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Stoichiometry:
Calculations with Chemical Formulas and Equations
Chemical Equations
 Chemical reactions are represented in a concise
method by a chemical equation.
 Ex) 2 H2(g) + O2(g)
2 H2O(l)
Reactants
Products
Chemical Equations
 Ex) 2 H2(g) + O2(g)
Phase Symbols
2 H2O(l)
Coefficient
Balancing an Equation
 A subscript in a chemical formula tells us how many of
each type of atom are in the compound.
 Ex) C6H12O6
 Subscripts cannot be altered!!!
 Atoms can be created nor destroyed in a chemical
reaction.
 Thus, we balance a reaction by adding coefficients in
front of each substance.
Balancing an Equation
 Balance by inspection.
 Use a tally sheet.
 Start with elements that occur once on each side.
 Combustion – do C, then H, then O.
 LEP #1.
Patterns of Reactivity
 Five basic types of reactions.
1 . Combination – two substances combine to make one
new one. Generic: A + B  C
Ex) 2 Mg(s) + O2(g)  2 MgO(s)
2. Decomposition – one substance decomposes to
several new ones. Generic: A  B + C
Ex) 2 NaN3(s)  2 Na(s) + 3 N2(g)
Patterns of Reactivity
3. Single Replacement – one element replaces the other.
Generic: A + BC  AC + B
Ex) 2 AgNO3(aq) + Cu(s)  Cu(NO3)2(aq) + 2 Ag(s)
4. Double Replacement (aka “Metathesis”) – trading
partners. Generic: AB + CD  AD + CD
Ex) Hg(NO3)2(aq) + 2 NaI(aq)  HgI2(s) + 2 NaNO3(aq)
5. Combustion – a rapid reaction with O2(g) producing a
flame.
Ex) CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
Quantitative Aspect
 A chemical formula also has a quantitative aspect.
 A Formula Weight for an element or compound is
found using the periodic table.
 Formula weights can refer to a single element’s weight or
an ionic compound.
 Molecular weight refers to a molecular compound’s
weight.
 Weights from periodic table should be rounded to the
nearest 0.1 amu at the bare minimum!
 LEP #2
Percent Composition
 A formula weight can be used to calculate the mass
percentage of any element in the formula by:
Mass % of A =
# atoms of A  atomic mass of A
 100
formula weight of compound
 This is one place to also test nomenclature!
 LEP #3
Moles
 It is not practical
to weigh things in
amu or think of
reactions in terms
of atoms or
molecules.
 The Mole is a
quantity used in
chemistry that is of
the size that we
can observe.
Molar Mass
 The molar mass of any compound is equal to the sum
of the atomic weights expressed in grams.
 Ex) The molar mass of CO2 is 44.0 grams.
 Thus, one mole of CO2 = 44.0 grams.
 One mole of anything will contain Avogadro’s Number
of particles.
 1 mol C = 6.02 x 1023 atoms
 1 mol CO2 = 6.02 x 1023 molecules
 1 mol NaCl = 6.02 x 1023 formula units
Mole Relationships
Mole Relationships
 These two concepts (molar mass, Avogadro’s number)
allow us to convert between mass, moles, and
molecules.
Examples – LEP #4
Empirical Formulas
 We can use moles to find an empirical (simplest)
formula from mass percentages by:
Assume a 100 gram sample (%  grams).
2. Convert grams of each element to moles use the formula
weights.
3. Divide each mole amount by the smallest one.
4. Using a multiplier to eliminate fractions like: 0.25, 0.33,
0.50, 0.67, and 0.75.
1.
Empirical Formulas
 LEP #5
 An empirical formula may not be the actual formula
since molecular formulas do not have to be the lowest
whole number subscripts.
 The multiplier, n, can be found if we know the overall
molecular weight of the compound.
 LEP #6
molecular weight
n=
empirical weight
Interpreting a Reaction
 A simple reaction like: N2(g) + 3 H2(g)  2 NH3(g),
can be interpreted on many levels.
 Molecular Level: one molecule of N2 plus three
molecules of H2 react to form two molecules of NH3
Interpreting a Reaction
 For this reaction, we can establish that:
1 molecule N2 = 3 molecules H2
1 molecule N2 = 2 molecules NH3
3 molecules H2 = 2 molecules NH3
LEP #7
Interpreting a Reaction
 The molecular level is really not practical as we cannot
do reaction on this scale.
 Rather, we can do them on a mole scale.
 Thus: one mole of N2 plus three moles of H2 react to
produce two moles of NH3.
 This means our relations can be shortened to moles.
LEP #7
Limiting Reactant
 If given amounts of both reactants, we may run out of
one of them first. This reactant limits how much can
be made.
 Analogy: Putting together a bicycle – parts on hand are
200 frames and 350 wheels. How many bicycles can
you make?
 Ex) 2 H2 + O2  2 H2O
 Suppose a vessel contained 10 molecules of H2 and 7
molecules of O2. How many water molecules are possible?
Limiting Reactant
 This also applies to mole amounts as well.
 LEP #8
Stoichiometry
 Pronounced: stoy-key-OM-uh-tree.
 Relating quantities in chemical reactions – in
particular – masses.
 Cannot use mole-to-mole ratios to convert mass of one
substance to mass of another by one single step.
 A mass-to-mass conversion must be done in three
steps.
Stoichiometry
Stoichiometry
 Can be used to find a mass of another reactant or a
product.
 Can be part of a limiting reactant where amounts of
both reactants are given.
 Can also be asked to find a percent yield.
Percent Yield =
Actual Mass Obtained
 100
Theoretical Mass
 Where the Theoretical Mass is the maximum amount
possible based on your limiting reactant.
 LEP #9 and #10