Question 13.10: Establishing a Molecular Formula with Freezing-Point Data Ni...
Establishing a Molecular Formula with Freezing-Point Data
Nicotine, extracted from tobacco leaves, is a liquid completely miscible with water at temperatures below 60 °C. (a) What is the molality of nicotine in an aqueous solution that starts to freeze at -0.450 °C? (b) If this solution is obtained by dissolving 1.921 g of nicotine in 48.92 g H_2O, what must be the molar mass of nicotine? (c) Combustion analysis shows nicotine to consist of 74.03% C, 8.70% H, and 17.27% N, by mass. What is the molecular formula of nicotine?
Analyze
(a) We can establish the molality of the nicotine by using equation (13.5) with the value of K_f for water listed in Table 13.2. (b) Once we know the molality, we can use the definition of molality, but with a known molality (from part a) and an unknown molar mass of solute (M). (c) To establish the empirical formula of nicotine, we need to use the method of Example 3-5.
\Delta T_f=-K_f \times m (13.5)
| TABLE 13.2 Freezing-Point Depression and Boiling-Point Elevation Constants^a | ||||
| Solvent | Normal Freezing Point,°C | K_f, °C m^{-1} | Normal Boiling Point, °C | K_b, °C m^{-1} |
| Acetic acid | 16.6 | 3.90 | 118 | 3.07 |
| Benzene | 5.53 | 5.12 | 80.10 | 2.53 |
| Nitrobenzene | 5.7 | 8.1 | 210.8 | 5.24 |
| Phenol | 41 | 7.27 | 182 | 3.56 |
| Water | 0.00 | 1.86 | 100.0 | 0.512 |
_{}^{a}\text{Values} correspond to freezing-point depressions and boiling-point elevations, in degrees Celsius, caused by 1 mol of solute particles dissolved in 1 kg of solvent in an ideal solution.
(a) Note that T_f = -0.450 °C, and that \Delta T_f = -0.450 – 0.000 °C = -0.450 °C.
molality = = \frac{\Delta T_f}{-K_f} = \frac{-0.450 ^{\circ} C }{-1.86 ^{\circ} C m ^{-1}} = 0.242 m
(b) Let’s represent the molar mass as x g/mol. The amount, in moles, contained in a 1.921 g sample is 1.921 g × (1 mol/ x g) = (1.921/ x) mol. Thus,
\text { molality } = \frac{1.921 x^{-1}}{0.04892 kg \text { water}} = 0.242 mol ( kg \text { water})^{-1}
x = 162
The molar mass is 162 g/mol.
(c) This calculation is left as an exercise for you to do. The result you should obtain is C_5H_7N. The formula mass based on this empirical formula is 81 u. The molecular mass obtained from the molar mass in part (b) is exactly twice this value−162 u. The molecular formula is twice C_5H_7N or C_{10}H_{14}N_2.
Assess
Using the freezing-point data is another experimental technique that can be used to obtain the chemical properties of a substance. In this example we were able to determine the molecular formula from the freezing-point depression and a known amount of substance dissolved in a solvent. Note that water was used as the solvent here; however, other solvents can be used in freezing-point experiments.