Quantifying Changes in Matter

1
Quantifying Changes in Matter Energy

• Once again whenever there is a change in matter
  there is a change in energy
• Both these changes can be measured with
  instruments
• Changes in matter are measured by balances
  volume measuring devices
• Changes in energy are measured by calorimeters
• Conservation of matter and energy apply
• The composition or location of matter has changed
  
• The type and location of the energy has changed
• Individual components and energy content of
  chemicals and relationships between reactants and
  products can be determined because
• Formulas and equations have EXACT numbers
• Energy content is proportional to the amount of
  matter

MOLES !!!!
2
The Dreaded Mole

• A mole
• A small blind rodent ? a pest
• An imperfection on skin ? sometimes unsightly
• 6.02 x 1023 of anything ? very useful
• When 1.00 mole of any collection of atoms
  organized as a formula is weighed, the mass in g
  the sum of the weighted atomic masses of the
  formula
• It is far more convenient to work with masses we
  can measure easily than to count large numbers of
  individual atoms
• The mole correlates numbers to masses

3
Some definitions

• Formula weight (FW) Sum of the atomic weights
  of all atoms in a formula
• If the substance is composed of atoms, the
  formula weight is also called the atomic weight.
• If the substance is composed of molecules, the
  formula weight is also called the molecular
  weight.
• Technically molecular weight should not be used
  to refer to the weight of ionic substances (not
  molecules) but it very often is
• Mole (mol) - An amount of a substance equal to
  its formula weight in grams.
• contains Avogadros number (6.02 x 1023) of
  formula units
• The formula weight is often called the molar mass.

4
Chemical Conversion Factors

• The mole can be used as a conversion factor in
  problem solving.
• 1 mol Cu 63.546 g Cu
• 1mol H2O 18.02 g H2O
• 1mol Ca(NO3)2 164.09 g
• These equalities can be changed into conversion
  factors

• Conversion factors can be flipped as needed

Recognizing when to flip is critically important
in problem solving
5
Stoichiometry - Switching Yard Part 1
g of species A
mole of A g of A
moles of species A
usually same dimension
molar mass
g of A mole of A
molar relationship between A and B
moles of B moles of A
moles of A moles of B
mole of B g of B
moles of species B
molar mass
g of B mole of B
g of species B
6
Some simple stoichiometry problems

• Determine the formula weight of Co(NO3)3
• How many moles are present in 65.56 g of CaCl2 ?
• How many g of N are there in 2.85 moles of
  Ca(NO3)2 ?

7
Some lessons learned

• Read formulas correctly
• Check the weight again account for all atoms in
  formula
• g to moles number is always smaller
• moles to g number is always larger
• Do not make the problem harder than it is
• Remember mole relationships

8
Relationships in Balanced Equations

• Since numbers of atoms in formulas are related to
  their weights by the mole, we can inter-relate
  both quantities in equations
• 3 Ca(NO3)2 (aq) 2 Na3PO4 (aq)? Ca3(PO4)2 (s)
  6 NaNO3 (aq)
• 3 moles of calcium nitrate reacts with 2 moles of
  sodium phosphate to give 1 mole of calcium
  phosphate and 6 moles of sodium nitrate
• And a bunch of other relationships!!!!!
• 3 moles (164.08 g Ca(NO3)2 /mol) will yield 6
  moles (84.99 g NaNO3 /mol)
• We can write other easy whole number
  relationships
• But how many g of Ca3(PO4)2 can we get from 27.55
  g of Ca(NO3)2 ?
• This is where we can use the switching yard most
  effectively

9
Stoichiometry - Switching Yard In operation
27.55 g of Ca(NO3)2
1mole of Ca(NO3)2 164.08 g of Ca(NO3)2
0.1679 moles of Ca(NO3)2
1 moles of Ca3(PO4)2 3 moles of Ca(NO3)2
0.0560 moles of Ca3(PO4)2
310.38 g of Ca3(PO4)2 1 mole of Ca3(PO4)2
17.37 g of Ca3(PO4)2
10
Two additional problems

• How many g of P in 48.25 g of Ca3(PO4)2 ?
• This could be a fertilizer problem how much
  calcium phosphate would we have to apply in order
  to make sure we added enough phosphorous to a
  soil that has been depleted in phosphorous
• We have to know the relationship between the P
  and the whole formula
• There are 2 Ps in the formula therefore there
  are 2 moles of P per mole of Ca3(PO4)2
• Now we can use the switching yard
• How many moles of H2 will be produced from 95.00
  g of Ca according to the following reaction
• Ca (s) 2 HCl (aq)? H2 (g) CaCl2 (aq)
• This could be used to make hydrogen gas to fill a
  cylinder since there is a relationship between
  moles and the pressure in a cylinder of known
  volume
• We can see that there is a one to one mole
  relationship once we get to moles of Ca that
  will be the same of moles of hydrogen gas but
  first we have to get to moles of Ca

11
Stoichiometry - Switching Yard In operation
48.25 g of Ca3(PO4)2
1 mole of Ca3(PO4)2 310.38 g Ca3(PO4)2
0.1554 moles of Ca3(PO4)2
2 moles of P 1 mole of
Ca3(PO4)2
0.3108 moles of P
30.97 g of P 1 mole of P
9.625 g of P
12

Stoichiometry - Switching Yard In operation
95.00 g of Ca
1 mole of Ca 40.08 g of Ca
2.37 moles of Ca
1 moles of H2 1 mole of Ca
2.37 moles of H2
Stop
13
Problem 5.49 p 203-04

• Ammonia and water react to form nitrogen and
  water
• 4 NH3 3 O2 ? 2 N2 6 H2O
• How many grams of O2 are needed to react with
  8.00 moles of NH3?
• How many grams of N2 can be produced when 6.50 g
  of O2 reacts?
• How many grams of water are formed from the
  reaction of 34.0g of NH3?

14
Limiting Reagents Limiting Reactant

• In an ideal chemical reaction, just the right
  amount of reactants are combined and the reaction
  goes to completion without any losses and we get
  100 conversion to products with 0 reactants
  left over
• GET REAL!
• Usually we put in a certain amount of the more
  expensive reactant and add an excess of the
  cheaper reactant in order to force the production
  of the most product based on the more expensive
  reactant
• When buying a kit to assemble a powered model car
  (or something similar) are there extra screws
  provided or are there extra motors provided!

15
Limiting Reagents Overview

• Consider this equation
• 3 NaOH (aq) H3PO4 (aq) ? 3 H2O (l) Na3PO4
  (aq)
• We need to have 3 moles of NaOH provided for
  every 1 mole of H3PO4
• What if there were only 2 moles of NaOH provided?
  
• We could only form 2/3 of the Na3PO4
• NaOH is the limiting reagent
• What if there were 4 moles of NaOH provided?
• We would get the full amount of Na3PO4 (aq) but
  there would be one mole of NaOH leftover
• H3PO4 is the limiting reagent
• We seldom are given values in moles!
• We have to do calculations from g to moles to
  determine what is the limiting reagent
• Then we often have to determine how much of a
  specific product will form

16
Limiting Reagents Calculations

• Consider this reaction by which iron is formed in
  a blast furnace
• Fe2O3 (s) 3 CO (g) ? 2 Fe (l) 3 CO2 (g)
• 1.14 g of CO is present to react with 1.00 g of
  Fe2O3.
• What is the limiting reactant?
• What is the theoretical yield of iron?
• If the actual yield is 0.612 g of iron, what is
  the percent yield?
• Two new concepts to work in
• Theoretical yield
• Percent yield
• Use switching yard

17
Stoichiometry - Switching Yard In operation
Fe2O3 (s) 3 CO (g) ? 2 Fe (l) 3 CO2 (g)
1.14 g CO
1mole of CO 28.01 g of CO
1/3 MOLES
0.0407 moles of CO
0.0136 moles of Fe2O3
HAVE
NEED
0.00626 moles of Fe2O3
0.0188 moles Of CO
HAVE
NEED
1 mole of Fe2O3 159.70 g of Fe2O3
1.00 g Fe2O3
3 MOLES
18
Using Limiting Reagent Information

• Determine the amount Fe formed theoretical
  yield
• Fe2O3 (s) 3 CO (g) ? 2 Fe (l) 3 CO2 (g)
• Fe2O3 was determined to be limiting reagent
  0.00626 moles
• If all of the Fe2O3 was used up, none would be
  left over but there would be some leftover CO
  (which we are not interested in)
• The number of moles of Fe that would form?
• 1 2 moles 2 0.0125 moles 3 0.00626 moles
  4 something else
• The number of g of Fe formed (theoretical yield)?
  
• 1 0.349 g 2 1.396 g 3 0.698 g 4
  something else
• Percent yield how much of what we expected did
  we get?
• Actual yield/theoretical yield x 100
• Percent yield will always be less than 100

19
Problem 5.74 p 213

• MODIFIED
• Propane gas, C3H8, a fuel for many barbecues,
  reacts with oxygen according to the following
  balanced equation
• C3H8 5 O2 ? 3 CO2 4 H2O
• If 5.50 g of propane and 11.0 g of oxygen are
  mixed together, how many grams of carbon dioxide
  can be produced?
• If the actual mass of carbon dioxide is 7.40 g,
  what is the percent yield?

20
Measuring changes in energy content

• Review
• Surroundings either accept energy from changes in
  matter OR provide energy to allow changes in
  matter
• PP chapter 2 pages 8 9
• Calorimeters surroundings that can measure
  heat transferred (produced or required) in
  changes of matter
• Heat transferred (Q)
• Q S.H. x mass x DT
• S.H. specific heat
• Every material has a different capacity to hold
  heat
• S.H. water 1.00 cal/oC-g 4.18 J/oC-g
• Mass ? the amount of heat that can be held
  depends on the amount of material
• DT ? the more heat added or removed, the
  larger the temperature change
• Specific heat (S.H.) of different materials can
  be determined by rearranging the equation

21
Changes in Heat (Q) and Heats of Reaction (DH)

• Q is heat for any experiment with any amounts of
  materials
• Values in cal, Joules, kcal, kJ depending on
  amount
• No sign for Q
• DH is heat for 1 mole of product(s) or
  reactant(s) produced or reacted (as defined)
• Also called enthalpy change
• Values always as kcal/mole or kJ/mole
• For exothermic reactions DH is for endothermic
  DH is
• Since g and moles of anything can be converted
  knowing mole to g relationships, Q DH can be
  interconverted

22
Stoichiometry - Switching Yard Connecting
Energy Content

Heat of reaction
kcal mole of A
mole of A kcal
kcal
23
Energy Calculations examples

• 1. The specific heat of copper metal is 0.0920
  cal/g-oC. How many kcal of heat are necessary to
  raise the temperature of a 800 g block of copper
  from 25.0oC to 88.5oC?
• Q S.H. x mass x DT
• Q 0.092 cal/g-oC x 800 g x (88.5-25.0)
  4673.6 cal 4.7 kcal
• Significant figures (SF)
• ? common sense
• ? either 2 or 3

2. (6.13) The equation for the formation of
silicon tetrachloride form silicon and chlorine
is Si 2 Cl2 ? SiCl4 DH -157 kcal How many
kcal are released when 125 g of Cl2 react with
silicon? 3. What is the heat of reaction for
C3H8 5 O2 ? 3 CO2 4 H2O if the combustion of
5.00 g of C3H8 produces 60.2 kcal of heat?
- 528 kcal
1 mole of Cl2 70.9 g ofCl2
1 mole of C3H8 44.0 g of C3H8
1.76 mole Cl2
0.114 mole C3H8
125 g Cl2
5.00 g C3H8
277 kcal
60.2 kcal
157 kcal mole Cl2
528 kcal mole C3H8