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QUANTITATIVE CHEMISTRY UNIT HL1

- IB Chemistry Gr.11
- IB Topic 1

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

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

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)

- 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

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!

3.7

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

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

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

- 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 )

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

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

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

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)

1.3 Chemical Equations

A chemical equation is a shorthand notation of a

chemical reaction.

- 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

Balancing Chemical Equations

- Write the correct formula(s) for the reactants

and the products.

Ethane reacts with oxygen to form carbon dioxide

and water

- 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.

- 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

- Balance those elements that appear in two or more

reactants or products.

7 oxygen on right

remove fraction multiply both sides by 2

- Check to make sure that you have the same number

of each type of atom on both sides of the

equation.

TOPIC

- 1.4 Mass and Gas Volume Relationships in

chemical reactions

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)

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

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

Limiting Reactants

What is the limiting reagent? NO Why?

Then, O2 is the excess reactant.

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

Use limiting reagent (Al) to calculate amount of

product that can be formed.

234 g Al2O3

124 g Al

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.

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

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)

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

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

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)

Avogadros Law

EXAMPLE 6

- Calculate the volume (in L) occupied by 7.40 g of

NH3 at STP. (1 L 1 dm3 )

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

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

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

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

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)

Solution Stoichiometry

The concentration of a solution is the amount of

solute present in a given quantity of solvent or

solution.

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

Preparing a Solution of Known Concentration

When ions (charged particles) are in aqueous

solutions, the solutions are able to conduct

electricity.

- Pure distilled water (nonconducting)
- Sugar dissolved in water (nonconducting) a

nonelectrolyte - NaCl dissolved in water (conducting) an

electrolyte

5.6

Chemistry in Action Metals from the Sea