroform? The boiling point of pure chloroform is 61.2 °C. The boiling point constant for chloroform is 3.85 °C/m

**Solution:**

1) Determine the molality of the menthol solution:

(29.3 g / 156.267 g/mol) / 0.0590 kg = 3.17796 m

2) We utilize this formula:

Δt = i K_{b}mx = (1) (3.85 °C kg mol¯

^{1}) (3.17796 mol/kg)x = 12.2 °C

61.2 + 12.2 = 73.4 °C <--- this is the bp of the solution

**Problem #8:** A solution was prepared by dissolving some acetamide, CH_{2}CONH_{2}, in 50.0 g
of pure water. The boiling point of the solution at 1 atm was 100.208 °C. How much acetamide, in grams, was dissolved to yield the solution.

Comment on question: note that the ebullioscopic constant for water is not given. This is sometimes done because the constants for water and other substances are widely available in reference sources.

**Solution:**

1) Determine the moles of acetamide dissolved:

Δt = i K_{b}m0.208 °C = (1) (0.52 °C kg mol¯

^{1}) (x/0.0500 kg)0.208 °C = (10.4 °C mol¯

^{1}) (x)x = 0.020 mol

2) Determine grams of acetamide:

(0.020 mol) (58.0596 g/mol) = 1.16 g (to three sig figs)

**Problem #9:** At a barometric pressure of 744 mmHg, the boiling point of water is 99.4 °C. What approximate mass of NaCl should be added to 3.50 kg of the boiling water to raise the boiling point to 100.0 °C?

**Solution:**

1) Calculate projected bp elevation:

100.0 − 99.4 = 0.6 °C

2) Determine moles of NaCl required:

Δt = i K_{b}m0.6 °C = (2) (0.52 °C kg mol¯

^{1}) (x/3.50 kg)0.6 °C = (0.297143 °C mol¯

^{1}) (x)Note: I used the theoretical value for sodium chloride's van 't Hoff factor.

x = 2.01923 mol

3) Determine grams of NaCl:

(2.01923 mol) (58.443 g/mol) = 118 g

**Problem #10:** 2.60 grams of a compound know to contain only indium and chlorine is dissolved in 50.0 g of tin(IV) chloride (K_{b} = 9.43 °C kg mol¯^{1}). The normal boilng point is raised from 114.1 °C for pure SnCl_{4} to 116.3 °C for the solution. What is the molecular weight and probable molecular formula for the solute?

**Solution**

1) Keep in mind that the units on molecular weight are grams per mole. We already have the grams, what we now need are the moles that 2.60 grams represent. We can get that by the molality (moles solute / kg solvent):

ΔT = i K_{b}m2.2 = 9.42 (x / 0.050); x = 0.01168 mol

2) Now we calculate the molecular weight:

2.60 g / 0.01168 mol = 222.6 g/mol

The probable molecular formula is InCl_{3}, but I will let you ponder on how I did it.

**Bonus Problem #1:** 5.00 g of an organic solid is dissolved in 100.0 g of benzene. The boiling temperature of this solution is 82.42 °C. The organic compound is 15.72% nitrogen, 7.92% hydrogen, 35.92% oxygen and the remainder is carbon. The boiling temperature of pure benzene is 80.1 °C; K_{b} = 2.53 °C kg/mol.

a. Determine the molecular weight of the organic solid.

b. Determine the molecular formula of the solid

c. Determine the mole fraction of the organic solid in the solution

d. If the density of this solution is 0.8989 g/mL , calculate the molarity of the solution

**Solution:**

1) Determine moles of solute using boiling point elevation data:

Δt = i K_{b}m1.42 °C = (1) (2.53 °C kg mol¯

^{1}) (x / 0.100 kg)1.42 °C = (25.3 °C mol¯

^{1}) (x)x = 0.0561265 mol

2) Use moles and 5.00 g to determine molecular weight (answer to part a):

5.00 g / 0.0561265 mol = 89.0845 g/mol89.1 g/mol (to 3 sig figs)

3) Use an on-line empirical formula calculator to determine the empirical formula from the percent composition:

C_{3}H_{7}NO_{2}

4) Determine the molecular formula (answer to part b):

the "empirical formula weight" of C_{3}H_{7}NO_{2}is 89.1 gthe molecular weight divided by the "EFW" yields 1

therefore, the molecular formula is:

Cthe same as the empirical formula_{3}H_{7}NO_{2}

5) Determine the mole fraction of the solute (answer to part c):

moles of solvent = 100.0 g / 78.1134 g/mol = 1.28019 molχ

_{solute}= 0.0561265 mol / (0.0561265 mol + 1.28019 mol)χ

_{solute}= 0.042

6) Determine molarity of solution (answer to part d):

i) our solution weighs 105.0 g; calculate its volume:0.8989 g/mL = 105.0 g / xii) calculate the molarity:x = 116.81 mL

0.0561265 mol / 0.11681 L = 0.48 M

**Bonus Problem #2:** A 0.255 g sample of an orange nonelectrolyte with the empirical formula of C_{10}H_{10}Fe is added to 11.12 g of benzene. The resulting boiling point is 80.26 °C. Boiling point of benzene is 80.10 °C and ebullioscopic constant of benzene is 2.53 °C/m. What is the molar mass and molecular formula of the orange compound?

**Solution:**

1) Determine the change in boiling point:

80.26 − 80.10 = 0.16 °C

2) We now utilize this formula:

Δt = i K_{b}m0.16 °C = (1) (2.53 °C kg mol¯

^{1}) (x / 0.01112 kg)0.16 °C = (227.52 °C mol¯

^{1}) (x)x = 0.000703235 mol

3) Calculate the molar mass:

0.255 g / 0.000703235 mol = 362.6 g/mol

4) Determine the weight of C_{10}H_{10}Fe, leading to the molecular formula:

186.034362.6 / 186.034 = 1.95

The molecular formula is C

_{20}H_{20}Fe_{2}

By the way, C_{10}H_{10}Fe is known as ferrocene. I did not know it could dimerize.

**Bonus Problem #3:** An organic compound contained: C, 18.3%; H, 0.51%; Br, 81.2%. When 0.793 g of the compound was dissolved in 14.80 mL of chloroform
(density = 1.485 g/mL), the solution boiled at 60.63 °C. Pure
chloroform boils at 60.30 °C and has K_{b} = 3.63 °C kg mol¯^{1}. What is the molecular formula of the organic compound?

**Solution:**

1) Kilograms of chloroform:

(1.485 g/mL) (14.80 mL) = 21.978 g = 0.021978 kg

2) Determine moles of the compound dissolved:

Δt = i K_{b}m0.33 °C = (1) (3.63 °C kg mol¯

^{1}) (x / 0.021978 kg)0.33 °C = (165.165 °C mol¯

^{1}) (x)Note: I used the value 165.165165165 (repeating) that was on the calculator to obtain the answer just below.

x = 0.001998 mol

3) Determine molecular weight of the compound:

0.793 g / 0.001998 mol = 396.9 g/mol

4) Determine empirical formula:

Assume 100 g of compound present. Now, determine moles of each element present:C ---> 18.3 g / 12.011 g/mol = 1.5236

H ---> 0.51 g / 1.008 g/mol = 0.5059524 mol

Br ---> 81.2 g / 79.904 g/mol = 1.01622 molDivide through by smallest:

C ---> 1.5236 mol / 0.5059524 mol = 3

H ---> 0.5059524 mol / 0.5059524 mol = 1

Br ---> 1.01622 mol / 0.5059524 mol = 2C

_{3}HBr_{2}

5) Determine molecular formula:

C_{3}HBr_{2}weighs 196.85396.9 / 196.85 = 2

C

_{6}H_{2}Br_{4}