The Mole Topic: Atomic Mass Objectives: Day 1 of 4 To learn how atomic mass is calculated using the...

The Mole

Topic: Atomic Mass

Objectives: Day 1 of 4

• To learn how atomic mass is calculated using the average natural abundance of isotopes

• To understand the quantity of a mole and Avogadro's number

Unit: The Mole

QuickwriteAnswer one of the questions below 1-2

sentences:• 1 light year (the distance light travels in a year) is equal to

9.5 trillion kilometers!!!!!!!!!! Why do you think scientists use light years to measure distances to nearby stars in light years and not kilometers?????

• Your pencil uses graphite (pure carbon) to write with; how many atoms do you think are in 12 grams of graphite or carbon????????

• How many items make up a dozen????? How many items make up a half dozen???? How many items are in two dozen??

• In 1811, an Italian scientist by the name of Avogadro, Amedeo (1776–1856) discovered that a mole of atoms is equal to the number 6.022 x1023

• In other words: 1 mole = 6.022 x1023

• Or, or if you have a mole of carbon atoms, then you have 6.022 x1023 atoms

The Mole

Element

Number of Atoms Present

Hydrogen 6.022 x1023

Helium 6.022 x1023

Carbon 6.022 x1023

Nitrogen 6.022 x1023

Oxygen 6.022 x1023

Aluminum 6.022 x1023

Sodium 6.022 x1023

How many atoms does A mole of hydrogen contain?

6.022 x1023

How many atoms does A mole of

helium contain?

How many atoms does

A mole Carbon contain?

How many atoms does a mole of nitrogen contain?

6.022 x1023

6.022 x1023

6.022 x1023

The point is, a MOLE of

anything has6.022 x1023 atoms!

What is a Mole?

• The amount of a substance that is equal to Avogadro's number, __________, which is the amount of atoms found in 12 grams of carbon

Write:

Answer Bank12.01Atoms

24Amount

6.022 x 1023(2)

1 Mole6.022 x 1023 atoms

12 grams of Carbon

6.022 x 1023atomsOr

12 grams of Carbon

1 MoleOr

The Mole• As it turns out, one mole of anything contains 6.022

x1023 units of that substance• Just as a dozen eggs is 12 eggs, a mole of eggs is

6.022 x1023 eggs• The mole is an incredibly large number to imagine -

602,000,000,000,000,000,000,000!!!!!!!!!!!!!!!!!• We use scientific notation to simply this number• We call this unbelievably large number Avogadro’s

number

The Mole• If I have a dozen eggs how many eggs do

I have?

• If I have 2 dozen eggs, how many eggs do I have?

• If I have a mole of eggs, how many eggs do I have?

• If I have a 2 mole of eggs, how many eggs do I have? (2) x (6.022 x 1023)

What is Avogadro’s Number?

• The amount of _____ in 1 mole of a substance which is ________

• Just as two dozen is (2) x (12), or ____eggs, 2 moles of atoms is equal to (2) x (6.022 x 1023 ) atoms

Answer Bank12.01Atoms

24Amount

6.022 x 1023(2)

The Mole• Take out your periodic table and find Carbon• Look Below the Atomic Symbol (C)(C)• What number do you see???• That’s right, 12.01 !!!!!!!!!!!!• Below every atomic symbol is number called

the ATOMIC MASS• A sample of carbon with a mass of 12.01 has

6.022 x1023 atoms!• So we can say that a mole of hydrogen (H)

atoms has a mass of 1.01g, a mole of oxygen, (O) has a mass of 16.00g, a mole of iron (Fe) is 55.80g, and so on

Element

Number of Atoms Present

Average Atomic Mass in grams

Hydrogen 6.022 x1023 1.008

Helium 6.022 x1023 4.00

Carbon 6.022 x1023 12.01

Nitrogen 6.022 x1023 14.01

Oxygen 6.022 x1023 16.00

Aluminum 6.022 x1023 26.98

Sodium 6.022 x1023 22.99

What is the MASS or weight of a MOLE of Hydrogen?

1.008 grams

What is the MASS or weight of a MOLE of

Helium?

What is the MASS or weight of a MOLE of

Carbon?

What is the MASS or weight of a MOLE of Nitrogen?

4.00 grams

12.01 grams

14.01 grams

The point is, a sample

Of any element that weighs A number of grams equal

To the average atomic mass of

that element contains

6.022 x1023 atoms!Or a MOLE!!!

Atomic Mass

• The average atomic mass for carbon is 12.01 amu

• Where does the 0.01 come from?• 0.01 is the percent abundance in nature of

the carbon isotopes• For example, if we weighed 12 grams of

carbon, 0.01% percent is the amount of Carbon 14 and Carbon 13 isotopes that exist in nature

Atomic Mass

• It is the average mass of an element containing 6.02 x 1023 atoms and calculated using the relative abundance of isotopes in a naturally-occurring element

• It is based on AMU’s and the natural abundance of an elements isotopes

What is atomic mass?• It is the _________ mass of an

element containing 6.02 x 1023 atoms and calculated using the relative abundance of isotopes in a naturally-occurring element

• For example a sample of Carbon containing 6.02 x 1023 atoms has a mass of ____grams

and a sample of Iron containing 6.02 x 1023 atoms has a mass of ____grams

Answer Bankaverage

1/12 simplifyCarbon

55.8abundance

Neutron12.01

The Mole• Consider the following sample of CARBON

atoms below (symbolized by red dots) which contains one mole (6.022 x 1023) of CARBON atoms

• Now consider another sample in which the number of CARBON atoms is unknown

Sample A = 12.01 grams Sample B = 6.005 grams

The Mole• We know sample A has 6.022 x1023 Carbon atoms • But how many atoms are in sample B?• We know the mass is 6.005 grams


The Mole• Let’s consider what we know• We know that 1 mol of Carbon atoms has a

mass of 12.01 grams• Sample B has a mass of 6.005 grams which is

exactly half the mass of a mole of Carbon atoms


The Mole• Let’s consider what we know• We know that 1 mol of CARBON atoms has a

mass of 12.01 grams• Sample B has a mass of 6.005 grams which is

exactly half the mass of a mole of CARBON atoms


The Mole• Let’s do the math!


Our conversion factor is:

1mol CARBON 12.01 grams

12.01 grams of carbon

1 mol carbon

= 0.50 molof Carbon in sample B

6.005 grams of carbon

We know we have 0.500 grams

of hydrogen

The Mole• Now that we have moles, we can compute the

number of atoms by using our conversion factor

Sample A = 12.01 gramsContains 6.022 x1023 Atoms

Sample B = 6.005 gramsContains ???? Atoms

1 mol of carbon

6.022 x1023 carbon atoms

= 3.0 x1023

carbon atoms in sample B

0.50 mol of carbon

We solved for molesIn the last example

Our conversion factor is:

6.022 x1023

1 mol

Practice:

• Your chicken laid 562 eggs. How many dozen eggs do you have?

12 eggs

1 dozen

= 46 dozen562 eggs

Practice:

• If you have 9.39 x1023 Aluminum Atoms, how many moles do you have?

• 1.56 mol of aluminum?

6.022 x1023 Al atoms

1 mol Al= 1.56 Moles of Al

9.39 x1023 lAl Atoms

number of molsNumber of atoms

Practice:• How many moles are in a 42 gram sample of

aluminum?

26.98 grams Al

1 mol aluminum= 1.56 molof Aluminum

42 grams Al

molesgrams

Practice:

• How many atoms are in a 18 gram sample of carbon?

12.01 grams C

6.022 x1023 C atoms18 grams C

= 9.07 x1023

Carbon Atoms

grams number of atoms

Summarize:

• If I have a _____of something I have_______ particles

• Avogadro number is ___________• A mole of carbon atoms weigh (mass)

______ grams and contains ______ atoms

• 2 moles of carbons atoms weighs _____ grams

• Review: An ______ is atom with a different amount of neutron than protons

Answer Bank12.01AtomsisotopeMole24.02

Amount6.022 x 1023(2)

Topic: Molar Mass (Molecular Weight) & Percent Composition


• To learn how to calculate Molar Mass

• To learn how to convert between moles and grams

• To learn how to calculate % composition

Unit: The Mole


sentences:• Let’s say you want to find the weight of your dog, which

is too big to stand on your bathroom scale; how could you find his weight???

• Together, you and your dog weigh 100 kilograms, you know that you weigh 75 kilograms, what percent by weight does your dog weigh????

• You know that a mole oxygen has a mass of 16grams and a mole of hydrogen has a mass of 1.0 grams, but how could you find the mass of 1 mole of water (H2O)

Molar Mass of a Compound

• Earlier we learned that a mole of any element contains 6.02 x1023 atoms

• But what about a compound or Molecule such as water (H2O)?

• A chemical compound such as water (H2O) is a collection of atoms

• Water contains 1 oxygen atom and 2 hydrogen atoms

• But how do we calculate the mass of one mole of water?

• In other words, what is the mass of 6.022 x1023 H2O Water molecules?

1 mole of H2O =6.02 x 1023

Molecules of H2O

• If we put one mole of water (H2O) on a scale and mass it, it will read 18.02 grams

• But how did we come up with that number?

• Well, we know water is made up of Oxygen and Hydrogen

• We also know the mass of one mole of Oxygen by looking on our periodic table 16.00 grams

• We also know the mass of one mole of Hydrogen by looking on our periodic table 1.01 grams

• Finally, we also know that water contains 2 Hydrogen atoms


1 mole of H2O =6.02 x 1023

Molecules of H2O

18.02 grams

• Because each water molecule (H2O) contains 1 Oxygen atom and 2 Hydrogen atoms, 1 mol (H2O) molecules consists of 1 mol Oxygen atoms and 2 mol of Hydrogen atoms

• So the mass of 1 mol of (H2O) is equal to: Mass of 1 mol of Oxygen (O) = 1 x 16.00g = 16.00 g

Mass of 2 mol of hydrogen (H) = 2 x 1.01 = 2.02g


1 mole of H2O =6.02 x 1023

Molecules of H2O

And contains 1 mole of Oxygen

Atoms And 2 moles of Hydrogen atoms

_______ Mass of 1 mol of (H2O) = 18.02 g

• Molar Mass is the mass in grams of 1 Mole

of a substance or compound • It is calculated by the SUM of the

atomic weights for every element that makes up the compound

• Ex: H2O

Mass of 1 mol of Oxygen (O) = 1 x 16.00g = 16.00 g

Mass of 2 mol of Hydrogen (H) = 2 x 1.01 = 2.02g _______ Mass of 1 mol of (H2O) = 18.02 g


What is molar mass or molecular weight?

• It is the mass in grams of 1 _____

of a substance or compound containing

______ molecules of that substance• It is calculated by the sum or ______up

the atomic mass for every element that makes up the compound

• Ex: H2O

Mass of 1 mol of Oxygen (O) = 1 x 16.00g = 16.00 g

Mass of 2 mol of Hydrogen (H) = 2 x 1.01 = 2.02g

Answer BankTotal

PercentElementAdding

6.02 x 1023

moleweight

_______ Mass of 1 mol of (H2O) = 18.02 g

Practice:

• Calculate the molar mass of sulfur dioxide(SO2):

Mass of 1 mol of sulfur (S) = 1 x 32.07 g = 32.07 gMass of 1 mol of sulfur (S) = 1 x 32.07 g = 32.07 g

Mass of 2 mol of oxygen (O) = 2 x 16.00 g = 32.00 g _______ Mass of 1 mol of (SO2) = 64.07 g/mol

Practice:• A sample of calcium carbonate (chalk) contains

4.86 mol. What is the mass in grams of this sample:

1 mol CaCO3

100.09 grams CaCO3 = 486 grams CaCO34.86 mol CaCO3

First calculate the molar mass of CaCO3:

Mass of 1 mol of Calcium (Ca)=1 x 40.08 g = 40.08 g

Mass of 1 mol of Carbon (C) =1x 12.01 g = 12.01 g

Mass of 3 mol of Oxygen (O) =3 x 16.00 g = 48.00 g _______ Mass of 1 mol of (CaCO3): = 100.09 g

20.56 gms H2O

18.016 gms H2O

Practice:

• A sample of water weighs 20.56 grams. How many water molecules are in a sample of water that weighs 20.56 grams?

=6.87 x 1023

molecules

6.022 x1023 H2O molecules

Percent Composition: part/whole

• Percent Composition is the percent by mass of an element in a compound or molecule

• % composition for each element is calculated as follows:

• For Example:

• Mass % is calculated by comparing the mass of a single element to the total mass (molar mass) of the compound

% Composition of an element =

Mass of the 1 mole of the element Mass of 1 mol of compound

% Composition =of Oxygen in H2O

Mass of the 1 mole of the Oxygen (O) Mass of 1 mol of Water (H2O)

• For example, lets consider methane (H2O):• We calculate the mass of each element present and the molar

mass of ethanol as follows:

1 mol O

16.00 grams O = 16.00 g O x 100 = 89.0 % O 18.02 g H2O

Mass of O = 1 mol O

1 mol H

1.01 grams H = 2.02 g H x 100 = 11.0% H 18.02 g H2O

Mass of H = 2 mol H

Mass % of O = 89.0 % O Mass % of H = 11.0 % H

• So we say the compound water H2O is 89.0 % Oxygen and 11.0% Hydrogen

100.00 % H2O

Percent Composition

What is Percent Composition?

• The ________by mass of an ________ in a compound or molecule

• Mass % is calculated by comparing the mass of a single element to the _____ mass (molar mass) of the compound

Answer BankTotal

PercentElementaddingmole

masses

_Part___ x 100 = % composition Whole

• Find the weight percent of Oxygen and Sulfur in SO2 Sulfur Dioxide:

1 mol S

32.07 grams S = 32.07g S32.07g S x 100 = 50.00 % S x 100 = 50.00 % S 64.07g SO2

Mass of S = 1 mol S

1 mol O

32.00 grams O = 32.00 g O x 100 = 50.00 % H 64.07g SO2

Mass of O = 2 mol O

Mass % ofMass % of S = 50.00 S = 50.00 % S% S Mass % ofMass % of OO = = 50.00 % O50.00 % O

• So we say the compound SO2is 50.00 % Sulfur 50.00 % Sulfur and 50.00% Oxygen

100.00 % SO2

Practice:

Summarize:

• Explain Molar mass in your own words:• Describe and explain how you would calculate %

compostion:• _____ _____is the mass in grams of one mole of a

compound which contains 6.02 x 1023 molcules • To calculate Molar mass, you would add up the ____

______for each element that make up a molecule• The percent by mass of an element in a compound is

called it’s ______ ______• Calculate the percentage of nitrogen in nitrogen dioxide

NO2

Topic: Molecular and Empirical Formulas


• To understand the difference between molecular and empirical formulas

• To learn how to calculate empirical formulas and molecular formulas given percent composition and mass

Unit: The Mole


sentences:

• Review: In iron (III) oxide, Fe2O3, how many iron atoms are present? How many oxygen atoms are present?

• How do think scientists determine the subscripts on chemical formulas such as Fe2O3? In other words how did they come up with the subscripts 2 and 3?

• Can hydrogen peroxide (H2O2) be simplified into a more basic chemical formula? If so, how?

Empirical Formulas

• The formula for a compound that is determined experimentally.

• A formula that represents the Smallest whole-number mole ratio of the different atoms in the compound.

• In other words, it is the simplest formula for a compound.

CHCH22OO CC22HH44OO22

CHCH33OOCHCH33OO

Empirical FormulaEmpirical FormulaMolecular Formula Molecular Formula A A formula formula based on the based on the actual numbers of atoms actual numbers of atoms of of

each type in the each type in the Empirical FormulaEmpirical Formula

A formula that gives the simplest whole-number A formula that gives the simplest whole-number ratioratio of the atoms of of the atoms of each element in a compound.each element in a compound.

Molecular Molecular FormulaFormula

Empirical FormulaEmpirical Formula

HH22OO22 HOHO

CHCH22OOCC66HH1212OO66

What is an Empirical Formula??

• A formula that represents the _____ whole-number ratio of the different atoms in the compound.

• In other words, it is the _____ formula for a compound.

• Example glucose

Answer BankSimplestNumbersoxygen

SmallestCH2O

Molecular Formula

Empirical Formula

C6H12O6??????

Practice: Practice: Write the empirical formula for NWrite the empirical formula for N22OO44

NN22OO44 NONO22

A formula that represents the A formula that represents the Using Using the smallest or lowest whole-the smallest or lowest whole-number ratio of number ratio of NN22OO4 4 we get….we get….

Steps for determining Empirical Formulas

1) Assume a 100 g sample when given percents. This makes 10.3 % = 10.3 g

2) Convert grams into moles for each element.

3) Divide all the moles by smallest number of moles to get the lowest whole number ratio.

4) Write the empirical formula.

A compound was found to contain 36.11 % calcium and 63.89 % chlorine by mass. What is its empirical formula? What assumption did you make?

36.11 % Ca = 36.11 g Ca

63.89 % Cl = 63.89 g Cl

= 0.9009 mol Ca

= 1.802 mol Cl

0.9009

0.9009

= 1 mol Ca

= 2 mol Cl

Therefore the empirical formula is CaCl2

Step 1 Assume a 100 g sample when given %

Step 2 Convert gramsinto moles for each element.

Step 3 Divide the all the moles by smallest number of moles

Step 4 Write the empirical formula

1 mol Ca

40.08g Ca

1 mol Cl

35.45g Ca

Problem: Write the Empirical Formula for a compound composed of: 72% iron and 27.6% oxygen by mass.

72.% Fe = 72.00 g Fe

27.6 % O = 27.60 g O

= 1.230 mol Fe

= 1.730 mol O

1.230

1.230

= 1 mol Fe

= 1.4 mol O

This gives us the empirical formula is Fe1O1.4

Step 1 Assume a 100 g sample when given %

Step 2 Convert gramsinto moles for each element.

Step 3 Divide all the moles by smallest number of moles

Step 4 Write the empirical formula

1 mol Fe

55.84g Fe

1 mol O

16.00g O

Since 1.4 atoms does not exist in nature, we need to multiply the compound by 2, so we get 2(Fe1O1.5) = Fe2O3

• A molecular formula is based on the actual number of atoms in each type of compound or molecule

• For example, consider glucose or sugar:

• The molecular formula tells us that it contains 2 Carbon atoms, 4 Hydrogen atoms, and 2 Oxygen atoms

Molecular Formulas

C2H4O2

A formula based on the actual _______ of atoms in each type of compound or molecule

Example: glucose C2H4O2 has 2 Carbon atoms, 4 Hydrogen atoms, and 2 ______ atoms

What is a Molecular Formula?

Answer BankSimplestNumbersoxygen

SmallestCH2O

Steps for determining Molecular Formulas

1. Find molar mass of the empirical formula

2. The molar mass of the molecule will be given.

3. Divide ___molar mass _of molecule___

molar mass of Empirical Formula 4. Multiply your answer from “step 3” by the

subscripts given in the empirical formula.

Practice: Find the molecular formula for a compound with an empirical formula of CH4N if the

molar mass of the molecule is 60.12 g/mole. 1. Find molar mass of the empirical formula

Molar mass of Empirical Formula – CH4NC = 1.0 x 12.0 = 12.0 g/moleH = 1.0 x 4 = 4.0 g/moleN = 1.0 x 14 = 14.0 g/mole

Molar mass of Empirical Formula = 30.0 g/mole

2. The molar mass of the molecule will be given.

Molar mass molecule (given) = 60.12 g/mole

Practice: Find the molecular formula for a compound with an empirical formula of CH4N if the

molar mass of the molecule is 60.12 g/mole.

4. Multiply your answer from the previous step by the subscripts given in the empirical formula.

2(CH4N) = C2H8N2

Therefore the Molecular Formula is C2H8N2

3. Divide __molar mass _of molecule___

molar mass of Empirical Formula ____Molar Mass _molecule____ = 60.12 g/mole = 2.00

Molar mass Empirical Formula 30.0 g/mole

Practice: Determine the molecular formula of a compound with an empirical formula of NH2 and Molecular or molar mass of 32.06

g/mole.

1. Find molar mass of the empirical formula Molar mass of Empirical Formula – NH2

N = 1 x14.0 = 14.0 g/moleH = 2.0 x 1 = 2.0 g/mole

Molar mass of Empirical Formula = 16.0 g/mole 2. The molar mass of the molecule will be given.

Molar mass molecule (given) = 32.06 g/mole

Practice: Determine the molecular formula of a compound with an empirical formula of NH2 and

Molecular or molar mass of 32.06 g/mole.

4. Multiply your answer from the previous step by the subscripts given in the empirical formula.

2(NH2) = N2H4

Therefore the Molecular Formula is N2H4

3. Divide ______molar mass _of molecule_______________

molar mass of Empirical Formula __ __ Molar Mass _molecule____ = 32.06 g/mole = 2.00

Molar mass Empirical Formula 16.0 g/mole

Summarize:• Compare and contrast the empirical

formula with the molecular formula:

• Can the empirical formula be the same as the molecular formula????

• What do you do if the subscript is not a whole number such as 1.4????

• Complete the table:Molecular Formula Empirical Formula

P4O6

C6H9

Topic: Moles in Chemical Reactions


• To learn how we go from moles of a reactant to moles of a product

• To learn how to calculate between moles of reactants to moles of products

Unit: The Mole

Quickwrite

Answer one of the questions below 1-2 sentences:

• A cake recipe requires 2 eggs, 2 cups of flour and 1 cup of sugar; you need to make 50 cakes for a friends birthday party, how many eggs, cups of flour and sugar should you buy????

• Using the recipe below:2 eggs + 2 cups of flour + 1 cup of sugar → 1 cake

To make one cake, a recipe requires how many cups of flour? How could you write this as a ratio?

Moles in Reactions• Chemistry is really all about reactions• Reactions involve the rearrangement of atoms• The calculation of the quantities of chemical

elements or compounds involved in chemical reactions is called Stoichiometry (our next unit)

• It is the coefficients in the balanced chemical equation that enables us to determine just how much product forms

• What we once called coefficients are now called moles!!!!

Moles in Reactions• To explore this idea, consider a non-

chemical analogy

• A particular cake recipe requires 2 eggs, 2 cups of flour, and 1 cup of sugar

• Or, you might represent this by:

2 eggs + 2 cups of flour + 1 cup of sugar → 1 cake

Moles in Reactions

• Now, lets say you need to make 50 cakes for a large party

• You will need ingredients to make 50 cakes• How do you figure out how much of each

ingredient you need to buy? • You could multiply the previous equation by 50:

50(2 eggs) +50 (2 cups of flour) + 50(1 cup of sugar) → 50 (1 cake)

100 eggs+ 100 cups of flour + 50 cups of sugar→ 50 cakes

Moles in Reactions

• Notice that the numbers 100:100:50 correspond to the ratio 2:2:1

• The equation for chemical reaction gives you the same type of information

• It indicates the relative numbers of reactant and product molecules required for the reaction to take place

Moles in Reactions

• To illustrate how this idea works, consider the reaction between gaseous carbon monoxide and hydrogen to produce liquid methanol

• The reactants and products are:• Unbalanced: CO(g) + H2(g) → CH3OH(l)

• Because atoms are just rearranged (not created or destroyed) in a chemical reaction, we must always balance the chemical equation

Moles in Reactions

Balanced: COCO(g)(g) ++ 2H2H2(g)2(g) →→ CHCH33OHOH(l)(l)

• It is important to recognize that the coefficients in a balanced equation give the relative number of molecules

• That is, we could multiply this balanced equation by any number and still have a balanced equation

• For example, we could multiply by 12,

• 12 [CO(g) + 2H2(g) → CH3OH(l) ] to obtain

• 12 CO(g) + 24H2(g) → 12 CH3OH(l)

Moles in Reactions

12 CO(g) + 24H2(g) → 12CH3OH(l)

• Is this still a balanced equation??? Yes!!!!

• Because 12 represents a dozen, we could even describe the reaction in terms of dozens:

1 dozen CO(g)+ 2 dozen H2(g) →1 dozen CH3OH(l)

• We could also multiply the original equation by a very large number, such as 6.022 x 1023

Moles in Reactions 6.022 x 1023 [ CO(g) + 2H2(g) → CH3OH(l) ]

• Which leads to the equation below: 6.022 x 1023 CO(g) + 2(6.022 x 1023 )H2(g) → 6.022 x 1023 CH3OH(l)

• Who wants to work with such a large number????• We also know this number 6.022 x 1023 is equal

to What????? 1 Mole!!!!!• Let’s replace 6.022 x 1023 with 1 Mole!!!!!

• Re-written in terms of moles we get: 1 mol CO(g) + 2 mol H2(g) → 1 mol CH3OH(l)

1 CO(g) + 2 H2(g) → 1 CH3OH(l)

Moles in Reactions• Think of the number of moles in a chemical reaction as

the amount of ingredients necessary for the reaction to take place

• Just like you need a certain amount ingredients to make a cake, you also need a certain amount of ingredients, or in this case molecules for a chemical reaction

• We can use an equation to predict the moles of products that a given number of moles of reactants will produce

• For example, consider the combination oxygen and hydrogen gas to synthesize water:

2H2(g) + O2(g) → 2H2O(l) • The equation tells us that 2 mol of H2 reacts and

requires 1 mol of O2 to create or produce 2 mol of H2O

Mole Ratio 2H2(g) + O2(g) → 2H2O(l) • Remember, the above equation tells us that 2 mol of H2 reacts

with 1 mol of O2 to create or yield 2 mol of H2O

• We can write this as a ratio:

• Or, we can use this ratio:

• This ratio is important because it allows us to go from moles of reactants to moles of products

• The mole ratio is defined as a conversion factor that allows us to go from moles of reactants to moles of products

moles of A on the reactant side moles B on the product

2 moles H2O or 2 moles H2__

2 moles H2 2 moles H2O

2 moles H2O or __1 moles O2___

1 moles O2 2 moles H2O

What is the mole ratio?

• A __________ factor that takes us from moles of ________ to moles of ________in a chemical reaction

Answer BankReactantsconversion

molesQuantitiesReactionsproducts

moles of A on the reactant side moles B on the product

Moles in Reactions• Remember, the equation tells us that 2 mol of H2 reacts and requires

1 mol of O2 to create or produce 2 mol of H2O

2H2(g) + O2(g) → 2H2O(l) • But we if double the reactants? That would give us:

22[2H2(g) + O2(g) → 2H2O(l)] or 4H2(g) + 2O2(g) → 4H2O(l)

• This equation tells us that 4 mol of H2 reacts and requires

2 mol of O2 to create or produce 4 mol of H2O• Or what if we Triple the reactant quantities? That would give us:

33[2H2(g) + O2(g) → 2H2O(l)] or 6H2(g) + 3O2(g) → 6H2O(l)

• This equation tells us that 6 mol of H2 reacts and requires

3 mol of O2 to create or produce 6 mol of H2O

Practice: First balance the equations below and determine the mole ratios

• NO2(g) + H2O (l) → HNO3(aq) + NO (g)

• Determine the mole ratio for NO2 and HNO3

• C6H6 (g) + H2 (g) → C6H12(g)

• Determine the mole ratio for H2 and C6H12

3 moles of NO2 or 2 moles of HNO3

2 moles of HNO3 3 moles of NO2

3 2

3

3 moles of H2 or 1 moles of C6H12

1 moles of C6H12 3 moles of H2

Mole Ratio• The mole ratio tells us how many moles of Reactant OR

Product will form in a chemical reaction• For example, consider the decomposition of water to

produce hydrogen gas and oxygen:

2H2O(l) → 2H2(g) + 1O2(g)

• Remember, equation tells us that 2 mole of H2O will produce 2 moles of H2 and 1 mole of O2

• But what if I have 4 moles of H2O????• Well, then I can produce 4 moles of H2 and 2

moles O2

• But what if I have 10 moles of H2O????• Well, then I can produce 10 moles of H2 and 5

moles O2

Mole Ratio• The mole ratio tells us how many moles of Reactant OR

Product will form in a chemical reaction• For example, consider the decomposition of water to form

hydrogen and oxygen gas :

2H2O(l) → 2H2(g) + 1O2(g)

• The equation tells us that 2 moles of H2O will produce 2 moles of H2 and 1 mole of O2

• In other words if I have 10 moles of H2O, then I can produce 10 moles of H2 and 5 moles of O2

Practice:• Using the equation: 2H2O(l) → 2H2(g)+ O2(g)

How many moles of O2 are produced by 4 moles of water???

• Using the equation: 2H2O(l) → 2H2(g)+ O2(g)

How many moles of O2 are produced by 1 mole of water???

Practice: (Record you solution in the answer bank)

• Using the equation below, what number of moles of O2 will be produced by the decomposition of 6.4 mol of water

2H2O(l) → 2H2(g) + O2(g)

1 mol O26.4 mol H2O

2 mol H2O

= 3.2 mol of O2


• Using the equation below, calculate the number of moles of NH3 that can be made from 1.3 mol H2

N2(g)+ 3H2(g) → 2NH3(g)

2 mol NH31.3 mol H2

3 mol H2

= 0.867 mol of NH3


• How many moles of N2 are needed to produce 8.5 moles of NH3?

• N2(g)+ 3H2(g) → 2NH3(g)

1 mol N28.5 mol NH3

2 mol NH3

= 4.25 mol of N2

Summarize:

If you have 2 moles of H2O

Then how many moles of…….

If you have 4 moles of H2O


If you have 1 mole of H2O


O2 would form?

____________

O2 would form?

____________

O2 would form?

____________

H2 would form?

____________

H2 would form?

____________

H2 would form?

____________

Consider the decomposition reaction of high quality H2O: 2H2O ---> 2H2(g) + O2(l)

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