February 28, 2006
Science A52 - Section 2 Handout
Molecular weight:
The molecular weight of a substance (abbreviated as MW) is the mass of one
mole of that substance. A mole is an arbitrary unit consisting of 6.02214199x1023
molecules. The number of molecules per mole is a convention adopted by chemists to
simplify equations and avoid carrying around 1023’s!
Molecular weights of some common elements:
Oxygen
16 grams/mole
Carbon
12 grams/mole
Hydrogen
1 gram/mole
Question: Find the molecular weight of C6 H12O6 :
MWC6 H12O6  6( MWCarbon )  12( MWHydrogen )  6( MWOxygen )
MWC6 H12O6  6(12 grams
)  12(1 grams
)  6(16 grams
)  180 grams
mole
mole
mole
mole
Molecular weight and moles:
The number of moles of a substance is given by the mass of a substance divided
by the molecular mass of the substance:
M subst .
N moles 
MWsubst .
The molecular weight can be used to convert grams to moles and moles to grams of an
element or compound. The molecular weight is analogous to the mass density where you
can go from mass to volume to mass.
Question: If you have 450 grams of C6 H12O6 , how many moles do you have?
M subst .
450grams
N moles 

 2.5moles
MWsubst . 180 grams
mole
Balancing equations:
A chemical equation has the general form: Reactants  Products.
We balance equations to ensure that the same number of elements appear on the reactant
side as the product side (we don’t want mass to inexplicably appear or disappear). A
balanced equation gives us some intuition about the reaction in that it is possible to
determine the limiting reactants and the amount of products.
Question: Balance the photosynthesis equation
a1CO2  a2 H 2O  a3C6 H12O6  a4O2
A simple approach to balancing follows:
Step 1: Start with the most complex molecule (in this case C6 H12O6 ) and set it’s
coefficient to 1 (a3=1).
Step 2: Next balance elements that only appear once on each side of the reactant side.
Balancing carbon gives a1=6. Balancing hydrogen gives a2=6.
Step 3: Next, balance oxygen. We find a4=6
Step 4: Put it all together. We find 6(CO2 )  6( H 2O)  1(C6 H12O6 )  6(O2 )
Oxidation / Reduction:
A method has been developed to quantify the oxidation / reduction state of
molecules. This convention assigns an oxidation number of -2 to oxygen atoms ( oxygen
carries a charge of -2 if it gains 2 electrons) and an oxidation number of +1 to hydrogen
atoms ( hydrogen carries a charge of +1 if it loses its electron ). The oxidation state of a
molecule is calculated by subtracting the sum of oxygen and hydrogen charges from the
overall charge of the particle. Here are two examples, first an uncharged molecule
(methane) and then a charged molecule (carbonate):
For methane:
C  4( H )  0
C40
CH4:
C  4
So we see the oxidation state of methane is -4.
For carbonate:
C  3(O)  2
2-
CO3 :
C  3(2)  2
C4
So we see the oxidation state of carbonate is +4.
Acids, Bases and pH:
The atmosphere contains many soluble gases. An acid is characterized by a
molecule containing hydrogen that readily dissociates when dissolved in water. This
dissociated hydrogen is released as a positively charged proton.
Here are some common acids and their dissociation:
H 2 SO4  H  (aq )  HSO4
(sulfuric acid)
HNO3  H  (aq)  NO3
(nitric acid)
Bases readily absorb protons in solution. Ammonia is a common example:
NH 3  H  (aq )  NH 4
Water can serve as both an acid and base H 2O  H  (aq)  OH  (aq)
The pH scale is a way of keeping track of H  (aq) and OH  (aq) concentrations in
solution. The pH scale is defined by the following equations:
[ H  (aq)][OH  (aq)]  1014
moles
2
liter

pH   log10 ([ H (aq)])
Manipulating the equations above we find:
[ H  (aq)]  10 pH
moles
liter
[OH  (aq)]  10 pH 14
moles
liter
Practice Problems:
[1] Coal-fired power plant:
A coal fired power plant uses 10,000 kg/hr of 3.00% sulfur coal. The sulfur is
oxidized by O2 as follows: S  O2  SO2 . How much SO2 does the plant produce per
hour in mass and moles? (Hint: the molecular mass of sulfur is 32 grams / mole)
First, let’s calculate mass of sulfur consumed:
kg
kg
M sulfur  .03*10,000 hour
 300 hour
From the balanced equation above we see that one mole of sulfur is consumed for each
mole of sulfur dioxide produced:
 MWsulfur _ dioxide 
M sulfur _ dioxide  M sulfur 

 MW
sulfur


grams
grams
MWsulfur _ dioxide  1(32 mole )  2(16 mole )  64 grams
mole
M sulfur _ dioxide  300
kg
hour
 64 grams

kg
mole
 grams   600 hour
 32 mole 
Next, let’s calculate the number of moles produced per hour. Notice that the number of
moles of sulfur consumed is equal to the number of moles sulfur dioxide produced:
kg
grams
300 hour
* 1000
M sulfur
kg
N sulfur _ dioxide 

 9375 moles
hour
MWsulfur
32 grams
mole
[2] Combustion of Octane:
When you drive around in a petroleum fueled car, you are tapping the chemical
energy of the hydrocarbons by oxidizing them to carbon dioxide and water. We’d like to
quantify the oxidation of octane in a combustion process.
Octane combustion is given by:
a1C8 H18  a2O2  a3CO2  a4 H 2O
(1) Balance the equation by finding appropriate values for the coefficients a1-4:
The best way to go about doing this is to set a value for moles of octane. Let’s start with
1 mole and see what we’ve got. Taking a1=1, we find a3=8 and a4=9 to balance the
carbon and hydrogen respectively. Accounting for the oxygen, we see that we will need
25 moles, so a2=12.5. So we’ve got:
1(C8 H18 )  252 (O2 )  8(CO2 )  9( H 2O)
However, it’s common practice to express all coefficients as integers, so we multiply
both sides by 2 to get:
2(C8 H18 )  25(O2 )  16(CO2 )  18( H 2O)
Both expressions are correct, but you will see integer expressions more frequently in the
literature.
(2) Calculate the initial oxidation state of octane (a1*C8H18) and the final oxidation state
of carbon dioxide (a2*CO2).
For Octane we find:
C  18( H )  0
C  18  0
C  18
Noxidation  a1 * C  36
For CO2 we find:
C  2(O)  0
C  2(2)  0
C4
Noxidation  a3 * C  32
[3] The ocean has a pH of 8.1. We’d like to calculate the concentrations of [ H  (aq)]
and [OH  (aq)] .
(a) Calculate the concentration of [ H  (aq)] and [OH  (aq)] .
From above:
[ H  (aq)]  10 pH
moles
liter
 108.1
moles
liter
 7.94*109
moles
liter
[OH  (aq)]  10 pH 14
moles
liter
 105.9
moles
liter
 1.26*106
moles
liter
(b) Is the ocean acidic or basic?
The concentration of [OH  (aq)] is more than 100 times greater than the
concentration of [ H  (aq)] , so the ocean is basic. However, CO2 is a weak acid that
dissolves into the ocean. The accumulation of CO2-acid over time is significantly
acidifying the ocean.