chemistry 101 : chap. 3 stoichiometry (1) chemical equations (2) some simple patterns of chemical...

Chemistry 101 : Chap. 3Chemistry 101 : Chap. 3

Stoichiometry

(1) Chemical Equations

(2) Some Simple Patterns of Chemical Reactivity

(3) Formula Weight

(4) Avogadro’s Number and the Mole

(5) Empirical Formulas from Analyses

(6) Quantitative Information from Balanced Equations

(7) Limiting Reactants

StoichiometryStoichiometry

A study of mass relationships that exist between substances consumed (reactants) and produced (products) in chemical reactions.

Stoichiometry is build upon an understanding of atomicmasses, chemical formulas and law of conservation of mass.

“… nothing is created: an equal quantity of matter

exists both before and after the experiment.” Antoine Lavoisier (1743 ~ 1794)

With the Dalton’s atomic theory, law of conservation of mass is known as law of conservation of atom

Balancing Chemical EquationsBalancing Chemical Equations

Chemical Equation:

2H2 + O2 = 2H2O

coefficientreactants products

+

Balanced Chemical Equation: Equations that represent the

correct amount of reactants and products in chemical reaction.

Balanced chemical equations satisfy

the law of conservation of atoms


Balancing Chemical Equations: Determining the coefficients

that provide equal numbers of each type of atom on

each side of the equation

(1) You only determine the coefficients, not the subscripts.

(2) Subscripts should never be changed in balancing equations. It changes the identities of reactants or products

(3) Balanced equation should contain the smallest possible whole-number coefficients

(4) It is usually best to balance first those elements that occur in the fewest chemical formulas on each side of equation


CO + O2

→ CO2

22

+ →

CH4

+ Cl2

→ CCl4

+ HCl

→+ +

4 4

Al + HCl → AlCl3

+ H2

2 23

23 x 3

NH4NO

3→ N

2 + O

2 + H

2O21

2x 2

C2H

4+ O

2→ CO

2+ H

2O2 23

2 Al + 6 HCl → 2 AlCl3

+ 3 H2

2 NH4NO

3→ 2 N

2 + O

2 + 4 H

2O



Example: Law of conservation of atom (or mass)

How many NH3 molecules should be shown in the

container on the right?

Three Basic Reactions TypesThree Basic Reactions Types

Combination Reactions: A + B C

2Mg (s) + O2 (g) 2MgO (s)

Two reactants combine to form a single product

NOTE: The symbols (s), (l), (g), (aq) specify the physical state of each substance

Examples of combination reactions [ Not balanced!]

(1) N2 + H2 NH3

(2) Fe + O2 Fe2O3 Iron (III) oxide

(3) S + O2 SO3


Decomposition Reactions: A B + C

2 HgO (s) 2 Hg (l) + O2 (g)

A single reactant decomposes to form two or more products

Examples of decomposition reactions [not balanced!]

(1) NaN3 (s) Na (s) + N2 (g)

(2) CaCO3 (s) CaO + CO2 (g)

NOTE: Over 22 million tons of limestone (CaCO3) are converted to

lime (CaO) each year for use in making glass, mortar, and for extracting iron from its ore


Combustion Reactions : Organic compounds (or hydrocarbons) is burned (combusted) in oxygen to produce CO2 and H2O

C3H8 (g) + O2 (g) CO2 (g) + H2O (g)3 45

NOTE: When balancing combustion reaction, it is best to balance the oxygen atoms last.

Examples of combustion reactions [not balanced]

(1) CH4(g) + O2 (g) CO2 (g) + H2O (g)

(2) C4H10 (g) + O2 (g) CO2 (g) + H2O (g)

Formula WeightFormula Weight

Chemical equations tell us the exact numbers of molecules and atomsinvolved in chemical reactions. But, in laboratory, what we usually measure is the weight of each compound. How can we relate them?

Formula Weight (FW) : The sum of the atomic weight of each atom in its chemical formula

FW of H2SO4 = 2 x (AW of H) + (AW of S) + 4 x (AW of O)

= 2 x 1.0 + 32.1 + 4 x 16.0 = 98.1 (amu)

NOTE: If the chemical formula is that of a molecule, the formula weight is also called molecular weight (MW).

NOTE: The masses used in the formula weight are from the periodic table (weighted average of all isotope’s masses)


Examples : Determined the formula weights of following substances

(1) FW of Mg(OH)2 =

(2) FW of Ca(NO3)2 =


Percentage Composition : The percentage by mass contributed by each element in the substance

100compound ofeight formular w

element of mass total element %

Example : Calculate the % weight of C, H and O in C12H22O11

The MoleThe Mole

The mole is the SI unit for amount (number) of substance. It was originally proposed by Wilhelm Ostward in 1896

Mole (mol): A quantity of objects equal to the number of atoms of carbon in exactly 12 g of 12C.

Wilhelm Ostward (1853 ~ 1932)He was awarded the novel prize forchemistry for his work on catalysisin 1909.

A mole of particles is an extremely large number of objects and it is approximately equal to 6.022 1023 particles. This number is referred to as Avogadro’s Number (NA) in honor of Amadeo Avogadro.

Amadeo Avogadro(1776 ~ 1856)

1 mol = 6.022 1023 particles

The MoleThe Mole

The mole is simply a unit of counting

1 dozen eggs = 12 individual eggs

½ dozen eggs = 6 individual eggs

1 mol of eggs = 6.022 1023 individual eggs

½ mol of eggs = 3.011 1023 individual eggs

1 mol of oxygen atoms = 6.022 1023 oxygen atoms (O)

1 mol of oxygen molecules = 6.022 1023 oxygen molecules (O2)

= 2 mol of oxygen atoms

The MoleThe Mole

Example : How many moles of eggs are in an egg carton holding 12 eggs?

Example : How many moles of oxygen atoms are in 1.5 moles of Ca(NO3)2 ?

Molar MassMolar Mass

Molar mass = mass of 1 mole of a substance in gram

(1) Molar mass is nothing to do with Molecule.

(2) 1 mole of He atoms includes the same number of particles

as 1 mole of Ne atoms. However, 1 mole of He atoms and

1 mole of Ne atoms have different weights because

He and Ne have different molar mass.

(3) 12 amu = mass of a single 12C atom (exact)

12 gram = mass of 1 mole of 12C atoms (exact)

molar mass of 12C = 12 g/mol

FW (or MW) in amu = molar mass in g/mol


1 mol of Cu = 6.02 1023 copper atoms

AW of Cu = 63.5 amu = average mass of naturally occurring

(FW or MW) single Cu atom

(This number is from the periodic table)

molar mass of Cu = 63.5 g/mol = mass of 6.02 1023 copper atoms

amount of a substance in gram

amount of a substance in mol

molar mass molar mass


Example: Calculate the mass of 0.433 mole of calcium nitrate.

Example: How many grams of oxygen are in 0.433 mole of Ca(NO3)2


Problem: How many hydrogen atoms are in 4.5g of CH3OH ?

1. Compute the molecular weight or molar mass of CH3OH:

2. Determine the amount of CH3OH in mol.

3. Determine the amount of H


Problem: How many grams of carbon are in 42g of CH3CH2OH ?

1. Compute the molecular weight or molar mass of CH3CH2OH:

2. Determine the amount of CH3CH2OH in mol.

3. Determine the amount of C

Molar Mass and Avogadro’s Number

Molar Mass and Avogadro’s Number

114.3 g of Copper

1.8 mol of Copper

1.08 1024 Copper atoms

molar mass Avogadro’s number

Empirical Formulas from AnalysesEmpirical Formulas from Analyses

Empirical Formula : Indicate the relative number of atoms

of each type in a molecule

If we know the relative number of atoms in mol, we can

determine the compound’s empirical formula

Example : A compound made of Hg and Cl has 73.9% mercury

(Hg) and 26.1% chlorine (Cl) by mass. What is the

empirical formula for this compound?


Example : A compound is found to contain by mass 47.7% C,

10.5% H and 42.1% O. What is the empirical formula

of the compound?


Example: A compound is found to have by mass 53.5% C, 11.1% H and 35.6% O. The experimentally determined molecular weight is 90 amu. What is the molecular formula of this compound?

2H2 + O2 2H2OTWO hydrogenmolecules

ONE oxygenmolecule

TWO watermolecules

2 mol H2 + 1 mol O2 2 mol H2O

6.021023

4g H2 + 32 g O2 36g H2O

molar mass

FOUR H atoms TWO O atoms FOUR H atoms + TWO O atoms

Conservationof mass

Conservation of atom

Quantitative Information from Balanced Chemical Equations




Example: How many moles of carbon dioxide would be produced

by burning 3 moles of carbon monoxide?

(1) Write down the reactants and products:

(2) Balance the chemical equation :

(3) Compute the amount of CO2 :



Example: How many grams of H2O are formed from the complete

combustion of 3.0 g of C2H6?

(1) Write down the reactants and products :

(2) Balance the chemical equation :

(3) Convert the mass (gram) to mol

(4) Compute the amount of water in mol :

(5) Convert the mol to mass (gram) :



Stoichiometric relations between compounds A and B



Example: Octane (C8H18), which is liquid in room temperature,

has a density of 0.692 g/ml at 20oC. How many grams

of O2 are required to completely burn 1.00 gal of octane?

(1) Balance the combustion reaction:

(2) Compute the mass of octane :

(3) Convert the mass of octane to mol

(4) Compute the amount of oxygen to react (in mol)

(5) Compute the amount of oxygen to react (in gram)

Limiting ReactantsLimiting Reactants

Limiting reactant : The reactant that is completely consumed in a reaction.

2H2 + O2 2H2O

Suppose we have a mixture of 10 mol H2 and 7 mol O2.

Then, how many moles of O2 will be used?

Since 2 moles of H2 consume 1 mole of O2, 10 moles of H2

will consume 5 moles of O2 to produce 10 moles of H2O

We have 7 mol – 5 mol = 2 mol of excess oxygens.

Hydrogen is the limiting reactant !


2H2 + O2 2H2O

Initial quantities : 10 mol 7 mol 0 mol

Changes : 10 mol 5 mol 10 mol

Final quantities : 0 mol 2 mol 10 mol


What if we had 4 moles of oxygens to start the same reaction?

2H2 + O2 2H2O

Initial quantities : 10 mol 4 mol 0 mol

Changes : 8 mol 4 mol 8 mol

Final quantities : 2 mol 0 mol 8 mol

Oxygens are completely used up.

Therefore, O2 is the limiting reactant


How many compete cars can be built from these parts?


Example : How many grams of P4O10 can be produced by the

reaction of 1.0g of phosphorous and 3.0g of oxygens?

P4 + 5O2 P4O10

MW=124 MW=32 MW=284

(1) Convert grams to mol :

(2) Find the limiting reactant :

(3) Set up the stoichiometic table (if you want)

(4) Convert mol to grams

YieldsYields

Theoretical Yield : The quantity of product that is calculated to

form when all of the limiting reactants react.

Actual Yield : The amount of product actually obtained in a

reaction.

This is from balanced chemical equations

This is from actual experiments

Percent Yield : Ratio between the theoretical yield and the actual yield

% 100yield ltheoretica

yield actual YieldPercent

Actual yield ≤ theoretical yield

YieldsYields

Example : 2.50 g of copper is reacted with excess sulfur and

3.00 g of copper(I) sulfide is produced. What is the

%yield of the reaction?

16Cu(s) + S8(s) 8Cu2S (s)

AW = 63.5 FW=159

(1)What is the limiting reactant?

(2) Calculate the theoretical yield of copper sulfide

(3) Calculate % yield :

YieldsYields

Example : Lithium and nitrogen react to produce lithium nitride. If 5.00 g of each reactant undergoes a reaction with a 88.5% yield, how many grams of Li3N are obtained?

6 Li (s) + N2 (g) 2 Li3N (s)

(1) Convert grams to mol:

(2) Determine the limiting reactant

(3) Compute the theoretical yield

(4) Compute the actual yield

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