Download Stoichiometry: Calculations with Chemical Formulas and Equations

Chapter 3
Stoichiometry:
Calculations with Chemical
Formulas and Equations
Stoichiometry
 2006, Prentice-Hall
Law of Conservation of Mass
“We may lay it down as an
incontestable axiom that, in all
the operations of art and nature,
nothing is created; an equal
amount of matter exists both
before and after the experiment.
Upon this principle, the whole art
of performing chemical
experiments depends.”
--Antoine Lavoisier, 1789
Stoichiometry
Chemical Equations
Concise representations of chemical
reactions
Stoichiometry
Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
Stoichiometry
Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
Reactants appear on the
left side of the equation.
CO2 (g) + 2 H2O (g)
Stoichiometry
Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
Products appear on the
right side of the equation.
CO2 (g) + 2 H2O (g)
Stoichiometry
Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
The states of the reactants and products
are written in parentheses to the right of
each compound.
Stoichiometry
Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
Coefficients are inserted to
balance the equation.
CO2 (g) + 2 H2O (g)
Stoichiometry
Subscripts and Coefficients Give
Different Information
• Subscripts tell the number of atoms of
each element in a molecule
Stoichiometry
Subscripts and Coefficients Give
Different Information
• Subscripts tell the number of atoms of
each element in a molecule
• Coefficients tell the number of
molecules
Stoichiometry
Reaction
Types
Stoichiometry
Combination Reactions
• Two or more
substances
react to form
one product
• Examples:
N2 (g) + 3 H2 (g)  2 NH3 (g)
C3H6 (g) + Br2 (l)  C3H6Br2 (l)
2 Mg (s) + O2 (g)  2 MgO (s)
Stoichiometry
2 Mg (s) + O2 (g)  2 MgO (s)
Stoichiometry
Decomposition Reactions
• One substance breaks
down into two or more
substances
• Examples:
CaCO3 (s)  CaO (s) + CO2 (g)
2 KClO3 (s)  2 KCl (s) + O2 (g)
2 NaN3 (s)  2 Na (s) + 3 N2 (g)
Stoichiometry
Combustion Reactions
• Rapid reactions that
produce a flame
• Most often involve
hydrocarbons
reacting with oxygen
in the air
• Examples:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
Stoichiometry
Formula
Weights
Stoichiometry
Formula Weight (FW)
• Sum of the atomic weights for the atoms
in a chemical formula
• So, the formula weight of calcium
chloride, CaCl2, would be
Ca: 1(40.1 amu)
+ Cl: 2(35.5 amu)
111.1 amu
• These are generally reported for ionic
compounds
Stoichiometry
Molecular Weight (MW)
• Sum of the atomic weights of the atoms
in a molecule
• For the molecule ethane, C2H6, the
molecular weight would be
C: 2(12.0 amu)
+ H: 6(1.0 amu)
30.0 amu
Stoichiometry
Percent Composition
One can find the percentage of the mass
of a compound that comes from each of
the elements in the compound by using
this equation:
(number of atoms)(atomic weight)
% element =
(FW of the compound)
x 100
Stoichiometry
Percent Composition
So the percentage of carbon in ethane
is…
(2)(12.0 amu)
%C =
(30.0 amu)
24.0 amu
x 100
=
30.0 amu
= 80.0%
Stoichiometry
Moles
Stoichiometry
Avogadro’s Number
• 6.02 x 1023
• 1 mole of 12C has a
mass of 12 g
Stoichiometry
Molar Mass
• By definition, these are the mass of 1
mol of a substance (i.e., g/mol)
– The molar mass of an element is the mass
number for the element that we find on the
periodic table
– The formula weight (in amu’s) will be the
same number as the molar mass (in g/mol)
Stoichiometry
Using Moles
Moles provide a bridge from the molecular scale
to the real-world scale
Stoichiometry
Mole Relationships
• One mole of atoms, ions, or molecules contains
Avogadro’s number of those particles
• One mole of molecules or formula units contains
Avogadro’s number times the number of atoms or
ions of each element in the compound
Stoichiometry
Finding
Empirical
Formulas
Stoichiometry
Calculating Empirical Formulas
One can calculate the empirical formula from
the percent composition
Stoichiometry
Calculating Empirical Formulas
The compound para-aminobenzoic acid (you may have
seen it listed as PABA on your bottle of sunscreen) is
composed of carbon (61.31%), hydrogen (5.14%),
nitrogen (10.21%), and oxygen (23.33%). Find the
empirical formula of PABA.
Stoichiometry
Calculating Empirical Formulas
Assuming 100.00 g of para-aminobenzoic acid,
C:
H:
N:
O:
1 mol
12.01 g
1 mol
5.14 g x
1.01 g
1 mol
10.21 g x
14.01 g
1 mol
23.33 g x
16.00 g
61.31 g x
= 5.105 mol C
= 5.09 mol H
= 0.7288 mol N
= 1.456 mol O
Stoichiometry
Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number
of moles:
C:
5.105 mol
0.7288 mol
= 7.005  7
H:
5.09 mol
0.7288 mol
= 6.984  7
N:
0.7288 mol
0.7288 mol
= 1.000
O:
1.458 mol
0.7288 mol
= 2.001  2
Stoichiometry
Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C7H7NO2
Stoichiometry
Empirical to molecular
• Since the empirical formula is the
lowest ratio, the actual molecule
would weigh more.
–By a whole number multiple.
• Divide the actual molar mass by the
empirical formula mass – you get a
whole number to increase each
coefficient in the empirical formula
Stoichiometry
Combustion Analysis
• Compounds containing C, H and O are routinely
analyzed through combustion in a chamber like this
– C is determined from the mass of CO2 produced
– H is determined from the mass of H2O produced
– O is determined by difference after the C and H have been
determined
Stoichiometry
Combustion Analysis
p. 94
• Isopropyl alcohol, a substance sold as
rubbing alcohol is composed of C, H, and
O. Combustion of 0.255 g of isopropyl
alcohol produces 0.561 g CO2 and 0.306
g H2O. Determine the empirical formula of
isopropyl alcohol.
We must calculate the grams of C from CO2
And the grams of H from H2O. To calculate
O we will subtract grams C and grams O
from the total.
Stoichiometry
Combustion Analysis
0.561 g CO2
1 mol CO2
44.01 g CO2
1 mol C
12.01 g C
=
1 mol C
1 mol CO2
0.153 g C
0.306 g H2O
1 mol H2O
18.02 g H2O
2 mol H
1 mol H2O
12.01 g H
1 mol H
=
0.0343 g H
Mass of O = mass of sample – (mass of C + mass of H)
0.255 g – (0.153 g + 0.0343 g) = 0.068 g O
Stoichiometry
Combustion Analysis
0.153 g C
1 mol C
Moles C =
= 0.0128 mol C
12.01 g C
0.0343 g H
1 mol H
Moles H =
= 0.0340 mol H
1.01 g H
0.068 g O
1 mol O
Moles O =
= 0.0043 mol O
16.00 g O
Stoichiometry
Combustion Analysis
C:
0.0128 mol
0.0043 mol
= 2.977  3
H:
0.0340 mol
0.0043 mol
= 7.907  8
O:
0.0043 mol
0.0043 mol
= 1.000
empirical formula = C3H8O
Stoichiometry
Combustion Analysis
• Other examples: p.107 #47
a. Combustion analysis of toluene, a common
organic solvent gives 5.86 g CO2 and 1.37 g of
H2O. If the compound contains only carbon and
hydrogen, what is the empirical formula?
C7H8
Stoichiometry
Combustion Analysis
b. Menthol, the substance we smell in
mentholated cough drops, is composed of C,
H, and O. A 0.1005 g sample of menthol is
combusted, producing 0.2829 g of CO2 and
0.1159 g of H2O. What is the empirical formula
for menthol?
emp. form. = C10H20O
If the compound has a molar mass of 156 g/mol,
what is its molecular formula?
mol. form. = C10H20O
Stoichiometry
Elemental Analyses
Compounds
containing other
elements are
analyzed using
methods analogous
to those used for C,
H and O
Stoichiometry
Hydrates
• Hydrates are solids with water trapped
in the solid in a repeating pattern.
• When a hydrate is heated, the water will
escape leaving an anhydrate (solid
without water).
–Copper(II) sulfate is a hydrate.
–Hydrated copper(II) sulfate is deep
blue in color.
Stoichiometry
How are Names of Hydrates Written?
• BaCl2•2H2O
• barium chloride dihydrate
• FeSO4•6H2O
• iron(II) sulfate hexahydrate
• Na2CO3•10H2O
• sodium carbonate decahydrate
• CuSO4•5H2O
• copper(II) sulfate pentahydrate
Stoichiometry
Hydrates
Stoichiometry
Hydrates
Stoichiometry
Hydrates
Stoichiometry
Stoichiometry
Greek for “measuring elements”
Stoichiometry
Quantitative Information from
Balanced Equations
• Balanced chemical equation gives number of
molecules that react to form products.
• Interpretation: ratio of number of moles of reactant
required to give the ratio of number of moles of
product.
• These ratios are called stoichiometric ratios.
NB: Stoichiometric ratios are ideal proportions
• Real ratios of reactants and products in the
laboratory need to be measured (in grams and
converted to moles).
Stoichiometry
Prentice Hall © 2003
Chapter 3
Stoichiometric Calculations
The coefficients in the balanced equation give
the ratio of moles of reactants and products
Stoichiometry
Stoichiometric Calculations
From the mass of
Substance A you can
use the ratio of the
coefficients of A and
B to calculate the
mass of Substance B
formed (if it’s a
product) or used (if
it’s a reactant)
Stoichiometry
Stoichiometric Calculations
C6H12O6 + 6 O2  6 CO2 + 6 H2O
Starting with 1.00 g of C6H12O6…
we calculate the moles of C6H12O6…
use the coefficients to find the moles of H2O…
and then turn the moles of water to grams
Stoichiometry
Let’s ride the Mole Train!
grams of
known
moles of
known
molar mass
of known
moles of
unknown
mole to
mole ratio
grams of
unknown
molar mass
of unknown
Stoichiometry
Stoichiometric Calculations
• p. 107 #55
Na2SiO3 + 8 HF  H2SiF6 + 2 NaF + 3H2O
a. How many moles of HF are needed to
react with 0.300 mol of Na2SiO3? 2.40 mol HF
b. How many grams of NaF form when
0.500 mol of HF reacts with excess
Na2SiO3? 5.25 g NaF
c. How many grams of Na2SiO3 can react
with 0.800 g of HF? 0.610 g Na2SiO3
Stoichiometry
Stoichiometric Calculations
• Additional examples #57 and #59
Stoichiometry
Limiting
Reactants
Stoichiometry
How Many Cookies Can I Make?
• You can make cookies
until you run out of one
of the ingredients
• Once this family runs
out of sugar, they will
stop making cookies
(at least any cookies
you would want to eat)
Stoichiometry
How Many Cookies Can I Make?
• In this example the
sugar would be the
limiting reactant,
because it will limit the
amount of cookies you
can make
Stoichiometry
Limiting Reactants
The limiting reactant
is the reactant
present in the
smallest
stoichiometric
amount
Stoichiometry
Limiting Reactants
• The limiting reactant is the reactant present in
the smallest stoichiometric amount
– In other words, it’s the reactant you’ll run out of first (in
this case, the H2)
Stoichiometry
Limiting Reactants
In the example below, the O2 would be the
excess reagent
Stoichiometry
Limiting Reactants
• The Limiting Reactant:
– The reactant that is used up first in a
reaction.
– It controls the amounts of the other
reactants that are used.
– It controls the amount of product produced
(maximum amount of product).
– It produces less product than the other
reactants possibly could produce. (***Use
this for problem solving).
Stoichiometry
Theoretical Yield
• The theoretical yield is the amount of
product that can be made
– In other words it’s the amount of product
possible as calculated through the
stoichiometry problem
• This is different from the actual yield,
the amount one actually produces and
measures
Stoichiometry
Percent Yield
A comparison of the amount actually
obtained to the amount it was possible
to make
Actual Yield
Percent Yield =
Theoretical Yield
x 100
Stoichiometry
Limiting Reactants
• p. 109 #69
• Sodium hydroxide reacts with carbon dioxide
as follows:
2 NaOH + CO2  Na2CO3 + H2O
• Which reagent is the limiting reactant when 1.70
mol NaOH and 1.00 mol CO2 are allowed to react?
• How many moles of Na2CO3 can be produced?
• How many moles of the excess reactant remain
after the completion of the reaction?
Stoichiometry
• To calculate the limiting reactant
1) Convert to moles
2) Compare – pick 1 reactant, solve for the
amount needed of the other and compare.
1.00 mol CO2 2 mol NaOH = 2.00 mol NaOH required
(given = 1.70 mol)
1 mol CO2
or
1.70 mol NaOH 1 mol CO2
= 0.85 mol CO2 required
2 mol NaOH
(given = 1.00 mol)
So: NaOH is the limiting reactant
Stoichiometry
How many moles of Na2CO3 can be produced?
1.70 mol NaOH 1 mol Na2CO3
= 0.85 mol Na2CO3
produced
2 mol NaOH
How many moles of the excess reactant remain
after the completion of the reaction?
1) use L.R. to determine quantity of excess reactant
used.
2)subtract to find quantity that remains.
1.70 mol NaOH 1 mol CO2
2 mol NaOH
= 0.85 mol CO2 used
Stoichiometry
1.00 mol given – 0.85 mol used = 0.15 mol CO2 remains