Stoichiometry
STOICHIOMETRY IS THE PART OF CHEMISTRY
THAT DEALS WITH THE AMOUNTS OF SUBSTANCES
INVOLVED IN CHEMICAL REACTIONS.
HENCE, IT IS THE STUDY OF THE QUANTITATIVE
RELATIONSHIPS IN CHEMICAL REACTIONS.
A BALANCED CHEMICAL EQUATION INDICATES THE
RELATIVE NUMBER OF MOLES OR “PARTICLES”
INVOLVED IN A CHEMICAL REACTION.
4 NO(g) + 6 H2O(g)
4 NH3(g) + 5 O2(g)
4 molecules 5 molecules
4 molecules 6 molecules
4 moles
5 moles
4 moles
6 moles
68.16 g
160.00 g
120.04 g
108.12 g
Recall the definition of a mole:
g-formula mass = 1 mole = 6.022 x 1023 particles
We can expand this definition to include gases.
Early chemists primarily worked with gases, and
devised many laws to explain the behavior(s) of these
gases.
In 1811 Avogadro postulated his own gas law.
He said that, “Equal volumes of gases contain equal
numbers of molecules.”
Thus, the volume is directly proportional to the
number of particles of the number of moles.
g-formula mass = 1 mole = 6.022 x 1023 particles
= 22.4 dm3 at STP*
*Standard Temperature and Pressure
0C and 1.0 atm
4 NH3(g) + 5 O2(g)
4 liters
5 liters
4 NO(g) + 6 H2O(g)
4 liters
6 liters
By writing balanced chemical equations and
incorporating mole conversions, you can calculate the
amount of reactants needed or the amount of products
produced. This is due to the fact that the equation
indicates the number of moles of reactants and
products.
TO SOLVE:
1. Write the balanced equation.
2. Find the number of moles of the given substance.
3. Inspect and use the balanced equation to
determine the ratio of moles of given
substance to moles of “wanted” substance.
4. Convert the amount of “wanted” substance into the
desired units.
THE TYPICAL FORMAT:
(given) grams
1 mole (given)
Step#2
g-formula mass (given)
moles
of wanted
Step#3
from equation
moles of given
g-formula mass
(wanted)
Step#4
1 mole (wanted)
Example: Sulfuric acid is neutralized by the addition of sodium hydroxide.
How many grams of water are produced when 15. grams of sodium
hydroxide react?
TO SOLVE:
1. Write the balanced equation.
H2SO4(aq) +
2 NaOH
(aq)
Na2SO4(aq) +
15 g
H O
2 HOH
2
(l)
??? g
2. Find the number of moles of the given substance.
15 g NaOH
1 mole NaOH
2 mole H2O
18.02 g H2O
40.00 g NaOH
2 mole NaOH
1 mole H2O
= 6.8 g H2O
3. Inspect and use the balanced equation to determine
the ratio of moles of given substance to moles of
“wanted” substance.
4. Convert the amount of “wanted” substance into the
desired units.
Example: How many moles of NaOH are required to neutralize
4.2 moles of H2SO4?
TO SOLVE:
1. Write the balanced equation.
2
H2SO4(aq) + NaOH(aq)
4.2 moles ??? moles
Na2SO4(aq) +
2
H2O(l)
2. Find the number of moles of the given substance.
4.2 moles H2SO4
2 moles NaOH
1 mole H2SO4
= 8.4 moles NaOH
3. Inspect and use the balanced equation to determine
the ratio of moles of given substance to moles of
“wanted” substance.
4. Convert the amount of “wanted” substance into the
desired units.
Example: How many grams of Na2SO4 are produced from 4.2 moles of H2SO4?
TO SOLVE:
1. Write the balanced equation.
H2SO4(aq) +
2 NaOH
(aq)
4.2 moles
Na2SO4(aq) +
??? g
2
H2O(l)
2. Find the number of moles of the given substance.
4.2 moles H2SO4
1 mole Na2SO4
142.04 gNa2SO4
1 mole H2SO4
1 mole Na2SO4
= 596.568 g Na2SO4 = 6.0 x 102 g Na2SO4
3. Inspect and use the balanced equation to determine
the ratio of moles of given substance to moles of
“wanted” substance.
4. Convert the amount of “wanted” substance into the
desired units.
So far you have worked stoichiometry problems in which the
given quantity of a reactant is consumed completely and we
use that quantity to figure out how much product is formed.
However, in most instances you mix different amounts of
various reactants and not all completely react. Lets look at the
following hypothetical reaction:
A
+ B
Start: 5 mol 3 mol
End: 2 mol 0 mol
C
0 mol
3 mol
As you can clearly see, all three moles of B were consumed but
since you had an excess of A, 2 moles were left over.
The reagent that is consumed 100% is referred to as the
limiting reagent or reactant (LR).
The LR allows you to calculate (just like you learned
previously) the amount of product formed and how
much of the other reactant was consumed.
This is a lot simpler to understand with an example.
Limiting reactant analogies….
For example,
bicycles require
one frame and
two wheels. If
you have 20
wheels but only 5
frames, it is clear
that the number
of frames will
determine how
many bicycles
can be made.
TO SOLVE LIMITING REACTANT (REAGENT)
PROBLEMS:
1. Write the balanced equation for the reaction.
2. Calculate the number of moles of each reactant.
3. Convert one of the reactants into the other. This tells you
how much of each you need to complete the reaction.
4. Compare the moles calculated in step #2 to the amounts in
#3. The one you have less of is the limiting reagent.
5. Use the limiting reagent as the given, and calculate the
desired quantity.
EXAMPLE: The Haber Process is used to convert nitrogen
and hydrogen gases to ammonia.
If 20. g N2 and 10. g H2 are mixed,
what is the limiting reagent?
1. Write the balanced equation for the reaction.
N2(g)
20. g
+
3H
2(g)
10. g
2 NH
3(g)
2. Calculate the number of moles of each reactant.
20. g N2
1 mole N2
28.02 g N2
= 0.71 moles N2(g)
10. g H2
1 mole H2
2.02 g H2
= 5.0 moles H2(g)
3. Convert one of the reactants into the other. This tells you how
much of each you need to complete the reaction.
.713775874375 moles N2
3 mole H2
1 mole N2
= 2.1 moles H2 needed
4. Compare the given amounts to the answer in #3. The one you
have less of is the limiting reagent.
= .71 N2 given and 2.1 moles need to react with it
•
You have 5 moles of H2…..more than enough…
N2 is the limiting reagent. There is an excess of H2, After the reaction has
completed there will be no N2 left, yet there will still be extra H2.
EXAMPLE: How much NH3 is produced?
Use the L R (N2 in this problem) to determine
amounts of products…..
N2(g)
+
3H
2 NH
2(g)
3(g)
From the last example, we know
.713775874375 moles N2
2 mole NH3
17.04 g NH3
1 mole N2
1 mole NH3
= 24 g NH3
EXAMPLE: How much excess is there?
Use the L R (N2 in this problem) to determine how
much of the other reactant (H2 in this problem) is
left over.
N2(g)
+
3H
2(g)
2 NH
3(g)
From the last example, we know
.713775874375 moles N2
3 mole H2
2.02 gH2
1 mole N2
1 mole H2
= 4.32548179871 g H2 reacted
10.
g H2 to start
- 4.325481 g H2 reacted
5.674518 g in excess
= 6 g H2 left over
EXAMPLE: How many grams of aluminum sulfide can form
from the reaction of 9.00 g of aluminum with
8.00 g of sulfur?
2 Al
(s)
9.00 g
1 mole Al
9.00 g Al
3S
Al2S3(s)
(g)
8.00 g
??? g
8.00 g S
26.98 g Al
1 mole S
32.07 g S
= 0.334 moles Al(S)
0.249 moles S
+
2 moles Al
3 moles S
= 0.249 moles S(s)
= 0.166 moles Al needed
There is more than enough aluminum
Use the L R (S in this problem) to determine
amounts of products…..
2 Al
+
(s)
9.00 g
0.249454318678 moles S
3S
Al2S3(s)
(g)
8.00 g
??? g
1 mole Al2S3
150.17 g Al2S3
3 mole S
1 mole Al2S3
= 12.5 g Al2S3 produced
EXAMPLE: How many grams of N2F4 can theoretically be
obtained from 4.00 g of NH3 and 14.0 g of F2
according to:
2 NH
3
4.00 g
4.00 g NH3
+
5F
N2 F 4
2
14.00 g
??? g
1 mole NH3
14.00 g F2
17.04 g NH3
6 HF
1 mole F2
38.00 g F2
= 0.3684 moles F2
= 0.235 moles NH3
0.235 moles NH3
+
5 moles F2
2 moles NH3
= .5875 moles F2 needed
There is NOT ENOUGH fluorine
Use the L R (F2 in this problem) to determine
amounts of products…..
2 NH
3
4.00 g
0.368421052632 moles F2
+
5F
2
14.00 g
N2 F 4
??? g
1 mole N2F4
104.02 g N2F4
5 mole F2
1 mole N2F4
= 7.665 g N2F4 produced
+
6 HF
Lets say you carry out the reaction described above in the lab
and you only obtain 4.80 g N2F4. What happened????
The quantities we calculate in mass-mass (Stoichiometry)
problems are THEORETICAL AMOUNTS; that is, the
maximum possible yield.
In reality, the actual amount of products is much less due to
human error, mechanical error, and possible side reactions.
By comparing the actual versus theoretical, scientists calculate
the percentage yield.
From the last problem
4.80 g N2F4 actual
7.665 g N2F4 theoretical
x 100 %
= 62.6% N2F4 yielded