CHEM 100A
Chemistry I
Exam III
Answers
1) Write the full set of four quantum numbers for:
a) The outermost electron in an Rb atom.
n = 5, l = 0, ml = 0, ms = +1/2
b) The electron gained when an S- ion becomes an S2- ion
n = 3, l = 1, ml = 1, ms = -1/2
c) The electron lost when a Cs atom ionizes
n = 6, l = 0, ml = 0, ms = +1/2
d) The electron lost when In ionizes to In+.
n = 5, l = 1, ml = -1, ms = +1/2
(Where acceptable, any other combination of l, ml, and ms was fine.)
2) One reason chemists study excited states of atoms is to gain information about the energies
of orbitals that are unoccupied in the atom’s ground state. Each of the following electron
configurations represents an atom in one of its excited states. Identify the element and write
its ground state electron configuration. You may use the condensed form if you wish.
a) 1s22s22p63s13p1 [Ne]3s2, Mg
b) 1s22s22p63s13p44s1 [Ne]3s23p4, S
c) 1s22s22p63s23p64s23d44p1 [Ar]4s23d5, Mn
d) 1s22s22p53s1 [He]2s22p6, Ne
e) 1s22s22p63s23p64s23d74p1 [Ar]4s23d8, Ni
3) Photochromic sunglasses, which darken when exposed to light, contain a small amount of
colorless AgCl(s) embedded in the glass. When irradiated with light, metallic silver atoms are
produced and the glass darkens: AgCl(s) → Ag(s) + Cl•. Escape of the chlorine atoms is
prevented by the rigid structure of the glass, and the reaction therefore reverses as soon as the
light is removed. If 310 kJ/mol of energy is required to make the forward reaction proceed,
what wavelength of light is necessary?
We need to convert energy from kJ/mol to J/molecule, then we can do the necessary
wavelength calculation. The given energy, 310 kJ/mol = (310000 J/mol)/(6.02 x 1023) = 5.15
x 10-19 J/molecule. This value we can substitute into our equation relating energy and
wavelength:
5.15 x 10-19 J =
(6.6261x10
−34
)(
J ⋅ s 2.9979 x10 8 m / s
λ
) which can be rearranged to
give:
λ=
(6.6261x10
−34
)(
)
J ⋅ s 2.9979 x10 8 m / s
= 3.86 x 10-7 m = 386 nm.
5.15 x10 −19 J
4) One possible use for all the cooking fat left over after making french fries is to burn it as a
fuel. Write the balanced equation for the burning of cooking fat, and then use it, and the
following data to calculate the energy released in kJ/(mL of fat) for the combustion of that fat.
Be careful of states. (Hint: express your states clearly!)
Cooking fat: C51H88O6
Density: 0.94 g/mL
∆H°f = -1310 kJ/mol
As I have said on many occasions, there is not a chemical reaction which is not helped by
writing the balanced equation. Since you needed to do this anyway for the problem, here it is:
∆Hf°
Mm
C51H88O6(l) + 70 O2(g) → 51 CO2(g) + 44 H2O(g) (liquid water was okay too, so long as
you clearly indicated it.)
-1310 kJ
0.0 kJ -393.5 kJ
-241.8 kJ
797.27 g/mol
The ∆Hrxn here is:[51(-393.5 kJ) + 44(-241.8 kJ)] –(-1310 kJ) = -29397.7 kJ/mole fat.
One approach converts g/mole of fat to mL/mole of fat: (797.27 g/mol)/(0.94 g/mL) = 848.16
mL/mole fat. Dividing the heat per mole by the mL per mole gives heat per mL:
− 29397.7kJ / mole
= -34.66 kJ/mL.
848.16mL / mole
A different approach converts kJ/mole of fat to kJ/g of fat, then multiplies by density to get
kJ/mL:
kJ/g =
− 29397.7kJ / mole
= -36.87 kJ/g (0.94 g/mL) = -34.66 kJ/mL fat.
797.27 g / mole
5) Often the alcohol (ethanol) we use in the chemistry lab is denatured i.e. it is mixed with a
certain amount of methanol to make it undrinkable. You take a sample of 100.0 g of
laboratory alcohol and burn it in excess oxygen. From this reaction, you determine that
2643.24 kJ of heat are evolved. Assuming that the alcohol is a mixture of ethanol, C2H6O, and
methanol, CH4O, calculate the percent methanol in the laboratory alcohol. (∆H°f(C2H6O) = 277.63 kJ; ∆H°f(CH4O) = -238.6 kJ) (Hint: treat all products as gases.)
Again, writing the separate, balanced equations gives a great deal of help.
CH4O(l) +
3
2
O2(g) → CO2(g) + 2H2O(g)
∆Hf° -238.6kJ 0.0 kJ -393.5kJ -241.8 kJ
∆Hrxn = (-393.5 +2(-241.8)) –(-238.6) = -638.5 kJ/mole CH4O.
C2H6O(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
∆Hf° -277.63kJ 0.0 kJ -393.5kJ -241.8 kJ
∆Hrxn = [2(-393.5) +3(-241.8)] –(-277.63) = -1234.77 kJ/mole C2H6O.
(Note that you really should not write one big balanced equation for the burning, as that
might cause assumptions to be made which are not necessarily valid!)
Let x = grams CH4O, and (100 – x) = grams C2H6O; then we can write:
x


 100 − x 

 (-638.5 kJ/mol) + 
 (-1234.77 kJ/mole) = -2643.24 kJ (Heat is evolved!)
 32.0424 
 46.0694 
(moles C2H6O)
(moles CH4O)
-19.927x + (-2680.239) + 26.802x = -2643.24 which leads to:
6.875x = 37.05, and x = 5.39 g CH4O and %CH4O = 5.39%.