Chapter Three:

STOICHIOMETRY

Chemical Stoichiometry

• Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

Atomic Masses

Atomic Masses

• Elements occur in nature as mixtures of isotopes

• Carbon = 98.89% 12C

1.11% 13C

<0.01% 14C Carbon atomic mass = 12.01 amu

Schematic Diagram of a Mass Spectrometer

Atomic Masses

An Atomic mass is the relative mass of an average atom of an element using the 12C isotope as a reference

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Relative masses of selected isotopes on the 12C atomic mass scale

6

1 grms 10 x 1.6606

amu 1

or grams10 x 1.6606 amu 1

24-

-24

“Average”Atom/ Weighted AveragesAtomic masses are weighted averages calculated from the following information:

1. The number of isotopes that exist.

2. The isotopic mass for each isotope on the 6C scale.

3. The percent abundance of each isotope.

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The element magnesium (Mg) has three stable isotopes with the following masses and abundances:

Isotope Mass (amu) Abundance24Mg 23.9850 78.99%25Mg 24.9858 10.00%26Mg 25.9826 11.01%

Calculate the average atomic mass (the atomic weight) of magnesium from these data.

IF ALL ATOMS ARE COMPOSED OF THE SAME COMPONENTS:

-- ELECTRONS

-- PROTONS

-- NEUTRONS

WHY DO DIFFERENT ATOMS HAVE DIFFERENT PROPERTIES?

CHEMICAL PROPERTIES OF ATOMS ARISE FROM THEIR ELECTRONS.

electron

neutron

proton

Moles, Molar Mass,

and Percent Composition

A MOLE

1 mole has exactly as many particles (atoms,

molecules, formulas) as in 1.200,000 grams of 6C.

This is Avogadro’s number, 6.022 X 1023

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CALCULATION OF FORMULA MASS

FORMULA MASS = Sum of all atomic masses in the compound formula

FORMULA MASS = Formula weight, f.w.

= Molecular mass

= Molecular weight, m.w.

Example: CaCl2 calcium chloride

Ca 1 atom X 40.078 amu / 1 atom = 40.078 amu, grams

Cl 2 atoms X 35.453 amu / 1 atom = 70.906 amu, grams

CaCl2 formula mass = 110.984 amu, grams

THE MASS OF A MOLE

Molar mass of an element is its atomic mass in grams.

Molar mass of a compound or molecule or formula is itsmolecular mass in grams.

Example: Ca 40.078 grams Ca = 1 mole of Ca

or 40.078 grams Ca = 1

1 mole CaCl2

CaCl2 110.984 grams CaCl2 = 1mole CaCl2

or 110.984 grams CaCl2 = 1 1 mole CaCl2

The Mole – A Chemist’s Counting Unit

A mole of anything is 6.022 x 1023 anything.

Avogadro’s number is 6.022 x 1023.

6.022 x 1023 atoms = 1 mole atom

Or

6.022 x 1023 molecules = 1 mole molecules

Or

Example:

110 x 6.022

molecules mole 123

1 atoms mole 1

atoms 10 x 022.6 23

224

2

223

2

CaCl molecules 10 x 2.186

CaCl mole 1

CaCl molecules10 x 6.022 x CaCl moles 3.631

USE DIMENSIONAL ANALYSIS ON EVERY OPERATION

Calculate the molar mass of each of the following substances:

a. I2 b. P4 c. CrCl3

d. C4H8 e. CrO2Cl2 f. CaF2

Calculate the number of atoms in each of the following quantities:

a. 36.1 g of argon

b. 5.00 g of chromium

c. 1.00 oz of silver

d. The 44 - carat Hope diamond, which is nearly pure

carbon (one carat equals 0.200 g)

e. 2.50 mL of mercury with a density of 13.6 g/mL

(Use Dimensional Analysis in every step)For calculation of

1. Number of PARTICLES.2. Number of MOLES.3. Number of GRAMS or MASS

All are related

ParticlesOf substance X

1 mole6.022 x 1023 particles = moles

Particles of substance

Avogadro’sNumber

Moles of substance

Grams ofsubstance X 1 mole

Molar mass of substance = molesGrams of substance

Moles of substance

Molar mass

Moles ofsubstance X Formula subscript, atoms

1 mole of compound = Moles of element

Moles of

compound

Moles of elementIn compound

formulasubscript

Determine the mass in grams of the following:

a. 3.00 X 1020 N2 molecules e. 2.00 X 10-15 mol N2

b. 3.00 X 10-3 mol N2 f. 18.0 picomoles of N2

c. 1.5 X 102 mol N2 g. 5.0 nanomoles of N2

d. A single N2 molecule

How many moles are represented by each of these samples?

a. 100 molecules (exactly) of H2O

b. 100.0 g H2O

c. 500 atoms (exactly) of Fe

d. 500.0 g Fe

e. 150 molecules (exactly) of O2

Particles of A

Molesof A

Particles of B

Moles of B

Grams of A

Grams of B

Molarmass

Molarmass

SCHEMATIC OF CALCULATIONS

(conversions)For Particles, Moles and Mass

Avogadro’s Avogadro’snumber number

formulasubscript

USE DIMENSIONAL ANALYSIS (UNITS)

UNITS WANTED = (HAVE) X (CONVERSION FACTORS) = (GIVEN UNITS) X (CONVERSION FACTOR UNITS)

Which of the following is closest to the average mass of one atom of copper?

a) 63.55 g

b) 52.00 g

c) 58.93 g

d) 65.38 g

e) 1.055 X 10-22 g

Which of the following 100.0 g samples contains the greatest number of atoms? a) Magnesium

b) Zinc

c) Silver

Rank the following according to number of atoms (greatest to least):

a) 100.0 g of silver

b) 62.0 g of zinc

c) 21.0 g of magnesium

Consider separate 100.0 gram samples of each of the following: 

H2O, N2O, C3H6O2, CO2

– Rank them from greatest to least number of oxygen atoms.

Percent Composition

Mass Percent

OR: USING DIMENSIONAL ANALYSIS:

Mass % = (g element. / 1 mol element.)(# moles element. / 1 mole subs) X 100%

(g substance / 1 mole substance)

Example: Find mass % Al in Al2O3

Mass % = (26.98 g Al / 1 mol Al)(2 mol Al / 1 molAl2O3) X 100%(101.96 g Al2O3/ 1 mole Al2O3)

= 52.92 % Al in Al2O3

100% X compound) mole 1 of wt.(molec.

mole)in atoms of element)(# of wt.(atomic

100% X compound of mole 1 of mass

mole 1in element of mass % Mass

CHEMICAL CALCULATIONSMole Concept

* a unit to count atoms, molecules, compounds,

* Avogadro’s number 6.022 X 1023

Chemical Formulas

* definite proportions of atoms to each other,

* calculation of formula mass

(also called formula weight, f.w.

or molecular weight, m.w.,

or molar mass)

* percent (%) composition

* empirical formulas

* molecular formulas

ALL OF THESE FORMULAS USE DIMENSIONAL ANALYSIS

Percent Composition

• Mass percent of an element:

• For iron in iron (III) oxide, (Fe2O3)

massm ass o f e lem en t in com pound

m ass o f com pound% 1 0 0 %

mass Fe%.

..

11 1 6 9

1 5 9 6 91 0 0 % 6 9 9 4 %

Determining the Formula of a Compound

EMPIRICAL and MOLECULAR FORMULAS

A MOLECULAR FORMULA gives the actual number of atoms present in a formula unit of a compound.An EMPIRICAL FORMULA gives the smallest whole number ratio of atoms in a formula unit of a compound.Molecular formula = whole number X empirical formula = (empirical formula)E

where E is the whole number multiplier.Examples:Compound name Empirical formula Molecular formula EDinitrogen tetrafluoride NF2 N2F4 2

hydrogenperoxide HO H2O2 2

sodium NaCl NaCl 1chloride

benzene CH C6H6 6

Aluminum metal is produced by passing an electric current

through a solution of aluminum oxide (Al2O3) dissolved in

molten cryolite (Na3AlF6). Calculate the molar masses of

Al2O3 and Na3AlF6.

Also for % composition m/m

-- Calculate the Mass % of Each Element in the Compound.

-- Let the Mass % = the Grams of the Element.

-- Calculate the Moles of Each Element.-- Divide Each Number of Moles by the Smallest Number. This represents the Empirical Formula Subscript for Each

Element in the Compound.

-- Multiply by an integer so Each Subscript is a Whole Number

e.g. 1.5 X 2 = 3 1.25 X 4 = 5

1.33 X 3 = 4 etc.

Calculate the percent of the first element in each of the following compounds.

a. BaCl2 b. Na2CO3 c. CuSO4 d. CoCl2

Determine the empirical formulas of the compounds with the following percent compositions.

a. 72.36% Fe , 27.64 % O

b. 58.53% C, 4.09 % H, 11.38 % N, 25.99 % O

c. 63.15 % C, 5.30 % H, 31.55 % O

d. 85.62 % C, 14.38 % H

One of the most commonly used white pigments in paint is a compound

of titanium and oxygen that contains 59.9 % Ti by mass. Calculate the

empirical formula of this compound.

A compound that contains only nitrogen and oxygen is 30.4 % N by

mass; the molar mass of the compound is 92 g/mol. What is the

empirical formula of the compound? What is the molecular formula of

the compound?

MOLECULAR FORMULA DETERMINATION

--Calculate the Empirical Formula.

--Compute the Empirical Formula Weight.

--Determine the Molecular Weight.

--Calculate:(g/mole)(empirical formula/g) = empirical formula / mole

Therefore:

(CaHbOc)E CaEHbEOcE

Where E = number of empirical formulas in one molecular formula or WHOLE NUMBER MULTIPLIER

Chemical Equations

Chemical Reaction:

A process in which at least one new substance

is created by chemical change.

Reactants:

substances present before reaction starts.

(written on the left side)

Products:

substances produced by chemical reaction.

(written on the right side)

Chemical EquationA written statement using symbols and formulas to describe a chemical reaction.

OR, it is... Chemist’s shorthand like: algebra to mathematiciansrecipes to a chefacronyms to the militarycomputer jargon

Format:1. Reactants on left, before reaction occurs.2. Products on right, after reaction occurs.3. Arrow points to products and means: “goes to”, “produces”,”forms”, “reacts to give.”4. + sign used between different reactants and products.Requirements:1. Must be consistent with experimental facts.2. Obeys conservation of mass.

LAW OF CONSERVATION OF MASS

Mass is neither created nor destroyed in an ordinary chemical reaction.

Or:

In a chemical reaction,

Total mass, reactants = total mass, products.

SYMBOLS USED IN EQUATIONSSymbol Meaning

(essential ones)

“to produce”, “goes to”, “forms”, “reacts to give”

+ “reacts with”, “and”, “plus”

(optional ones)

(s) solid

(l) liquid

(g) gas

(aq) aqueous solution(dissolved in water)

Information Conveyed by the Balanced Equation for the Combustion of Methane

Reactants Products

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

1 molecule CH4 1 molecule CO2

+ 2 molecules O2 + 2 molecules H2O

1 mole CH4 molecules 1 mole CO2 molecules

+ 2 mole O2 molecules + 2 mole H2O molecules

6.022 X 1023 CH4 molecules 6.022 X 1023 CO2 molecules

+ 2 (6.022 X 1023) O2 molecules + 2(6.022 X 1023) H2O molecules

16 g CH4 + 2(32 g) O2 44 g CO2 + 2(18 g) H2O 80 grams reactants 80 grams products

GRAM

MOLE

MOLECULES

ATOMS

DIVIDE BY MW

X 6.023 X 1023

X # ATOMS

EQUATIONS AND THE MOLE CONCEPT

moles of equation moles of

substance A coefficients substance B

Coefficients give the numerical relationships among formula units consumed and/or produced in a chemical reaction.

P4O10 + 6 H2O 4 H3PO4

3 mole to mole relationships:

1 mole P4O10 produces 4 moles of H3PO4

6 moles of H2O produces 4 moles of H3PO4

1 mole of P4O10 produces 6 moles of H2ONow write dimensional analysis factors:

1 mole P4O10 and 4 moles H3PO4

4 moles H3PO4 1 mole P4O10

or -------

Give a word interpretation of the balanced equation

CS2 + 3 O2 CO2 + 2 SO2

in terms of (a) molecules and (b) moles.

Write the twelve mole to mole conversion factors that can be derived from the balanced equation.

3 HNO2 2 NO + HNO3 + H2O

Write the twelve mole-to-mole conversion factors that can be derived from the balanced equation.

N2H4 + 2 H2O2 N2 + 4 H2O

BALANCING CHEMICAL EQUATIONS

A balanced equation has an equal number of each element in reactants as in products.

coefficients -- number of moles of a substancesubscripts -- number of atoms of an element or polyatomic ions in

one mole of the substance (compound)coefficient - multiplier for H and O

2H2Osubscript - multiplier for H onlycoefficient - multiplier for Ca , P and O

3 Ca3(PO4)2

subscripts - 3 multiplier for Ca- 4 multiplier for O- 2 multiplier for both P, O

STEPS TO BALANCE AN EQUATION1. Coefficients are the smallest set of whole numbers.

2. Consider polyatomic ions as single entities (use ( ) to keep them together).

3. Subscripts in a compound may not be changed during balancing, only coefficients may be.

4. Product compounds of a reaction will be given or predicted from known principles.

5. Balancing “by inspection” is the method for this class. Certain reactions which involve changes in oxidation states require more detailed methods.

6. All chemical equations are the result of experimental data and may be verified by experiment.

Balance these equations:

C3H8 + O2 CO2 + H2O

BaCl2 + Na3PO4 Ba3(PO4)2 + NaCl

AgNO3(aq) + Al(s) Al(NO3)3 (aq) + Ag(s)

Which of the following correctly balance the chemical equation given below? There may be more than one correct balanced equation. If a balanced equation is incorrect, explain what is incorrect about it.

CaO + C CaC2 + CO2

I. CaO2 + 3C CaC2 + CO2

II. 2CaO + 5C 2CaC2 + CO2

III. CaO + (2.5)C CaC2 + (0.5)CO2

IV. 4CaO + 10C 4CaC2 + 2CO2

Which of the following are true concerning balanced chemical equations? There may be more than one true statement.– The number of molecules is conserved.

– The coefficients tell you how much of each substance you have.

– Atoms are neither created nor destroyed.

– The coefficients indicate the mass ratios of the substances used.

– The sum of the coefficients on the reactant side equals the sum of the coefficients on the product side.

SUMMARY

• The number of atoms of each type of element must be the same on both sides of a balanced equation.

• Subscripts must not be changed to balance an equation.

• A balanced equation tells us the ratio of the number of molecules which react and are produced in a chemical reaction.

• Coefficients can be fractions, although they are usually given as lowest integer multiples.

• Trial and error is a valid method to balance a chemical equation.

Stoichiometric Calculations:Amounts of

Reactants and Products

Calculations Based on Chemical Equations - StoichiometryChemical stoichiometry is the study of the quantitative relationships among

reactants and products in a chemical reaction

Particles of A

Moles ofA

Avogadro’s number

Particles of B

Grams of B

Grams of A

Moles ofB

equationcoefficients

Avogadro’s number

Molarmass

Molarmass

Always use dimensional analysis

STOICHIOMETRYStoichiometry is the relationship of quantities of reactants consumed and products made in all chemical reaction.

MoleA mole is the number of atoms in exactly 12 grams of carbon 12.

One mole of any substance contains 6.022 X 1023 atoms (Avogadro’s number) of that substance.

The mass of one mole of an element is the weighted average atomic mass of all isotopes of that element OR the total of all the element masses in a compound or chemical formula.

STOICHIOMETRIC CALCULATIONS

Of Masses/Moles Use Dimensional Analysis:

Balance the reaction equation.

Convert Known masses to moles.

Set up mole ratios from the equation

Calculate Moles of desired product or reactant.

Convert From moles to mass as required.

Given the equation:

4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g)

(a) How many moles of O2 are needed to produce 1.34 moles of NO?

(b) How many moles of H2O will be produced from 0.789 mole of NH3?

(c) How many moles of NH3 are needed to react with 3.22 moles of O2 ?

(d) How many moles of NO are produced when 0.763 mole of H2O is

produced?

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g)

(a) How many moles of O2 are needed to produce 2.38 moles of H2O?

(b) How many moles of CO2 will be produced from 0.57 moles of C3H8?

(c) How many moles of C3H8 are needed to react with 1.45 moles of O2 ?

(d) How many moles of CO2 are produced when 1.11 moles of H2O are

produced?

The reusable booster rockets of the U.S. space shuttle employ a mixture of aluminum and ammonium perchlorate for fuel. A possible equation for this reaction is

3 Al (s) + 3 NH4ClO4 (s) Al2O3 (s) + AlClO3 (s) + 3 NO (g) + 6 H2O (g)

What mass of NH4ClO4 should be used in the fuel mixture for every kilogram of Al?

Anthraquinone (C14H8O2), an important intermediate in the dye industry, is produced from the reaction of benzene (C6H6), followed by dehydration with sulfuric acid. The overall reaction is

C8H4O3 (s) + C6H6 (l) C14H8O2 (s) + H2O (l)

A. What mass of benzene reacts completely with 2.00 X 103 g phthalic anhydride?

B. What masses of anthraquinone and of water are produced assuming 100% yield?

C. If 1.96 X 103 g anthraquinone is actually obtained, what is the percentage yield of

the reaction?

Elixirs such as Alka-Seltzer use the reaction of sodium bicarbonate with citric acid in aqueous solution to produce a fizz:

2 NaHCO3 (aq) + C6H8O7 (aq) 3 CO2 (g) + 3 H2O (l) + Na3C6H5O7 (aq)

A. What mass of C6H8O7 should be used for every 1.0 X 102 mg NaHCO3?

B. What mass of CO2 (g) could be produced from such a mixture?

Nitric acid is produced commercially by the Ostwald process, represented by the following equations:

4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g)

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

3 NO2 (g) + H2O (l) 2 HNO3 (aq) + NO (g)

What mass of NH3 must be used to produce 1.0 X 106 kg. HNO3 by the Ostwald process, assuming 100% yield in each reaction?

SIMULATANEOUS ANDCONSECUTIVE REACTIONS

When two or more reactions are involved in stoichiometric calculations, each molar relationship must be taken into account.

Example:

In steelmaking, a three-step process leads to conversion of Fe2O3 into metallic Fe.

1. 3 Fe2O3 + CO 2 Fe3O4 + CO2

2. Fe3O4 + CO 3 FeO + CO2

3. FeO + CO Fe + CO2

How many grams Fe are produced from 500.0 grams of Fe2O3?

WANT = HAVE X CONVERSIONS

Simultaneous ReactionsA mixture of composition 60.0% ZnS and 40.0% CuS are heated in air until the sulfides are completely converted to oxides as shown by the following equations.

2 ZnS + 3 O2 2 ZnO + 2 SO2

2 CuS + 3 O2 2 CuO + 2 SO2

How many grams of SO2 are produced from reaction of 82.5 g of the sulfide mixture?

A mixture of composition 50.0% H2S and 50.0% CH4 is reacted with oxygen, producing SO2, CO2, and H2O. The equations for the reactions are

2 H2S + 3 O2 2 SO2 + 2H2O

CH4 + 2 O2 CO2 + 2 H2O

How many grams of H2O are produced from the reaction of 65.0 g of mixture?

Consecutive Reactions

How many grams of SO2 can be obtained from 50.0 g of KClO3 by the following two-step chemical process?

2 KClO3 2KCl + 3 O2

S + O2 SO2

Limiting Reactants

Consider the reaction

Mg (s) + I2 (s) MgI2 (s)

Identify the limiting reagent in each of the reaction mixtures below:

a. 100 atoms of Mg and 100 molecules of I2

b. 150 atoms of Mg and 100 molecules of I2

c. 200 atoms of Mg and 300 molecules of I2

d. 0.16 mole Mg and 0.25 mole I2

e. 0.14 mole Mg and 0.14 mole I2

f. 0.12 mole Mg and 0.08 mole I2

g. 6.078 g Mg and 63.455 g I2

h. 1.00 g Mg and 2.00 g I2

I. 1.00 g Mg and 20.00 g I2

Consider the reaction

2 H2 (g) + O2 (g) 2 H2 O (g)

Identify the limiting reagent in each of the reaction mixture given below:

a. 50 molecules of H2 and 25 molecules of O2

b. 100 molecules of H2 and 40 molecules of O2

c. 100 molecules of H2 and 100 molecules of O2

d. 0.5 mole H2 and 0.75 mole O2

e. 0.80 mole H2 and 0.75 mole O2

f. 1.0 g H2 and 0.25 mole O2

g. 5.00 g H2 and 56.00 g O2

LIMITING REACTANTSSTOICHIOMETRIC CALCULATIONS

Of Masses/Moles

USE DIMENSIONAL ANALYSISBalance the reaction equation.

Convert Known masses to moles.

DETERMINE THE LIMITING REACTANT

Set up mole ratios from the equation.

Calculate Moles of desired product or reactant.

Convert from moles to mass as required.

Nitrogen (N2) and hydrogen (H2) react to form ammonia (NH3). Consider the mixture of N2 ( ) and H2 ( ) in a closed container as illustrated below.

Assuming the reaction goes to completion, draw a representation of the product mixture. Explain how you arrived at this representation.

SUMMARY

• Use the coefficients in the balanced equation to decide the amount of each reactant that is used, and the amount of each product that is formed.

• The coefficients in the balanced equation have nothing to do with the amount of each reactant that is given in the problem. The balanced equation represents a ratio of reactants and products, not what actually “happens” during a reaction.

• Reactants are only placed on the left side of the arrow, products are only placed on the right side of the arrow.

• Methane (CH4) reacts with the oxygen in

the air to produce carbon dioxide and water.

• Ammonia (NH3) reacts with the oxygen in

the air to produce nitrogen monoxide and water.

• What mass of ammonia would produce the same amount of water as 1.00 g of methane reacting with excess oxygen?