Stoichiometry PowerPoint

Stoichiometry
Connecting Mathematics to Balanced
Chemical Equations

Balanced Equations
• You already have experience balancing
equations. We have used this simple one
for the formation of water before.
• 2H 2 + O 2 2 H 2 O

Understanding Stoichiometry
• Hydrogen atoms are tiny, with a mass of
only 1 amu per atom, and 2 amu per
molecule. Oxygen atoms are much larger.
1 1 1
• 1 H atom
1 H 2 molecule
16
• O atom
16 16
O 2 molecule

Understanding Stoichiometry
• The trick to understanding stoichiometry is
to realize that the ratio of the parts (atoms or
molecules) is NOT the same as the ratio of
the masses of the parts. In our example,
reactant and product parts occur in a 2:1:2
ratio; 2 hydrogen molecules and 1 oxygen
molecule make 2 water molecules.

Understanding Stoichiometry
• However, the ratio of the masses is
definitely NOT 2:1:2 because H and O
atoms have very different masses. What is
the mass ratio for this reaction?
1
1
1
1
16 16
1
1 1
1
16
16

Understanding Stoichiometry
• If you said 4 amu H 2 : 32 amu O 2 : 36 amu
H 2 O, you are correct.
1
1
1
1
16 16
1
1 1
1
16
16
• On a more practical level, we can also say
that 4 g H 2 and 32 g O 2 give 36 g H 2 O.

The Magic of Avogadro’s
Number
• Remember what you learned earlier about
Avogadro’s magical number. Our way of
counting atoms and molecules is to take
their masses on the atomic level (amu) and
translate them directly into grams. The
gram atomic or gram molecular mass of an
element or compound = 1 mole of that
substance or 6.02 x 10 23 particles.

Conservation of Mass
• Remember also that atoms and molecules
cannot appear or disappear, so the total
masses of all the reactants must equal the
total masses of all the products.
• 4g H 2 + 32g O 2 36g H 2 O
• 36 g reactants 36g products

Conservation of Mass
• All of the atoms present at the start are still present
at the end; they have simply been rearranged into
new combinations.
• It’s sort of like the world’s coolest Lego set. You
have some basic building blocks, and you can
make all sorts of great stuff.
• This is true for all chemical reactions and
processes.
Formative assessment: Unit 9, 5 clicker questions.

Other Examples
• Remember this drawing?
1
1
1
1
16 16
1
1 1
1
16
16
• Can you make a similar drawing for the
reaction: Mg + F 2 MgF 2 ?
• What are the particle and mass ratios for
this reaction?

Mg + F 2 MgF 2
• Did your drawing look like this?
• 24.3 19 19
19 24.3 19
•Particle ratio: 1:1:1
•Mass ratio: 24.3amu : 38.0amu : 62.3amu
•Or 24.3g : 38.0g : 62.3g

Other Examples
• Try this one. Again, use a diagram to
determine particle and mass ratios.
• 4 Al + 3 O 2 2 Al 2 O 3
27
27
16 16
16
27
16
27
16
27
27
16 16
16 16
16
27
16
27
16

4 Al + 3 O 2 2 Al 2 O 3
• Particle ratio: 4 : 3 : 2
• Mass ratio: 108amu : 96amu : 204amu
• Or 108g : 96g : 204 g
(4 mol : 3 mol : 2 mol)

Balanced Equation Coefficients
• By now you have noticed that the ratio of
the particles is the same as the coefficients
in the balanced equation.
• The coefficients in any balanced chemical
equation represent both particles and moles
of the substances in the reaction.

Mass Ratios
• Determining mass ratios without pictures is
a simple matter of converting moles to
grams.
• From the 4 Al + 3 O 2 2 Al 2 O 3 example,
the math looks like this:
• 4 mol Al x (27.0g/mol) = 108g Al
• 3 mol O 2 x (32g/mol) = 96g O 2
• 2 mol Al 2 O 3 x (102g/mol) = 204g Al 2 O 3

Now The Fun Begins
• Solving stoichiometry problems is easy if
you know
• The mole ratio of the reactants and
products from the balanced equation
and
• The molar mass of each reactant and
product.

Now The Fun Begins
• Once you know these 2 things, you
can be given a starting amount of any
material in the problem and calculate
how much of everything else will be
needed or produced.

Looking Back
• If we go back to our first example:
1
1
1
1
16 16
1
1 1
1
16
16
• What if there were 6 H 2 molecules (or moles)
rather than 2? How many O 2 molecules (or
moles) would be required, and how many water
molecules (or moles) could we make?

Mole Ratio
• Three times as much hydrogen requires 3
times as much oxygen. 3 molecules or
moles of oxygen are required.
• Also 3 times as much water will be
produced. 6 molecules or moles of water.
• 6 H 2 + 3 O 2 6 H 2 O
• Notice that the mole ratio (2:1:2) from the
balanced equation has been maintained.

Masses Involved
• What will be the mass of 6 mol of H 2 ?
• 6 mol (2.0g/mol) = 12g H 2
• According to our picture, for every 4g of
H 2 , 32g of O 2 are required. We need 3
times as much oxygen: 3 x 32g = 96g O 2
needed.
• The original reaction made 36g of water, so
starting with 6g of hydrogen makes 108g
H 2 O (3 times as much.)

Solving Stoichiometry Problems
• The heart of every stoichiometry problem is
the ratio of moles from the balanced
equation.
• The number of moles of any material in the
problem determines how many moles of
everything else there will be.

Mole Ratios
• If we use our basic water example again:
• 2H 2 + O 2 2 H 2 O
• If we start with 16 moles of Hydrogen we
need 8 moles of oxygen and make 16 moles
of water.
• If we start with 0.0018 moles of hydrogen,
we need 0.0009 moles of oxygen and make
0.0018 moles of water.

Solving Stoichiometry Problems
• Once you understand that relationship, the
rest of the stoichiometry problem is easy.
• Chemists don’t measure moles directly.
They generally measure grams for solids,
liters (of volume) for gases and mL or L for
solutions. (We will deal with solutions
later.)
• Sometimes # of molecules will be involved.

Solving Stoichiometry Problems
• So solving stoichiometry problems involves
converting between these starting units to
moles, and then from moles back to the
units required at the end (mole map stuff.)
• The typical sequence looks like this:
• Amount known mol known
mol unknown amount unknown
• Each arrow represents a calculation.

Solving Stoichiometry Problems
• Remember the screws, nuts and washers?
• This is exactly the same kind of problem.
• Grams of screws/mass of 1 screw = # screws.
• # screws x 1 = # of nuts
• # nuts x mass of 1 nut = g of nuts OR
• # screws x 2 = # of washers
• # washers x mass or 1 washer = g of washers.

Solving Stoichiometry Problems
• For chemical problems:
• g of Known Substance ÷ mass of 1 mole =
moles of Known Substance
• Use moles of Known Substance and ratio
from equation to get moles of Unknown
Substance.
• Moles Unknown Substance x mass of 1
mole = g of Substance 2.

Examples
• Let’s try a couple of full-blown stoichiometry
problems, but we will start simple.
• CH 4 + O 2 CO 2 + H 2 O
• First, balance the equation.
• CH 4 + 2 O 2 CO 2 + 2 H 2 O
• What is the mole ratio of the reactants and
products?
• 1:2:1:2

Examples
• Next, calculate the molar masses of each
material.
• CH 4 = 16.0 g/mol
• O 2 = 32.0 g/mol
• CO 2 = 44.0 g/mol
• H 2 O = 18.0 g/mol

Examples
• If we start with 50.0grams of methane, how
many grams of oxygen will be required? These
problems are as easy as 1, 2, 3.
• 1. Convert grams methane to moles.
• 50.0g x (1 mol CH 4 /16.0g) = 3.12 mol CH 4
• 2. Convert mol of methane to mol of oxygen.
• 3.12 mol CH 4 x (2 mol O 2 /1 mol CH 4 ) = 6.24 mol O 2
• 3. Convert mol O 2 to grams.
• 6.24 mol O 2 x (32.0g O 2 /mol) = 200. g O 2

Examples
• If we start with 50.0grams of methane, how many
grams of water will be produced?
• Again the procedure is the same.
• 50.0g CH 4 x (1 mol/16.0g) = 3.12 mol CH 4
• 3.12 mol CH 4 x (2 mol H 2 O/1 mol CH 4 ) = 6.24 mol H 2 O
• 6.24 mol H 2 O x (18.0g/1 mol) = 112 g H 2 O
• The 3-step sequence is the same:
• gknown mol known mol unknown g unknown

Curve Ball
• Starting with the same 50.0g of CH 4 , how
many liters of CO 2 gas at STP will be
produced?
• The problem starts out just the same as
before: Convert g known to moles.
• 50.0g CH 4 x (1 mol/16.0g) = 3.12 mol CH 4
• Next, use the mole ratio from the equation to
convert mols known to mols unknown.
• 3.12 mol CH 4 x (1 mol CO 2 /1 mol CH 4 ) = 3.12
mol CO 2

Curve Ball
• The last step is just a little different.
• This time, we were not asked for grams of
CO 2 , but for liters of gas.
• We simply need to adjust the last
calculation a little. Do you remember how
many liters of gas at STP are in a mole?
• 3.12 mol CO 2 x (22.4L/1 mol) = 69.9 L

Try One on Your Own
• Fe 2 O 3 + 3 CO(g) 2 Fe + 3 CO 2
• How many grams of Fe 2 O 3 will be needed to
make 10.0g of Fe?
• 10.0g Fe (1 mol/55.8g) = 0.179 mol Fe
• 0.179 mol Fe (1 mol Fe 2 O 3 /2 mol Fe) =
0.0895 mol Fe 2 O 3
• 0.0895 mol Fe 2 O 3 (159.6g/mol) = 14.3g Fe 2 O 3

Try One on Your Own
• How many liters of CO gas will be required
to make the same 10.0g of Fe?
• 10.0g Fe (1 mol/55.8g) = 0.179 mol Fe
• 0.179 mol Fe (3 mol CO/2 mol Fe) = 0.268
mole CO
• 0.268 mol CO (22.4 L/mol) = 6.00 L of CO

Simplify
• As you become more practiced at the 3 step
stoichiometry problems, you can combine
all the steps into 1 equation.
• The last problem looks like this:
• 10.0g Fe 1 mol Fe 3 mol CO 22.4 L CO =
55.8 g Fe 2 mol Fe 1 mol CO
• 6.00 L CO

Simplify
• 10.0g Fe 1 mol Fe 3 mol CO 22.4 L CO =
55.8 g Fe 2 mol Fe 1 mol CO
• All three steps are here:
• 1. Convert g mol of known
• 2. Convert mol known mol unknown
with ratio from equation.
• 3. Convert mol unknown liters of
unknown