STOICHIOMETRY
• I CAN solve a stoichiometry (mass – mass)
problem using a chemical equation and mass
data.
What is Stoichiometry?
• Stoichiometry is the
quantitative study of
reactants and products in
a chemical reaction.
What You Should Expect To Do?
• Given : Amount of reactant(s)
• Question: How much product(s) can be
formed?
• Example
• XA +
YB
?C
• “Given X grams of A and Y grams of B,
how many grams of C can be produced?”
What Skills Do You Need?
Some skills you will need to use:
• Molar Ratios (come from coefficients of a BALANCED equation).
• Molar Masses (Formula Weight calculations).
• Balancing Equations
• Conversions between grams and moles
Note: This type of problem is often called "mass-mass."
Steps Involved in Solving MassMass Stoichiometry Problems
• Balance the chemical equation
correctly.
• Convert GRAMS to MOLES of the
starting substance.
• Set up a MOLE RATIO between starting
and ending substances.
• Convert MOLES to GRAMS of ending
substance.
Mole Ratios
A mole ratio COMPARES moles of
one compound in a balanced
chemical equation to moles of
another compound (look at their
coefficients).
Example
In the reaction between magnesium and
oxygen to form magnesium oxide, what
is the mole ratio for the substances?
2 Mg
2 moles
+
:
O2
1 mole
:
2 MgO
2 moles
Steps to Solve a
Stoichiometry Problem
•
•
•
•
Make sure the equation is BALANCED.
Label the QUANTITIES GIVEN.
DRAW a DIMENSIONAL ANALYSIS
FRAME:
Grams
starting
X
1 mole
starting
Moles
ending
FW
ending, g
FW
starting, g
Moles
starting
1 mole
ending
• Fill in the frame with the MOLES and FWs as
indicated on the frame.
• Complete the math across the top of the
frame, then across the bottom.
• Finally divide the TOP by BOTTOM and ROUND
to SIGNIFICANT FIGURES.
Practice Problem
• If 75.2 g of KClO3 decomposes, how
many grams of O2 will be produced?
2 KClO3
75.2 g
2 KCl + 3 O2
?g
✔
balanced
FW O2
75.2g KClO3
X
1 mole
KClO3
122.55 g
KClO3
FW
KClO3
3 mole
O2
31.998 g
O2
2 mole
KClO3
1 mole
O2
Mole
Ratio
=
______ g O2
75.2g KClO3
X
1 mole
KClO3
122.55 g
KClO3
3 mole
O2
31.998 g
O2
2 mole
KClO3
1 mole
O2
=
7218.7488 g O2
____________________
245.1
=29.45225 g
≈ 29.5 g O
2
MORE PRACTICE
• #1) If 75.2 grams of Calcium Sulfate are
reacted with an unlimited amount of Sodium
Phosphate, what mass of Calcium Phosphate
will be formed according to the reaction:
• CaSO4 + Na3PO4  Ca3(PO4)2 + Na2SO4
• #2) How many grams of Potassium are
needed to form 125.6 grams of Potassium
Carbonate according to the reaction below?
•
K
+
Li2CO3  K2CO3 + Li