Download Electron configuration

Stoichiometry
Chapter 3
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.
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….
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.
Example:
CH4 + 2O2  CO2 + 2H20
• 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?
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 1023
atoms, molecules, or particles.
• e.g. 1 mole of silver = 6.02 x 1023 atoms of silver
Particles in a Mole
Amadeo
Avogadro
Amedeo Avogadro (1766-1856)
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
(1776 – 1856)
chemistry teacher, in 1895.
?
quadrillions
thousands
trillions
billions
millions
1 mole = 602213673600000000000000
or 6.022 x 1023
There is Avogadro's number of particles in a mole of any substance.
Analogy
Mole
other
1 mole of Na = 6.02 x 1023 atoms of 1 dozen of eggs = 12 pieces of eggs
Na
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
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).
Molar Mass Examples
• carbon
12.01 g/mol
• aluminum
26.98 g/mol
• zinc
65.39 g/mol
Molar Mass Examples
• water
 H2O
 2(1.01) + 16.00 = 18.02 g/mol
• sodium chloride
 NaCl
 22.99 + 35.45 = 58.44 g/mol
Molar Mass Examples
• sodium bicarbonate
 NaHCO3
 22.99 + 1.01 + 12.01 + 3(16.00)
• sucrose
= 84.01 g/mol
 C12H22O11
 12(12.01) + 22(1.01) + 11(16.00)
= 342.34 g/mol
Mole Conversion
٪ NA
Particles
x NA
Mole
22,4L ٪
MM
NA
STP
= Molar Mass
= 6.02 x 1023
= Standard
Temperature
Pressure
X MM
٪ MM
X 22,4L
Volume
(STP)
Mass
Stoichiometry has 5 basic steps
1.
2.
3.
4.
5.
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)
What is a mole ratio?
• Mole ratio is based on the coefficient of the
balanced chemical equation.
• e.g.
CH4 + 2O2  CO2 + 2H2O
• Remember :
“The ratio of coefficient = the ratio of mole”
• The mole ratio = 1 : 2 : 1 : 2
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?
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?
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?
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
2 C57H110O6(s) + 163 O2(g)  114 CO2(g) + 110 H2O(l)
takes place. What mass of water can be made from 1.0 kg of fat?
Rocket Fuel
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).
Chemical equation
B2H6 + O2
Balanced chemical equation B2H6 + 3 O2
10 kg
xg
B2O3 + H2O
B2O3 + 3 H2O
Lithium Hydroxide Scrubber
Modified by Apollo 13 Mission
Astronaut John L. Swigert holds the
jury-rigged lithium hydroxide scrubber
used to remove excess carbon dioxide
from the damaged Apollo 13 spacecraft.
Percentage composition
Formula:
n = number of element
Percentage composition tell you the percent of
mass of the element which made up the
compound.
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)
= 12 + 16 + 28 + 4
= 60
% C = (1 x 12) x 100% = 20%
60
% N = (2 x 14) x 100% = 46.67%
60
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
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)
Limiting 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
Limiting Reactants
• Limiting 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
Limiting Reagents
6 green used up
6 red left over
3.9
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
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:
10.0 g Al
1 mol Al
27.0 g Al
2 mol AlCl3 133.5 g AlCl3
2 mol Al
1 mol AlCl3
= 49.4g AlCl3
• Now Cl2:
35.0g Cl2
1 mol Cl2
71.0 g Cl2
2 mol AlCl3 133.5 g AlCl3
3 mol Cl2
1 mol AlCl3
= 43.9g AlCl3
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
Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3
Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al
mol Al
mol Fe2O3 needed
g Fe2O3 needed
OR
g Fe2O3
124 g Al x
mol Fe2O3
1 mol Al
27.0 g Al
x
Start with 124 g Al
mol Al needed
1 mol Fe2O3
2 mol Al
160. g Fe2O3
=
x
1 mol Fe2O3
g Al needed
367 g Fe2O3
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
3.9
Use limiting reagent (Al) to calculate amount of product that
can be formed.
g Al
mol Al
mol Al2O3
2Al + Fe2O3
124 g Al x
1 mol Al
27.0 g Al
x
g Al2O3
Al2O3 + 2Fe
1 mol Al2O3
2 mol Al
102. g Al2O3
=
x
1 mol Al2O3
234 g Al2O3
3.9
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.
Molecular
formula
Empirical
Formula
the true or actual ratio of
the atoms in a compound
the simplest whole number
ratio of the atoms in a
compound
Example
C6H12O6
CH2O
37
Learning Check
Timberlake LecturePLUS
1. What is the empirical formula for
C4H8?
A ) C 2 H4
B) CH2
C) CH
2. What is the empirical formula for
C8H14?
A) C4H7
B) C6H12
C) C8H14
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
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
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)
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)
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.
The Answer
• 1. – Convert the grams to mol for each element
N = mass
= 42 g = 3 mol
Molar mass
H = mass
Molar mass
14 g/mol
=9g
1 g/mol
= 9 mol
- Write the number of mol as a subscript in a
chemical formula
- 3 mol of N
- 9 mol of H
N 3 H9
- Divide each number by the lowest number.
NH3
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)
1.
Pretend that you have a 100 gram sample of the
compound.
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.
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
1gH
•
47 g N
= 3.4 mole N
14 g N
• 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
Example
• The ratio is 3.1 mol C
3.1 mol C
• The ratio is 16 mol H
3.1 mol C
• The ratio is 3.4 mol N
3.1 mol C
=
=
=
1 mol C
1 mol C
5 mol H
1 mol C
1 mol C
1 mol C
• C1H5N1 is the empirical formula or CH5N
Empirical
Molecular
6. Caffeine has a molar mass of 194
g. what is its molecular formula?
•
Find x if
molar mass
x
empirical formula mass
194 g
97 g
2X
C4H5N2O1
C8H10N4O2
=2
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
98.96 g. What is its molecular formula?