Chapter 3 Stoichiometry - Lamar University · Chapter 3 Stoichiometry STOICHIOMETRY : The chemical...

Chapter 3

Stoichiometry

STOICHIOMETRY: The chemical

arithmetic used to relate the amount of

products and reactants to each other

1st Write Chemical Equation

2nd Balance Equation

3rd Interpret Equation


REACTANTS �� PRODUCTS

A2 + B2 � A2B


A balanced chemical equation has the

same type and number of ______ in

the reactants as in the products.

A2 + B2 � A2B

= A =

= B =


2 A2 + B2 � 2 A2B

Two units of A2 _

React with One unit of B2 __

Forming Two units of A2B__

Magnesium reacts with “air”

forming Magnesium Oxide



Mg(s) + O2(g) → MgO(s)

2 Mg + O2 →→→→ 2 MgO

2 = Mg = 22 = O = 2


2 Mg + O2 →→→→ 2 MgO

Some Balanced Chemical Reactions

Combination & Decomposition reactions

In a combination reaction two or more substances form a single compound.

__N2(g) + __H2(g) → __NH3(g)

In a decomposition reaction, a single compound forms two or more new substances.

__KClO3(s) → __KCl(s) + __O2(g)

1 23

2 2 3

CHEMICAL REACTIONS

(YOU SHOULD KNOW)

1. COMBUSTION (Of a Hydrocarbon)

2. NEUTRALIZATION (Acid + Base)

3. ACID + ACTIVE METAL

4. FORMATION Reaction

CHEMICAL REACTIONS

(YOU SHOULD KNOW)

1. COMBUSTION (Of a Hydrocarbon)

The combustion of a Hydrocarbon

produces CO2 and H2O :

CH4

+ O2

→→→→ CO2

+ H2O

Combustion of Butane

C4H10 + 6 ½ O2 →→→→ 4 CO2 + 5 H2O

Is This Reaction Balanced ?

With Fractions ?

CHEMICAL REACTIONS

(YOU SHOULD KNOW)

2. NEUTRALIZATION (ACID+ BASE)

HCl (aq) + NaOH →→→→ H2O + A SALT

In this case the salt is NaCl Sodium Chloride

CHEMICAL REACTIONS

(YOU SHOULD KNOW)

3. ACID + ACTIVE METAL

HCl (aq) + Zn (s) →→→→ H2

(g) + A SALT

In this case the salt is

CHEMICAL REACTIONS

(YOU SHOULD KNOW)

4. FORMATION Reaction

Zn (s) + Cl2 (g) →→→→ ZnCl

2(s)

Reactants are in their “natural” state

Which of the following are

Balanced ?

C4H10 + 6 ½ O2 →→→→ 4 CO2 + 5 H2O

4 C4H10 + 26 O2 → 16 CO2 + 20 H2O

Uniquely balanced equation

C4H10 + 6 ½ O2 →→→→ 4 CO2 + 5 H2O

4 C4H10 + 26 O2 → 16 CO2 + 20 H2O

The ONLY Uniquely balanced equation is:

2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

(MOLECULAR) WEIGHTS

The FORMULA WEIGHT of a

substance is the sum of the atomic

weights of each atom in its chemical

formula.

Calculate the Formula Weight of

Hydrogen Sulfide

Nickel II Carbonate

Magnesium Acetate

Ammonium Sulfate

Potassium Phosphate

Iron III Oxide

Diphosphorus pentasulfide

1st you must know the formula

Calculate the Formula Weight of

H2S

NiCO3

Mg(C2H3O2)2

(NH4)2SO4

K3PO4

Fe2O3

P2S5

Avogadro’s Number & The Mole

Avogadro’s number is chosen so that

1 mole of 12C atoms has a mass of exactly

12 grams.

1 mole 12C atoms = 6.02x1023 12C atoms

6.02x1023 12C atoms = 12 g

Mascots

Lamar University

University of Texas

LSU

Rice

Oregon

Nederland

CARDINALS

LONGHORNS

TIGERS

OWLS

DUCKS

BULLDOGS

Biology

Department

Mascot

Chemistry

Department

Mascot

Rabbits (& other animals)

If you had a mole of

rabbits, how many

rabbits would you

have ?

How many rabbit ears

would you have?

How many rabbit feet?

Interpreting Chemical Formulas

If you had a mole of water ,

how many molecules of

water would you have ?

How many Hydrogen atoms

would you have?

How many Oxygen atoms?

• The mole is just a conversion factor

that allows us to accurately work

with atoms and molecules.

• One mole of any substance contains

6.02 x 1023 units of that substance.

Grams, Moles & Avogadro

Conversion Factors

Use Molecular Weight

To Convert Grams to Moles

or

To Convert Moles to Grams

Grams to Moles

3.2 Grams of Oxygen = ? Moles

Moles ? Grams

Mole

?

1 x ? =Grams

Moles 0.10 Grams

Mole

32

1 x Grams 2.3 =

Moles to Grams

1.5 Moles of Methane = ? Grams

Grams ? Mole

Grams

1

? x Moles ? =

Grams 24 Mole

Grams

1

16 x Moles 1.5 =

Use AVOGARDRO’S Number

• For NUMBER of Atoms

• Or NUMBER of Molecules

• 32 Grams of Oxygen = ? Molecules

• 32 Grams of Oxygen = ? Atoms

• 1 Molecule of Oxygen = ? Grams

Interpretation of Chemical Reactions

Using Stoichiometry

Example 1

Nitrogen monoxide reacts with oxygen to

produce Nitrogen dioxide

1st Write the reaction

2nd balanced reaction

3rd Interpret Reaction

How many moles of oxygen gas are required

to react completely with 2.0 mole NO?

2 NO (g) + O2 (g) �� 2 NO2 (g)

(a) 0.5 mol O2

(b) 1.0 mol O2

(c) 1.5 mol O2

(d) 2.0 mol O2


to react completely with 1.0 mole NO?

(a) 0.5 mol O2

(b) 1.0 mol O2

(c) 1.5 mol O2

(d) 2.0 mol O2

(e) 2.5 mol O2

2 NO (g) + O2 (g) 2 NO2 (g)


to react completely with 2.50 moles NO?

(a) 0.5 mol O2

(b) 1.00 mol O2

(c) 1.25 mol O2

(d) 1.50 mol O2

(e) 2.00mol O2

2 NO (g) + O2 (g) 2 NO2 (g)


to react completely with 10 moles NO?

(a) 5 mol O2

(b) 10 mol O2

(c) 15 mol O2

(d) 20 mol O2

(e) 25 mol O2

2 NO (g) + O2 (g) 2 NO2 (g)

If 6.0 moles of NO are reacted with 3.0 mole O2,

how many moles NO2 are produced?

(a) 2.0 mol NO2

(b) 6.0 mol NO2

(c) 10.0 mol NO2

(d) 16.0 mol NO2

(e) 32.0 mol NO2

2 NO (g) + O2 (g) 2 NO2 (g)

If 10.0 moles of NO are reacted with 5.0 mole

O2, how many moles NO2 are produced?

(a) 2.0 mol NO2

(b) 6.0 mol NO2

(c) 10.0 mol NO2

(d) 16.0 mol NO2

(e) 32.0 mol NO2

2 NO (g) + O2 (g) 2 NO2 (g)

If 10.0 moles of NO are reacted with 6.0 moles


(a) 2.0 mol NO2

(b) 6.0 mol NO2

(c) 10.0 mol NO2

(d) 16.0 mol NO2

(e) 32.0 mol NO2

2 NO (g) + O2 (g) 2 NO2 (g)

• Chemical Reactions do not always go

the way we expect them to

• Using stoichiometry we can calculate

the theoretical (Maximum) amount

of product formed in a reaction.

.

If the actual amount of product formed in a

reaction is less than the theoretical amount

we can calculate a percentage yield.

100% yieldproduct lTheoretica

yieldproduct Actual yield% ×=

The Limiting Reactant

A reaction stops when one reactant

is totally consumed.

This is the limiting reactant.

The other reactants are excess reactants.

The Limiting Reactant

• How many bikes can be made

from 10 frames and 16 wheels ?

1 frame + 2 wheels →→→→ 1 bike

What is the limiting “reagent” ?

Excess “reagent” ?

If 10.0 moles of NO are reacted with 6.0 moles


(a)What LIMITS the reaction?

(b)What is in excess ?

2 NO (g) + O2 (g) 2 NO2 (g)

If 10.0 moles of NO are reacted with 6.0

moles O2, how many moles of the excess

reagent remain?

• 1.0 mol O2

• 5.0 mol O2

• 4.0 mol NO

• 8.0 mol NO

2 NO (g) + O2 (g) 2 NO2 (g)

Example 2:

Hydrogen reacts with Nitrogen

to form Ammonia

1st Write Reaction

Next Balance Equation

Interpret Balanced Equation

Interpretation of a Chemical Reaction

3 H2 + 1 N2 →→→→ 2 NH3

Three moles One mole Two moles

of Hydrogen of Nitrogen of Ammonia====================================================

Mole Ratio MUST Always be 3 : 1 : 2

3 H2 + 1 N2 →→→→ 2 NH3

3/3 H2 + 1/3 N2 →→→→ 2/3 NH3

3/2 H2 + 1/2 N2 →→→→ 1 NH3

3/4 H2 + 1/4 N2 →→→→ 2/4 NH3

3/8 H2 + 1/8 N2 →→→→ 2/8 NH3

All are 3 : 1 : 2

How many moles of Nitrogen are needed to

react with 3 moles of Hydrogen ?

3 H2 + 1 N2 → 2 NH3

(a) ½ (b) 1 (c)1 ½ (d) 2 (e) 2 ½ (f) 3


react with 1 ½ moles of Hydrogen ?

3 H2 + 1 N2 → 2 NH3

(a) ½ (b) 1 (c)1 ½ (d) 2 (e) 2 ½ (f) 3

Chemical Reactions are

Interpreted on the

Mole basis but chemicals are weighed

in the laboratory in Grams

Use

Molecular Weight

To Convert Grams to Moles

or

To Convert Moles to Grams

Units, Units, Units

Units, Units, Units

___moles Grams

moles

?

1 x ___Grams =

___Grams moles

Grams

1

? x ___moles =


react with Three grams of Hydrogen ?

3 g x 1 mole = 1 ½ moles of H22 g

3 H2 + 1 N2 → 2 NH3

(a) ½ (b) 1 (c)1 ½ (d) 2 (e) 2 ½ (f) 3

3 H2 + 1 N2 →→→→ 2 NH3

How many moles of Ammonia ( NH3 ) are

produced from

(a) 3 grams of H2 and ½ mole of N2?

1 mole = (1 mole)x(17 g/mole) grams of NH3

(b) 3 grams of H2 and 28 grams of N2?

1 mole NH3

with 14 g of Nitrogen in excess

3 H2 + 1 N2 →→→→ 2 NH3

With a 50 % Yield, How many moles of NH3

are produced from

(a) 3 grams of H2 and ½ mole of N2?

½ mole = (½ mole)x(17 g/mole) grams of NH3

(b) 3 grams of H2 and 28 grams of N2?

½ mole NH3

with 14 g of Nitrogen in excess

DESCRIBING COMPOUNDS

Percent Composition

Empirical Formulas

Molecular Formulas

Two Approaches to Chemical Formulas

1. Given Formula

Determine Percent Composition of

Each Element in Compound

2. Given Percent Composition of Each

Element in Compound

Determine Formula

A student obtains the following data

Mass of crucible and cover = 28.288 gms

Mass crucible, cover & K = 28.709 gms

Mass crucible, cover & = 28.793 gms

potassium oxide

Determine the EMPIRICAL formula

First determine % by Weight

Weight K = 28.709 – 28.288 = 0.421

Weight O = 28.793 – 28.709 = 0.084

Total = 0.505

83.36633 100 x 0.505

0.421 K by Wt % ==

83.4 % by wt K and 16.6 % by wt O

Choose any total weight 100 grams is convenient

Construct the following work sheet

Element % Wt moles ratio

K 83.4 83.4 g 83.4 g / 39 = 2.14 = 2

O 16.6 16.6 g 16.6 g / 16 = 1.04 = 1

Therefore formula is

Chapter 3 Stoichiometry - Lamar University · Chapter 3 Stoichiometry STOICHIOMETRY : The chemical...

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