Ch. 6 – Chemical QuantitiesThe Mole

• What is a mole? It is a unit for _________in chemistry.

• It is similar to a dozen, except instead of 12 things, it’s

602 billion trillion things… (602,000,000,000,000,000,000,000)

• ___________ (in scientific notation)

• This number is named in honor of AmedeoAmedeo _________ (1776 – (1776 –

1856)1856), who studied quantities of gases and discovered that equal

volumes of gases under the same conditions of temperature and

pressure contained the same number of particles.

Avogadro

6.02 x 10 23

counting

• A mole is a term for a certain ______________ of objects.

1 mole = 6.02 x 1023 objects

*Other words that represent quantities:

1 pair = __ objects; 1 dozen = __ objects

1 gross = ____ objects; 1 _____ = 24 cans of soda

• Since this value is so huge, it is used to measure very small objects like ___________ and _______________.

The Mole Concept

number

2 12

144 case

atoms molecules

Just How Big is a Mole?Just How Big is a Mole?

• It’s enough soft drink cans to cover the surface of the earth to a depth of over 200 miles.

• If you had Avogadro's number of un-popped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles.

• If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.

• 1 dozen cookies = ___ cookies• 1 mole of cookies = ___________ cookies

• 1 dozen cars = ___ cars• 1 mole of cars = __________ cars

• 1 dozen Al atoms = ___ Al atoms• 1 mole of Al atoms = __________ atoms

• Mole is abbreviated ______ .

The MoleThe Mole12

6.02 x 1023

126.02 x 1023

126.02 x 1023

mol

Gram-Formula Mass

• The # of grams that 6.02 x 1023 particles, (or ___ mole), weighs is called the gram-formula mass. The mass is found from the atomic masses of the elements on the ____________ __________.

• This quantity has other nicknames too!

________ ______or _________ ________ or Formula Weight

1

periodic table

Practice Problems:

Calculate the gram-formula mass of each compound.a) CaCO3 b) (NH4)2SO4

c) N2O5

Molar Mass Formula Mass

Ca = 40.1C = 12.03 O’s =3 x 16.0 = 48.0

Add them up!

100.1 g/mol2 N’s = 2 x 14.0 = 28.08 H’s = 8 x 1.0 = 8.0S = 32.14 O’s = 4 x 16.0 = 64.0

Add them up!

132.1 g/mol

2 N’s = 2 x 14.0 = 28.05 O’s = 5 x 16.0 = 80.0

Add them up!

108.0 g/mol

3) Convert 835 grams of SO3 to moles.

4) How many molecules of CH4 are there in 18 moles?

5) How many grams of helium are there in 5.6 x 1023 atoms of helium?

6) How many molecules are there in 3.7 grams of H2O?

80.1 g SO3

1 mole SO3835 g SO3

x = 10.4 moles of SO3

1 mole CH4

6.02 x 1023 molecules CH418 moles CH4 x = 108 x 1023 molecules CH4

4.0 grams He5.6 x 1023 atoms He

x 3.72 grams He

18.0 grams H2O3.7 grams H2O

x = 1.24 x 1023 molecules H2O

or 1.08 x 1025 molecules CH4

6.02 x 1023 atoms He=

6.02 x 1023 molecules H2O

Calculating Percent Composition by MassStep 1: Find the molar mass of the compound by adding the

individual masses of the elements together.

Step 2: Divide each of the individual masses of the elements by the molar mass of the compound.

Step 3: Convert the decimal to a % by multiplying by 100.

Practice Problems:

(1) Find the % composition of the elements in each compound.

a) Na3PO4 b) SnCl4

3 Na’s = 3 x 23.0 = 69.0

P = 31.0

4 O’s = 4 x 16.0 = 64.0+164

÷ 164

÷ 164

÷ 164

= 0.421 = 42.1%

= 0.189 = 18.9%

= 0.390 = 39.0%

Sn = 118.7

4 Cl’s = 4 x 35.5 = 142.0+

260.7

÷ 260.7

÷ 260.7

= 45.5%

= 54.5%

Elements in the Universe: % Composition by Mass

Earth’s Crust: % Composition by Mass

Entire Earth (Including Atmosphere): % Composition by Mass

Human Body: % Composition by Mass

Determining the Empirical Formula for a Compound

• The empirical formula for a compound is the simplest __________ number __________ of the atoms in the compound.

Examples: H2O is the empirical formula for water.

_______ is the empirical formula for glucose, C6H12O6.

Step 1: Divide the % composition data by the atomic mass of the element. This will give you a ratio of the # of atoms in the formula.

Step 2: Divide each of these answers by the smallest ratio.

Step 3: If there is still a decimal, multiply each answer by the denominator of the “freak”, (i.e. -- multiply all the ratios by the denominator of the ratio that is still a decimal.)

[1/2= 0.5 1/3≈ 0.33 2/3≈ 0.67 3/4= 0.75]

wholeratio

Helpful Rhyme: % to mass, mass to mole, divide by small, times ’til whole.

C1H2O1

Practice Problems: 1) An unknown compound is composed of 81.8% carbon and 18.2%

hydrogen. Determine the empirical formula for the compound.

2) An unknown compound is composed of 42.9% carbon and the rest of the compound is oxygen. Determine the empirical formula for the compound.

C = 81.8% = 81.8 gH = 18.2% = 18.2 g

(mass to moles) 81.8 g C ÷ 12.0 = 6.82 moles18.2 g H ÷ 1.0 = 18.2 moles

(÷ by small)C6.82H18.2

6.82 6.82

C1H2.67

(x ‘til whole)

x 3 x 3

C3H8

C = 42.9% = 42.9 gO = 57.1% = 57.1 g

(mass to moles) 42.9 g C ÷ 12.0 = 3.575 moles57.1 g O ÷ 16.0 = 3.569 moles

(÷ by small)C3.575O3.569

3.569 3.569

C1.0O1.0 CO

2) An unknown compound is composed of 42.9% carbon and the rest of the compound is oxygen. Determine the empirical formula for the compound.

C = 42.9% = 42.9 gO = 57.1% = 57.1 g

(mass to moles) 42.9 g C ÷ 12.0 = 3.575 moles C57.1 g O ÷ 16.0 = 3.569 moles O

(÷ by small)C3.575O3.569

3.569 3.569

C1.0O1.0 CO

To check your answer, you can simply find the % composition by mass of your formula…

% C = 12.0/28 = 42.9%

%O = 16.0/28 = 57.1%

CO = 28 g/mole

That matches up with the original problem!!

Determining the Molecular Formula for a Compound• The molecular formula for a compound is either the same as the

empirical formula ratio or it is a “_________ _________ of this ratio. It represents the true # of atoms in the molecule.

Examples: 1) H2O is the empirical & molecular formula for water. 2) CH2O is the empirical formula for sugar, ethanoic

acid, and methanol. The molecular formula for glucose is C6H12O6, (___times the empirical ratio!)

Step 1: Determine the empirical formula for the compound. (See the previous steps in the notes.)

Step 2: Calculate the empirical formula mass of the compound.

Step 3: Determine the “whole # multiple” by dividing the molecular formula mass (given in the problem) by the empirical

formula mass. Multiply each of the empirical ratios by this whole number.

whole # multiple

6

Practice Problems:

1) An unknown compound is composed of 58.8% carbon and 9.8% hydrogen and 31.4% oxygen. The molecular formula mass is 204 g. Determine the molecular formula for the compound.

C = 58.8% = 58.8 gH = 9.8% = 9.8 g

(mass to moles) 58.8 g C ÷ 12.0 = 4.9 moles C9.8 g H ÷ 1.0 = 9.8 moles H

(÷ by small)C4.9H9.8O1.96

1.96 1.96

C2.5H5O1

(x ‘til whole)

x 2 x 2 x 2

C5H10O2

O = 31.4% = 31.4 g 31.4 g O ÷ 16.0 = 1.96 moles O

1.96

Our formula mass = 102 g(Compare)

(204 ÷ 102 = 2)[C5H10O2 ]x 2 = C10H20O4

Practice Problems:

2) An unknown compound is composed of 40% carbon, 6.6% hydrogen, and 53.4% oxygen. Determine the molecular formula for the compound if the mass of one mole of the compound is 120 g.

C = 40% = 40 gH = 6.6% = 6.6 g

(mass to moles) 40 g C ÷ 12.0 = 3.3 moles C6.6 g H ÷ 1.0 = 6.6 moles H

(÷ by small)C3.3H6.6O3.3

3.3 3.3

C1H2O1

O = 53.4% = 53.4 g 53.4 g O ÷ 16.0 = 3.3 moles O

3.3

Our formula mass = 30 g(Compare)

(120 ÷ 30 = 4) [C1H2O1 ]x 4 = C4H8O4