Luana C.
Ch. 11 & 3 intermolecular force & stoichiometry
1. Using concepts involving intermolecular forces, describe the basis of absorption
chromatography. [10 points]
Due to the difference in adhesiveness, the two dyes in the mobile phase sticks differently to
the stationary phase. The dye that is more adhesive attaches itself thus becoming immobile.
The dye that is less adhesive remains in the mobile phase and passes through the stationary
phase. The more adhesive dye eventually comes out.
2. Describe the basis of London dispersion intermolecular force. [10 points]
London dispersion intermolecular force occurs when a nonpolar molecule becomes
temporary positive on one side and temporary negative thereby forming a temporary polar
molecule which could induce a nonpolar molecule to become a polar molecule. Therefore,
you would have IMF involving these two temporary polar molecules. London IMF also
depends on the polarizability. The more electrons in the compound, the higher likelihood of
forming a temporary dipole since there is a higher probability for an asymmetric electron
distribution. London dispersion intermolecular force is formed due to changes in the
compound’s dipole moment, which could occur in either polar or nonpolar compounds.
3. Vapor pressure versus temperature curve [15 points]
a. sketch a vapor pressure versus temperature curve for methanol, methane, and
ethane. Label each curve with the appropriate chemical.
b. justify / rationalize your answer.
Methanol: polar, contains London IMF and H-bond
Methane: nonpolar, contains London IMF only
Ethane: nonpolar, contains London IMF only
Methanol has the lowest vapor pressure because the IMF strength of Methanol, due to
hydrogen bonding, is the largest.
Larger IMF (greater size)  harder to separate liquid molecules  ↓ evaporation  ↓ gas  lowest vapor pressure
Vapor pressure of nonpolar molecules depends on the size. Since methane is smaller than
ethane:
CH4 = small  ↓ polarizability  ↓ IMF  easier to separate liquid molecules  ↑ evaporation  ↑ gas  highest vapor pressure
4. Balance the following chemical reactions using the lowest ratio of integers and put a “1” as
appropriate; an empty blank will be considered as an error. [15 points]
a. __1__ Cu + __4__ HNO3  __1__ Cu(NO3) 2 + __2__ NO2 + __2__ H2O
b. __1__ S4 + __4__ BCl3 + __6__ H2  __12__ HCl + __4__ BS
c. __2__ CH3(CH2)2CH3 + __13__ O2  __8__ CO2 + __10__ H2O
5. Determine the % yield in the following chemical reaction: [15 points]
2.5 g CH3CH3 + excess O2  CO2 + 2.5 g H2O
2𝐶𝐻3 𝐶𝐻3 + 7𝑂2 → 4𝐶𝑂2 + 6𝐻2 𝑂
The limiting reactant is CH3CH3, since there are excess O2.
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝐻3 𝐶𝐻3 = 2(12 + 3) = 30𝑔/𝑚𝑜𝑙𝑒
1 𝑚𝑜𝑙𝑒 𝐶𝐻3 𝐶𝐻3
2.5 𝑔 𝐶𝐻3 𝐶𝐻3 𝑥
30 𝑔 𝐶𝐻3 𝐶𝐻3
= 0.083 𝑚𝑜𝑙𝑒𝑠 𝐶𝐻3 𝐶𝐻3
6 𝑚𝑜𝑙𝑒𝑠 𝐻 𝑂
18 𝑔 𝐻 𝑂
0.083 𝑚𝑜𝑙𝑒𝑠 𝐶𝐻3 𝐶𝐻3 × 2 𝑚𝑜𝑙𝑒𝑠 𝐶𝐻 2𝐶𝐻 × 1 𝑚𝑜𝑙𝑒 𝐻2
3
3
2𝑂
= 4.5 𝑔 𝐻2 𝑂
𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
% 𝑦𝑖𝑒𝑙𝑑 = 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 × 100%
2.5 𝑔 𝐻 𝑂
= 4.5 𝑔 𝐻2 𝑂 × 100%
2
= 56% 𝑦𝑖𝑒𝑙𝑑 𝑜𝑓 𝐻2 𝑂
6. A compound contains:
32.73 % carbon
63.64 % chlorine
3.64 % hydrogen
Determine the empirical formula of the compound. [15 points]
Assume there is 100 g sample.
1 𝑚𝑜𝑙𝑒 𝐶
𝑀𝑜𝑙𝑒𝑠 𝐶 = 32.73 𝑔 𝐶 × 12 𝑔 𝐶 = 2.73 𝑚𝑜𝑙𝑒 𝐶
1 𝑚𝑜𝑙𝑒 𝐶𝑙
𝑀𝑜𝑙𝑒𝑠 𝐶𝑙 = 63.64 𝑔 𝐶𝑙 × 35.45 𝑔 𝐶𝑙 = 1.80 𝑚𝑜𝑙𝑒 𝐶𝑙
𝑀𝑜𝑙𝑒𝑠 𝐻 = 3.64 𝑔 𝐻 ×
1 𝑚𝑜𝑙𝑒 𝐻
1𝑔𝐻
= 3.64 𝑚𝑜𝑙𝑒 𝐻
Carbon : Chlorine : Hydrogen
2.73 :
1.8 :
3.64
1.5 :
1 :
2 b/c 𝐶:
3 :
2 :
4
The empirical formula is C3H4Cl2
2.73
1.8
= 1.5
𝐶𝑙:
1.8
1.8
=1
𝐻:
3.64
1.8
=2