The Mole Chapter 11Chemistry

RiverDell High SchoolMs. C. Militano

•What is a mole in chemistry?

•What conversion factors are associated with the mole?

•Types of conversions involving mole equalities

I. What is a Mole? A. SI base unit that measures amount of a

substance

B. 1 mol = Avogadro Number of particles (particles can be atoms, molecules or ions) 6.02 x 1023 is Avogadro's Number

C. Molar Mass – mass of one mole of atoms

of an element

1. ex. C = 12.0amu N = 14.0 amu

D. Mole Equalities

- 1 mole = molar mass

- 1 mole = 6.02 x 1023 particles

II. Mole Conversions [mass-mole-atoms] A. Type of Problem Equality

1. MOLES MASS

2. MASS MOLES 1 mole= molar mass (g)

3. MOLES ATOMS

4. ATOMS MOLES 1 mole = 6.02 X 1023 atoms

5. MASS ATOMS 1 mole = 6.02 x 1023 atoms

6. ATOMS MASS 1 mole = molar mass (g)

Making Conversions – use quantities in the Atom and Mass box as conversion factors

6.02x1023 molar mass ATOMS MOLE MASS(g)

1 mole 1 mole

B. Using Factor Label to Make Mole Conversions

1 mole molar mass

6.02 x 1023 1 mole

PARTICLES <----> MOLES <----> MASS

6.02 x 1023 1 mole

1 mole molar mass

C. Solving Mole Problems

EXAMPLES

1. 1.00 mole of He = 4.00 g.

2. 2.00 mole of He = _____g

2.00 mol He X 4.00g He =

1 1 mole He

8.00 g He

EXAMPLES

3. 1.00 mole He = 6.02 X 1023 atoms

4. 2.00 mole He = ________atoms He

2.00 mole He x 6.02 x 1023 atoms He = 12.04 x 1023

1 1 mole He 1.20 x 1024 atoms He

5. 16.00g He = _____ moles He

16.0 g He x 1 mole He = 1 4.00 g He 4.00g He

EXAMPLES

6. 3.01 X 1023 atoms He = _____ moles

3.01 x 1023 atoms He x 1 mole He =

1 6.02 x 1023 atoms He

.500 mol He

7. 8.00g He =______atoms He

8.00 g He x 1 mole He x 6.02 x 1023 atoms He =

1 4.00g He 1 mole He

12.04 x 1023 atoms He = 1.20 x 1024 atoms He

Sample Problems – More Practice

1. Moles to mass.

Find the mass of 3.50 moles of carbon.

2. Mass to Moles

How many moles of carbon are contained in

60.0 g of carbon?

3. Moles to Atoms

How many atoms of carbon are found in 4.00 moles of carbon?

Sample Problems More Practice

4. Atoms to Moles How many moles of carbon are represented by 1.806 x 1024 atoms of carbon?

5. Mass to Atoms How many carbon atoms are found in 36.0g of carbon?

6. Atoms to Mass What is the mass of 1.204 x 1024 atoms of

carbon?

Answers to Sample Problems

1. 42.0g C

2. 5.00 mol C

3. 24.1 x 1023 = 2.41 x 1024 atoms C

4. .300 x 10-1 = 3.00 mol C

5. 18.06 x 1023 = 1.81 x 1024 atoms C

6. 24.0 g C

More Sample Problems2.00 moles of Cu = atoms of Cu

60.0 grams of C = moles of C

3.00 x 1023 atoms He = moles of He

2.50 moles Al = grams of Al

28.0 grams N = atoms of N

1.80 x 1023 atoms Mg = grams of Mg

Answers to More Sample Problems

1. 12.04 x 1023 = 1.20 x 1024 atoms Cu

2. 5.00 mol C

3. .498 mol He or 4.98 x 10-1 mol He

4. 67.5 g Al

5. 12.04 x 1023 = 1.20 x 1024 atoms N

6. 7.27 g Mg

Compounds and Diatomic Molecules

1. Convert 4.00 mol NaOH to grams.

a. Na = 23.0 O = 16.0 H = 1.00

formula mass = 40.0

b. 4.00mol NaOH x 40.0g NaOH = 160.g NaOH

1 1 mol NaOH

2. How many molecules are found in 2.00 mol of H2SO4? (sulfuric acid)

2.00 mol H2SO4 x 6.02 x 1023 molecules H2SO4

1 1 mol H2SO4

12.04 x 1023 =

1.20 x 1024 molecules H2SO2

3. How many moles are found in 64.0 g

of oxygen gas (O2)?

64.0g O2 x 1mol O2 = 2.00 mol O2

1 32.0g O2

4. How many formula units are found in 117.0g of sodium chloride (NaCl)?

117.0g NaCl x 1 mol NaCl x 6.02x1023 units NaCl

1 58.5 g NaCl 1 mol NaCl

12.04 x 1023 = 1.20 x 1024 formula units NaCl

III. Percent Composition

A. Procedure

1. Determine total mass for each element

2. Determine the molar mass (formula

mass)

3. Divide mass of each element by the

molar mass (formula mass)

4. Multiply by 100%

B. Problem Solving

1. Determine the percent composition of each element in carbon dioxide (CO2)

C = 12.0 2O = 2 x 16.0 = 32.0

Sum = 44.0 (formula mass)

Carbon 12.0/44.0 = .273 = 27.3%

Oxygen 32.0/44.0 = .727 72.7%

B. Problem Solving

2. Determine the percent composition of each element in calcium hydroxide Ca(OH)2

Ca = 40.1

O = 16.0 x 2 = 32.0

H = 1.00 x 2 = 2.00 sum is 74.1(molar mass)

% Ca = 40.1/74.1 = 54.1%

% O = 32.0/74.1 = 43.2%

% H = 2.00/74.1 = 2.70%

B. Problem Solving

3. Determine the precent composition of each element in TNT (trinotrotoluene) C7H5(NO2)3

7C = 7(12.0) = 84.05H = 5(1.00) = 5.00 3N = 3(14.0) = 42.06O = 6(16.0) = 96.0 sum = 227 (molar mass)

% C = 84.0/227 = 37.0% % H = 5.00/227 = 2.20%% N = 42.0/227 = 18.5%% O = 96.0/227 = 42.3%

IV. Hydrates

A. Definitions

1. hydrate – compound that has a specific number of water molecules in its crystal (solid state)

2. water of hydration – water molecules that are part of the crystal (solid state)

3. anhydride – compound without water of

hydration

B. Naming Hydrates

1. CaCl2 2H2O calcium chloride dihydrate

2. NaC2H3O2 3H2O sodium acetate trihydrate

3. CuSO4 5H2O copper sulfate pentahydrate

4. MgSO4 7H2O magnesium sulfate heptahydrate

5. Na2CO3 10H2O sodium carbonate decahydrate

C. Heating a Hydrate

∆

Hydrate anhydride + water ∆

Na2CO3 10H2O Na2CO3 + 10 H2O

∆

MgSO4 7H2O MgSO4 + 7 H2O

∆

Na2CO3 10H2O Na2CO3 + 10H2O

D. Problem Solving - % Water in Hydrates

1. Procedure

a. determine formula mass of the compound

b. divide mass of only water molecules

by the formula mass of the compound

c. multiply answer by 100%

Examples – Find % water in the hydrate2a. CaCl2 2 H2O Ca = 40.1 Cl = 35.5 (2) = 71.0 H = 1.00(4) = 4.00 O = 16.0(2) = 32.0 Total mass is 147.1

Mass of water 2(18.0) = 36.0

% H2O mass water = 36 = .2447 = mass of compound 147.1 24.5%

2b. Find the % water in the hydrate

MgSO4 7H2O

Mg = 24.3

S = 32.1

O(4) = 64.0

H2O(7) = 18(7) = 126 sum = 246.4

% water = 126/246.4 = 51.1%

V. Empirical Formulas

A. lowest whole number ratio of subscriptsB. How to Determine the Empirical Formula 1. if given % composition write % as a number of grams without the % sign 2. divide # grams by molar mass to get the number of moles 3. Divide # moles for each element by smallest 4. Round to nearest whole number when possible 5. Multiply by 2,3,or 4 to get whole numbers- if necessary 6. Use resulting numbers as subscripts in the formula

C. Sample Problems1. Find the empirical formula of a compoundcontaining 19.55 g of potassium (39.08) and2.00g of oxygen (16.00).

Determine # moles of each element# moles Oxygen – 2.00/ 16.00 = .125# moles Potassium - 19.55/ 39.08 = .500

Divide # moles of each by the smallest .125/.125 = 1 .500/.125 = 4.00

Formula is K4O

2. Find the empirical formula of a compound

containing 5.41g Fe (55.85), 4.64g Ti (47.88), and

4.65g O (16.00).

# moles Iron(Fe) - 5.41/55.85 = .0968

# moles Titanium(Ti) - 4.64/47.88 = .0969

# moles Oxygen(O) - 4.65/16.00 = .291

Fe = .0968/.0968 = 1.00 Ti = .0969/.0968 = 1.00

O = .291/.0968 = 3.01

Formula is FeTiO3

3. Find the empirical formula of methyl acetate which contains 48.64% C, 8.16% H, 43.20% O.

# moles C = 48.64/12.00 = 4.053

# moles H = 8.16/1.00 = 8.16

# moles O = 43.20/16.00 = 2.700

Oxygen – 2.700/2.700 = 1.000 x 2 = 2.000

Hydrogen – 8.16/2.700 = 3.02 x 2 = 6.04

Carbon - 4.053/2.700 = 1.500 x 2 = 3.000

Formula is C3H6O2

VI. Determining Molecular Formulas

A. Procedure

1. Determine empirical formula

2. Calculate empirical formula mass

3. Divide actual formula mass to get “X”

empirical formula mass

4. Multiply subscripts in the empirical

formula by “X”

B. Determine Molecular Formula

1.Determine the molecular formula of succinic acid.(Molar mass is 118.1g) C = 40.68% H = 5.05%, O =54.24%

Carbon = 40.68/12.0 = 3.39Hydrogen = 5.08/1.00 = 5.08 # of molesOxygen = 54.24/16.0 = 3.39

C = 3.39/3.39 = 1.0 O = 3.39/3.39 = 1 mole ratio H = 5.08 / 3.39 = 1.49 Multiply by 2 to get the smallest whole number ratio

Empirical formula = C2H3O2

Calculate empirical formula mass

Empirical formula mass =

2(12.00) + 3(1.00) + 2(16.00) = 59.00g

Divide formula mass by empirical formula mass

118.1/59.0 = 2.00 = “X”

Multiply each subscript in the empirical formula by X

The molecular formula is C4H6O4

2. Determine the molecular formula for styrene.

C = 92.25%, H = 7.75%. (Molar mass – 104.00g)

92.25/ 12.0 = 7.69 mol C 7.75/1.00 = 7.75 mol H

7.69/7.69 = 1.00 7.75/7.69 = 1.01

Empirical formula is CH

Empirical formula mass is 12.0 + 1.00 = 13.0

104.00/13.0 = 8 ( “X”)

Formula is C8H8

3. Determine the molecular formula for ibuprofen.

C=75.7%, H=8.80%, O=15.5%. Molar Mass-206.00g

Determine the # of moles

75.7/12.0 = 6.31 mol C 8.80/ 1.00 = 8.80 mol H

15.5/16.0 = .969 mol O

Divide by the smallest number of moles

.969/.969 = 1.00 6.31/.969 = 6.51 8.80/.969 = 9.08 Multiply by 2 to get whole number ratio

Empirical Formula is C13H18O

Empirical Formula Mass 13(12.0)+18(1.00)+2(16.0) = 206.0

Divide formula mass by empirical formula mass to get X

206/206 = 1.00 (“X”)

Molecular Formula is C13H18O2

4. Given the molecular formula, determine the empirical formula for the following.

a. C6H6 (benzene)

b. C2H6 (ethane)

c. C10H8 (naphthalene)

d. C8H10N4O2 (caffeine)

e. C14H18N2O5 (aspartame)

Answers a) CH b) CH3 c) C5H4

d) C4H5N2O e) C14H18N2O5

5. Determine the molecular formula if the empiricalformula is CH and the molar mass is 78.00g.Empirical formula mass is 12.0 + 1.00 = 13.O78.00/13.0 = 6.00 (“X”)

The molecular formula is C6H6

6. Determine the molecular formula for butane if

the empirical formula is C2H5 and the molar massis 58.00g. 2(12.0) + 5(1.00) = 29.0 (empirical formula mass)58.0/29.0 = 2.00 (“X”)

The molecular formula is C4H10