Transcript Chapter 3 PPT - Richsingiser.com

Daniel L. Reger
Scott R. Goode
David W. Ball
http://academic.cengage.com/chemistry/reger
Chapter 3
Equations, the Mole,
and Chemical Formulas
Chemical Equation
• Stoichiometry is the study of the
quantitative relationships involving the
substances in chemical reactions.
• A chemical equation describes the
identities and relative amounts of
reactants and products in a chemical
reaction.
• Just like a chemical formula, a chemical
equation expresses quantitative relations.
Chemical Reactions
•
A chemical equation is a shorthand
notation to describe a chemical reaction.
2Mg + O2

2MgO
magnesium react to form magnesium
and oxygen
oxide
Definitions
•
•
•
•
Reactants are the substances
consumed.
Products are the substances formed.
Coefficients are numbers before the
formula of a substance in an equation.
A balanced equation has the same
number of atoms of each element on
both sides of the equation.
Writing Balanced Equations
•
•
Write the correct formula for each
substance.
H2 + Cl2  HCl (unbalanced)
Add coefficients so the number of atoms
of each element are the same on both
sides of the equation.
H2 + Cl2  2HCl (balanced)
Balancing Chemical Equations
•
•
•
•
•
Write the correct formula for each substance.
C5H12 + O2  CO2 + H2O
Assume one molecule of the most complicated
substance. Adjust the coefficient of CO2 to
balance C.
C5H12 + O2  5CO2 + H2O
Adjust the coefficient of H2O to balance H.
C5H12 + O2  5CO2 + 6H2O
Adjust the coefficient of O2 to balance O.
C5H12 + 8O2  5CO2 + 5H2O
Check the balance by counting the number of
atoms of each element.
Test Your Skill
•
Balance the equation
C4H9OH + O2  CO2 + H2O
Test Your Skill
•
Balance the equation
C4H9OH + O2  CO2 + H2O
•
Answer:
C4H9OH + 6O2  4CO2 + 5H2O
Balancing Equations
•
Sometimes fractional coefficients are
obtained.
C5H10
C5H10
C5H10
C5H10
•
+
+
+
+
O2  CO2 + H2O
O2  5CO2 + H2O
O2  5CO2 + 5H2O
15/2O2  5CO2 + 5H2O
Multiply all coefficients by the
denominator.
2C5H10 + 15O2  10CO2 + 10H2O
Neutralization Reactions
•
Neutralization is the reaction of an acid
with a base to form a salt and water.
•
•
•
An acid is a compound that dissolves in
water to produce hydrogen ions.
A base dissolves in water to produce
hydroxide ions. It is usually a soluble
metal hydroxide.
A salt is an ionic compound consisting of
the cation of a base and the anion of an
acid.
Examples of Neutralization
HCl
+ NaOH 
H2SO4 + 2KOH 
2HClO4 + Ca(OH)2 
acid
base

NaCl +
H2O
K2SO4 + 2H2O
Ca(ClO4)2 + 2H2O
salt
water
Balancing Neutralization Reactions
•
The number of hydrogen ions provided
by the acid must equal the number of
hydroxide ions provided by the base.
H3PO4 + 3NaOH  Na3PO4 + 3H2O
3H+ + 3OH- 
3H2O
Combustion Reactions
•
A combustion reaction is the process
of burning.
• Most combustion reactions are the
combination of a substance with oxygen.
• When an organic compound burns in
oxygen, the carbon is converted to CO2,
and the hydrogen forms water, H2O.
Test Your Skill
•
Identify the type of the reaction, then
balance it.
(a) (C2H5)2O + O2  CO2 + H2O
(b) H2SO4 + Ca(OH)2  CaSO4 + H2O
Oxidation and Reduction
• Oxidation is the loss of electrons by a
chemical process.
• When sodium forms a compound, Na+ is
formed. Sodium is oxidized.
• Reduction is the gain of electrons by a
chemical process.
• When Cl- ions are formed from elemental
chlorine, chlorine is reduced.
Oxidation-Reduction (“Redox”)
• An oxidation-reduction reaction, or
redox reaction, is one in which electrons
are transferred from one species to
another.
• In every redox reaction, at least one species
is oxidized and at least one species is
reduced.
• 2Na(s) + Cl2(g) → 2NaCl(s) is a redox
reaction because Na is oxidized and Cl is
reduced.
Oxidizing and Reducing Agents
• An oxidizing agent is the reactant that
accepts electrons, causing an oxidation to
occur.
• The oxidizing agent is reduced.
• A reducing agent is the reactant that
supplies electrons, causing a reduction to
occur.
• The reducing agent is oxidized.
• In the reaction of sodium with chlorine, Na is
the reducing agent and Cl2 is the oxidizing
agent.
Half-reactions
• In a half-reaction, either the oxidation or
reduction part of a redox reaction is
given, showing the electrons explicitly.
• Half-reactions emphasize the transfer of
electrons in a redox reaction.
• For 2Na(s) + Cl2(g) → 2NaCl(s) :
• Na → Na+ + 1e• Cl2 + 2e- → 2Cl-
oxidation half-reaction
reduction half-reaction
Oxidation States
• The oxidation state is the charge on the
monatomic ion, or the charge on an atom
when the shared electrons are assigned
to the more electronegative atom.
• Electron pairs shared by atoms of the same
element are divided equally.
• In CaCl2, an ionic compound:
• calcium has an oxidation state of +2.
• chlorine has an oxidation state of -1.
Rules for Assigning Oxidation Numbers
1. Oxidation numbers for atoms in their
elemental form are 0.
2. The oxidation number of a monatomic
ion is equal to the charge on the ion.
3. In compounds, F is always -1. Other
halogens are also -1 unless they are
combined with a more electronegative
element (O or a halogen above it in the
periodic table).
Rules for Assigning Oxidation Numbers
(cont’d)
4. In compounds, O is -2 except for
peroxides (where it is -1) or when
combined with F.
5. In compounds, H is +1 except in metal
hydrides, where it is -1.
6. The sum of all the oxidation numbers of
the atoms in a substance must sum to
the charge on the substance.
Assigning Oxidation Numbers
• Assign oxidation numbers for each atom
in K2CrO4.
•
•
•
•
K = +1 by rule 2.
O = -2 by rule 5.
Cr = +6 by rule 6.
2(+1) + 6 + 4(-2) = 0, the overall charge on
the substance.
Test Your Skill
• Assign oxidation numbers to each atom
in the following substances.
(a) PF3
(b) CO (c) NH4Cl
Balancing Redox Equations
• Determine oxidation numbers that are
changing and write the skeleton halfreactions.
• Balance each half-reaction separately.
•
•
•
•
•
Balance element being oxidized or reduced.
Balance all other elements except H and O.
Balance O by adding H2O as needed.
Balance H by adding H+ as needed.
Balance charges by adding e- as needed.
Balancing Redox Equations (cont’d)
• Multiply one or both reactions by an
integer so that the number of electrons in
both half-reactions is the same.
• Add the two half-reactions, canceling out
the electrons and any other species that
appears on both sides of the equation.
Balancing Redox Equations (cont’d)
• Balance the following equation.
• Cu + NO3- → Cu2+ + NO
• Cu is oxidized from 0 to +2
• Write a skeleton equation for Cu:
• Cu → Cu2+
• No other step necessary except to
balance charges with electrons:
• Cu → Cu2+ + 2e-
Balancing Redox Equations (cont’d)
• N is reduced from +5 to +2.
• Write a skeleton equation for N:
• NO3- → NO
• Balance O by adding water:
• NO3- → NO + 2H2O
• Balance H by adding H+:
• 4H+ + NO3- → NO + 2H2O
• Balance charge by adding e-:
• 3e- + 4H+ + NO3- → NO + 2H2O
Balancing Redox Equations (cont’d)
• Make number of electrons in both
reactions the same by multiplying to
least common multiple (6, in this case):
• 3×(Cu → Cu2+ + 2e-)
• 2×(3e- + 4H+ + NO3- → NO + 2H2O)
• Combine the two reactions and cancel
as appropriate:
• 3Cu + 8H+ + 2NO3- → 3Cu2+ + 2NO + 4H2O
• Reaction is balanced.
Balancing Redox Reactions
• In basic solutions, add OH- to each side
of the reaction to eliminate any H+ by
combining them to make H2O.
Test Your Skill
• Balance the following equation in acid
solution:
• Cr2O72- + C2H5OH → Cr3+ + CO2
Test Your Skill
• Balance the following equation in basic
solution:
• Zn + ClO- → Zn(OH)42- + Cl-
The Mole
•
•
One mole is the amount of substance
that contains as many entities as the
number of atoms in exactly 12 grams of
the 12C isotope of carbon.
Avogadro’s number is the
experimentally determined number of 12C
atoms in 12 g, and is equal to
6.022 x 1023.
One Mole of Several Elements
Argon in green balloons
H
g
Cu
Na
Fe
Al
Converting Moles and Entities
•
One mole of anything contains
6.022 x 1023 entities.
•
•
•
•
1 mol H = 6.022 x 1023 atoms of H
1 mol H2 = 6.022 x 1023 molecules of H2
1 mol CH4 = 6.022 x 1023 molecules of
CH4
1 mol CaCl2 = 6.022 x 1023 formula units of
CaCl2
Moles to Number of Entities
Avogadro’s number allows the interconversion
of moles and numbers of atoms or molecules.
Moles of
substance
Avogadro’s
number
Number of
atoms or
molecules
Example: Convert Moles and Entities
•
•
How many atoms are present in 0.35 mol
of Na?
How many moles are present in 3.00 x
1021 molecules of C2H6?
Molar Mass
•
The molar mass (M) of any atom,
molecule or compound is the mass (in
grams) of one mole of that substance.
• The molar mass in grams is numerically
equal to the atomic mass or molecular mass
expressed in u.
• Molar mass converts from mass (in grams)
to amount (in moles) or the reverse.
Molar and Atomic Masses
Atomic Scale
Lab Scale
Substance
Name
Mass
Molar Mass
Ar
C2H6
atomic mass
molecular
mass
formula mass
39.95 u
30.07 u
39.95 g/mol
30.07 g/mol
41.99 u
41.99 g/mol
NaF
Converting Moles and Mass
Moles of
substance
Molar mass
of substance
Mass of
substance
Molar Mass Conversion
•
What is the mass of 0.25 moles of CH4?
•
Molar mass of CH4 = 16.0 g/mol.
Example: Molar Mass Conversions
•
•
How many moles of ethylene (C2H4,
M = 28.0 g/mol) are present in 16 g of
that compound?
What is the mass, in grams, of 0.178
moles of Fe?
Test Your Skill
•
What mass of compound must be
weighed out to have a 0.0223 mol
sample of H2C2O4 (M = 90.04 g/mol)?
Mass Percentage from Formula
• Use the formula of the compound to
calculate the mass of each element in
the compound and use those numbers to
calculate percentage composition.
Example: Percentage Calculation
•
What is the mass percentage
composition of H2C2O4?
A Combustion Experiment
The sample burns in excess O2: the H2O
is trapped by the CaCl2 and CO2 is trapped by
the NaOH.
Carbon and Hydrogen Content
• The masses of C and H in the sample are
calculated from the masses of H2O and
CO2 formed in the combustion reaction.
Example: Mass C, H and O in Sample
•
A compound contains only C, H, and O.
A 0.1000 g-sample burns completely in
oxygen to form 0.0930 g of water and
0.227 g of CO2. Calculate the mass of
each element in this sample.
Calculate Empirical Formula
Calculate Empirical Formula
•
What is the empirical formula of a
compound that contains 0.799 g C and
0.201 g H in a 1.000 g sample?
Example: Empirical Formula
•
What is the empirical formula of a
chromium oxide that is 68.4% Cr and
31.6% O?
Test Your Skill
•
What is the empirical formula of a
compound that is 59.9% Ti and 40.1%
O?
Molecular Formula
• The molecular formula must be a whole
number multiple of the empirical formula.
• If the empirical formula is CH2, the molecular
formula is (CH2)n where
molar mass of compound
n
molar mass of empirical formula
• The molecular mass must be measured
experimentally.
Example: Molecular Formula
• An empirical formula calculation based
on a combustion analysis experiment
shows a new compound has the
empirical formula C4H8O. A mass
spectrometry experiment determines that
the molar mass of the compound is 216
g/mol. What is the molecular formula?
Mole Relationships in Equations
Guidelines for Reaction Stoichiometry
•
•
•
•
Write the balanced equation.
Calculate the number of moles of the
species for which the mass is given.
Use the coefficients in the equation to
convert the moles of the given
substance into moles of the substance
desired.
Calculate the mass of the desired
species.
Reaction Stoichiometry
Example: Stoichiometry
•
What mass of SO3 forms from the
reaction of 4.1 g of SO2 with an excess of
O2?
Test Your Skill
•
Given the equation
4FeS2 + 11O2 2Fe2O3 + 8SO2
what mass of SO2 is produced from
reaction of 3.8 g of FeS2 with excess
oxygen?
Theoretical Yield
•
The previous calculation showed that
given the equation
4FeS2 + 11O2 2Fe2O3 + 8SO2
4.1 g SO2 is produced from reaction of
3.8 g of FeS2 with excess oxygen.
•
The 4.1 g SO2 is the theoretical yield –
the maximum quantity of product that
can be obtained from a chemical
reaction, based on the amounts of
starting materials.
Limiting Reactant
• Limiting reactant: the reactant that is
completely consumed when a chemical
reaction occurs.
Limiting Reactant
• In the below reaction of Cl2 (green) and Na
(purple), Cl2 is the limiting reactant. Once
the limiting reactant is consumed the
reaction stops and no more product forms.
The formed NaCl and the excess Na are
present at the end of the reaction.
Example: Limiting Reactant
•
Calculate the mass of the NH3 product
formed (theoretical yield) when 7.0 g of
N2 reacts with 2.0 g of H2.
Strategy for Limiting Reactant
Mass of A
(reactant)
Mass of B
(reactant)
Molar mass of A
Molar mass of B
Moles of A
Moles of B
Coefficients in the equation
Moles of
Product
Choose
smaller
amount
Moles of
Product
Molar mass of product
Mass of
product
Example: Limiting Reactant
•
Calculate the mass of the NH3 product
formed (theoretical yield) when 7.0 g of
N2 reacts with 2.0 g of H2.
Actual Yield, Percent Yield
•
•
As shown in the picture,
in many reactions not all
of the product formed
can be isolated.
Actual yield: mass of
product isolated in a
reaction.
Yields are generally reported as a percent:
Actual yield
Percent yield 
 100%
Theoretical yield
Example: Calculating Percent Yield
Given the reaction,
PCl3 + Cl2  PCl5
what is the percent yield when 50.0 g of
PCl3 reacts with 35.0 g Cl2 and a
chemist isolated 61.3 g of PCl5.?