The Mole

Chapter 11

By: Kathryn Johnson 2008

& BuggLady 2006

What is a mole?

Not that one….

A Mole is….. (p.310)

… an SI unit of measure used for large numbers of particles

Like the following: 1 dozen = 12 1 gross = 144 1 ream = 500 1 thousand = 1,000 1 billion = 1,000,000,000

A Mole (mol) =

= 602,213,670,000,000,000,000,000 particles

= 6.02x1023 particles (scientific notation)

= 6.02E23 particles (calculator notation)

Particles - atoms, molecules, formula units, ions,…

This number is also called ‘Avogadro's number’

Why is the number so large?

Because atoms are so small!

Why do we use the mole?

Because it is easier to talk of 4 moles than to say we have 2,410,000,000,000,000,000,000,000 or 2.41x1024 molecules of something

We use it the same way we talk of

miles instead of feet when we travel

Remember…Conversion Factors (Chapter 2)

They allow us to decide whether to divide or multiply by a value

Also called: Dimensional Analysis Unit Cancellation Factor Label

How to set one up…

Set up a fraction Place the units first

the unit you have on the bottom the other unit from the relationship on top

Scratch off (Cancel) units that are on both the top and the bottom

Repeat until you only have the units you want Then place the numbers from the relationships

into the fractions

Last…Do the Math

Multiply all the numbers on the top together

Divide by all the numbers on the bottom

Round off to correct significant figures

USE YOUR CALCULATOR!

Mole ↔ Molecules (p.311)

Set up a conversion factor!

Moles x Molecules = Molecules

Moles

Molecules x Moles = Moles

Molecules

1 mol = 6.02E23 molecules

6.02E23 molecules

1 mol

multiplication

1 mol .

6.02E23 molecules

division

Be Careful!

This will be easy at first and you will want to skip steps….

But we will be setting up lots of these fractions and you have to keep them straight

Now, lets add another step!

Molar Mass is

the mass (g)

of one mole (mol)

of a substance

Molar Mass of Elements (p.313)

The mass (g) of a mole (mol) of a particular element is equal to its Atomic Mass on the Periodic Table Use two decimal places

Carbon: 1 mol C = 12.01 g C

12.01 g/mol

Molar Mass of Compounds (p.322)

The mass (g) of a mole (mol) of a molecule is found by adding the atomic masses of all the elements, multiply by the subscripts

Water… H2O

2(1.01) + 16.00 = 18.02 g/mol or 1 mol H2O = 18.02 g

Be Careful….

Polyatomic ions with parenthesis complicate the math!

Ammonium Phosphate… (NH4)3PO4

= 3(14.01+4(1.01)) + 30.97 + 4(16.00)= 3(14.01+4.04) + 30.97 + 64.00

= 3(18.05) + 30.97 + 64.00 = 54.15 + 30.97 + 64.00 = 149.12 g/mol

Another way to do it…

Ammonium Phosphate… (NH4)3PO4

N: 3(14.01) = 42.03

H: 4∙3(1.01) = 12.12

P: (30.97) = 30.97

O: 4(16.00) = 64.00

Molar Mass = 149.12 g/mol

Mass ↔ Mole (p.314)

Now we have a new conversion factor!

Mass x Mole = Moles

Mass

Moles x Mass = Mass

Moles

1 mol = (Molar Mass) g

(Molar Mass) g

1 mol

multiplication

1 mol .

(Molar Mass) g

division

Mass ↔ Molecules (p.316)(the 2 step!)

Now we set up two conversion factors! You always have to go to Moles first!

Mass x Mole x Molecules = Molecules Mass Mole

Molecules x Mole x Mass = Mass Molecules Mole

Ratio of Moles of an Elementin 1 mole of a Compound (p. 320)

equal to the subscript beside the element

Water… H2O

2 mol H and 1 mol O .

1 mol H2O 1 mol H2O

It gets complicated…

Watch out for those polyatomic parenthesis!

Ammonium Phosphate… (NH4)3PO4

3 mol N 12 mol H .

1 mol (NH4)3PO4 1 mol (NH4)3PO4

and

1 mol P 4 mol O .

1 mol (NH4)3PO4 1 mol (NH4)3PO4

Mass Compound ↔ Element (the 3 step!)

Now we set up many conversion factors! Always go to moles!

Mass compound x Mole compound x Mole element… Mass compound Mole

compound

… x Mass element = Mass element

Mole element

Yes… it keeps going!

You can set up these conversion factors in infinite combinations to transform from one value to another

The only trick is… to go from one ‘thing’ (compound) to another ‘thing’ (element in that compound) you have to go through the mole ratio based on the subscript

... Don’t Skip Steps!

Moles of Element

Mass of Element Atoms of Element

Moles of Compound

Mass of Compound Molecules of Compound

(subscript) mol of Element = 1 mol of Compound

(atomic mass) g = 1 mol 1 mol = (6.02E23) atoms

(molar mass) g = 1 mol 1 mol = (6.02E23) molecules

And we still have it easy…

wait until we start dealing with

Chemical Reactions!(Chapter 12)

Percent Composition (p.328)

percent (%) by mass of each element in a compound Review: Name → Formula (Chapter 8&9) Watch out for those polyatomic parenthesis!

% Element = (subscript)x(Atomic Mass)x100Molar Mass Compound

All the % should add up to 100!

Example: Water…. H2O

% H = 2(1.01) g H x 100 = 11.2%

18.02 g H2O

% O = 16.00 g O x 100 = 88.8%

18.02 g H2O

Check your work:

11.2% + 88.8% = 100%

Another way to do it…

Ammonium Phosphate… (NH4)3PO4

N: 3 (14.01) = 42.03 → 28.2%

H: 12 (1.01) = 12.12 → 8.1%

P: (30.97) = 30.97 → 20.8%

O: 4 (16.00) = 64.00 → 42.9%

Molar Mass = 149.12 g/mol

Practice

Determine the percent by mass of each element in calcium chloride.

Calculate the percent composition of sodium sulfate.

Which has the larger percent by mass of sulfur: H2SO4 or H2S2O8?

Empirical Formula (p.331)

The smallest whole-number mole ratio of the elements in a compound

Derived from %comp of the elements Turn Percent Composition (%) into Mass (g) Convert Mass→Mole with the Molar Mass Divide by the smallest value to get the whole

number ratios

Like a %comp in reverse

Example: 59.95% O & 40.05% S

0: 59.95 g x 1 mol = 3.747 mol = 3

16.00 g 1.249 mol S: 40.05 g x 1 mol = 1.249 mol = 1

32.07 g 1.249 mol

Therefore: SO3… Sulfur trioxide!

If the ratios are still decimals, multiply them by a small factor to make them whole (EX: 0.5 = ½, multiply by 2) (EX: 0.33 = 1/3, multiply by 3) (EX: 0.25 =1/4, multiply by 4)

Example: 48.64% C, 8.16% H, & 43.20% O

0: 43.20 g x 1 mol = 2.700 mol = 1 x 2 = 2

16.00 g 2.700 mol C: 48.64 g x 1 mol = 4.050 mol = 1.5 x 2 = 3

12.01 g 2.700 mol H: 8.16 g x 1 mol = 8.10 mol = 3 x 2 = 6

1.01 g 2.700 mol Therefore: C3H6O2

Two or more substances with different properties can have the same % composition and

empirical formula!

The simplest ratio in the empirical formula does not

indicate the actual number of elements in the compound

Molecular Formula (p.333)

• The actual numbers of atoms in a compound• Identifies the compound

• A multiple of the empirical formula• Molecular Formula = (Empirical Formula) n

n = Measured molar mass of compound_ _. Calculated molar mass of empirical formula

Example: 46.68% N & 53.32% OMeasured Molar Mass = 60.01 g/mol

0: 53.32 g x 1 mol = 3.333 mol = 1

16.00 g 3.332 mol N: 46.68 g x 1 mol = 3.332 mol = 1

14.01 g 3.332 mol Empirical Formula: NO … nitrogen monoxide

Molar Mass = 14.01 + 16.00 = 30.01 g/mol But, 60.01 g/mol = 2 therefore…

30.01 g/mol Molecular Formula: N2O2 … dinitrogen dioxide!

Review

Mol ↔ Molecules/Atoms (Avogadro's #) Mol ↔ Mass (Molar Mass) Mass ↔ Molecules/Atoms (The 2 step) Element ↔ Compound (Subscript, 3 step) % Composition Empirical/Molecular Formula

Practice…

Practice…

Practice…

Work Smart, Not Hard!

You have to be fast enough

to finish the test!

Mole Ratios

Coefficients indicate the number of moles Mole Ratios – the ratio between the numbers of moles of any

2 substances in a balanced equation

Ex: 2 Al + 3 Br2 2 AlBr3

All possible mole ratios: 2 mole Al : 3 mole Br2 2 mole Al : 2 mole AlBr3 3 mole Br2 : 2 mole Al 3 mole Br2 : 2 mole AlBr3 2 mole AlBr3 : 2 mole Al 2 mole AlBr3 : 3 mole Br2

Interpreting Mole Conversions

4 Fe + 3 O2 2 Fe2O3

You can interpret any balanced equation in terms of moles, mass, and molecules

Molecules: 4 atoms of iron react with 6 atoms of oxygen to produce 2 molecules of iron (III) oxide

Moles: 4 moles of iron react with 3 moles of oxygen to produce 2 moles of iron (III) oxide

Mass: convert to grams first

Practice Mole Ratios

For the following equations:Balance the equations

Interpret each in terms of moles, molecules, and mass

Determine all possible mole ratios 1. CaC2 + N2 CaNCN + C 2. NH3 + CuO Cu + H2O + N2

3. MnO2 + HCl MnCl2 + Cl2 + H2O 4. P4O10 + C P4 + CO 5. ZrI4 Zr + I2 6. PbS + O2 PbO + SO2

Mole to Mole Conversions Ex: Determine the number of moles of H2 produced when 0.040 moles

of K is used. Steps for solving: Step 1. Write the balanced equation:

2 K + 2 H2O 2 KOH + H2 Step 2. Identify the known (given) substances and the unknown

substances:Known: .040 moles KUnknown: moles of H2

Step 3. Determine the mole ratio of unknown substance to known substance:

1 mole H2 (use coefficients from balanced equation)2 mole K

Step 4. Convert using the known substance and the mole ratio:0.040 moles K x 1 mole H2 = 0.02 mole H2 produced

2 mole K

Practice

Handout: Mole to Mole Stoichiometry Problems

Mole to Mass Conversions Ex: Determine the mass of NaCl produced when 1.25 moles of Cl2 reacts with

sodium Steps: 1. Write the balanced equation:

2 Na + Cl2 2 NaCl 2. Identify the known (given) substances and the unknown substances:

Known: 1.25 moles Cl2Unknown: mass of NaCl

3. Determine the mole ratio of unknown substance to known substance:2 moles NaCl (use coefficients from balanced equation)1 mole Cl2

4. Convert using the known substance and the mole ratio:1.25 moles Cl2 x 2 moles NaCl = 2.5 moles NaCl

1 mole Cl2 5. Convert moles of unknown to grams of unknown: (use molar mass)

2.5 mole NaCl x 58.44 g = 146 grams NaCl 1 mole

Example: Mole to Mass

Reactants: 4 mole Fe x 55.85 g Fe = 223.4 g Fe 1 mole Fe

3 mole O2 x 32.00 g O2 = 96.0 g O2

1 mole O2

Product: 2 mole Fe2O3 x 158.7 g Fe2O3 = 319.4 gFe2O3

1 mole Fe2O3

Example, cont.

** Law of Conservation of Mass – the sum of the reactants should ALWAYS equal the sum of the product – check your work!!!

223.4 grams of Fe react with 96.0 grams of O2 to produce 319.4 grams of Iron (III) oxide

Mass to Mass Conversions Ex: Determine the mass of H2O produced from the decomposition of 25.0

grams of NH4NO3 Steps: 1. Write the balanced equation:

NH4NO3 N2O + 2H2O 2. Identify the known (given) substances and the unknown substances:

Known: 25.0 grams of NH4NO3

Unknown: mass of H2O 3. Convert grams of known to moles of known:

25.0 g NH4NO3 x 1 mole NH4NO3 = 0.312 mole NH4NO3 80.04 g

4. Determine the mole ratio of unknown substance to known substance:2 moles H2O (use coefficients from balanced equation)1 mole NH4NO3

5. Convert using the known substance and the mole ratio:0.312 mole NH4NO3 x 2 moles H2O = 0.624 mole H2O

1 mole NH4NO3 6. Convert moles of unknown to grams of unknown: (use molar mass)

0.624 mole H2O x 18.0 g = 11.2 grams H2O 1 mole

Hydrates (p.338)

Compound that has a specific number of water molecules (H2O) bound to its atoms

Solids in which water molecules are trapped Like a sponge, different amounts of water can

be trapped

Formulas and Names

1 formula unit of compound with a dot and then the attached water moleculesEx: Na2CO3∙10H2O

Name the molecule Use the prefixes in front of the

word ‘hydrate’Ex: sodium carbonate decahydrate

1 Mono-

2 Di-

3 Tri-

4 Tetra-

5 Penta-

6 Hexa-

7 Hepta-

8 Octa-

9 Nona-

10 Deca-

Anhydrous

“without water” Water is removed from the hydrate by

heating Desiccants are used to absorb excess

moisture in packages Calcium chloride Calcium sulfate

How much water attached?

How many moles of water attached to one mole of the molecule (BaCl2∙xH2O)

compare the mass before and after heating Mass hydrate – Mass anhydrous = Mass water

Convert the Mass water and Mass anhydrous to moles using the molar mass of each

Calculate the ratio x = mol H2O / mol BaCl2