Download Chapter 5 Atomic Structure

Topic 1: Stoichiometric Relationships
1.1-1.3
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Honors Chemistry
Mrs. Peters
2014-15
1.1: Introduction to the particulate nature of
matter and chemical change
Applications:
1.
Deduction of chemical equations when
reactants and products are specified.
2.
Application of the state symbols (s), (l), (g),
and (aq) in equations.
*Understandings 1-3, Application 3 were covered in Unit 1
with Topic 11.
1.1 A1. Deduce chemical equations when all
reactants and products are specified
What is a chemical reaction?
Rearrangement of atoms forming new substances
Reactants  Products
Atoms (matter) are not created or destroyed, they
are rearranged! (Law of conservation of matter)
1.1 A1. Deduce chemical equations when all
reactants and products are specified
Some reactions are desirable…
Glucose + oxygen  Carbon dioxide + water
…some are not.
Iron + oxygen  iron (III) oxide
1.1 A.2 Application of the state
symbols (s), (l), (g) and (aq)
Symbols used in chemical reactions
+
Used to separate two reactants or products
“Yields” separates products from reactants
 Used in place of  for reversible reactions
1.1 A.2 Application of the state
symbols (s), (l), (g) and (aq)
Symbols used in chemical reactions
*(s)
*(l)
*(g)
*(aq)

Designates a solid reactant or product
Designates a liquid reactant or product
Designates a gaseous reactant or product
Designates an aqueous reactant or product
Indicates that heat is supplied to the reaction
* Required
1.1 A1. Deduce chemical equations when all
reactants and products are specified.
Balancing chemical equations
The Law of Conservation of Matter: In a chemical (non-
nuclear) reaction, atoms are neither created nor destroyed.
For an equation to be balanced the number of atoms of each element is
the same on both sides of the equation.
H2(g) + O2(g)  H2O (l)
2H2(g) + O2(g)  2H2O (l)
(unbalanced)
(balanced)
K(s) + H2O(l)  KOH (aq) + H2(g)
2K(s) + 2H2O(l)  2KOH (aq) + H2(g)
(unbalanced)
(balanced)
C6H12O6 + O2  CO2 + H2O
C6H12O6 + 6O2  6CO2 + 6H2O
(unbalanced)
(balanced)
1.1 A1. Deduce chemical equations when
all reactants and products are specified.
Rules for Balancing Equations
1)
Write the correct formulas for the reactants on the left
side of the yield sign and products on the right side.
2)
Count the number of atoms of each element in the
products and the reactants.
3)
Balance the elements one at a time by using coefficients.
Do not change the subscripts in the chemical formulas.
4)
Check each atom or polyatomic ion to make sure the
equation is balanced.
5)
Make sure all coefficients are in the lowest possible ratio.
1.1 A1. Deduce chemical equations when all
reactants and products are specified.
N2 + H2  NH3
Rules for Balancing Equations
1) Write the correct formulas for the
reactants on the left side of the
yield sign and products on the
right side.
2) Count the number of atoms of
each element in the products and
the reactants.
3) Balance the elements one at a time
by using coefficients. Do not
change the subscripts in the
chemical formulas.
4) Check each atom or polyatomic ion
to make sure the equation is
balanced.
5) Make sure all coefficients are in the
lowest possible ratio.
1) Formulas given
2) Reactants
Products
2N 2H  1N 3H
3) Balance N by putting 2 in front of NH3
N2 + H2  2NH3
Balance H by putting 3 in front of H2
N2 + 3H2  2NH3
4)
2N
6H  2N
6H
5)
Lowest ratio of coefficients
1.1 A1. Deduce chemical equations when all
reactants and products are specified.
Rules for Balancing Equations
1) Write the correct formulas for the
reactants on the left side of the
yield sign and products on the right
side.
2) Count the number of atoms of each
element in the products and the
reactants.
3) Balance the elements one at a time
by using coefficients. Do not
change the subscripts in the
chemical formulas.
4) Check each atom or polyatomic ion
to make sure the equation is
balanced.
5) Make sure all coefficients are in the
lowest possible ratio.
KClO3  KCl + O2
1) Formulas given
2) Reactants
Products
1K 1Cl 3O  1K 1Cl 2O
3) Balance O by putting 2 in front of
KClO3
and 3 in front of O2
2KClO3  KCl + 3O2
Balance K & Cl by putting 2 in front
of KCl
2KClO3  2KCl + 3O2
4)
5)
2K 2Cl 6O  2K 2Cl 6O
Lowest ratio of coefficients
1.1 A1. Deduce chemical equations when
all reactants and products are specified.
Balance the following equations

Fe(s) + O2(g)  Fe2O3(s)

CaCO3(s)  CaO(s) + CO2(g)

Al(s) + Fe2O3(s)  Al2O3(s) + Fe(s)

H2SO4(aq) + NaCN(aq)  Na2SO4(aq) + HCN(g)

C2H5OH(l) + O2(g)  CO2(g) + H2O(g)
1.1 A1. Deduce chemical equations when
all reactants and products are specified.
Balance the following equations

4Fe(s) + 3O2(g)  2Fe2O3(s)

CaCO3(s)  CaO(s) + CO2(g)

2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(s)

H2SO4(aq) + 2NaCN(aq)  Na2SO4(aq) + 2HCN(g)

C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(g)
1.1 A1. Deduce chemical equations when
all reactants and products are specified.
The 7 Diatomics (KNOW THESE)
Seven elements are diatomic, meaning they have to have two atoms
when alone:
H2, N2, O2, F2, Cl2, Br2, I2
1.1 A1. Deduce chemical equations when all
reactants and products are specified.
Write the equation, then balance:
Aluminum bromide + chlorine yield
aluminum chloride + bromine
AlBr3 + Cl2  AlCl3 + Br2
1.1 A1. Deduce chemical equations when all
reactants and products are specified.
Write the equation, then balance:
Aluminum bromide + chlorine yield
aluminum chloride + bromine
2AlBr3 + 3Cl2  2AlCl3 + 3Br2
1.1 A1. Deduce chemical equations when all
reactants and products are specified.
Write the equation, then balance:
Copper + oxygen produces copper(I) oxide
Cu + O2  Cu2O
1.1 A1. Deduce chemical equations when all
reactants and products are specified.
Write the equation, then balance:
Copper + oxygen produces copper(I) oxide
4Cu + O2  2Cu2O
1.1 A1. Deduce chemical equations when
all reactants and products are specified.
Write the equations, then balance:
1.
Sodium chlorate decomposes to sodium chloride and
oxygen gas
2.
3.
Aluminum nitrate plus sodium hydroxide yields
aluminum hydroxide plus sodium nitrate
Ethane (C2H6) burns in oxygen to produce carbon
dioxide and water vapor
1.1 A1. Deduce chemical equations when
all reactants and products are specified.
1. Sodium chlorate decomposes to sodium chloride and oxygen
gas
NaClO3  NaCl + O2
2. Aluminum nitrate plus sodium hydroxide yields aluminum
hydroxide plus sodium nitrate
Al(NO3)3 + NaOH  Al(OH)3 + NaNO3
3. Ethanol (C2H5OH) burns in oxygen to produce carbon
dioxide and water vapor
C2H5OH + O2  CO2 + H2O
1.1 A1. Deduce chemical equations when
all reactants and products are specified.
1. Sodium chlorate decomposes to sodium chloride and oxygen
gas
2NaClO3  2NaCl + 3O2
2. Aluminum nitrate plus sodium hydroxide yields aluminum
hydroxide plus sodium nitrate
Al(NO3)3 + 3NaOH  Al(OH)3 + 3NaNO3
3. Ethanol (C2H5OH) burns in oxygen to produce carbon
dioxide and water vapor
C2H5OH + 3O2  2CO2 + 3H2O
1.1 A1. Deduce types of reactions
Synthesis or Combination Reaction: Two or more
substances combine to form a single substance.
R + S  RS


2Na(s) + Cl2(g)  2NaCl(s)
CaO(s) + H2O(l)  Ca(OH)2(aq)
1.1 A1. Deduce types of reactions
Decomposition Reactions: A single substance is broken
down into two or more substances.
RS  R + S


CaCO3(s)  CaO(s) + CO2(g)
2H2O(l)  2H2(g) + O2(g)
1.1 A1. Deduce types of reactions
Single-Replacement Reactions: One element replaces
another in a compound.
R + ST  RT + S


Mg(s) + Zn(NO3)2(aq)  Mg(NO3)2(aq) + Zn(s)
2K(s) + 2H2O(l)  2KOH (aq) + H2(g)
1.1 A1. Deduce types of reactions
Double-Replacement Reactions: Ions of two reacting
compounds trade places.
RS + TU  RU + TS
(cation1 anion1 + cation2 anion2  cation1 anion2 + cation2 anion1)


Na2CO3(aq) + CaCl2(aq)  CaCO3(s) + 2NaCl(aq)
Na2S(aq) + Cd(NO3)2(aq)  CdS(s) + 2NaNO3(aq)
1.1 A1. Deduce types of reactions
Combustion Reactions: An element or compound reacts with
oxygen. Products are usually carbon dioxide and water.
CxHy + O2  CO2 + H2O


CH4(g) + O2(g)  CO2(g) + H2O(g)
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
1.1 A1. Identify the types of
reactions to predict products
Combination Reaction: R + S  RS
Decomposition Reaction: RS  R + S
Single-Replacement Reaction: T + RS  TS + R
Double-Replacement Reaction:
R+S- + T+U-  R+U- + T+SCombustion Reaction: CxHy + O2  CO2 + H2O
1.1 Identify the mole ratio of any two
species in a chemical reaction.
How do you calculate the quantity of reactants and
products in chemical reactions?
The calculation of quantities in chemical reactions is called stoichiometry.
Use the chemical equation to predict ratio of reactants and products
N2
+
3H2

2NH3
1 molecule N2 reacts with 3 molecules H2 producing 2 molecules NH3
1 mole N2 reacts with 3 moles H2 producing 2 moles NH3
The balanced equations gives us the ratio in particles or moles (not mass) of the
chemicals involved in the reaction.
The ratio of N2 to NH3 is 1:2
The ratio of NH3 to H2 is 2:3
The ratio of H2 to N2 is 3:1
1.1 Identify the mole ratio of any
two species in a chemical reaction.
Coefficients are in particles or in moles
2 NaClO3  2 NaCl + 3 O2
We will use moles because we measure moles in grams
The coefficients give us mole ratios
Mole ratio NaClO3:NaCl is 2:2
Mole ratio NaCl:O2 is 2:3
Mole ratio O2:NaCl is 3:2
1.1 Identify the mole ratio of any
two species in a chemical reaction
Consider the following equation
2K2Cr2O7 + 2H2O + 3S  4KOH + 2Cr2O3 + 3SO2
What
What
What
What
is the KOH:S mole ratio?
is the K2Cr2O7:Cr2O3 mole ratio?
is the mole ratio between sulfur dioxide and water?
species will give mole ratios of 4:2?
1.1 Identify the mole ratio of any
two species in a chemical reaction.
Consider the following equation
2K2Cr2O7 + 2H2O + 3S  4KOH + 2Cr2O3 + 3SO2
What is the KOH:S mole ratio? 4:3
What is the K2Cr2O7:Cr2O3 2:2
What is the mole ratio between sulfur dioxide and water?
3:2
What species will give mole ratios of 4:2?
KOH:K2Cr2O7 KOH:H2O KOH:Cr2O3
1.1 Identify the mole ratio of any two
species in a chemical reaction.
Consider the reaction:
2C8H18 + 25O2  16CO2 + 18H2O
What
What
What
What
is
is
is
is
the
the
the
the
ratio
ratio
ratio
ratio
of
of
of
of
O2 to CO2? 25:16
H2O to C8H18? 18:2
CO2 to H2O?
O2 to C8H18?
Remember these ratios are in particles or moles. We will be
using moles.
1.2: The Mole Concept
Understandings:
1.
The mole is a fixed number of particles and refers
to the amount ,n, of a substance.
2.
Masses of atoms are compared on a scale relative
to 12C and are expressed as relative atomic mass
(Ar) and relative formula/molecular mass (Mr)
3.
Molar Mass (M) has the units g mol-1
4.
The empirical formula and molecular formula of a
compound give the simplest ratio and the actual
number of atoms present in a molecule
respectively.
1.2: The Mole Concept
Applications and Skills:
1. Calculation of the molar masses of atoms, ions, molecules
and formula units.
2. Solutions of problems involving the relationships between
the number of particles, the amount of substance in moles
and the mass in grams.
3. Inter conversion of the percentage composition by mass
and the empirical formula
4. Determination of the molecular formula of a compound
from its empirical formula and molar mass
5. Obtaining and using experimental data for deriving
empirical formulas from reactions involving mass changes.
1.2 The Mole
1 mole of any substance contains 6.02 x 1023
particles of that substance


This is called Avogadro’s number (named after a scientist).
Applies to all kinds of particles such as:
atoms, molecules, formula units, ions, electrons





34
How
How
How
How
How
many
many
many
many
many
atoms are in 1 mol of Au?
molecules are in 1 mol of CO?
formula units are in 1 mol of NaCl?
ions are in 1 mol of S2-?
electrons are in 1 mol of H atoms?
6.02
6.02
6.02
6.02
6.02
X 1023 atoms
x 1023 molecules
x 1023 f.u.
x 1023 ions
x 1023 electrons
1.2 U2. Relative atomic mass (Ar)
and relative molecular mass (Mr).

The relative atomic mass (Ar) of an element is the
weighted mean mass of all of the naturally occurring
isotopes of an element relative to one twelfth of the
mass of a carbon-12 atom.
 This explains why the relative atomic masses given in
data books are not exactly whole numbers.
 Relative atomic masses have no units.
 (This is review from the Unit on Atomic Structure and
Relative Atomic Mass calculations)
35
1.2 U2. Relative atomic mass (Ar)
and relative molecular mass (Mr).

The relative molecular mass or formula mass (Mr) of a
compound is the sum of the relative atomic masses of
the elements in the compound.
36
1.2 U3. Molar Mass


Molar Mass = mass of one mole of a substance
Units: g per mole or g mol-1
37
1.2 Molar Mass (g mol-1)
Moles
1 mole of a substance contains the number of particles in the
formula mass measured in grams (molar mass)
1 particle of oxygen has a relative mass of 16.0 so
1 mole of oxygen particles has a mass of 16.0 g
1 particle of water has a relative mass of 18.0 so
1 mole of water particles has a mass of 18.0 g
38
1.2 A1. Molar Mass
Molar Mass (all masses round to the hundredths place)

The relative atomic mass of an element can be found on the periodic
table (e.g. Na = 23.00).

The relative molecular mass or formula mass of a compound is the sum
of the atomic masses of the elements in the compound.
Examples:
Compound
Oxygen
Water
Sodium phosphate
Iron(III) nitrate
39
Formula Molar Mass Calculation
O2
2(16.00) = 32.00
H2 O
2(1.00) + 16.00 = 18.00
Na3PO4 3(23.00) + 31.00 + 4(16.00) = 164.00
Fe(NO3)3 55.80 + 3(14.00) + 9(16.00) = 241.80
1.2 A1. Molar Mass Calculations
Molar Mass
(all formula masses round to the hundredth place)

Find the molar mass of Cl2O7

Find the molar mass of Mg(MnO4)2

Find the molar mass of sulfur trioxide

Find the molar mass of copper (II) carbonate
40
1.2 A1. Molar Mass Calculations
Molar Mass
(all formula masses round to the hundredths place)
Find the molar mass of Cl2O7
2(35.50) + 7(16.00) = 183.00
 Find the molar mass of Mg(MnO4)2
1(24.30) + 2(54.90) + 8(16.00) = 262.10
 Find the molar mass of sulfur trioxide (SO3)
1(32.10) + 3(16.00) = 80.10
 Find the molar mass of copper (II) carbonate (CuCO3)
1(63.50) + 1(12.00) + 3(16.00) = 123.50

41
1.2 A1. Molar Mass (g mol-1)
MOLES
Scientists need to measure amounts of substances in grams, not amu.
1 mole of a substance contains the number of particles found
in the formula mass measured in grams (gram-formula mass)
1 mole of Na3PO4 particles has a mass of 164.0 grams
1 mole of Fe(NO3)3 particles has a mass of 241.8 grams
What is the mass in grams of 1 mole of K2CrO4 particles?
What is the mass in grams of 1 mole of CH3OH particles?
42
1.2 A2. Problems involving the relationship between
the number of particles, amount of substance in
moles, and mass
Converting Moles to Grams
Multiply the mole amount given by the molar mass
to find grams
What is the mass in grams of 1 mole of K2CrO4 particles? 194.2 g
What is the mass in grams of 1 mole of CH3OH particles? 32.0 g
What would be the mass of 1.72 moles of K2CrO4?
1.72 mol x 194.2 g/mol = 334 g
What would be the mass of 0.308 mol CH3OH?
0.308 mol x 32.0 g/mol = 9.86 g
43
1.2 A2. Problems involving the relationship between
the number of particles, amount of substance in
moles, and mass
Converting Moles to Grams
Multiply the mole amount given by the molar mass
to find grams
What is the mass in grams of 1 mole of K2CrO4 particles? 194.2 g
What is the mass in grams of 1 mole of CH3OH particles? 32.0 g
What would be the mass of 1.72 moles of K2CrO4?
1.72 mol x 194.2 g/mol = 334 g
What would be the mass of 0.308 mol CH3OH?
0.308 mol x 32.0 g/mol = 9.86 g
44
1.2 A2. Problems involving the relationship between
the number of particles, amount of substance in
moles, and mass
Converting Grams to Moles
Divide the gram amount given by the molar mass to
find moles
What is the mass in grams of 1 mole of K2CrO4 particles? 194.2 g
What is the mass in grams of 1 mole of CH3OH particles? 32.0 g
How many moles are in 124 g of K2CrO4?
124 g / 194.2 g/mol = 0.639 mol
If you had 1.93 g of CH3OH, how many moles would you have?
1.93 g / 32.0 g/mol = 0.0603 mol
45
1.2 A2. Problems involving the relationship between
the number of particles, amount of substance in
moles, and mass
Review




Moles  Grams:
Grams Moles:
Moles  Particles:
Particles  Moles:
Grams
mol x molar mass
grams / molar mass
mol x 6.02 x 1023
particles / 6.02 x 1023
/ Formula mass
X Formula mass
Grams
Moles
46
Particles
/ 6.02 E23
X 6.02 E23
Particles
1.2 A2. Problems involving the relationship between
the number of particles, amount of substance in
moles, and mass
Grams to Particles

How many particles are in 6.91 grams BaSO4?
 Convert grams to moles: 6.91 / 233.4 = 0.0296 mol
 Convert moles to particles : 0.0296 x 6.02 E23 = 1.78 x 1023 f.u.
Grams
/ Formula mass
X Formula mass
Grams
Moles
47
Particles
/ 6.02 E23
X 6.02 E23
Particles
1.2 A2. Problems involving the relationship between
the number of particles, amount of substance in
moles, and mass
Particles to Grams

How many grams are in 7.04 x 1023 molecules N2O?
 Convert molecules to moles: 7.04 x1023 / 6.02 x 1023 = 1.17 mol
 Convert moles to grams : 1.17 x 44.0 = 51.5 g
Grams
/ Formula mass
X Formula mass
Grams
Moles
48
Particles
/ 6.02 E23
X 6.02 E23
Particles
1.2 A2. Problems involving the relationship between the
number of particles, amount of substance in moles, and
mass
Practice

How many grams are in 1.91 x 1024 f.u. Na2CO3

How many molecules are in 154 g PCl3?
Grams
/ Formula mass
X Formula mass
Grams
Moles
49
Particles
/ 6.02 E23
X 6.02 E23
Particles
1.2 A2. Problems involving the relationship between the
number of particles, amount of substance in moles, and
mass
Practice


How many grams are in 1.91 x 1024 f.u. Na2CO3
 336 g
How many molecules are in 154 g PCl3?
 6.74 x 1023
Grams
/ Formula mass
X Formula mass
Grams
Moles
50
Particles
/ 6.02 E23
X 6.02 E23
Particles
1.2 A3. Percentage
Composition
Determine the % composition of a compound.
Find molar mass (formula mass)
1.
Divide the individual atom masses by the molar mass
and multiply by 100.
2.
51
1.2 A3. Percentage
Composition
Determine the % composition of ammonium nitrate.
1.
2.
Find formula mass
NH4NO3 = 14.00 + 4(1.00) + 14.00 + 3(16.00) =
80.00
Divide the individual masses by the formula mass and
multiply by 100%
Percent nitrogen
= (2(14.00) / 80.00) * 100
= 35.00 % N
Percent hydrogen = (4(1.00) / 80.0) * 100 = 5.00 % H
Percent oxygen
= (3(16.00) / 80.0) * 100
= 60.00 % O
52
1.2 A3. Percentage
Composition
Determine the % composition of sodium chloride.
1.
2.
53
Find formula mass
Formula mass of NaCl is 23.0 + 35.5 = 58.5
Divide the individual masses by the formula mass and
multiply by 100%
%Na = (23.0 / 58.5) x 100 = 39.3 %
% Cl = (35.5 / 58.5) x 100 = 60.7 %
1.2 A3. Percentage
Composition
Determine the % of nitrogen in urea (NH2)2CO.
Formula mass of (NH2)2CO is 2(14.0) + 4(1.0) + 12.0 + 16.0 = 60.0
% N = 28.0/60.0 * 100% = 46.7% N
Determine the mass of nitrogen in 85.8 g urea
85.8 g x 0.467 = 40.1 g

Determine the percentage composition of Mg(ClO3)2.

Determine the percentage of Fe in FeCl3.

Determine the mass of Fe found in 51.3 grams FeCl3.
54
1.2 A3. Percentage Composition



Determine the percentage composition of Mg(ClO3)2.
24.3 + 2(35.5) + 6(16.0) = 191.3
% Mg = 24.3 / 191.3 * 100% = 12.7%
% Cl = 71.0 / 191.3 * 100% = 37.1%
% O = 96.0 / 191.3 * 100% = 50.2%
Determine the percentage of Fe in FeCl3.
55.9 + 3(35.5) = 162.4
%Fe = 55.9 / 162.4 * 100% = 34.4%
Determine the mass of Fe found in 51.3 grams FeCl3.
51.3g x .344 = 17.6 g
55
1.2 U4. Empirical formula and Molecular
formula.
Empirical Formula
Smallest whole number ratio of
atoms in a compound
Molecular Formula
Actual ratio of atoms in a
compound
Sometimes the empirical
formula and molecular
formula are the same
Molecular
Formula
N2O4
Empirical
Formula
NO2
CO2
CO2
C6H12O6
CH2O
C3H6O3
CH2O
56
1.2 A5. Determine Empirical Formula
Find the empirical formula of a compound
1.
2.
3.
Assume 100 g of the compound and
change % to grams.
Convert the grams to moles
Divide each mole quantity by the
smallest number of moles
57
1.2 A5. Determine Empirical Formula
What is the empirical formula of a compound that
contains 40.0% C, 6.7% H, and 53.3% O?
1.
Assume 100 g of the compound and change % to grams.
2.
Convert the grams to moles
40.0 g C / 12.0 = 3.33 mol C
6.7 g H / 1.0 = 6.7 mol H
53.3 g O / 16.0 = 3.33 mol O
3. Divide each mole quantity by the smallest number of moles
3.33 mol C / 3.33 = 1 C
6.7 mol H / 3.33 = 2 H
3.33 mol O / 3.33 = 1 O
4. Empirical formula is CH2O
58
1.2 A5. Determine Empirical Formula
What is the empirical formula of a compound that contains 26.2% N,
7.53% H and 66.4% Cl?

Assume 100 g of the compound and change % to grams.

Convert the grams to moles

Divide each mole quantity by the smaller number of moles

Empirical formula is
59
1.2 A5. Determine Empirical Formula
What is the empirical formula of a compound that contains 26.2% N,
7.53% H, and 66.4% Cl?

Assume 100 g of the compound and change % to grams.

Convert the grams to moles

Divide each mole quantity by the smallest number of moles

Empirical formula is NH4Cl
26.2 g N / 14.0 = 1.87 mol N
7.53 g H / 1.01 = 7.46 mol H
66.4 g Cl / 35.5 = 1.87 mol Cl
1.87 mol N / 1.87 = 1 mol N
7.46 mol H / 1.87 = 4 mol H
1.87 mol Cl / 1.87 = 1 mol Cl
60
1.2 A5. Determine Empirical Formula
What is the empirical formula of a compound that
contains 1.67 g Ce and 4.54 g I?



Convert the grams to moles
1.67 g Ce / 140.1 = .0119 mol Ce
4.54 g I / 126.9 = .0358 mol I
Divide each mole quantity by the smallest number of moles
.0119 mol Ce / .0119 = 1 Ce
.0358 mol I / .0119 = 3.01 = 3 I
The empirical formula is CeI3
61
1.2 A5. Determine Empirical Formula
What is the empirical formula of a compound that contains
2.74 g Na, 0.120 g H, 1.43 g C and 5.71 g O?
62
1.2 A5. Determine Empirical Formula
What is the empirical formula of a compound that contains
2.74 g Na, 0.120 g H, 1.43 g C and 5.71 g O?

Convert the grams to moles
2.74g Na / 23.0 = 0.119 mol Na1.43 g C / 12.0 = 0.119 mol C
0.120 g H / 1.0 = 0.12 mol H
5.71 g O / 16.0 = 0.357 mol O

Divide each mole quantity by the smallest number of moles
0.119 mol Na / 0.119 = 1 Na
0.12 mol H / 0.119 = 1 H

0.119 mol C / 0.119 = 1 C
0.357 mol O / 0.119 = 3 O
The empirical formula is NaHCO3
63
1.2 A4. Determine Molecular Formula
To find molecular formula you need the empirical
formula and the formula (molar) mass



A substance has an empirical formula of CH and a molar
mass of 78.0. What is the molecular formula?
Find the mass of the empirical formula
Mass of CH is 12.0 + 1.0 = 13.0
Divide the molar mass by the empirical formula mass
78.0 / 13.0 = 6
Multiply each subscript in the EF by the number
(CH)6 = C6H6
64
1.2 A4. Determine Molecular Formula
To find molecular formula you need the empirical
formula and the formula (molar) mass
A substance consists of 42.9% C and 57.1% O with a molar mass of
56.0. What is the molecular formula?
65
1.2 A4. Determine Molecular Formula
To find molecular formula you need the empirical
formula and the formula (molar) mass
A substance consists of 42.9% C and 57.1% O with a molar mass of
56.0. What is the molecular formula?
Find Empirical Formula:
42.9 g C / 12.0 = 3.58
EF is CO
57.1 g / 16.0 = 3.57
Mass of CO: (12.0 + 16.0) = 28.0.
56.0 / 28.0 = 2 2(CO): MF = C2O2
66
1.3: Reacting Masses and
Volumes
Understandings:
1.
Reactants can be either limiting or excess.
2.
The experimental yield can be different from
the theoretical yield
*Understandings 4-6, will be covered in IB Chem 2
1.3: Reacting Masses and
Volumes
Applications and Skills:
1.
Solution of problems relating to reacting
quantities, limiting and excess reactants,
theoretical, experimental, and percentage
yields
*Applications and Skills 2-8, will be covered in IB Chem 2
1.3 U1. Limiting Reactants
Stoichiometry: the quantitative method of
examining the relative amounts of reactants and
products.
Reactants can limit the amount of product made or
reactants can be left over when a specific amount
of product is made
1.3 A1. Theoretical Yield Calculations
The Four Steps to Stoichiometric
Enlightenment
Step 1: Write the balanced chemical equation.
Step 2: Convert the known chemical amount to moles
If amount in grams: Divide by formula weight
If amount in particles: Divide by 6.02x1023
Step 3: Ratio the unknown to the known molar quantities.
From Problem
From Balanced Equation
X moles of known X
Coefficient of Unknown = moles unknown
Coefficient of Known
Step 4: Convert molar quantity from step 3 to amount asked for
If amount in grams: Multiply by formula weight
If amount in particles: Multiply by 6.02x1023
1.3 A1. Theoretical Yield Calculations
Step 3: Ratio the unknown to the known molar
quantities.
From Problem
X moles of known
From Balanced Equation
X Coefficient of Unknown = moles unknown
Coefficient of Known
1.3 A1. Theoretical Yield Calculations
Step 4: Convert molar quantity from step 3 to amount
asked for
If amount in grams: Multiply by formula weight
If amount in particles: Multiply by 6.02x1023
1.3 A1. Theoretical Yield Calculations
How many grams of hydrogen
are produced from the
decomposition of 1,234 g of
water?
How many grams of hydrogen
are produced from the
decomposition of
1,234 g of water?
Step 1: Write the balanced equation.
Step 2: Convert grams to moles.
Step 3: Ratio unknown to known.
From Problem
From Balanced
Equation
moles of known X Coefficient of Unknown =
Coefficient of known
= moles of unknown
Step 4: Convert moles to grams.


2H2O  2H2 + O2
1,234 g / 18.0 g/mol = 68.56
mol H2O
From Problem From Equation:
68.56 Moles H2O X 2 Moles H2
2 Moles H2O
= 68.56 moles H2


Mass H2 = 68.56 mol H2 (2.02
g/mol) = 138 g (3 s.f.)
1.3 A1. Theoretical Yield
Calculations
How many grams of sodium carbonate are
needed to precipitate the calcium from 254 g
of calcium chloride?
How many grams of sodium carbonate
are needed to precipitate the calcium
from 254 g of calcium chloride?
Step 1: Write the balanced equation.
Step 2: Convert grams to moles.
Step 3: Ratio unknown to known.
From Problem
From Balanced Equation
moles of known x Coefficient of Unknown =
Coefficient of Known
= moles of unknown
Step 4:
Convert moles to grams.

CaCl2 + Na2CO3  CaCO3 + 2NaCl

254 g / 111.1 g/mol = 2.29 mol CaCl2

From Problem
From Equation:
2.29 Moles CaCl2 x 1 mol Na2CO3 =
1 mole CaCl2
= 2.29 moles Na2CO3

Mass Na2CO3 = 2.29 mol Na2CO3 (106.0
g/mol) = 243 g (3 s.f.)
1.3 A1. Theoretical Yield
Calculations
How much oxygen does it take to burn 100.0 grams of
octane?
Step 1: 2C8H18 + 25O2  16CO2 + 18H2O
1.3 A1. Theoretical Yield
Calculations
How much oxygen does it take to burn 100.00 grams of
octane?
Step 1: 2C8H18 + 25O2  16CO2 + 18H2O
Step 2: moles of octane = 100.00 g / 114.00 g/mol = 0.8772
mol
1.3 A1. Theoretical Yield
Calculations.
How much oxygen does it take to burn 100.0 grams of octane?
Step 1: 2C8H18 + 25O2  16CO2 + 18H2O
Step 2: moles of octane = 100.0 g / 114.0 g/mol = 0.8772 mol
Step 3: Moles of O2 required:
0.8772 mol C8H18 x
25 O2
= 10.97 mol O2
2 C8H18
1.3 A1. Theoretical Yield
Calculations
How much oxygen does it take to burn 100.0 grams of
octane?
Step 1:
2C8H18 + 25O2  16CO2 + 18H2O
Step 2:moles of octane = 100.0 g / 114.0 g/mol = 0.8772 mol
Step 3: Moles of O2 required:
0.8772 C8H18 x 25 O2
2 C8H18
= 10.97 mol O2
Step 4: Mass of O2 required: 10.97 mol (32.00 g/mol) =
351.0 g O2
1.4.1 Calculate theoretical yields from
chemical equations.
In a spectacular reaction called the thermite reaction,
iron(III) oxide reacts with aluminum producing iron and
aluminum oxide. How many grams of iron will be produced
from 43.7 grams of aluminum?
1.3 A1. Theoretical Yield
Calculations
In a spectacular reaction called the thermite reaction, iron(III)
oxide reacts with aluminum producing iron and aluminum
oxide. How many grams of iron will be produced from 43.7
grams of aluminum?
Step 1:
Fe2O3 + 2Al  2Fe + 2Al2O3
Step 2: The moles of Al = 43.7 g / 27.0 g/mol = 1.62 mol
Step 3: Moles of Fe produced:
1.62 Al x 2 mol Fe
= 1.62 mol Fe
2 mol Al
Step 4: Mass of Fe produced: 1.62 mol (55.8 g/mol) =
90.4 g Fe
1.3 A1. Determining Limiting
Reactants
Limiting Reactants
You are given amounts for two reactants and one reactant will run out first.
This is called the limiting reactant. It limits the amount of product that
can be made. The reactant that is left over is called the excess
reactant.
A strip of zinc metal weighing 2.00 g is placed in a solution containing 2.50
g of silver nitrate causing the following reaction to occur:
Zn + 2AgNO3  2Ag + Zn(NO3)2
How many grams of Ag will be produced?
1.3 A1. Determining Limiting Reactants
1) Balanced equation: Zn + 2AgNO3  2Ag + Zn(NO3)2
2) Moles of starting substances:
2.00 g Zn/ 65.39 g/mol = .03059 mole Zn and
2.50 g AgNO3 / 169.88 g/mol = .01472 mol AgNO3
3) For each value above, use mole ratio for substance you are trying to find.
.03059 Zn X 2 mole Ag
= .06118 mol Ag
1 mole Zn
.01472 AgNO3 x 2 Ag
2 AgNO3
= .01472 mol Ag
4) Choose the smaller value from step 3 to calculate the grams.
.01472 mol Ag x 107.87 = 1.58 g Ag (3 sig figs)
AgNO3 is the limiting reactant because it will produce the least amount of Ag.
Zn is the reactant in excess. There will be Zn left over after the reaction
1.3 A1. Determining Limiting Reactants
How many grams of Cu2S will be produced when 80.0 g Cu
reacts with 25.0 g S?
1.3 A1. Determining Limiting Reactants
How many grams of Cu2S will be produced when 80.0 g Cu
reacts with 25.0 g S?
1)
2Cu + S  Cu2S
1.3 A1. Determining Limiting Reactants
How many grams of Cu2S will be produced when 80.0 g Cu
reacts with 25.0 g S?
1)
2)
2Cu + S  Cu2S
80.0 g Cu/63.55 = 1.259 molCu
25.0 g S/32.06 = .7798 mol S
1.3 A1. Determining Limiting
Reactants
How many grams of Cu2S will be produced when 80.0 g Cu reacts with 25.0
g S?
1)
2)
1)
2Cu + S  Cu2S
80.0 g Cu/63.55 = 1.259 mol
25.0 g S/32.06 = .7798 mol
1.259 Cu x 1 mole Cu2S = 0.6295 mol Cu2S
2 moles CU
0.7798 S x 1mol Cu2S = 0.7798 mol Cu2S
1 mole S
1.3 A1. Determining Limiting Reactants
How many grams of Cu2S will be produced when 80.0 g Cu
reacts with 25.0 g S?
1)
2Cu + S  Cu2S
2)
80.0 g Cu/63.55 = 1.259 mol
25.0 g S/32.06 = .7798 mol
3) 1.259 Cu x 1 mole Cu2S = 0.6295 mol Cu2S
2 moles CU
0.7798 S x 1mol Cu2S = 0.7798 mol Cu2S
1 mole S
4)
0.6295 mol Cu2S x 159.16 = 100 g or 1.00 x 102 g (3 sig
figs)
1.3 A1. Solve problems involving theoretical,
experimental and percentage yield.
The theoretical yield from a chemical reaction is the yield
calculated by assuming the reaction goes to completion.
In practice we often do not obtain as much product from a
reaction mixture as theoretically possible due to a number
of factors:

Many reactions do not go to completion

Some reactants may undergo two or more reactions
simultaneously forming undesired products

Not all of the desired product can be separated from the
rest of the products
The amount of a specified pure product actually obtained from
a given reaction is the experimental or actual yield.
1.3 A1. Solve problems involving theoretical,
experimental and percentage yield.
Percentage yield indicates how much of a
desired product is obtained from a reaction
Percentage yield = (experimental yield of a product) X 100
(theoretical yield of a product)
1.3 A1. Solve problems involving theoretical,
experimental and percentage yield.
Consider the reaction: P4 + 5O2  P4O10
If 272 g of phosphorus reacts, the percentage yield of
tetraphosphorus decoxide is 89.5%. What mass of P4O10 is
obtained?
Theoretical amt:

272 g/124.0 g/mol = 2.194 mol P4

2.19 P4 x 1 mole P4O10 = 2.194 mol P4O10
1 mole P4

2.194 mol P4O10 x 284.0 = 623 g
Experimental amt:

623 g x .895 = 558 g
1.3 A1. Solve problems involving theoretical,
experimental and percentage yield.
Consider the reaction: P4O10 + 6H2O  4H3PO4
If 558 g of P4O10 reacts, the experimental yield of H3PO4 is 746
g. What is the percentage yield of H3PO4?
Theoretical amt:

558 g/284 g/mol = 1.965 mol P4O10

1.965 P4O10 x 4 H3PO4 = 7.86 mol H3PO4
1 P4O10

7.86 mol H3PO4 x 98.0 = 770. g H3PO4
Percentage yield:

746 g x 100% = 96.9%
770 g
1.3 A1. Solve problems involving theoretical,
experimental and percentage yield.
Consider the reaction: Fe2O3 + 3CO  2Fe + 3CO2
When 84.8 g iron(III) oxide reacts with an excess of carbon
monoxide, 54.3 g of iron is produced. What is the
percentage yield of this reaction?
1.3 A2. Solve problems involving theoretical,
experimental and percentage yield.
Consider the reaction: Fe2O3 + 3CO  2Fe + 3CO2
When 84.8 g iron(III) oxide reacts with an excess of carbon
monoxide, 54.3 g of iron is produced. What is the
percentage yield of this reaction?
Theoretical amt:

84.8 g/159.2 g/mol = 0.5327 mol Fe2O3

0.5327 mol Fe2O3 x 2 mol Fe
= 1.065 mol Fe
1 mol Fe2O3

1.065 mol Fe x 55.8 g.mol = 59.4 g Fe
Percentage yield:

54.3 g x 100% = 91.4%
59.4 g
1.3 A2. Solve problems involving theoretical,
experimental and percentage yield.
If 50.0 g silicon dioxide is heated with an excess of carbon, the
percentage yield of silicon carbide 76.0%. What mass of
silicon carbide will actually be produced?
SiO2 + 3C  SiC + 2CO
1.3 A1. Solve
If 50.0 g silicon dioxide is heated with an excess of carbon, the
percentage yield of silicon carbide 76.0%. What mass of silicon
carbide will actually be produced? SiO2 + 3C  SiC + 2CO
Theoretical amt:

50.0 g/60.1 g/mol = 0.8319 mol SiO2

0.8319 mol SiO2 x 1 mol SiC
= 0.8319 mol SiC
1 mol SiO2

0.8319 mol SiC x 40.1 g.mol = 33.4 g SiC
Experimental yield:

33.4 g x .760% = 25.4 g SiC