1.1.1 Apply the mole concept to substances.
Apply: Use an idea, equation principle, theory or law in a new situation.
IB Core
Size matters
Atoms are VERY small, how small? Take a ping pong ball an
placed it in the center of a soccer stadium. This is the atoms nucleus.
Then take a pin and stand on the very edge of the stadium. The pin head is the electron.
Dealing with VERY small things To count small things we need a large number
1 dozen atoms is only 12 atoms
We need a new number. Avogadro's Number
The mole: 6.02 x 10 23 1 mole atoms = 6.02 x 10 23 atoms 3 moles of atoms = 3 (6.02 x 10 23 ) atoms
1.1.1 Apply the mole concept to substances.
1.1.1 Apply the mole concept to substances.
Understanding the number• If you ordered a mole of donuts instead of a
dozen, that would be enough to give every person in the world over 9 trillion donuts.
• The Earth is estimated to be about 4.54 billion years old. That is approximately 1.43 x 1017 seconds, still much less than a mole.
• Spreading a mole of marbles over the Earth would produce a layer almost five kilometers thick
• If you traveled a mole in inches, you would be able to cross the galaxy and back eight times, or go around the Earth 300 trillion times.
1.1.1 Apply the mole concept to substances.
The Sahara Desert is about 1500 km north to south and 4000 km east to west.
If the average depth of sand throughout is 10 m, and there were about 1000 grains of sand in each cm3, then the Sahara Desert would contain about 6 x 1022 grains of sand.
This is still one tenth of a mole. This means there are more atoms of
carbon in a 12 g sample (one mole) then there are grains of sand in the Sahara Desert
IB Core
1.1.2 Determine the number of particles and the amount of substance (in moles).
Determine: Find the only possible answer. (Obj. 3)
Using Avogadro's Number How many Hydrogen and Oxygen atoms are
there in 1 mol of water? What is the total # of atoms?
H2O Ratio 1:2: There are 2 Hydrogen atoms for
every water molecule. 1 mol of water molecules has 2 moles of hydrogen
atoms 2 mol Hydrogen x (6.02 x 10 23 )atoms/mol = 12.04 x 10 23 hydrogen atoms.
Ratio 1:1 There is 1 Oxygen atom for ever water molecule. 1 mol oxygen 6.02 x 10 23 atoms/mol = 6.02 x 10 23 oxygen atoms / mol
1.1.2 Determine the number of particles and the amount of substance (in moles).
Using 1 mol of substance, how many atoms? HCl
O3
CH4
There are 2 mols of atoms: 1.20*1024 atoms
There are 3 mols of atoms: 1.81*1024 atoms
There are 5 mols of atoms: 3.01*1024 atoms
Click for answers
1.1.2 Determine the number of particles and the amount of substance (in moles).
IB Core1.2.1 Define the terms relative
atomic mass (Ar) and relative molecular mass (Mr).
Define: Give the precise meaning of a word, phrase or physical quantity. (Obj 1)
Relative Atomic Mass & Relative Molecular Mass
Relative Atomic Mass (Ar): Weighted mean mass of all naturally occurring isotopes
of an element. The number is the ratio of the average mass per atom to 1/12 of an atom of the C-12 isotope.
Relative Molecular Mass (Mr) The mass of a molecule. Indicates how heavy the
molecule is compared to the C-12 isotope.
Relative things have NO UNITS!
But!! Atomic Mass does! (g/mol) or (g mol-1)
You find these values on the Periodic Table
To get this value, just use the Ar from
above and add g/mol to the end
IB Core1.2.2 Calculate the mass of one
mole of a species from its formula.
Calculate: Find a numerical answer showing the relevant stages in the working (unless instructed not to do so).
Molar Mass
Molar mass (M) is the mass of one mol of atoms (g/mol)
For compounds or Elements with multiple atoms: Add the atomic mass
of all the atoms in the compound.
What is the Molar mass of H2O?
H: 1.008 g/mol O: 16.00 g/mol
M = 2(1.008)g/mol + 1(16.00) g/mol =18.02 g/mol
Number of Atoms
involved
Atomic Mass(On P-table)
Units...Don’t be lazy!
Final Molar Mass of Water
Sig. Figs
IB Core
1.2.3 Solve problems involving the relationship between the amount of substance in moles, mass and molar mass.
Solve: Obtain an answer using algebraic and/or numerical methods.
1.2.3 Solve problems involving the relationship between the amount of
substance in moles, mass and molar mass.Stoichiometry Time!
Time to learn what stoichiometry is all about!
Yee-haw!!Definition of stoichiometry: The
relationships among the quantities of reactants and products involved in chemical reactions.
1.2.3 Solve problems involving the relationship between the amount of
substance in moles, mass and molar mass.
unit desired unitgiven
unit desiredunit x Given
• When you have something to start with, and you know what unit youare after, you can solve the problem!!• Be sure to cancel out units as you go through.
You may also see it done this way...
Formula n= m
M n = # of moles (mol)
m = mass (g)
MM = Molar Mass (g/mol)
Oxygen8O
15.999
1.2.3 Solve problems involving the relationship between the amount of substance in moles, mass, and molar mass.
Sample problems:
If you have 4.904 g of sulfuric acid, how many moles would you have?
You have 2.00 mols of carbon dioxide. How many grams of carbon dioxide do you have?
Answers to previous slide
Sulfuric acid is H2SO4 . So the molar mass would be (2 x 1.01) + (32.07) + (4 x 16.00) = 98.09 g mol-1 4.904 g H2SO4 x = 0.04999 mols
Carbon dioxide is CO2. So 12.01 + (2 x 16.00) = 44.01 g mol-1.
2.0 mol CO2 x = 88 g CO2
H2SO4 g 98.09
H2SO4 mol 1
CO2 mol 1
CO2 g01.44
1.2.3 Solve problems involving the relationship between the amount of substance in moles, mass, and molar
mass.
The combustion of Hydrogen gas and Oxygen. 1) Determine the product and write the
chemical equation for the reaction 2) Balance the equation 3) What is the ratio of Oxygen to Hydrogen?
If I react 24 g of Hydrogen gas with oxygen, 1) How many grams of oxygen will I need? 2) How many grams of product will I make?
Answer 2H2 + O2 → 2H2O
Ratio:
You will need 380 grams of oxygen.
You will make 214 grams (210 using sig figs) of product (H2O)
H2 mol 2
O2 mol 1
IB Core Objective
1.2.4 Distinguish between the terms empirical formula and molecular formula.
Distinguish: Give the difference between two or more different terms.
1.2.4 Distinguish between the terms empirical formula and molecular formula.
Empirical formula States the elements present in the compound. Simplest whole number ratio of these
elements. Example: Empirical formula of glucose=CH2O
Molecular formula• The elements present in the compound• Actual number of atoms of the elements in one
molecule• Example: Molecular formula of
glucose=C6H12O6
IB Core Objective 1.2.5 Determine the empirical
formula from the percentage composition or from other experimental data.
Determine: Find the only possible answer (Obj. 3)
1.2.5 Determine the empirical formula from the percentage composition or from other experimental data.An analogy
A test is worth 100% and there are 3 sections:
Section A = 20 marks Section B = 50 marks Section C = 30 marks
What % of the test is Section A? What would you make the test out of to make
mark calculations easy? The easiest way to mark this test is to make the total
100.
1.2.5 Determine the empirical formula from the percentage composition or
from other experimental data. You need to determine the empirical formula for a compound of
phosphorous and oxygen that contains 43.64% phosphorus by mass. You are not given a mass, so make up one. Remember, 100 is an easy
number, so let’s start there…. In 100g, there would be 43.64 g of P, and 56.36g of O (100-
43.64=56.36). Now mole it out (find the moles!!) Amount of phosphorus =
= 1.409 mols Amount of oxygen =
= 3.523 molsRatio = 1.409 : 3.523Divide 3.523 by 1.409= 2.5, so 1:2.5. But we need whole numbers!!!!So multiply both sides by two= 2:5 ratio. What is the empirical formula?
1-mol 97.30
64.43
g
g
1-mol g 16.00
56.36g
1.2.5 Determine the empirical formula from the percentage composition or
from other experimental data. Empirical formula with 2:5 ratio would be: P2O5
Finding The Empirical Formula
A compound was found to have a % composition of 38.67 % Carbon 16.22 % Hydrogen 45.11 % Nitrogen
What is the Empirical formula? Click for answer
Answer: CH5N
1.2.5 Determine the empirical formula from the percentage
composition or from other experimental data. You run an experiment where you start with a
compound containing hydrogen and carbon. You use 2.80 g of this compound, and after it
combusts, you have 8.80 g of carbon dioxide and 3.60 g of water.
Use your stoichiometry skills, mole it out, follow the mole ratios, and find the empirical formula.
IB Core Objective 1.2.6 Determine the
molecular formula when given the empirical formula and experimental data.
Determine: Find the only possible answer (Obj. 3)
Molecular Formula Molecular formula is related to the molar
mass of the compound
Sometimes the Molecular Formula is the same as the Empirical Formula.
Ratio = Relative Molecular Mass (Mr)
Empirical Relative Molecular Mass
Multiply the Empirical Formula by the ratio found
Empirical and Molecular Formulas
A compound was found to contain 71.65 % Cl 24.27 % C 4.07 % H The molar mass is known to be 98.96g/mol
What is the empirical and molecular formula?
Remember, assume a 100g sample
IB Core Objective
1.3.1 Deduce chemical equations when all reactants and products are given.
Deduce: Reach a conclusion from the information given.
Tips on Balancing
1) To start, balance all the metals
2) Next balance all the non metals EXCEPT for Oxygen and Hydrogen AND that are not in a complex ion.
3) Next balance all complex ions (you must be able to recognize these!!!)
4) Finally balance Oxygen and Hydrogen, if one of the 2 is in its elemental state balance it last.
1.3.1 Deduce chemical equations when all reactants and products are given.
You already have practice and knowledge of chemical equations.
Practice balancing these equations:
CH4 + O2 CO2 + H2O
C2H5OH(l) + O2(g) CO2(g) + H2O(g)
(NH4)2Cr2O7(s) Cr2O3(s) + N2(g) + H2O(g)
Balancing Practice CH4 + O2 CO2 + H2O
C2H5OH(l) + O2(g) CO2(g) + H2O(g)
(NH4)2Cr2O7(s) Cr2O3(s) + N2(g) + H2O(g)
Solid lithium hydroxide is used as a carbon dioxide absorber on the space shuttle. The products formed are solid lithium carbonate and liquid water. What mass and volume of carbon dioxide will be absorbed if 1.00kg of lithium hydroxide is used at S.T.P.? (HW)
Don’t forget the State
subscripts!
IB Core Objective 1.3.2 Identify the mole ratio of
any two species in a chemical equation.
Identify: What is the definition? Your turn….
Identify: Find an answer from a given number of possibilities. (Obj 2)
1.3.2 Identify the mole ratio of any two species in a chemical equation.You already have practice with this as well.
Yeahh!Practice makes perfect….
2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O (l)
What is the ratio of sodium hydroxide to sodium sulfate?
What is the ratio of sodium hydroxide to sulfuric acid?
What is the ratio of moles of oxygen atoms in sulfuric acid to moles of sulfuric acid?
What is the ratio of moles of oxygen atoms in water to moles of water?
IB Core Objective 1.3.3 Apply the state symbols
(s), (l), (g), and (aq).
Apply: Use an idea, equation, principle, theory, or law in a new situation.
1.3.3 Apply the state symbols (s), (l), (g), and (aq).
(s) = solid(l) = liquid(g) = gas(aq) = aqueous solution
1.3.1 Deduce chemical equations when all reactants and products are given.
1.3.3 Apply the state symbols (s), (l), (g), and (aq). For a reaction occurring in aqueous solution forming a precipitate (solid), or part of an acid-base or redox reaction, it is often better to write the ionic equation.
For example: Pb(NO3)2(aq) + 2NaCl(aq) PbCl2(s) + 2NaNO3(aq)
What are the ions which would be in solution (non-solid) for each of these molecules? Write them down.
Then put the ions and solid in an equation, just like above. This is the ionic equation.
1.3.1 Deduce chemical equations when all reactants and products
are given.1.3.3 Apply the state symbols (s),
(l), (g), and (aq). The previous ionic equation is often referred to as
the complete ionic equation. However, we can reduce this to the net ionic equation.
To do this, we get rid of the spectator ions. This would be any ions that are not involved in forming the main product.
Pb2+(aq) + 2NO3
-(aq) + 2Na+
(aq) + 2Cl- (aq)
PbCl2(s) + 2Na+(aq) + 2NO3
-(aq)
Which ones are the spectator ions? (Which appear twice?). Take these out, re-write, and you have the net ionic equation.
1.3.1 Deduce chemical equations when all reactants and products are given.1.3.3 Apply the state symbols (s), (l),
(g), and (aq).Answer
Net ionic equation:Pb2+
(aq) + 2Cl- (aq)
PbCl2(s)
IB Core Objective
1.4.1 Calculate theoretical yield from chemical equations.
Calculate: Find a numerical answer
showing the relevant stages in the working (unless instructed not to do so).
1.4.1 Calculate theoretical yield from chemical equations.
You already have practice doing this with mass to mass calculations!
Practice: What mass of sodium hydrogencarbonate (NaHCO3) must be heated to give 8.80g of carbon dioxide (CO2)?
2NaHCO3 → Na2CO3 + CO2 + H2O
Answer: 33.6 g
IB Core Objective
1.4.2 Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given.
Determine: Find the only possible answer.
1.4.2 Determine the limiting reactant and the reactant in
excess when quantities of reacting substances are given.
We sometimes don’t always have perfect amounts of each reactant.
We can only make as much product as the limiting reactant will allow.
We must determine which reactant is limiting
1.4.2 Determine the limiting reactant and the reactant in
excess when quantities of reacting substances are given.Finding The Limiting Reactant
1) Balance the chemical equation
2) Find the number of mol’s of each reactant
3) Divide the # of moles by the coefficient found in the balanced equation.
4) The chemical that has the smallest value is the limiting reactant.
1.4.2 Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given.
In the reaction 2NH3 + CO2 → (NH2)2CO + H2O
You have 25.5 g of ammonia (NH3) and 22.0 g of carbon dioxide (CO2).
Which is the limiting reactant? What is the mass of each product? How much reactant is left over?
Answers: Limiting: carbon dioxide Mass of each product: (NH2)2CO = 30.0g H2O =
9.01 g. NH3 left over= 8.52 g
1.4.2 Determine the limiting reactant and the reactant in
excess when quantities of reacting substances are given.
2.50 g of methane is mixed with 3.00 g of water 1) Which reactant is the limiting one? 2) What is the mass of each product? 3) What is the mass of reactant left?
CH4(g) + H2O(g) CO(g) + H2(g)
IB Core Objective
1.4.3 Solve problems involving theoretical, experimental and percentage yield.
Solve: Obtain an answer using algebraic and/or numerical methods.
1.4.3 Solve problems involving theoretical, experimental and percentage yield.
Theoretical yield: Value that you should get mathematically.
Experimental yield: The value you actually get from an experiment.
Percentage yield= x 100% yield lTheoretica
yield alexperimentor Actual
1.4.3 Solve problems involving theoretical, experimental and percentage yield.
Put it all together: limiting reactants and yield.For the reaction H2O2(aq) + 2KI(aq) + H2SO4(aq) →
I2(s) + K2SO4(aq) + 2H2O(l)
If you add 100.00 g of KI to a solution of 12.00 g of hydrogen peroxide (H2O2) and 50.00 g of sulfuric acid (H2SO4), what would the theoretical yield of iodine (I2) be?
Answer: 76.44 g
In the experiment you performed, you produced an experimental yield of 62.37 g of iodine (I2). What was the percentage yield?
Answer: 81.59%
Question
What is the composition by mass of 1 mol of Iron(III) Oxide? Click for answer:
M = 159.7 g/molFe = 69.9%O = 30.1%
If it doesn’t add up to 100% something is
wrong!
IB Core Objective
1.5.1 Distinguish between the terms solute, solvent, solution and concentration (g dm-3 and mol dm-3)
Distinguish: Give the difference between two or more different items.
1.5.1 Distinguish between the terms solute, solvent, solution and concentration (g dm-3 and mol dm-3)
Solution: Homogenous mixture of two or more substances.
Solvent: Liquid in which the dispersion occurs.
Solute: The substance dissolved in the solvent.
So….a solution contains a solute dissolved in a solvent.
Concentration: Amount of substance contained within a given volume
1.5.1 Distinguish between the terms solute, solvent, solution and concentration (g dm-3 and mol dm-3)
Concentration C = Concentration or Molarity (mol/L or
mol/dm3 or M) Molarity = moles of solute
Liters of solvent V = volume (L, dm3)
Q: How many moles are in a 25cm3 of 10M sulphuric acid (H2SO4)?
A: 0.25 moles
IB Core Objective
1.5.2 Solve problems involving concentration, amount of solute and volume of solution.
Solve: Obtain an answer using algebraic and/or numerical methods.
1.5.2 Solve problems involving concentration, amount of solute and volume of solution.
Q: Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough water to make 26.8ml(cm3) of solution.
A: 1.60M HCl
Click for answer
1.5.2 Solve problems involving concentration, amount of solute and volume of solution.
Calculate the number of moles of each ion in 1.75L of 1.0 x 10-3 M ZnCl2
A: 1.8*10-3 mol Zn2+
3.5*10-3 mol Cl-
Click for answer
1.5.2 Solve problems involving concentration, amount of solute and volume of solution.
Dilutions Sometimes a concentrated solution will
have solvent added to produce a more dilute one. # of moles does not change so, ‘n’ is common
to both!!!
C1 x V1 = C2 x V2
1.5.2 Solve problems involving concentration, amount of solute and volume of solution.
Q: What volume of 16M sulfuric acid must be used to prepare 1.5L of a 0.10M H2SO4 solution? Click for answer
A: 9.4 *10-3dm3 or 9.4 cm3 (mL)
Special note: Always add acid to water not the other way around!!
The Challenger
Shuttle1986
Mistakes in Stoichiometry
1999 Mars Orbital Lander
IB Core Objective
1.4.4 Apply Avogadro’s law to calculate reacting volumes of gases.
Apply: Use an idea, equation, principle, theory, or law in a new situation.
1.4.4 Apply Avogadro’s law to calculate reacting volumes of gases.
At a constant temperature and pressure, a given volume of any gas always contains the same number of particles.
In other words, equal amounts of gases at the same temperature and pressure occupy the same volume!
1.4.4 Apply Avogadro’s law to calculate reacting volumes of gases.
Using the following reaction C2H2(g) + 2H2(g) → C2H6(g)
If you had 10cm3 of ethyne (C2H2), what volume of hydrogen (H2) would you need to fully react with it?
What is the volume of ethane (C2H6) that would be produced?
Answer: 20cm3 H2
10cm3 of ethane
IB Core Objective
1.4.5 Apply the concept of molar volume at standard temperature and pressure in calculations.
Apply: Use an idea, equation, principle, theory or law in a new situation. (Obj. 2)
1.4.5 Apply the concept of molar volume at standard temperature and pressure in
calculations.
Molar volume: The volume of any gas that contains one mole.
If temperature and pressure are specified, then molar volume can be calculated.
At Standard Temperature and Pressure (STP): 1 mol of gas occupies 22.4dm3
0oC = 273.15K, 1atm = 760 mm Hg = 760 torr = 101.3kPa
1.4.5 Apply the concept of molar volume at standard temperature and pressure in
calculations.
How many moles of oxygen (O2) are there in 5.00 dm3 of oxygen?
Answer: 0.223 mol
IB Core Objective
1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas.
Solve: Obtain an answer using algebraic and/or numerical methods. (Obj. 3)
1.4.6 Solve problems involving the relationship between temperature, pressure and volume for a fixed mass of an ideal
gas.
Ideal gas: Particles have negligible volume, there are no attractive forces between particles, and kinetic energy is proportional to absolute temperature.
Boyle’s Law: volume is inversely proportional to pressure at a constant temperature.
i.e. When volume decreases, pressure increases. When volume increases, pressure decreases.
1.4.6 Solve problems involving the relationship between
temperature, pressure and volume for a fixed mass of an ideal gas.
Charles’s Law: Volume is proportional to temperature at constant pressure. (This is for variable volume only, such as a balloon) Therefore, doubling the temperature doubles the volume.
Gay-Lussac’s Law: The pressure is proportional to the absolute temperature of the gas.Therefore, increasing the temperature increases the pressure, and vice versa.How do we put this all together….?
1.4.6 Solve problems involving the relationship between temperature,
pressure and volume for a fixed mass of an ideal gas.
P1V1/T1 = P2V2/T2
The one refers to the initial measurement, the two refers to the second measurement.
If one of the values is constant, then you can just remove that value!
1.4.6 Solve problems involving the relationship between temperature,
pressure and volume for a fixed mass of an ideal gas.
A sample of methane gas that has a volume of 3.8L at 5oC is heated to 86 oC at constant pressure. Calculate its new volume.
Remember to convert Celsius to Kelvins!
Answer: 4.9L
Click for answer
IB Core Objective
Solve problems using the ideal gas equation, PV=nRT
Solve: Obtain an answer using algebraic and/or numerical methods. (Obj. 3)
1.4.7 Solve problems using the ideal gas equation, PV=nRT
Universal Gas Equation: PV=nRT
P = Pressure V = Volume N = # of mols T = Temperature (K) R = Universal Gas Constant = 8.314 JK-1 mol-1
OR =0.08206 (L•atm)(K•mol)-1
1L = 1000cm3
1L = 1dm3
1000cm3 = 1dm3
1.4.7 Solve problems using the ideal gas equation, PV=nRT
A sample of hydrogen gas has a volume of 8.56dm3 at a temperature of 0.00C and a pressure of 152.0 kPa. Calculate the moles of hydrogen gas present in this gas sample
P = V = n = T = R = 0.08206 (L•atm)(K•mol)-1
Or = 8.314 JK-1 mol-1
PV = nRT
Answer: 0.57 mol
Click for answer
IB Core Objective
1.4.8 Analyse graphs relating to the ideal gas equation.
Analyse: Interpret data to reach conclusions. (Obj. 3)