stoichiometry - college of dupage€¦ · stoichiometry with equations: the roadmap equations are...
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Learning objectives
Use chemical equations to predict quantity
of one substance from given quantity of
another
Determine percentage yield
Identify limiting reactant
The chemical equationaA + bB = cC + dD
The Law of Conservation of Matter states that matter
is neither created nor destroyed
All atoms on left must be same as those on right
Reactant
side
Product
side
coefficient ELEMENT or
COMPOUND
Working with equations:
STOICHIOMETRY
Predict how much product is obtained from
given amount of reactant
Predict how much reactant is needed to give
required amount of product
Predict how much of one reactant is
required to give optimum result with given
amount of another reactant
Stoichiometry with equations:
The roadmap
Equations are in moles, but we measure in grams
Three conversions required:
A is given substance (reactant or product); B is target
substance (reactant or product)
1. Must convert grams A to moles A using molar mass
2. Use coefficients in equation to get moles B from moles A
3. Convert moles B to grams B using molar mass
Mass/molar mass Mole:mole ratioMoles x molar
massMass A Moles A Moles B Mass B
Mole:mole ratio: the central step
Tells us molar ratio of substances in
balanced chemical equation
xX + yY = cC + dD
Mole:mole ratio (target:given) B:A = b/a
Mass/molar mass Mole:mole ratioMoles x molar
massMass A Moles A Moles B Mass B
Types of problems:
Moles A → moles B
(A is given; B is target)xX + yY = cC + dD
Single step
Require mole:mole ratio:
Always target/given (B/A)
In balanced equation a mol A ≡ b mol B
Mass/molar mass Mole:mole ratioMoles x molar
massMass A Moles A Moles B Mass B
moles B b mol B=
moles A a mol A
Mole:mole ratio problems
CH4 + 2O2 = CO2 + 2H2O
How many moles of H2O are produced from 5 moles of CH4?
Identify given substance: A = CH4
Identify target substance: B = H2O
Mole:mole ratio is
Identify coefficients: CH4 a = 1; H2O b = 2
moles B b mol B=
moles A a mol A
4
2
4
2
CH mol 1
OH mol 2
CH moles
OH moles
Moles A → mass BxX + yY = cC + dD
Two steps on road map
1. Convert moles A → moles B:
Mole:mole ratio (target/given):
2. Convert moles B → mass B using molar mass B
Mass/molar mass Mole:mole ratioMoles x molar
massMass A Moles A Moles B Mass B
moles B b mol B=
moles A a mol A
mass B (g) = moles B (mol) x molar mass B (g/mol)
Mole:mass/mass:mole problems
CH4 + 2O2 = CO2 + 2H2O
What mass of H2O is produced from burning 5 moles CH4?
How many moles of CO2 is produced when 64 g O2 is consumed?
Mass A → mass BxX + yY = cC + dD
All three steps
1. Mass A → moles A using molar mass A
2. Moles A → moles B using mole:mole ratio
3. Moles B → mass B using molar mass B
Mass/molar mass Mole:mole ratioMoles x molar
massMass A Moles A Moles B Mass B
moles B b mol B=
moles A a mol A
mass B (g) = moles B (mol) x molar mass B (g/mol)
mass A (g)moles A (mol) =
molar mass A (g/mol)
2 NaOH + Cl2 = NaOCl + NaCl + H2O What mass NaOH required to react with 142 g Cl2?
Molar mass Cl2 = 35.5 x 2 = 71.0 g/mol
Molar mass NaOH = 23.00 + 16.00 + 1.01 = 40.01 g/mol
Summary of stoichiometry problems
Maximum of three conversions required1. Must convert grams A to moles A using molar mass
2. Use coefficients in equation to get moles B from moles A
3. Convert moles B to grams B using molar mass
Maximum of three pieces of information required
1. Molar mass of given substance (maybe)
2. Molar mass of target substance (maybe)
3. Balanced chemical equation (always)
Work this example
CH4 + 2O2 = CO2 + 2H2O
What mass of CO2 is produced by the
complete combustion of 16 g of CH4
Atomic weight H = 1, C = 12, O = 16
44 g
Do stoichiometry exercises
Reaction Yield
The actual yield from a chemical reaction is
normally less than predicted from the
stoichiometry. Incomplete reaction
Product lost in recovery
Competing side reactions
Percent yield is:
actual yield% yield=
theoretical yield
Worked example
Actual yield of product is 32.8 g after reaction of 26.3 g of C4H8 with excess CH3OH to give C5H12O.
What is theoretical yield? Use stoichiometry to get mass of product:
convert mass moles moles mass
Theoretical yield = 41.4 g
Percent yield = 32.8/41.4 x 100 %
Mass/molar mass Mole:mole ratioMoles x molar
massMass A Moles A Moles B Mass B
CH4 + 2O2 = CO2 + 2H2O What is percent yield if 40 g of CO2 is
produced by the complete combustion of 16
g of CH4?
First calculate theoretical yield as previously
Theoretical yield CO2 = 44 g
Actual yield CO2 = 40 g
Percent yield = 40 g/44 g x 100% = 91%
actual yield% yield=
theoretical yield
Limiting reactant
Balanced equations involve precisely the
right amounts of each reactant. None is left
over
Real-life situations usually involve an excess
of one reactant
Burning CH4 in excess O2
Reacting Zn with excess HCl
Identifying the limiting reactant
In balanced equations
each reactant will give
the same amount of
products
When one reactant is
limiting (deficient) the
amount of product will
be determined by its
amount