TOPIC 1 TOPIC 1 STOICHIOMETRIC STOICHIOMETRIC RELATIONSHIPS RELATIONSHIPS

TOPIC 1 TOPIC 1

STOICHIOMETRIC STOICHIOMETRIC

RELATIONSHIPS RELATIONSHIPS

1.2 1.2

THE MOLE CONCEPT THE MOLE CONCEPT

By: Merinda Sautel Alameda Int’l Jr/Sr High School Lakewood, CO msautel@jeffco.k12.co.us

ESSENTIAL IDEA ESSENTIAL IDEA

The mole makes it possible to The mole makes it possible to correlate the number of particles correlate the number of particles

with the mass that can be with the mass that can be

measured.

measured.

NATURE OF SCIENCE (2.3) NATURE OF SCIENCE (2.3)

Concepts – the concept of the mole Concepts – the concept of the mole developed from the related concept of developed from the related concept of

“equivalent mass” in the early 19

“equivalent mass” in the early 19 ^{th th} century. century.

INTERNATIONAL-MINDEDNESS INTERNATIONAL-MINDEDNESS

The SI system (Systeme International d’Unites) The SI system (Systeme International d’Unites)

refers to the metric system of measurement refers to the metric system of measurement

based on seven base units.

based on seven base units.

The International Bureau of Weights and The International Bureau of Weights and

Measures (BIPM according to its French initials) Measures (BIPM according to its French initials) is an international standards organization, which is an international standards organization, which aims to ensure uniformity in the application of SI aims to ensure uniformity in the application of SI

units around the world.

units around the world.

THEORY OF KNOWLEDGE THEORY OF KNOWLEDGE

The magnitude of Avogadro’s constant is The magnitude of Avogadro’s constant is

beyond the scale of our everyday beyond the scale of our everyday experience. How does our everyday experience. How does our everyday

experience limit our intuition?

experience limit our intuition?

UNDERSTANDING/KEY IDEA UNDERSTANDING/KEY IDEA

1.2.A 1.2.A

The mole is a fixed number of The mole is a fixed number of

particles and refers to the particles and refers to the

amount, n, of substance.

amount, n, of substance.

A chemical species may be an atom, a A chemical species may be an atom, a

molecule or an ion.

molecule or an ion.

This is the same as using the terminology This is the same as using the terminology

that “particles” refer to atoms, molecules and that “particles” refer to atoms, molecules and

formula units. (Note that formula units are formula units. (Note that formula units are

made up of ions.)

made up of ions.)

The exact definition of a mole is the amount The exact definition of a mole is the amount

of a substance that contains the same of a substance that contains the same

number of chemical species as there are number of chemical species as there are

atoms in exactly 12g of the isotope atoms in exactly 12g of the isotope

carbon-12.

carbon-12.

Avogadro’s constant (L) has the value Avogadro’s constant (L) has the value

6.02x10

6.02x10 ^{23 23} particles per mole. particles per mole.

The mole is the chemist’s counting unit.

The mole is the chemist’s counting unit.

It contains 6.02 x 10

It contains 6.02 x 10

particles of a particles of a substance.

substance.

Example:

Example: 1 dozen = 12 of something 1 dozen = 12 of something

1 mole = 6.02 x 10 1 mole = 6.02 x 10

of of something

something

What are species (particles)?

What are species (particles)?



Atoms – smallest species of an element Atoms – smallest species of an element



Molecules – smallest species of a covalent cmpd Molecules – smallest species of a covalent cmpd



Ions – smallest species of an ionic cmpd Ions – smallest species of an ionic cmpd

When we want to count atoms or molecules, When we want to count atoms or molecules, we use Avogadro’s # as a conversion factor.

we use Avogadro’s # as a conversion factor.

There are 6.02 x 10

There are 6.02 x 10

number of atoms in one number of atoms in one mole of an element.

mole of an element.

There are 6.02 x 10

There are 6.02 x 10

number of molecules in number of molecules in a covalent compound.

a covalent compound.

UNDERSTANDING/KEY IDEA UNDERSTANDING/KEY IDEA

1.2.B 1.2.B

Masses of atoms are Masses of atoms are

compared on a scale relative compared on a scale relative to to _{12 12} C and are expressed as C and are expressed as

relative atomic mass (A

relative atomic mass (A _{r r} ) and ) and relative formula/molecular

relative formula/molecular mass (M

mass (M _{r r} ). ).

1 1 The The relative atomic mass (A relative atomic mass (A _{r r} ) ) is the is the average mass of an atom, taking into average mass of an atom, taking into

account the relative abundances of all the account the relative abundances of all the

naturally occurring isotopes of the element.

naturally occurring isotopes of the element.

2 2 The The relative molecular mass (M relative molecular mass (M _{r r} ) ) of a of a

covalent or molecular compound is the sum covalent or molecular compound is the sum

of all the atomic masses of the elements in of all the atomic masses of the elements in

the compound.

the compound.

3 3 The The relative formula mass (M relative formula mass (M _{f f} ) ) of an ionic of an ionic compound is the sum of all the atomic

compound is the sum of all the atomic masses of the ions in the formula.

masses of the ions in the formula.

 Notice that A Notice that A _{r r} , M , M _{r r} and M and M _{f f} have no units have no units because they are “relative” terms.

because they are “relative” terms.

UNDERSTANDING/KEY IDEA UNDERSTANDING/KEY IDEA

1.2.C 1.2.C

Molar mass (M) has the Molar mass (M) has the

units g/mol.

units g/mol.

The mass of one mole of a species is The mass of one mole of a species is

called the

called the molar mass molar mass (M) and it has the (M) and it has the units of g/mol or g mol

units of g/mol or g mol ^{-1 -1} . .

APPLICATION/SKILLS APPLICATION/SKILLS

Be able to calculate the molar Be able to calculate the molar

masses of atoms, ions, molecules masses of atoms, ions, molecules

and formula units.

and formula units.

To calculate molar mass, add up the To calculate molar mass, add up the

atomic masses of the separate atoms atomic masses of the separate atoms

from the periodic table.

from the periodic table.

CO CO _{2 2} 1-C = 1 x 12 = 12 1-C = 1 x 12 = 12 2-O = 2 x 16 = 2-O = 2 x 16 = 32 32

44 g/mol 44 g/mol

CH CH _{4 4} 1-C = 1 x 12 = 12 1-C = 1 x 12 = 12 4-H = 4 x 1 =

4-H = 4 x 1 = 4 4 16 g/mol

16 g/mol

1. NaHCO 1. NaHCO _{3 3}

2. Sr(CN) 2. Sr(CN) _{2 2}

3. Al

3. Al _{2 2} (SO (SO _{3 3} ) ) _{3 3} 4. C

4. C _{12 12} H H _{22 22} O O _{11 11} 5. (NH

5. (NH _{4 4} ) ) _{2 2} CO CO _{3 3}

FIND THE FOLLOWING MOLAR MASSES

FIND THE FOLLOWING MOLAR MASSES

APPLICATION/SKILLS APPLICATION/SKILLS

Be able to solve problems involving Be able to solve problems involving

the relationships between the the relationships between the

number of particles, the amount of number of particles, the amount of

substance in moles and the mass substance in moles and the mass

in grams.

in grams.

Converting from moles to particles Converting from moles to particles

or particles to moles or particles to moles

To convert from moles to particles or To convert from moles to particles or

particles to moles, use Avogadro’s number particles to moles, use Avogadro’s number

as the conversion factor.

as the conversion factor.

VOLUME (L) MOLE MASS

(g)

PARTI CLES

ATOMS

MOLECULES

FORMULA UNITS

DIVIDE BY 6.02X10²³ PART MOL MULTIPLY BY

6.02X10²³ PART MOL

Mole/Particle Conversion Problems Mole/Particle Conversion Problems

 1. 1. How many atoms are in 3.0 mol of Sn? How many atoms are in 3.0 mol of Sn?

 2. 2. How many formula units are in 0.40 mol KCl? How many formula units are in 0.40 mol KCl?

 3. 3. How many molecules are in 7.50 mol SO How many molecules are in 7.50 mol SO

? ?

 4 4 . How many formula units are in 4.80x10 . How many formula units are in 4.80x10

mole NaI? mole NaI?

 5. 5. How many moles are in 1.50x10 How many moles are in 1.50x10

molecules of NH molecules of NH

? ?

 6. 6. How many moles are in 1.0x10 How many moles are in 1.0x10

molecules of O molecules of O

? ?

 7. 7. How many moles are in 6.02x10 How many moles are in 6.02x10

molecules of Br molecules of Br

? ?

 8. 8. How many moles are in 4.81x10 How many moles are in 4.81x10

atoms of Li? atoms of Li?

Converting from moles to mass or Converting from moles to mass or

mass to moles mass to moles

To convert from moles to mass or mass to To convert from moles to mass or mass to

moles, use the molar mass as the moles, use the molar mass as the

conversion factor.

conversion factor.

VOLUME (L) MOLE MASS

(g)

PARTI CLES

ATOMS

MOLECULES

FORMULA UNITS

DIVIDE BY MOLAR MASS

MULTIPLY BY MOLAR MASS

MASS/MOLE CONVERSIONS MASS/MOLE CONVERSIONS

 1. What is the mass of 2.40 mol N 1. What is the mass of 2.40 mol N

? ?

 2. How many moles is 5.00g H 2. How many moles is 5.00g H

? ?

 3. What is the mass of 3.32 mol K? 3. What is the mass of 3.32 mol K?

 4. How many moles is 11.0g CH 4. How many moles is 11.0g CH

? ?

 5. What is the mass of 0.160 mol H 5. What is the mass of 0.160 mol H

O O

? ?

 6. How many moles is 847g (NH 6. How many moles is 847g (NH

) )

CO CO

? ?

 7. What is the mass of 5.08 mol Ca(NO 7. What is the mass of 5.08 mol Ca(NO

) )

? ?

 8. How many moles is 333g SnF 8. How many moles is 333g SnF

? ?

 9. What is the mass of 10.0 mol Cr? 9. What is the mass of 10.0 mol Cr?

 10. How many moles is 0.000264g Li 10. How many moles is 0.000264g Li

HPO HPO

? ?

UNDERSTANDING/KEY IDEA UNDERSTANDING/KEY IDEA

1.2.D 1.2.D

The empirical formula and The empirical formula and

molecular formula of a molecular formula of a

compound give the simplest compound give the simplest

ratio and the actual number of ratio and the actual number of

atoms present in a molecule atoms present in a molecule

respectively.

respectively.

Empirical formula

Empirical formula – the simplest whole – the simplest whole number ratio of the atoms it contains.

number ratio of the atoms it contains.

Molecular formula

Molecular formula – multiple of the empirical – multiple of the empirical formula showing the actual number of atoms formula showing the actual number of atoms

of each element present.

of each element present.

APPLICATION/SKILLS APPLICATION/SKILLS

Be able to determine the empirical Be able to determine the empirical

formula when given the percentage formula when given the percentage

composition by mass.

composition by mass.

Rules for empirical formula Rules for empirical formula

1. Find the % composition.

1. Find the % composition.

2. Assume 100 g and convert %’s to g’s.

2. Assume 100 g and convert %’s to g’s.

3. Convert the grams to moles by dividing 3. Convert the grams to moles by dividing

by the molar mass.

by the molar mass.

4. Divide all mole quantities by the smallest 4. Divide all mole quantities by the smallest

number of moles found in step 3.

number of moles found in step 3.

5. If the numbers found in step 4 are not 5. If the numbers found in step 4 are not

whole numbers, multiply by a factor to make whole numbers, multiply by a factor to make

whole numbers.

whole numbers.

Sample Problem Sample Problem

A hydrocarbon contains 85.7% by mass of carbon. What is A hydrocarbon contains 85.7% by mass of carbon. What is the empirical formula?

the empirical formula?



Step 1: Find %’s Step 1: Find %’s

• 85.7 % C 85.7 % C

• 100-85.7 = 14.3% H (Hydrocarbons contain H and C) 100-85.7 = 14.3% H (Hydrocarbons contain H and C)



Step 2: Convert to grams Step 2: Convert to grams

• 85.7 g C 85.7 g C

• 14.3 g H 14.3 g H



Step 3: Convert to moles Step 3: Convert to moles

• 85.7 g C x mol/12.01 g C = 7.14 mol C 85.7 g C x mol/12.01 g C = 7.14 mol C

• 14.3 g H x mol/1.01 g H = 14.16 mol H 14.3 g H x mol/1.01 g H = 14.16 mol H



Step 4: Divide by the smallest number of Step 4: Divide by the smallest number of moles.

moles.

• 7.14 mol / 7.14 mol = 1 7.14 mol / 7.14 mol = 1

• 14.16 mol / 7.14 mol = 1.98 = 2 14.16 mol / 7.14 mol = 1.98 = 2



Step 5: The numbers in Step 4 are whole Step 5: The numbers in Step 4 are whole numbers so this is the end of the problem.

numbers so this is the end of the problem.

• The empirical formula is CH The empirical formula is CH

APPLICATION/SKILLS APPLICATION/SKILLS

Be able to determine the molecular Be able to determine the molecular

formula of a compound from its formula of a compound from its

empirical formula and molar mass.

empirical formula and molar mass.

 Hint: You have to be given the Hint: You have to be given the

experimental molar mass in order to solve a experimental molar mass in order to solve a

molecular formula problem.

molecular formula problem.

Rules for molecular formula Rules for molecular formula

1. Solve for empirical formula.

1. Solve for empirical formula. (It may be (It may be given to you or you may have to solve for it.) given to you or you may have to solve for it.)

2. Find the empirical molar mass.

2. Find the empirical molar mass.

3. Divide the given molecular or molar mass 3. Divide the given molecular or molar mass

by the empirical molar mass – this answer by the empirical molar mass – this answer

should be a whole number.

should be a whole number.

4. Multiply the empirical formula by the 4. Multiply the empirical formula by the

number found in step 3.

number found in step 3.

Sample problem Sample problem

The compound with the empirical formula CH

The compound with the empirical formula CH

is analyzed is analyzed by a mass spectrometer and found to have a relative

by a mass spectrometer and found to have a relative molecular mass of 42.09. Deduce its molecular formula.

molecular mass of 42.09. Deduce its molecular formula.



Step 1: The empirical formula is CH Step 1: The empirical formula is CH

. .



Step 2: Find the empirical molar mass. Step 2: Find the empirical molar mass.

• CH CH

has a molar mass of 14.03 g/mol. has a molar mass of 14.03 g/mol.



Step 3: Divide the given molar mass by the Step 3: Divide the given molar mass by the empirical molar mass.

empirical molar mass.

• 42.09 / 14.03 = 3 42.09 / 14.03 = 3



Step 4: Multiply the empirical formula by the Step 4: Multiply the empirical formula by the number found in Step 3.

number found in Step 3.

• (CH (CH

) )

= C = C

H H

APPLICATION/SKILLS APPLICATION/SKILLS

Be able to obtain and use Be able to obtain and use

experimental data for deriving experimental data for deriving

empirical formulas from reactions empirical formulas from reactions

involving mass changes.

involving mass changes.

EMPIRICAL FORMULAS FROM EMPIRICAL FORMULAS FROM

COMBUSTION COMBUSTION

1. Sometimes you are not given the percentages, but 1. Sometimes you are not given the percentages, but

have to calculate them from a combustion problem.

have to calculate them from a combustion problem.

2. The problems usually give you an amount of 2. The problems usually give you an amount of

original substance and amounts of CO

original substance and amounts of CO

and H and H

O that O that are formed from the combustion reactions.

are formed from the combustion reactions.

3. Make these assumptions:

3. Make these assumptions:



All of the carbon in the original sample is converted All of the carbon in the original sample is converted completely to CO

completely to CO

. .



All of the hydrogen in the original sample is converted All of the hydrogen in the original sample is converted completely to H

completely to H

O. O.

EXAMPLE EXAMPLE

You have a compound composed of carbon, hydrogen and You have a compound composed of carbon, hydrogen and nitrogen. When .1156g of this compound is burned, .1638g nitrogen. When .1156g of this compound is burned, .1638g of CO

of CO

and .1676g of H and .1676g of H

O is produced. Determine the O is produced. Determine the empirical formula of the compound.

empirical formula of the compound.

Citations Citations

International Baccalaureate Organization. Chemistry Guide, International Baccalaureate Organization. Chemistry Guide, First assessment 2016. Updated 2015.

First assessment 2016. Updated 2015.

Brown, Catrin, and Mike Ford.

Brown, Catrin, and Mike Ford. Higher Level Chemistry Higher Level Chemistry . 2nd ed. . 2nd ed.

N.p.: Pearson Baccalaureate, 2014. Print.

N.p.: Pearson Baccalaureate, 2014. Print.

ISBN 978 1 447 95975 5 ISBN 978 1 447 95975 5 eBook 978 1 447 95976 2 eBook 978 1 447 95976 2

Most of the information found in this power point comes directly Most of the information found in this power point comes directly from this textbook.

from this textbook.

The power point has been made to directly complement the The power point has been made to directly complement the Higher Level Chemistry textbook by Brown and Ford and is Higher Level Chemistry textbook by Brown and Ford and is used for direct instructional purposes only.

used for direct instructional purposes only.