[11%] Compound A contains only carbon, fluorine, and chlorine. — Gas Laws, Titration, Stoichiometry, and Organic Structure Chemistry Question
Problem Context
[11%] Compound A contains only carbon, fluorine, and chlorine.
The vapor density of A at 1.00 bar and 25.0 °C is 6.895 g L-1. What is the molar mass of A?
Model Answer
PV = nRT, so n = PV/RT. 1.000 L of gas will have n = (1.000 bar)(1.000 L)/(298.15 K)(0.08314 L bar mol-1 K-1)
n = 0.04034 mol
molar mass = (6.895 g L-1)/(0.04034 mol L-1) = 170.9 g mol-1
Chlorofluorocarbons such as A react with sodium metal to convert all the chlorine present into sodium chloride. The products from the reaction of sodium with 100.0 mL of gaseous A (at 1.00 bar and 25.0 °C) are dissolved in water and a few drops of sodium chromate solution are added. This solution is then titrated with 0.3540 M AgNO3 solution until a bright red precipitate appears, which requires 22.79 mL of the titrant. How many chlorine atoms are there in a molecule of A?
Model Answer
0.02279 L titrant × 0.3540 mol L-1 = 8.068 × 10-3 mol chloride
100.0 mL gas has n = (1.000 bar)(0.1000 L)/(298.15 K)(0.08314 L bar mol-1 K-1) = 4.034 × 10-3 mol A
(8.068 × 10-3 mol chloride)/(4.034 × 10-3 mol A) = 2.000 mol Cl/mol A
What is the molecular formula of A? Justify your answer.
Model Answer
Since the molar mass of A is 170.9 and there are 2 mol Cl per mol A, the carbons and fluorines in A must have a combined mass of 170.9 – 2(35.45) = 100.0 amu. If A has only one carbon, it must have (100.0 – 12.01)/19.00 = 4.63 F atoms, which is impossible (it must have an integer number of each type of atom per formula unit, plus over six halogens per carbon atom is not chemically reasonable). Two carbons in A would imply (100.0 – 2×12.01)/19.00 = 4.00 atoms of F, which is possible. Each additional carbon in the formula would result in 0.63 fewer F atoms, which will not be especially close to an integer from 3-7 carbon atoms, at which point there would need to be fewer than one F atom. Thus the only possibility is C2F4Cl2.
There is only one known isomer of A, compound B. The two compounds have nearly identical boiling points, but can be distinguished by 19F NMR spectroscopy, a technique that is sensitive to small differences in the chemical environments of fluorine atoms in compounds. When analyzed by this method, compound A exhibits only a single type of chemical environment for fluorine, while compound B shows two distinct environments for its fluorine atoms. Draw structural formulas for compounds A and B that are consistent with these observations.
Model Answer
all F atoms in the same environment (each bonded to a carbon with 2F's and 1 Cl) Isomer A
Two different F environments (one F bonded to a carbon with 2 Cl's, three F's bonded to a C with no Cl's) Isomer B