Download Changing Matter

Changing Matter
• Matter can be changed two ways:
–Physically
• Physical reaction
• Physical change
–Chemically
• Chemical reaction
• Chemical change
Physical Changes
• Do NOT CHANGE THE TYPE OF
MATTER
– Nothing new or different is formed
– Could be a change in:
•
•
•
•
•
•
Mass
Size
Volume
Density
Change in state
Color
Shape
Examples of Physical Changes
•
•
•
•
Boiling, vaporization…. Any state change
Dissolving
Breaking
Making a mixture
– 2 or more types of matter (substances) mixed
together
• Not in specific amounts
• Can be separated physically
Chemical Changes
• Atoms have electrons arranged in energy levels
or energy shells
• Electrons in the last (outermost) shell are called
valence electrons
• Valence electrons let atoms bond with other
atoms
– Ionic bonding
• Gaining or losing electrons
– Covalent bonding
• Sharing electrons
Chemical Changes
• Atoms that bond form molecules
– May be the same type (nonmetals) of atom or,
– Different types (metal + nonmetal) of atoms
• Different types  compounds
Chemical Changes
• Molecules can bond and “unbond”
– Atoms can re-arranged in different
combinations
– For example:
CaCO3 (1 atom Ca, 1 atom C, 3 atoms O)
Add heat to re-arranged the atoms:
CaO
CO2
Chemical Changes
• Evidence of a chemical reaction
– Formation of gas
– Formation of precipitate
– Change in color
– Change in energy
• Endothermic
– Absorbs heat energy (gets cold)
• Exothermic
– Releases heat energy (gets hot)
Chemical Changes
• Chemical reactions can be represented by
equations
CaCO3  CaO + CO2
Reactants
Products
Chemical Changes
• Atoms are re-arranged, NOT created or
destroyed
– Law of Conservation of Matter
– Law of Conservation of Mass
Chemical Changes
• Matter is conserved  type of atoms
does not change
– Nothing is created or destroyed
• Mass is conserved  amount of atoms
cannot change
– Nothing is created or destroyed
Chemical Changes
• To show conservation of mass  Balance
equations
– Make sure there are the same number of
each type of atom in the products and in the
reactants
Balancing Equations
The equation for the burning of
methane gas in oxygen is:
CH4 + 2 O2 → CO2 + 2 H2O
Subscript
Coefficient
Shows # of atoms
Shows # of molecules
Balancing Equations
• No subscript or coefficient is understood to
be 1
CH4 + 2 O2 → CO2 + 2 H2O =
C 1H 4 + 2 O 2 → C 1 O 2 + 2 H 2O 1
1C
4H
4O



1C
4H
4O
Periodic Properties Post Lab
• Periodic Trends affect properties because
of Zeff
– Ionization energy
• Energy required to remove an electron
– Electronegativity
• Ability to attract electrons
– Atomic Radius
• Distance of the valence electrons to the nucleus
Other Periodic Trends
• Groups have similar properties
– Valence electrons
• Members of the same representative family have
the same number of valence electrons
– Reactivity
• Because they have the same number of electrons,
they react similarly.
– CaCl2, MgCl2, SrCl2, etc.
– Density = mass/volume
• % error = I actual – theoretical I
theoretical
Sn = 7.265
Pb = 11.34
Si = 2.33
Ge = 5.323
Solubility Trends
• Common in Double Replacement Rxns
Double Replacement Rxns
• Two compounds react to form two new
compounds
– Metal replaces metal
– Remember basic formula writing rules
Ex:
Ca(NO3)2 +
Na2CO3 
CaCO3 + 2 NaNO3
ppt
Some Types of Chemical Reactions
Type of
Reaction
Definition
Synthesis
Two or more elements or
compounds combine to
make a more complex
substance
Decomposition
Compounds break down
into simpler substances
Single
Replacement
Occurs when one element
replaces another one in a
compound
Double
Replacement
Occurs when different
atoms in two different
compounds trade places
 Equation
A + B → AB
AB → A + B
AB + C → AC + B
AB + CD → AC + BD
Identifying Chemical Reactions
S = Synthesis
____ P +
D = Decomposition
O2 → P4O10
____ HgO → Hg
____ Cl2 +
+
SR = Single Replacement
DR = Double Replacement
____ Mg + O2
O2
NaBr → NaCl + Br2
→
____ Al2O3
MgO
→ Al + O2
____ H2 + N2 → NH3
MOLAR MASS
Molar Mass is shown below the element
symbol on the Periodic Table.
Units: grams
mole
Use molar mass to convert between:
Number of Moles
Mass (grams)
Number of Molecules or Atoms
Molar Mass Examples
• sodium bicarbonate
 NaHCO3
 22.99 + 1.01 + 12.01 + 3(16.00)
• sucrose
= 84.01 g/mol
 C12H22O11
 12(12.01) + 22(1.01) + 11(16.00)
= 342.34 g/mol
Divide the mass (in
grams) by Molar Mass
to get the number of
moles.
MASS in
GRAMS
MOLAR
MASS
grams/mol
Multiply the moles
by the Molar Mass
to get the mass
(in grams).
MOLES
Divide the number of
molecules or atoms by
Avogadro’s Constant
to get the number of moles.
AVOGADRO’S
CONSTANT
6.022x1023
particles/mol
Multiply the number of
moles by Avogadro’s
Constant to get the
number of atoms or
molecules.
MOLECULES
ATOMS
Molar Conversion Examples
• How many moles of carbon are
in 26 g of carbon?
26 g C 1 mol C
= 2.2 mol C
12.01 g C
Molar Conversion Examples
• How many molecules are in
2.50 moles of C12H22O11?
23
6.02

10
2.50 mol
molecules = 1.51  1024
molecules
1 mol
C12H22O11
Molar Conversion Examples
• Find the mass of 2.1  1024
molecules of NaHCO3.
2.1  1024
molecules
1 mol
84.01 g
6.02  1023 1 mol
molecules
= 290 g NaHCO3
Sample Problems –
Moles and Atoms
• Determine the number of atoms
present in 2.50 moles of strontium.
• Convert 5.01 x 1024 atoms of
strontium to moles of strontium.
Sample Problems –
Moles and Mass
• Determine the mass in grams of 2.50
moles of strontium.
• Determine the number of moles
represented by 943.5 grams of
strontium.
Sample Problems –
Moles, Atoms, Mass
• Determine the mass in grams of
(exactly) 5 atoms of strontium.
• Determine the number of atoms
represented by 43.5 grams of
strontium.
Percentage Composition
• the percentage by mass of each
element in a compound
mass of element
% composition 
 100
total mass
Percentage Composition
• Find the % composition of Cu2S.
127.10 g Cu
79.852%
%Cu =
 100 =
Cu
159.17 g Cu2S
%S =
32.07 g S
159.17 g Cu2S
 100 = 20.15%
S
Percentage Composition
• Find the percentage composition
of a sample that is 28 g Fe and
8.0 g O.
28 g
 100 = 78% Fe
%Fe =
36 g
%O =
8.0 g
36 g
 100 = 22% O
Percentage Composition
• How many grams of copper are in
a 38.0-gram sample of Cu2S?
Cu2S is 79.852% Cu
(38.0 g Cu2S)(0.79852) = 30.3 g Cu
Percentage Composition
• Find the mass percentage of
water in calcium chloride
dihydrate, CaCl2•2H2O?
%H2O =
36.04 g
 100 = 24.51%
H2O
147.02 g
Empirical Formula
• Smallest whole number ratio of
atoms in a compound
C 2H 6
reduce subscripts
CH3
Empirical Formula
1. Find mass (or %) of each element.
2. Find moles of each element.
3. Divide moles by the smallest # to
find subscripts.
4. When necessary, multiply
subscripts by 2, 3, or 4 to get
whole #’s.
Empirical Formula
• Find the empirical formula for a
sample of 25.9% N and 74.1% O.
25.9 g 1 mol
= 1.85 mol N
=1N
1.85 mol
14.01 g
74.1 g 1 mol
= 4.63 mol O
= 2.5 O
16.00 g
1.85 mol
Empirical Formula
N1O2.5
Need to make the subscripts whole
numbers  multiply by 2
N2O5
Molecular Formula
• “True Formula” - the actual number
of atoms in a compound
empirical
formula
CH3
?
molecular
formula
C2H6
Molecular Formula
1. Find the empirical formula.
2. Find the empirical formula mass.
3. Divide the molecular mass by the
empirical mass.
4. Multiply each subscript by the
answer from step 3.
MF mass
n
EF mass
EF n
Molecular Formula
• The empirical formula for ethylene
is CH2. Find the molecular formula
if the molecular mass is
28.1 g/mol?
empirical mass = 14.03 g/mol
28.1 g/mol
14.03 g/mol
= 2.00
(CH2)2  C2H4
Using the Basics…
The empirical formula of a
compound is found to be P2O5.
The molar mass of the compound
is 284 grams/mole.
What is the molecular formula for the
compound?
Proportional Relationships
• Stoichiometry
– mass relationships between
substances in a chemical reaction
– based on the mole ratio
• Mole Ratio
– indicated by coefficients in a balanced
equation
2 Mg + O2  2 MgO
N2 + 3H2  2NH3
1 mol
3 mol
2 mol
1 mol N 2
3 mol H 2
43
N2 + 3H2  2NH3
1 mol
3 mol
2 mol
3 mol H 2
2 mol NH 3
44
Stoichiometry Steps
1. Write a balanced equation.
2. Identify known & unknown.
3. Line up conversion factors.
•– Mole ratio - moles
molesmoles
moles
– Molar mass - moles  grams
– Molarity moles  liters soln
– Molar volume - moles  liters gas
Core step in all stoichiometry problems!!
4. Check answer.
Molar Volume at STP
1 mol of a gas=22.4 L
at STP
Standard Temperature & Pressure
0°C and 1 atm
Molar Volume at STP
LITERS
OF GAS
AT STP
Molar Volume (22.4 L/mol)
MASS
IN
GRAMS
Molar Mass
MOLES
(g/mol)
6.02 
1023
particles/mol
Molarity (mol/L)
LITERS
OF
SOLUTION
NUMBER
OF
PARTICLES
Mole-Mole
Calculations
48
Phosphoric Acid
• Phosphoric acid (H3PO4) is one of the
most
widely
produced
industrial
chemicals in the world.
• Most of the world’s phosphoric acid is
produced by the wet process which
involves the reaction of phosphate rock,
Ca5(PO4)3F, with sulfuric acid (H2SO4).
Ca5(PO4)3F(s) + 5H2SO4  3H3PO4 + HF + 5CaSO4
49
Calculate the number of moles of phosphoric acid
(H3PO4) formed by the reaction of 10 moles of
sulfuric acid (H2SO4).
Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4
1 mol
5 mol
3 mol
1 mol
5 mol
Step 1 Moles starting substance: 10.0 mol H2SO4
Step 2 The conversion needed is
moles H2SO4  moles H3PO4
Mole Ratio
3 mol H 3PO4
10 mol H 2SO4 x
= 6 mol H3PO4
5 mol H 2SO4
50
Calculate the number of moles of sulfuric acid
(H2SO4) that react when 10 moles of Ca5(PO4)3 react.
Ca5(PO4)3F + 5H2SO4  3H3PO4 + HF + 5CaSO4
1 mol
5 mol
3 mol
1 mol
5 mol
Step 1
The starting substance is 10.0 mol Ca5(PO4)3F
Step 2
The conversion needed is
moles Ca5(PO4)3F  moles H2SO4
Mole Ratio
5 mol H 2SO4
10 mol Ca 5 (PO4 )3F x
= 50 mol H 2SO4
1 mol Ca 5 (PO4 )3F
51
Stoichiometry Problems
• How many moles of KClO3 must
decompose in order to produce 9
moles of oxygen gas?
2KClO3  2KCl + 3O2
? mol
9 mol O2 2 mol KClO3
3 mol O2
9 mol
= 6 mol
KClO3
Mole-Mass
Calculations
53
Calculate the number of moles of H2SO4
necessary to yield 784 g of H3PO4.
Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4
Method 1 Step by Step
Step 1 The starting substance is 784 grams of H3PO4.
Step 2 Convert grams of H3PO4 to moles of H3PO4.
 1 mol H 3PO4 
784 g H 3PO4 
 = 8.00 mol H 3PO4
 98.0 g H 3PO4 
Step 3 Convert moles of H3PO4 to moles of H2SO4 by
the mole-ratio method.
 5 mol H 2SO4 
8.00 mol H 3PO4 
 = 13.3 mol H 2SO4
 3 mol H 3PO4 
54
Mole Ratio
Calculate the number of moles of H2SO4
necessary to yield 784 g of H3PO4
Ca5(PO4)3F+ 5H2SO4  3H3PO4 + HF + 5CaSO4
Method 2 Continuous
The conversion needed is
grams H3PO4  moles H3PO4  moles H2SO4
Mole Ratio
 1 mol H 3PO4 
784 g H 3PO4

 98.0 g H 3PO4 
 5 mol H 2SO4 

 = 13.3 mol H 2SO4
 3mol H 3PO4 
55
Stoichiometry Problems
• How many grams of KClO3 are req’d to
produce 9.00 L of O2 at STP?
2KClO3  2KCl + 3O2
?g
9.00 L
9.00 L
O2
1 mol
O2
2 mol 122.55
KClO3 g KClO3
22.4 L
O2
3 mol
O2
1 mol
KClO3
= 32.8 g
KClO3
Mass-Mass
Calculations
57
Calculate the number of grams of NH3 formed by
the reaction of 112 grams of H2.
N2 + 3H2  2NH3
Method 1 Step by Step
Step 1 The starting substance is 112 grams of H2.
Convert 112 g of H2 to moles.
grams  moles
 1 mol H 2 
112 g H 2 
  55.4 moles H 2
 2.02 g H 2 
Step 2 Calculate the moles of NH3 by the mole ratio
method.
 2 mol NH 3 
55.4 moles H 2 
 = 36.9 moles NH 3
 3 mol H 2 
58
Calculate the number of grams of NH3 formed by
the reaction of 112 grams of H2.
N2 + 3H2  2NH3
Method 1 Step by Step
Step 3 Convert moles NH3 to grams NH3.
moles  grams
 17.0 g NH 3 
36.9 moles NH 3 
 = 629 g NH 3
 1 mol NH 3 
59
Calculate the number of grams of NH3 formed by
the reaction of 112 grams of H2.
N2 + 3H2  2NH3
Method 2 Continuous
grams H2  moles H2  moles NH3  grams NH3
 1 mol H 2   2 mol NH 3   17.0 g NH 3 
112 g H 2 
 = 629 g NH 3


 2.02 g H 2   3 mol H 2   1 mol NH 3 
60
Stoichiometry Problems
• How many grams of silver will be
formed from 12.0 g copper?
Cu + 2AgNO3  2Ag + Cu(NO3)2
12.0 g
?g
12.0 1 mol 2 mol 107.87
g Cu Cu
Ag
g Ag
= 40.7 g
63.55 1 mol 1 mol
Ag
g Cu
Cu
Ag
Stoichiometry Problems
• How many grams of Cu are required
to react with 1.5 L of 0.10M AgNO3?
Cu + 2AgNO3  2Ag + Cu(NO3)2
?g
1.5L
0.10M
1.5 .10 mol 1 mol 63.55
L AgNO3
Cu
g Cu
1L
= 4.8 g
2 mol 1 mol
Cu
AgNO3 Cu
Limiting Reactant
63
• The limiting reactant is one of the
reactants in a chemical reaction.
• It is called the limiting reactant
because the amount of it present is
insufficient to react with the
amounts of other reactants that are
present.
• The limiting reactant limits the
amount of product that can be
formed.
64
How manyFrom
bicycles
From
eight three
wheels
From
pedal
four
fourassemblies
frames four
can be assembled
bikes three
can be
bikes
constructed.
bikes
cancan
be be
constructed.
constructed.
from the parts
shown?
The limiting part is the
number of pedal
assemblies.
65
9.2
H2 + Cl2  2HCl
+
4 molecules Cl2 can form 8
molecules HCl
7 molecules H2 can form
14 molecules HCl
9.3

Cl is the limiting
reactant
3 molecules of H2 remain
2
H2 is in excess
66
Steps Used to Determine the
Limiting Reactant
67
1. Calculate the amount of product (moles
or grams, as needed) formed from each
reactant.
2. Determine which reactant is limiting.
(The reactant that gives the least amount
of product is the limiting reactant; the
other reactant is in excess.
3. Calculate the amount of the other
reactant required to react with the
limiting reactant, then subtract this
amount from the starting quantity of the
reactant. This gives the amount of the
substance that remains unreacted.
68
Examples
69
How many moles of HCl can be produced by
reacting 4.0 mol H2 and 3.5 mol Cl2? Which
compound is the limiting reactant?
H2 + Cl2 → 2HCl
Step 1 Calculate the moles of HCl that can form
from each reactant.


2
mol
HCl
4.0 mol H 2 
  8.0 mol HCl
 1 mol H 2 
3.5 mol Cl2  2 mol HCl   7.0 mol HCl
 1 mol Cl 2 
Step 2 Determine the limiting reactant.
The limiting reactant is Cl2 because it
70
produces less HCl than H2.
How many moles of silver bromide (AgBr) can be formed
when solutions containing 50.0 g of MgBr2 and 100.0 g of
AgNO3 are mixed together? How many grams of the excess
reactant remain unreacted?
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 1 Calculate the grams of AgBr that can
form from each reactant.
The conversion needed is
g reactant → mol reactant → mol AgBr → g AgBr
 1 mol MgBr2   2 mol AgBr   187.8 g AgBr 
 102 g AgBr
50.0
g
MgBr

 
 

2 
 184.1 g MgBr2   1 mol MgBr2   1 mol AgBr 
 1 mol AgNO3   2 mol AgBr  187.8 g AgBr 
110.5 g AgBr
100.0 g AgNO3 



71
 169.9 g AgNO3   2 mol AgNO3  1 mol AgBr 
How many moles of silver bromide (AgBr) can be
formed when solutions containing 50.0 g of MgBr2
and 100.0 g of AgNO3 are mixed together? How
many grams of the excess reactant remain
unreacted?
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 2 Determine the limiting reactant.
The limiting reactant is MgBr2
because it
forms less Ag Br.
 1 mol MgBr2   2 mol AgBr   187.8 g AgBr 
 102 g AgBr
50.0
g
MgBr

 
 

2 
 184.1 g MgBr2   1 mol MgBr2   1 mol AgBr 
 1 mol AgNO3   2 mol AgBr  187.8 g AgBr 
110.5 g AgBr
100.0 g AgNO3 



72
 169.9 g AgNO3   2 mol AgNO3  1 mol AgBr 
How many grams of the excess reactant (AgNO3)
remain unreacted?
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 3 Calculate the grams of unreacted AgNO3.
First calculate the number of grams of
AgNO3 that will react with 50 g of MgBr2.
The conversion needed is
g MgBr2 → mol MgBr2 → mol AgNO3 → g AgNO3
 1 mol MgBr2  2 mol AgNO3   169.9 g AgNO3 
50.0 g MgBr2 
  92.3 g AgNO3


 184.1 g MgBr2  1 mol MgBr2   1 mol AgNO3 
The amount of MgBr2 that remains is
100.0 g AgNO3 92.3 g AgNO3 = 7.7 g AgNO733
Reaction Yield
74
The quantities of products calculated
from
equations
represent
the
maximum yield (100%) of product
according to the reaction represented
by the equation.
75
Many reactions fail to give
a 100% yield of product.
This occurs because of side reactions
and the fact that many reactions are
reversible.
76
• The theoretical yield of a reaction is the
calculated amount of product that can
be obtained from a given amount of
reactant.
• The actual yield is the amount of
product finally obtained from a
given amount of reactant.
77
• The percent yield of a reaction is the ratio
of the actual yield to the theoretical yield
multiplied by 100.
actual yield
x 100 = percent yield
theoretical yield
78
Silver bromide was prepared by reacting 200.0 g of
magnesium bromide and an adequate amount of
silver nitrate. Calculate the percent yield if 375.0 g
of silver bromide was obtained from the reaction:
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq
Step 1 Determine the theoretical yield by
calculating the grams of AgBr that can
be formed.
The conversion needed is
g MgBr2 → mol MgBr2 → mol AgBr → g AgBr
 1 mol MgBr2 
200.0
g
MgBr


2 
184.1
g
MgBr
2 

 2 mol AgBr   187.8 g AgBr 
408.0 g AgBr



 1 mol MgBr2   1 mol AgBr 
79
Silver bromide was prepared by reacting 200.0 g of
magnesium bromide and an adequate amount of
silver nitrate. Calculate the percent yield if 375.0 g
of silver bromide was obtained from the reaction:
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq
Step 2 Calculate the percent yield.
must have same units
actual yield
percent yield =
x 100
theoretical yield
must have same units
375.0 g AgBr
x 100 = 91.9%
percent yield =
408.0 g AgBr
80