1. Stoichiometry
Chapter 3

2. Cookery and Chemistry Chefs have recipes, chemists have recipes.
Recipes in chemistry can be seen on chemical equation.
Instead of using cups and teaspoons, chemists use moles.
Instead of eggs, butter, sugar, etc. Chemists use chemical compounds as ingredients.

3. How to make
chocolate chips cookies?
Ingredients:
a cup of butter
a half cup of sugar
a cup of brown sugar
a teaspoon of vanilla
2 pieces of eggs
2,5 cups of flour
a teaspoon of baking soda
a teaspoon of salt
2 cups of chocolate chips
How to make a soap?
Ingredients:
75 grams of texapon
30 grams of coconut oil
30 grams of glycerin
50% of Salt solution
50% of Citric acid
and so on….

4. The Relation between cookies and chemistry: STOICHIOMETRY
Reaction equation tells us about what you need to react (reactant) to get a product. (like the cookies recipe)
STOICHIOMETRY is derived from Greek languages: stoicheion (element) and metron (measure)
Usage: STOICHIOMETRY is used to measure the amount of substances involved in chemical reactions.

5. Example:
CH O2  CO H20
This reaction tells us that by mixing 1 mole of methane with 2 moles of oxygen we will get 1 mole of carbon dioxide and 2 moles of water.
If we want to get 10 moles of water, how many moles of methane and oxygen is needed? How many grams of CO2 is produced?

6. What is a Mole?
The unit of measurement which is used to count the number of atoms, molecules, or particles.
1 mole of any substance = NA = 6.02 x atoms, molecules, or particles.
e.g. 1 mole of silver = 6.02 x 1023 atoms of silver

7. Amadeo Avogadro (1776 – 1856) Particles in a Mole
Amedeo Avogadro ( )
never knew his own number;
it was named in his honor by a
French scientist in 1909.
its value was first estimated
by Josef Loschmidt, an Austrian
chemistry teacher, in 1895. ?
A more precise value of Avagadro’s number = E23.
How could you determine the number of sugar crystals in a bowl of sugar?
Is it possible to get a reasonably close estimate?
quadrillions
thousands
trillions
billions
millions
1 mole =
or x 1023
There is Avogadro's number of particles in a mole of any substance.

8. Analogy Mole other 1 mole of Na = 6.02 x 1023 atoms of Na
1 dozen of eggs = 12 pieces of eggs
1 mole water = 6.02 x 1023 molecules of water
A pair of shoes = 2 pieces of shoes
1 mole of HCl = 22,4 L HCl (STP)
1 rims of paper = 500 sheets of paper

9. Molar Mass (MM)
When we measure one mole of a substance on a balance, it called “molar mass” and the unit is g/mol (gram per mole).

10. Molar Mass Examples carbon aluminum zinc 12.01 g/mol 26.98 g/mol

11. Molar Mass Examples H2O 2(1.01) + 16.00 = 18.02 g/mol NaCl
water
sodium chloride
H2O
2(1.01) = g/mol
NaCl
= g/mol

12. Molar Mass Examples NaHCO3
sodium bicarbonate
sucrose
NaHCO3
(16.00) = g/mol
C12H22O11
12(12.01) + 22(1.01) + 11(16.00) = g/mol

13. Mole Conversion 22,4L ٪ MM = Molar Mass NA = 6.02 x 1023
Particles
Mass
Volume
(STP)
٪ NA
X MM
x NA
٪ MM
X 22,4L
22,4L ٪
MM = Molar Mass
NA = 6.02 x 1023
STP = Standard
Temperature
Pressure

14. Stoichiometry has 5 basic steps
Write and balance the equation
Write down all given information
Convert everything into moles
Use mole ratio to solve the problem
Convert everything into the required unit (Mass, particles, volume)

15. What is a mole ratio?
Mole ratio is based on the coefficient of the balanced chemical equation.
e.g CH O2  CO H2O
Remember :
“The ratio of coefficient = the ratio of mole”
The mole ratio = 1 : 2 : 1 : 2

16. Example of stoichiometric problem
H2 + O2  H2O (not balance)
Question:
If 3 moles of oxygen are completely react with hydrogen, how many grams off water produced?
If 72 grams of water are produced, how many moles of oxygen are needed?

17. Example of stoichiometric problem
N2 + H2  NH3 (not balance)
Question:
How many molecules of ammonia are produced when 2 grams of nitrogen is reacted with hydrogen ?
How many grams of oxygen are needed to produce 10 grams of ammonia?

18. Stoichiometry in Real Life :
Ethane gas (C2H6) is burnt at STP by following a reaction below:
C2H6 + O2  CO2 + H2O (not balance)
Question:
How many liters of oxygen are needed to burn 12 moles of C2H6?
How many liters of CO2 are produced if we burn 14 moles of O2?

19. Water from a Camel
Camels store the fat tristearin (C57H110O6) in the hump. As well as
being a source of energy, the fat is a source of water, because when
it is used the reaction
takes place.
2 C57H110O6(s) O2(g)  114 CO2(g) H2O(l)
What mass of water can be made from 1.0 kg of fat?

20. Rocket Fuel B2H6 + O2 B2O3 + H2O B2H6 + O2 B2O3 + H2O 3 3
The compound diborane (B2H6) was at one time
considered for use as a rocket fuel. How many grams
of liquid oxygen would a rocket have to carry to burn
10 kg of diborane completely?
(The products are B2O3 and H2O).
B2H6 + O2
B2O3 + H2O
Chemical equation
Balanced chemical equation
B2H O B2O H2O 3 3
10 kg x g

21. Lithium Hydroxide Scrubber Modified by Apollo 13 Mission
Interior view of the Apollo 13 Lunar Module (LM) as the astronauts jerry-rig a system to use the Command Module lithium hydroxide canisters to purge carbon dioxide from the LM. Photo: NASA/JSC
Astronaut John L. Swigert holds the
jury-rigged lithium hydroxide scrubber
used to remove excess carbon dioxide
from the damaged Apollo 13 spacecraft.

22. Water in Space
In the space shuttle, the CO2 that the crew exhales is removed from the
air by a reaction within canisters of lithium hydroxide. On average, each
astronaut exhales about 20.0 mol of CO2 daily. What volume of water will
be produced when this amount of CO2 reacts with an excess of LiOH?
(Hint: The density of water is about 1.00 g/mL.)
CO2(g) LiOH(s)  Li2CO3(aq) + H2O(l)
20.0 mol
excess
x g
Water is NOT at STP!

23. Percentage composition
Formula: n = number of element Percentage composition tell you the percent of mass of the element which made up the compound.

24. Example
Calculate the percentage composition of C and N in urea, CO(NH2)2. Answer: Molar mass of urea = Ar C + Ar O + (2 x Ar N) + (4 x Ar H) = = 60 % C = (1 x 12) x 100% = 20% 60 % N = (2 x 14) x 100% = 46.67%

25. Exercise 1. Calculate the percentage composition of nitrogen in :
(NH4)2SO4
NH4NO3
2. Determine the mass of nitrogen in:
- 100 grams of Ca(NO3)2
- 200 grams of (C2H5)2NH

26. Exercise
3. Calculate the percentage composition of carbon and oxygen in :
C6H12O6
CH3OCH3
4. Chlorophyll contains 4.8% of magnesium. Assume that each molecules of chlorophyll contain 1 atom Mg. determine the relative molecular mass (Mr) of chlorophyll. (Ar Mg=24)

27. Limiting Reactants Limiting Reactant Excess Reactants
Available Ingredients
4 slices of bread
1 jar of peanut butter
1/2 jar of jelly
Limiting Reactant
bread
Excess Reactants
peanut butter and jelly

28. Limiting Reactants Limiting Reactant Excess Reactant
used up in a reaction
determines the amount of product
Excess Reactant
added to ensure that the other reactant is completely used up
cheaper & easier to recycle
If one or more of the reactants is not used up completely but is left over when the reaction is completed, then the amount of product that can be obtained is limited by the amount of only one of the reactants
A limiting reactant is the reactant that restricts the amount of product obtained. The reactant that remains after a reaction has gone to completion is present in excess.

29. Limiting Reagents
6 red left over
6 green used up
3.9

30. Method 1 Pick A Product Try ALL the reactants
The lowest answer will be the correct answer
The reactant that gives the lowest answer will be the limiting reactant

31. Limiting Reactant: Method 1
10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced?
2 Al + 3 Cl2  2 AlCl3
Start with Al:
Now Cl2:
10.0 g Al 1 mol Al mol AlCl g AlCl3
27.0 g Al mol Al mol AlCl3
= 49.4g AlCl3
35.0g Cl mol Cl mol AlCl g AlCl3
71.0 g Cl mol Cl mol AlCl3
= 43.9g AlCl3

32. Method 2 Convert one of the reactants to the other REACTANT
See if there is enough reactant “A” to use up the other reactants
If there is less than the GIVEN amount, it is the limiting reactant
Then, you can find the desired species

33. Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al
mol Al
mol Fe2O3 needed
g Fe2O3 needed OR
g Fe2O3
mol Fe2O3
mol Al needed
g Al needed
1 mol Al
27.0 g Al x
1 mol Fe2O3
2 mol Al x
160. g Fe2O3
1 mol Fe2O3 x =
124 g Al
367 g Fe2O3
Start with 124 g Al
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
3.9

34. 1 mol Al 27.0 g Al x 1 mol Al2O3 2 mol Al x 102. g Al2O3 1 mol Al2O3 x
Use limiting reagent (Al) to calculate amount of product that
can be formed.
g Al
mol Al
mol Al2O3
g Al2O3
2Al + Fe2O Al2O3 + 2Fe
1 mol Al
27.0 g Al x
1 mol Al2O3
2 mol Al x
102. g Al2O3
1 mol Al2O3 x =
124 g Al
234 g Al2O3
3.9

35. Types of Formulas Empirical Formula
The formula of a compound that expresses the smallest whole number ratio of the atoms present.
Ionic formula are always empirical formula
Molecular Formula
The formula that states the actual number of each kind of atom found in one molecule of the compound.

36. Molecular formula Empirical Formula
the true or actual ratio of the atoms in a compound
C6H12O6
the simplest whole number ratio of the atoms in a compound
Example
CH2O

37. Learning Check 1. What is the empirical formula for C4H8?
Timberlake LecturePLUS
1. What is the empirical formula for C4H8?
A) C2H4 B) CH C) CH
2. What is the empirical formula for C8H14?
A) C4H7 B) C6H C) C8H14

38. Calculating Empirical
Just find the lowest whole number ratio
and
It is not just the ratio of atoms, it is also the ratio of moles of atoms

39. In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen
In one molecule of CO2 there is 1 atom of C and 2 atoms of O

40. Learning Check
1. Find the empirical formula of a compound that contains 42 g of nitrogen and 9 g of hydrogen. (Ar N = 14 Ar H = 1)

41. Learning Check
2. Find the empirical formula of a compound containing 20 g of calcium, 6 g of carbon and 24 g of oxygen. (Ar Ca = 20 Ar C = 12 Ar O = 16)

42. Convert the grams to mol for each element
Write the number of mol as a subscript in a chemical formula
Divide each number by the lowest number.
Multiply the result to get rid of any fractions.

43. 1. – Convert the grams to mol for each element
The Answer
1. – Convert the grams to mol for each element
N = mass = 42 g = 3 mol
H = mass = 9 g = 9 mol
Molar mass
14 g/mol
1 g/mol
Molar mass

44. - Divide each number by the lowest number.
Write the number of mol as a subscript in a chemical formula
3 mol of N
- 9 mol of H
N3H9
- Divide each number by the lowest number.
NH3

45. 4. Calculate the empirical formula of a compound composed of 37 % C, 16 % H, and 47 % N. (Ar C = 12 Ar H = 1 Ar N = 14)
Pretend that you have a 100 gram sample of the compound. 1. 2.
change the % to grams 3.
Convert the grams to mol for each element 4.
Write the number of mol as a subscript in a chemical formula 5.
Divide each number by the lowest number. 6.
Multiply the result to get rid of any fractions.

46. Example
Calculate the empirical formula of a compound composed of 37 % C, 16 % H, and 47 %N.
Assume 100 g so
37 g C = 3.1 mol C
12 g/mol
16 g H = 16 mol H
1 g H
47 g N = 3.4 mole N
14 g N

47. C3.1H16N3.4 If we divide all of these by the smallest
3.1 mol C
16 mol H
3.4 mol N
C3.1H16N3.4
If we divide all of these by the smallest
one It will give us the empirical formula

48. Example The ratio is 3.1 mol C = 1 mol C 3.1 mol C 1 mol C
The ratio is mol H = mol H mol C mol C
The ratio is mol N = mol C mol C mol C
C1H5N1 is the empirical formula or CH5N

49. Empirical
Molecular

50. 6. Caffeine has a molar mass of 194 g. what is its molecular formula?

51. Find x if
194 g
97 g
= 2
2 X
C4H5N2O1
C8H10N4O2

52. Learning Check
A compound is known to be composed of 71 % Cl, 25 % C and 4 % H. Its molar mass is known (from gas density) is known to be g. What is its molecular formula?