Download Binnie Chapter 3

Chapter 3: Stoichiometry
Formula Weight
• A formula weight is the 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
• Formula weights are generally reported for ionic compounds.
Calculate the formula weight for:
• Aluminum hydroxide
• Calcium nitrate
© 2009, Prentice-Hall, Inc.
Molecular Weight (MW)
• A molecular weight is the 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
© 2009, Prentice-Hall, Inc.
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)
x 100
% element =
(FW of the compound)
© 2009, Prentice-Hall, Inc.
Percent Composition
• Calculate the percentage of carbon in ethane (C2H6)
• Calculate the percentage of oxygen in water
© 2009, Prentice-Hall, Inc.
The Mole
Avogadro’s Number
• 6.02 x 1023 things (atoms, particles, molecules…)
• 1 mole of 12C has a mass of 12 g.
• How many carbon atoms in .350 mol of glucose?
© 2009, Prentice-Hall,
Molar Mass
• By definition, a molar mass is 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).
• What is the mass of 5 moles of water?
© 2009, Prentice-Hall, Inc.
Using Moles
Number of
particles
Moles provide a bridge from the molecular scale
to the real-world scale.
© 2009, Prentice-Hall,
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.
© 2009, Prentice-Hall,
Practice
• Calculate the number of moles of sulfur
dioxide in 130 g.
• Calculate the mass in grams of .6 moles of
calcium nitrate.
Empirical Formulas from Analyses
One can calculate the empirical formula from the
percent composition.
Helpful Rhyme: % to mass, mass to mole, divide by small, times ’til whole.
© 2009, Prentice-Hall,
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.
remember: % to mass, mass to mole, divide by small, times ’til whole.
% to mass: assume 100 g
mass to mole:
divide by small:
times ’til whole
© 2009, Prentice-Hall, Inc.
para-aminobenzoic acid
C7H7NO2
© 2009, Prentice-Hall, Inc.
Practice
Ascorbic acid, vitamin C, contains 40.92% C, 4.58% H, and 54.50% O.
What is the empirical formula for ascorbic acid?
Molecular from Empirical
Molecular formulas are a whole number multiple of the empirical formula.
The empirical formula for ascorbic acid was C3H4O3. It’s molecular weight
is 176. What is the molecular formula?
© 2009, Prentice-Hall, Inc.
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.
© 2009, Prentice-Hall,
Steps for Combustion Analysis Problems
1. Calculate the moles of CO2 produced (same as mols of C)
2. Calculate the moles of H2O produced Multiply this number by
two to calculate the number of moles of H atoms present in the
original compound.
3. Calculate the mass of C and H
4. If there is another element present (typically O) in the
combusted substance then calculate its mass by subtracting
the mass of C and H from the total mass of the combusted
sample.
5. Once you have determined the mass of each piece of the
original compound the problem can be solved just like an
empirical formula problem.
EXAMPLE 1:
A 0.255g sample of isopropyl alcohol was burned in a combustion
train and 0.561g CO2 and 0.306g H2O were formed. Determine
the empirical formula of ascorbic acid.
Steps to Follow:
(1)
Calculate the moles of Carbon and Hydrogen in the compound.
(2)
Calculate the moles of Oxygen in the compound.
(3)
The mole ratios are the subscripts of the empirical formula. Use the “Empirical Formula
Rhyme” to complete the rest of the problem.
Combustion Analysis
• A sample of a common alcohol with a mass of
4.599 g, containing C, H and O, was
combusted in excess oxygen to yield 8.802 g of
CO2 and 5.394 g of H2O.
• The alcohol contains only carbon, hydrogen,
and oxygen. What is the empirical formula of
the alcohol?
EXAMPLE 2:
The combustion of 62.63 grams of a compound which contains
only C,H and O yields 134.43 grams of CO2 and 13.75 grams of
H2O. What is the empirical formula of the compound?
–
If the compound’s molecular weight is 156 g/mol, determine the molecular
formula.
EXAMPLE 3:
The combustion of 40.10 g of a compound which contains only C,
H, Cl and O yields 58.57 g of CO2 and 14.98 g of H2O. Another
sample of the compound with a mass of 75.00 g is found to contain
22.06 g of Cl. What is the empirical formula of the compound?
–
(Hint: Use the mass data to calculate the % composition of each element in
the compound.)
Reactions
Chemical equations are concise
representations of chemical reactions.
© 2009, Prentice-Hall, Inc.
Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
© 2009, Prentice-Hall, Inc.
Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
Reactants appear on the left
side of the equation.
© 2009, Prentice-Hall, Inc.
Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
Products appear on the
right side of the equation.
© 2009, Prentice-Hall, Inc.
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.
© 2009, Prentice-Hall, Inc.
Subscripts and Coefficients Give Different
Information
• Subscripts tell the number of atoms of
each element in a molecule.
© 2009, Prentice-Hall,
Subscripts and Coefficients Give Different
Information
• Subscripts tell the number of atoms of
each element in a molecule
• Coefficients tell the number of molecules.
© 2009, Prentice-Hall,
Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
Coefficients are inserted to
balance the equation.
© 2009, Prentice-Hall, Inc.
Combustion Reactions
• These are generally rapid
reactions that produce a
flame.
• Most often involve
hydrocarbons reacting
with oxygen in the air.
• Excess oxygen  CO2
• Insufficient oxygen  CO
• Examples:
– CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
– C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
© 2009, Prentice-Hall, Inc.
Combination Reactions
• In this type of
reaction two
or more
substances
react to form
one product.
• Examples:
– 2 Mg (s) + O2 (g)  2 MgO (s)
– N2 (g) + 3 H2 (g)  2 NH3 (g)
– C3H6 (g) + Br2 (l)  C3H6Br2 (l)
© 2009, Prentice-Hall, Inc.
Decomposition Reactions
• In a decomposition 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)
© 2009, Prentice-Hall,
Categories of Decomposition (and Composition ) Reactions
a) carbonates  metallic oxide + CO2
CaO + _____
CO2
CaCO3  _____
b) chlorates  metallic chloride + O2
NaCl + _____
O2
NaClO3  _____
c) hydroxides  metallic oxide + H2O
MgO + _____
H2O
Mg(OH)2  _____
d) oxy acids  nonmetal oxide + H2O
SO3 + _____
H2O
H2SO4  _____
e) binary compounds  2 elements
Na + _____
Cl2
NaCl  _____
•
•
Notice that in carbonates, hydroxides, and oxy acids, the oxygen
combines with both the metal AND the nonmetal!!
Every time you try to write the formula for a new ionic compound,
charges of the ions and ___________
cross
you must look up the ___________
them if they are different!!
Quantitative Information from
Balanced Equations
The coefficients in the balanced equation give the
ratio of moles of reactants and products.
You can only change the coefficients to balance
(never subscripts)
You can only use whole numbers to balance
© 2009, Prentice-Hall,
Balancing Hints:
–Balance the metals first.
–Balance the ion groups next.
–Balance the other atoms.
–Save the non ion group oxygen and
hydrogen until the end.
• Balance the following equations;
__Ca + __H2O → __H2 + __ Ca(OH)2
__ Cu + __ O2 → __ CuO
__ Na + __ O2 → __ Na2O
__ Fe + __ HCl → __ FeCl2 + __ H2
__ Fe + __ Br2 → __ FeBr3
__ C4H8 + __ O2 → __ CO2 + __ H2O
Practice writing reactions
Write the balanced chemical equation for:
• The combination reaction the occurs between lithium
metal and fluorine gas.
• The reaction that occurs when solid sodium azide (NaN3)
decomposes rapidly into it’s constituent elements as an
airbag is inflated.
• Si2H6 burns when exposed to air
Stoichiometric Calculations
Starting with 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).
© 2009, Prentice-Hall,
• Balance this equation showing the burning of
methane below, to answer the following
questions.
• __ CH4 + __ O2 → __ CO2 + __ H2O
How many grams of oxygen are required to
produce 2.23 g of carbon dioxide?
How many grams of water will be produced
when 34.0 g of methane is burned?
Stoichiometric Calculations
• How many grams of water are produced in
the combustion of 1 g of glucose?
© 2009, Prentice-Hall,
3.7 Limiting Reactants
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).
© 2009, Prentice-Hall,
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.
© 2009, Prentice-Hall,
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).
© 2009, Prentice-Hall,
Limiting Reactants
In the example below, the O2 would be the excess
reagent.
© 2009, Prentice-Hall,
Let’s Try This Problem Together
• H2(g) + O2(g) → H2O(l)
a. Calculate the stoichiometric ratio of moles H2 to moles O2
b. Calculate the moles of water created when 1.50 mol H2 is
mixed with 1.00 mol O2
c. Calculate the limiting reactant (H2 or O2) for the mixture in
(b)
Let’s Try This Problem Together
• For the reaction: Na2O + H2O → NaOH
What weight of NaOH could be made from 12.4
g of Na2O and 42.1 g of H2O?
What would be the limiting reagent if 100 g
each of Na2O and H2O were allowed to react?
Practice
• Part of the SO2 that is introduced into the atmosphere by combustion
of sulfur containing compounds ends up being converted to sulfuric
acid according to the following reaction:
SO2 (g) + O2 (g) + H2O(l)  H2SO4 (aq)
How much H2SO4 can be formed from 5.0 moles of SO2, 2.0 moles of O2,
and an unlimited supply of H2O?
Theoretical Yield
• The theoretical yield is the maximum 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, which is
the amount one actually produces and
measures.
© 2009, Prentice-Hall, Inc.
Percent Yield
One finds the percent yield by comparing
the amount actually obtained (actual yield)
to the amount it was possible to make
(theoretical yield).
Actual Yield
Percent Yield =
x 100
Theoretical Yield
© 2009, Prentice-Hall, Inc.