Stoichiometry
n
Stoichiometry: the study of the quantitative, or measurable relationships that exist
in chemical formulas and chemical reactions
n
Coefficients in the balanced equation tell:
–
The relative number
of particles (atoms, formula units, etc)
–
The number of moles
–
The relative volume of gases at content
temperature and pressure
How to Interpret
Coefficients to solve Stoichiometry Problems
2H2 + O2 à 2H2O
n Two
moles of hydrogen gas react with one mole of oxygen gas to produce two moles of
water.
Mass-mass relationships
Use the factor label method to solve:
–
Write the grams of known
product and place this value over one.
–
Use molar mass (of the known
substance) to convert the grams of the known substance to moles
•
The moles should be on
top to cancel grams.
–
Use the coefficients of
the balanced chemical equation to convert from the moles of the known substance
to moles of the unknown substance
•
The moles of unknown
compound should be on top so the known moles will cancel out.
–
Use molar mass (of the unknown
substance) to convert the moles of the unknown substance to grams
•
Grams should be on top
as that is the unit you want
Mole to Mole Example
NH4NO3 ® N2O
+ 2H2O
n How
many moles of N2O and H2O are produced from 2.25 moles of
NH4NO3?
2.25 moles NH4NO3 x
1 mol N2O = 2.25 mol N2O
1 mol NH4NO3
2.25 moles NH4NO3 x
2 mol H2O = 4.5 mol H2O
1 mol NH4NO3
Mass-mass relationships cont.
n
Multiply by all numbers
on top and divide by all numbers on the bottom
n
Check for sigfigs by
looking at the number of sigfigs in the original question.
n
The 2nd and 4th
step are just the mass to mole and mole to mass conversions done in the last
unit
n
The coefficients from
the balanced equation are the numbers of moles of each compound in that
particular equation
Mass-Mass Example
NH4NO3 ® N2O
+ 2H2O
n How
many grams of N2O and H2O are produced from 10.0 g of NH4NO3?
10.0 g NH4NO3 x
1 mol NH4NO3 x
1 mol N2O
x 44.0 g N2O = 5.50 g N2O
80.1 g NH4NO3 1 mol NH4NO3 1 mol N2O
10.0 g NH4NO3 x
1 mol NH4NO3 x 2 mol H2O x
18.02 g H2O =
4.50 g N2O
80.1 g NH4NO3 1 mol NH4NO3 1 mol H2O
The Law of Conservation of Mass
n Stoichiometry can be used to verify the law of
conservation of mass
NH4NO3
® N2O + 2H2O
n We calculated that 10.0 g NH4NO3
of decomposed to form 5.50 g N2O of and 4.50 g of H2O.
10.0
g = 5.50 g + 4.50 g
–
Mass was neither created
nor destroyed!!!
Mass-Volume Problems
n Use
the factor label method to solve:
– Use molar mass to
convert the grams of the known substance to moles
– Use the
coefficients of the balanced chemical equation to convert from the moles of the
known substance to moles of the unknown substance
– Use the conversion
factor 22.4 L/mol to convert from mols of the known substance to L
Mass – Volume Example
2 NaN3 ® 2 Na + 3 N2
n How
many liters of N2 gas are produced from 125 g of NaN3?
125 g NaN3 x
1 mol NaN3
x 3 mol N2 x
22.4 L N2 =
64.6 L N2 (STP)
65.0 g NaN3 2 mol NaN3 1 mol N2
Volume – Volume Problems
n Volume
– Volume Problems: are much
like mole-mole problems
– 1 mole of any gas
occupies the same volume as 1 mole of any other gas at STP
– Thus, the ratio
of volumes also represents the ratio of
moles
Volume – Volume Example
N2 + 3 H2 ® 2 NH3
Calculate the volume of H2 gas that reacts with 15.5 liters of N2
(at STP).
15.5 L N2 x 3 L H2 = 46.5 L H2
1
L N2
Limiting Reactant Problems
n In real life, quantities of reactants are not usually
available in the exact ratio described by the balanced equation.
–
There may be too little
of one reactant to fully react with the other reactants.
•
Limiting
Reactant: the reactant that limits the amount of product formed
in a chemical reaction.
–
The quantities of
products formed in a reaction are always determined by the limiting reactant
•
Excess
Reactant: is the reactant that is leftover after the reaction
has ended.
Limiting Reactant Example
n Wood
burning in air. The oxygen in the air
is considered the reactant in excess and the wood is considered the limiting
reactant. When the wood is all burned,
there will no longer be a reaction occurring (fire).
Limiting Reactant Problems
n
First write the balanced equation.
n
Next determine which reactant produces the least
product. This will be the limiting
reactant.
n
The amount of product produced can be calculated in
grams, liters, or moles.
Limiting Reactant Example Problem
3.5 g Cu is added to a solution containing 6.0 g of silver (I) nitrate.
Determine the limiting reactant.
Cu + 2 AgNO3 ®
Cu(NO3)2 + 2 Ag
3.5 g Cu x 1 mol Cu x
2 mol Ag = 0.11 mol Ag
63.5 g Cu 1 mol
Cu
6.0 g AgNO3 x
1 mol AgNO3 x 2 mol Ag = 0.035 mol Ag
169.9 g AgNO3 2
mol AgNO3
The limiting reactant is silver (I) nitrate.
Limiting Reactant Example 2
n Consider
the following reaction:
3
NaHCO3 + H3C6H5O7 ® 3 CO2 + 3 H2O + Na3C6H5O7
n Suppose 2.0 g of NaHCO3 and 0.5 g of H3C6H5O7
are present. Which is the limiting
reactant and what volume of CO2 will be produced?
2.0 g NaHCO3
x 1 mol NaHCO3 x 3 mol CO2 x 22.4 L CO2 =
0.53 L CO2
84 g NaHCO3 3 mol NaHCO3 1 mol CO2
0.5 g H3C6H5O7
x 1 mol H3C6H5O7 x 3
mol CO2 x 22.4
L CO2 = 0.2 L
CO2
192 g H3C6H5O7 1 mol H3C6H5O7 1 mol CO2
n The limiting reactant is H3C6H5O7
and 0.2 L of CO2 are produced.
Percent Yield
n Expected
Yield: the amount of product that
should be obtained based on the calculations
n Percent
Yield: the percent of the
expected yield that was actually obtained
Percent yield = actual yield x 100%
expected
yield
Solving Percent Yield Problems
n
Write the balanced chemical equation
n
Calculate the expected yield
n
Calculate the percent yield
Percent Yield Example Problem
n A
5.00 g piece of Cu is placed in excess silver (I) nitrate. 15.2 g of silver metal is produced. What is the percent yield?
Cu + 2 AgNO3 ®
Cu(NO3)2 + 2 Ag
5.0 g Cu x 1 mol Cu x
2 mol Ag x 107.9 g Ag = 17.0 g Ag
63.5 g Cu 1 mol
Cu 1 mol Ag
Percent Yield = 15.2
g Ag x 100 % = 89.4%
17.0 g Ag