Names (please print): KEY
CHEM 141L – Chemical Principles I Recitation: IMFs and Stoichiometry
Fall 2012
Chapter 11: IMFs
1. Draw the Lewis structure of each of the following and identify the strongest type of IMFs that exist
in the liquid state of each molecule. [Hint: You must first identify if the molecule is polar or
nonpolar.]
(a) SO2
(b) OF2
(c) PCl5
(d) CH3OH
Polar
Dipole-dipole
Polar
Dipole-dipole
Nonpolar
Dispersion (or LDF)
Polar
Hydrogen-bonding
2. Using IMFs alone, predict which of the molecules above will exhibit the following properties.
(a) Highest vapor pressure – easy to vaporize, meaning weak IMF, so PCl5, with only weak LDF
interactions
(b) Lower boiling point – same reason as in 2a, so PCl5
(c) Highest boiling point – strong IMF, difficult to vaporize and boil, so CH3OH, with strong Hbonding
Chapter 3: Stoichiometry
NOTE: You must show all calculation work in the space provided, and use proper units and significant
figures to get credit.
1. The chemical industry is replacing CFCs with hydrofluorocarbons (HFCs). The best replacement
HFC is called HFC-134a (C2H2F4). HFC-134a is made by the British company ICI using the
unbalanced chemical equation shown below:
C2HCl3
+
4 HF
C2H2F4
+ 3 HCl
(a) Balance the above equation by writing the proper coefficient in each blank.
(b) Assume that a reaction was carried out using 455.3 g of C2HCl3 and 122.1 g of HF. Calculate the
theoretical yield in grams of C2H2F4. First you need to calculate the molar mass of each species in
question.
Molar mass (MM):
C2HCl3 =131.39 g/mol
HF = 20.006 g/mol
C2H2F4 = 102.03 g/mol
WORK:
Calculate yield from each reactant, then pick the smaller of the two yields. The reactant it came from is
the limiting reactant.
g C2H2F4 from C2 HCl3 = 3.465 mol C2HCl3 x
1 mol C2H2F4
x
102.03 g C2H2F4
1 mol C2HCl3
1 mol C2H2F4
(stoichiom. ratio)
= 355.53 g
g C2H2F4 from HF = 6.103 HF x
1 mol C2H2F4
4 mol HF
x
102.03 g C2H2F4
1 mol C2H2F4
= 155.67 g
= 1.526 mol C2H2F4
The limiting reactant is: HF (it gave the lower of the two yields above)
The theoretical yield (in moles and in grams) of C2H2F4 is 1.526 mol C2H2F4
Moles = 1.526
grams = 155.67
Alternate work: (My preferred approach)
Calculate yield (theoretical) from the limiting reactant:
g C2H2F4 from HF = 6.103 HF x
1 mol C2H2F4
4 mol HF
x
102.03 g C2H2F4
= 155.67
1 mol C2H2F4
(c) If you ran this reaction in the lab and obtained 92.78 g of C2H2F4, then what is the % Yield?
WORK:
% yield of C2H2F4 =
actual gram yield
x 100
theoretical gram yield
=
% Yield = 59.60
92.78 g
x 100 = 59.60 %
155.67 g