Download Stoichiometry: Understanding Balanced Chemical Equations and Calculations and more Study notes Stoichiometry in PDF only on Docsity!
12.1 – The Arithmetic of Equations
- OBJECTIVES:
- Explain how balanced equations apply to both
chemistry and everyday life.
- Interpret balanced chemical equations in terms
of: a) moles, b) representative particles, c) mass,
and d) gas volume (Liters) at STP.
- Identify the quantities that are always conserved
in chemical reactions.
Chapter 12 “Stoichiometry”
Stoichiometry is…
• Greek for “measuring elements”
Pronounced “stoy kee ahm uh tree”
• Defined as: calculations of the
quantities in chemical reactions,
based on a balanced equation.
• There are 4 ways to interpret a
balanced chemical equation
#1. In terms of Particles
- An Element is made of atoms
- A Molecular compound (made of only nonmetals)
is made up of molecules
- Ionic Compounds (made of a metal and nonmetal
parts) are made of formula units
+ O
2
→ 2H
2
O
- Two molecules of hydrogen and one molecule of
oxygen form two molecules of water
O
3
4Al + 3O 2
O 3
form four atoms Al and three
molecules of oxygen
#2. In terms of Moles
- The coefficients tell us how many moles of each
substance
2 Al 2
O
3
4 Al + 3 O 2
2 Na + 2 H 2
O 2 NaOH + H 2
- Remember: A balanced equation is a Molar Ratio
#3. In terms of Mass
- The Law of Conservation of Mass applies to rxns
- We can check mass by using moles
2H
2
+ O
2
2H
2
O
2 moles H 2
2.02 g H 2
1 mole H 2
= 4.04 g H 2
1 mole O 2
32.00 g O 2
1 mole O 2
= 32.00 g O 2
Reactants total mass = 4.04 g + 32.00 g
= 36.04 g
So 36.04 g of water are produced
2 moles H 2
O
18.02 g H 2
O
1 mole H 2
O
36.04 g H 2
O
36.04 grams reactant = 36.04 grams product
#4. In terms of Volume
- At STP, 1 mol of any gas = 22.4 L
2H
2
+ O
2
2H
2
O
(2 x 22.4 L H 2 ) + (1 x 22.4 L O 2 ) (2 x 22.4 L H 2 O)
NOTE: mass and atoms are ALWAYS conserved -
however, molecules, formula units, moles, and
volumes will not necessarily be conserved!
67.2 Liters of reactant ≠ 44.8 Liters of product!
Practice:
- Show that the following equation
follows the Law of Conservation of
Mass (show the atoms balance, and
the mass on both sides is equal)
2Al 2
O 3
4Al + 3O 2
12.2 – Chemical Calculations
- OBJECTIVES:
- Construct “mole ratios” from balanced
chemical equations, and apply these ratios in
mole-mole stoichiometric calculations.
- Calculate stoichiometric quantities from
balanced chemical equations using units of
moles, mass, representative particles, and
volumes of gases at STP.
Mole to Mole conversions
2Al
2
O
3
Al + 3O
2
- each time we use 2 moles of Al 2
O 3
we will also
make 3 moles of O 2
2 moles Al 2
O
3
3 mole O 2
or
2 moles Al 2
O
3
3 mole O 2
These are the two possible conversion
factors to use in the solution of the problem.
Mole to Mole conversions
are
produced when 3.34 moles of
Al 2
O 3
decompose?
2Al 2
O 3
Al + 3O 2
3.34 mol Al 2
O
3
2 mol Al 2
O
3
3 mol O 2
= 5.01 mol O 2
If you know the amount of ANY chemical in the reaction,
you can find the amount of ALL the other chemicals!
Conversion factor from balanced equation
Practice:
2C
2
H
2
+ 5 O
2
4CO
2
+ 2 H
2
O
H
2
are burned, how
many moles of O
2
are needed?
(9.6 mol)
H
2
are needed to
produce 8.95 mole of H
2
O?
(8.95 mol)
H
2
are burned, how
many moles of CO 2
are formed? (4.94 mol)
Steps to Calculate Stoichiometric
Problems
1. Correctly balance the equation.
2. Convert the given amount into
moles.
3. Set up mole ratios.
4. Use mole ratios to calculate moles of
desired chemical.
5. Convert moles back into final unit.
“Limiting” Reagent
- If you are given one dozen loaves of bread, a
gallon of mustard, and three pieces of salami,
how many salami sandwiches can you make?
- The limiting reagent is the reactant you run out
of first.
- The excess reagent is the one you have left over.
- The limiting reagent determines how much
product you can make
Limiting Reagents - Combustion
How do you find out which is limited?
- The chemical that makes the least
amount of product is the “limiting
reagent”.
- You can recognize limiting reagent
problems because they will give you 2
amounts of chemical
- Do two stoichiometry problems, one
for each reagent you are given.
- If 10.6 g of copper reacts with 3.83 g
sulfur, how many grams of the product
(copper (I) sulfide) will be formed?
2Cu + S Cu 2
S
10.6 g Cu
63.55g Cu
1 mol Cu
2 mol Cu
1 mol Cu 2
S
1 mol Cu 2
S
159.16 g Cu 2
S
= 13.3 g Cu 2
S
3.83 g S
32.06g S
1 mol S
1 mol S
1 mol Cu 2
S
1 mol Cu 2
S
159.16 g Cu 2
S
= 19.0 g Cu 2
S
= 13.3 g Cu 2
S
Cu is the
Limiting
Reagent, since
it produced less
product.
Another example:
- If 10.3 g of aluminum are reacted
with 51.7 g of CuSO 4
how much
copper (grams) will be produced?
2Al + 3CuSO 4
→ 3Cu + Al 2
(SO 4
) 3
the CuSO 4
is limited, so Cu = 20.6 g
- How much excess reagent will
remain?
Excess = 4.47 grams
What is Yield?
- Yield is the amount of product made in a
chemical reaction.
- Actual yield- what you actually get in the lab
when the chemicals are mixed
- Theoretical yield- what the balanced equation
tells should be made
- Percent yield = Actual
Theoretical
x 100
Example:
- 6.78 g of copper is produced when 3.92 g
of Al are reacted with excess copper (II)
sulfate.
2Al + 3 CuSO
4
Al
2
(SO
4
3
+ 3Cu
- What is the actual yield?
- What is the theoretical yield?
- What is the percent yield?
= 6.78 g Cu
= 13.8 g Cu
= 49.1 %
Details on Yield
- Percent yield tells us how “efficient” a
reaction is.
- Percent yield can not be bigger than
- Theoretical yield will always be larger
than actual yield!
- Why? Due to impure reactants; competing side
reactions; loss of product in filtering or
transferring between containers; measuring
