StoichiometryChapter 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

Amadeo Avogadro(1776 – 1856)

1 mole = 602213673600000000000000or 6.022 x 1023

thousandsmillionsbillionstrillions

quadrillions?

There is Avogadro's number of particles in a mole of any substance.

Particles in a MoleParticles in a Mole

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

chemistry teacher, in 1895.

AnalogyMole 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

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

•aluminum

•zinc

12.01 g/mol

26.98 g/mol

65.39 g/mol

Molar Mass Examples•water

•sodium chloride

H2O

2(1.01) + 16.00 = 18.02 g/mol

NaCl 22.99 + 35.45 = 58.44 g/mol

Molar Mass Examples•sodium bicarbonate

•sucrose

NaHCO3

22.99 + 1.01 + 12.01 + 3(16.00)

= 84.01 g/mol

C12H22O11

12(12.01) + 22(1.01) + 11(16.00)

= 342.34 g/mol

Mole Conversion

X MM

٪ MM

٪ NA

x NA

22,4L ٪ X 22,4L

MM = Molar MassNA = 6.02 x 1023

STP = Standard Temperature

Pressure

Stoichiometry has 5 basic steps1. Write and balance the equation2. Write down all given information3. Convert everything into moles4. Use mole ratio to solve the problem5. 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

takes place.

2 C57H110O6(s) + 163 O2(g) 114 CO2(g) + 110 H2O(l)

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).

B2H6 + O2Chemical equation

Balanced chemical equation

10 kg x g

B2O3 + H2O

3 3B2H6 + O2 B2O3 + H2O

Lithium Hydroxide ScrubberModified 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.

ExampleCalculate 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

Exercise3. 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 IngredientsAvailable Ingredients

▫4 slices of bread▫1 jar of peanut butter▫1/2 jar of jelly

• Limiting ReactantLimiting Reactant– bread

• Excess ReactantsExcess Reactants– peanut butter and jelly

Limiting Reactants• Limiting ReactantLimiting Reactant

▫used up in a reaction▫determines the amount of product

• Excess ReactantExcess Reactant▫added to ensure that the other reactant

is completely used up▫cheaper & easier to recycle

6 green used up

6 red left over

Limiting Reagents

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:

•Now Cl2:

10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g AlCl3

27.0 g Al 2 mol Al 1 mol AlCl3= 49.4g AlCl3

35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g AlCl3

71.0 g Cl2 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

ORg Fe2O3 mol Fe2O3 mol Al needed g Al needed

124 g Al1 mol Al

27.0 g Alx

1 mol Fe2O3

2 mol Alx

160. g Fe2O3

1 mol Fe2O3

x = 367 g Fe2O3

Start with 124 g Al need 367 g Fe2O3

Have more Fe2O3 (601 g) so Al is limiting reagent3.9

Use limiting reagent (Al) to calculate amount of product thatcan be formed.

g Al mol Al mol Al2O3 g Al2O3

124 g Al1 mol Al

27.0 g Alx

1 mol Al2O3

2 mol Alx

102. g Al2O3

1 mol Al2O3

x = 234 g Al2O3

2Al + Fe2O3 Al2O3 + 2Fe

3.9

Types of FormulasTypes of Formulas• Empirical FormulaEmpirical Formula

The formula of a compound that The formula of a compound that expresses the expresses the smallest whole number smallest whole number ratioratio of the atoms present. of the atoms present.

Ionic formula are always empirical Ionic formula are always empirical formulaformula

• Molecular FormulaMolecular FormulaThe formula that states the The formula that states the actualactual

number of each kind of atom found in number of each kind of atom found in one one moleculemolecule of the compound. of the compound.

Molecular formula

the true or actual ratio of the atoms in a compound

C6H12O6

Empirical Formula

the simplest whole number ratio of the atoms in a compound

Example

CH2O

Timberlake LecturePLUS

37

1. What is the empirical formula for C4H8?

A) C2H4 B) CH2 C) CH

2. What is the empirical formula for C8H14?

A) C4H7 B) C6H12 C)

C8H14

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Just find the lowest whole number ratio

andIt is not just the ratio of atoms, it is also the

ratio of moles of atoms

Calculating Empirical

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

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)

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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

H = mass = 9 g = 9 molMolar mass

14 g/molMolar mass

1 g/mol

- 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

Pretend that you have a 100 gram sample of the compound.

change the % to grams

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.

1.

2.

3.

4.

5.

6.

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)

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

•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 = 1 mol C 3.1 mol C 1 mol C

• The ratio is 16 mol H = 5 mol H 3.1 mol C 1 mol C

• The ratio is 3.4 mol N = 1 mol C 3.1 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 massformulaempirical

massmolarx

194 g

97 g= 2

C4H5N2O1

C8H10N4O2

2 X

•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?

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