QUANTITATIVE CHEMISTRY UNIT HL1

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QUANTITATIVE CHEMISTRY UNIT HL1

• IB Chemistry Gr.11
• IB Topic 1

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• Topics
• 1.1 The mole and Avogadros Constant
• 1.2 Formulas

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IB STANDARDS

• Upon completion of this unit the SWBAT
• 1.1.1 Apply the mole concept to substances
  (5.2.12.B.3)
• 1.1.2 Determine the number of particles and the
  amount of substance (in moles) (5.2.12.B.3,
  9.1.12.A.1)
• 1.2.1 Define the terms relative atomic mass and
  relative molecular mass (5.2.12.B.3)
• 1.2.2 Calculate the mass of one mole of a species
  from its formula (5.2.12.B.3)
• 1.2.3 Solve problems involving the relationship
  between the amount of substance in moles, mass,
  and molar mass

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1.1 The Mole and Avogadros Constant.
The mole (mol) is the amount that contains the
same number of chemical species as there are
atoms in exactly 12.00 grams of 12C.

• Avogadros constant (L)

1 mol L 6.022x 1023

• The relative atomic mass of an element is the
  average mass of an atom of the element taking
  into account all its isotopes and their relative
  abundance, compared to one atom of carbon12
  expressed in atomic mass units (amu)

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• Molar mass is the mass of 1 mole of a species
  expressed in grams and has units of
  grams/mole

1 mole 12C atoms 6.022 x 1023 atoms 12.00 g 1
12C atom 12.00 amu
1 mole 12C atoms 12.00 g 12C 1 mole lithium
atoms 6.941 g of Li
Number of particles (N) number of moles
(n) x Avogadros Constant (L) N nL
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Keep these relationships in mind
use molar mass
molecules
grams
use Avogadros number
moles
Remember the critical link between moles and
grams of a substance is the molar mass.
ITS SIMPLE THINK IN TERMS OF PARTICLES!
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Example 1 How many H atoms are in 72.5 g of
propanol, C3H8O ?
Solution 1 mol C3H8O (3 x 12.0) (8 x 1.0)
16.0 60 g C3H8O
1 mol C3H8O molecules 8 mol H atoms 1 mol H
6.022 x 1023 atoms H
Step 1 Convert mass to moles
1.21 moles C3H8O
72.5 g C3H8O
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STEP 2 CONVERT MOLES C3H8O TO MOLES H ATOMS
9.68 moles H atoms
1.21 moles C3H8O
Step 3 Convert moles H atoms to atoms H
9.68 moles H atoms
5.83 x 1024 atoms H
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Example 2 How many atoms are in 0.551 g of
potassium (K) ?
1 mol K 39.10 g K
1 mol K 6.022 x 1023 atoms K
0.551 g K
8.49 x 1021 atoms K
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• IB Standards
• 1.2.4 Distinguish between the terms empirical
  formula and molecular formula
• 1.2.5 Determine the empirical formula from the
  percentage composition or from other experimental
  data (9.1.12.A.1)
• 1.2.6 Determine the molecular formula when given
  both the empirical formula and experimental data
  (5.3.B1, 5.6.A1, 9.1.12.A.1 )

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1.2 CHEMICAL FORMULA

• Molecular Formula gives the actual number of
  atoms of each element present in a molecule of a
  compound.
• Empirical formula gives the ratio of the atoms
  of different elements in a compound. It is the
  molecular formula expressed as its simplest
  ratio.
• What is the empirical formula of glucose,
• C6 H12O6?
• Answer The simplest ratio CH2O

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Formula Mass is the sum of the atomic masses (in
amu) in a formula unit of an ionic compound.
For any ionic compound formula mass (amu)
molar mass (grams)
1 formula unit NaCl 58.5 amu 1 mole NaCl 58.5
g NaCl
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Do NOW What is the formula mass of Ca3(PO4)2 ?
1 formula unit of Ca3(PO4)2
310.3 amu
What is the molar mass of Ca3(PO4)2 ? Answer
310.3 grams
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TOPIC1.3 CHEMICAL EQUATIONS

• IB Standards
• Deduce chemical equations when all reactants and
  products are given
• Identify the mole ratio of any two species in a
  chemical equations
• Apply the state symbols (s), (l), (g), and (aq)

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1.3 Chemical Equations
A chemical equation is a shorthand notation of a
chemical reaction.
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• Types of Reactions
• 1. Combination A B ? AB
• CO2 H2O ? H2CO3
• 2. Decomposition AB ? A B
• CaCO3 ?CO2 CaO 3. Combustion (Burning
  fuels) CH4 2 O2 ? CO2 2 H2O4. Single
  Replacement
• A BC? AC B
• 2Al Fe2O3 ? Al2O3 2Fe
• 5. Double Replacement
• AB CD ? AD CB
• HCl NaOH ? NaCl H2O

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Balancing Chemical Equations

1. Write the correct formula(s) for the reactants
   and the products.

Ethane reacts with oxygen to form carbon dioxide
and water

1. Change the coefficients of the chemical formulas
   to make the number of atoms of each element equal
   on both sides of the equation. Do not change the
   subscripts.

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1. Start by balancing those elements that appear in
   only one reactant and one product.

start with C or H but not O
multiply CO2 by 2
multiply H2O by 3
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1. Balance those elements that appear in two or more
   reactants or products.

7 oxygen on right
remove fraction multiply both sides by 2
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1. Check to make sure that you have the same number
   of each type of atom on both sides of the
   equation.

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TOPIC

• 1.4 Mass and Gas Volume Relationships in
  chemical reactions

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IB STANDARDS

• Calculate theoretical yields from chemical
  equations
• Determine the limiting reactant and the reactant
  in excess when quantities of reacting substances
  are given
• Solve problems involving theoretical,
  experimental, and percentage yield (9.1.12.A.1)

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1.4 Mass and Gaseous Volume Relationships in
Chemical Reactions
Quantities of reactants and products in a
balanced chemical equation has many applications.
Example 3 How many moles of O2 are needed to
react with 4.26 moles of H2? 2
H2 O2 ? 2 H2O 4.26 mol H2 X ( 1 mole O2 / 2
mole H2) 2.13 mole O2
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Example 4 Methanol burns in air according to the
equation
If 209 g of methanol are used up in the
combustion, what mass of water is produced?
molar mass CH3OH
molar mass H2O
coefficients chemical equation
209 g CH3OH
235 g H2O
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Limiting Reactants
What is the limiting reagent? NO Why?
Then, O2 is the excess reactant.
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Example 5 In one process, 124 g of Al are
reacted with 601 g of Fe2O3 ,
Calculate the mass of Al2O3 formed.
367 g Fe2O3
124 g Al
Have more Fe2O3 (601 g) so Al is limiting reagent
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Use limiting reagent (Al) to calculate amount of
product that can be formed.
234 g Al2O3
124 g Al
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Reaction Yield
Theoretical Yield is the maximum amount of
product That can be produced.
Actual Yield is the amount of product actually
obtained from a reaction.
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Class Exercise

• If the theoretical yield of iron was 30.0 g and
  the actual yield was 26.8 grams, calculate the
  per cent yield.
• 2Al Fe2O3 ? Al2O3 2Fe

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IB STANDARDS

• Apply Avogadros Law to calculate reacting
  volumes of gases (9.1.12.A.1)
• Apply the concept of molar volume at standard
  temperature and pressure in calculations
• Solve problems using the relationships between
  temperature, pressure, and volume for a fixed
  mass of an ideal gas (9.1.12.A.1)
• Solve problems using the ideal gas equation
• Analyze graphs relating to the ideal gas equation
  (9.1.12.A.1, 9.1.12.B.1)

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Molar Volume of a Gas STP standard temperature
and pressure 0 0C and 100 kPa. Previous
standard is 1 atm 101.3 kPa. Molar Volume at
STP 22.4 dm3
1 dm3 1 L
Molar Volume at RTP 24 dm3
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The Gas Laws Boyles Law
P a 1/V
Constant temperature Constant amount of gas
P x V constant
P1 x V1 P2 x V2
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Relationship of Gas Volume (V) with
Temperature at Constant Pressure
Charless Law V a T
Temperature must be in Kelvin
T (K) t (0C) 273.15
Relationship Between Gas Temperature And
Pressure (P) Gay-Lussacs Law P a
T Temperature in Kelvins P 0 at absolute
zero (- 273 oC)
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Avogadros Law
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EXAMPLE 6

• Calculate the volume (in L) occupied by 7.40 g of
  NH3 at STP. (1 L 1 dm3 )

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Ideal Gas Equation
Charless law V a T (at constant n and P)
Avogadros law V a n (at constant P and T)
R is the gas constant
PV nRT
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COMBINED GAS LAW P1V1/ T1 P2V2/T2

• REMINDER
• Use SI units when using the ideal gas equation.
• R 8.31 J/Kmol
• P in Pa (pascal)
• V in m3
• T in Kelvin

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Density (p) Calculations
m is the mass of the gas in g
M is the molar mass of the gas P is the pressure
of the gas
Molar Mass (M) of a Gaseous Substance
p is the density of the gas in g/L
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TOPIC1.5 SOLUTIONS AND SOLUTION
CONCENTRATION

• IB Standards
• Distinguish between the terms solute, solvent,
  solution, and concentration
• Solve problems involving concentration, amount of
  solute, and volume of solution

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A solution is a homogenous mixture of 2 or more
substances.
The solute is (are) the substance(s) present in
the smaller amount(s).
The solvent is the substance present in the
larger amount.
Soft drink (l)
H2O
Sugar, CO2
Air (g)
N2
O2, Ar, CH4
Pb
Sn
Soft solder (s)
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Solution Stoichiometry
The concentration of a solution is the amount of
solute present in a given quantity of solvent or
solution.
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How to prepare a 1.00 M NaCl solution
mol solute
M
L of solution
Note you do NOT add 58.5 g NaCl to 1.00 L of
water. The 58.5 g will take up some volume,
resulting in slightly more than 1.00 L of
solution and the molarity would be lower.
5.5
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Preparing a Solution of Known Concentration
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When ions (charged particles) are in aqueous
solutions, the solutions are able to conduct
electricity.

1. Pure distilled water (nonconducting)
2. Sugar dissolved in water (nonconducting) a
   nonelectrolyte
3. NaCl dissolved in water (conducting) an
   electrolyte

5.6
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Chemistry in Action Metals from the Sea