Chapter 11

The Mole

Sec. 11.1

Measuring Matter

The Mole (mol):

SI base unit used to measure the amount of a substance.

1 mole = 6.02 x 10²³ representative particles such as atoms, molecules, formula units, electrons, or ions.

One mole of anything has 6.02 x 10²³ number of representative particles!

Avogadro’s Number:

6.02 x 10²³

602 000 000 000 000 000 000 000

Because Avogadro’s number is so enormous, it is used to count extremely small particles!

Converting moles to particles:

If one mole of copper contains 6.02 x 10²³ copper atoms, then how many copper atoms does two moles of copper contain?

Answer:

2 x (6.02 x 10²³) = ? Copper atoms

Complete practice problems p. 311

Converting particles to

moles:

Question:

If one mole of iron contains 6.02 x 10²³, how many moles of iron contain 2.50 x 1020 contain?

Answer:

2.50 x 10 ²º/ 6.02 x 10²³

= ?

Complete practice problems p. 312

Sec 11.2Mass and the

Mole

Would a dozen limes have the same mass as a dozen eggs?

No, because they have different size and composition.

Would the mass of one mole of copper differ from the mass of one mole of carbon?

Yes, because carbon atoms differ from copper atoms.

Molar Mass: The mass in grams of one mole of any pure substance.

The molar mass of any element is numerically equal to its atomic mass and has the units g/mol

What is the atomic mass of manganese?

54.94amu

What is the molar mass of manganese?

54.94 g/mol

What is the molar mass of:

Carbon= Zinc= Calcium=Iron=

Using molar mass:

Mole=mass (g)/molar mass

(g/mol)

Calculate the mass in grams of 0.0450 moles of chromium.

By rearranging the previous equation:

Mass=mole x molar mass

0.0450 mol Cr x 52.00g/mol Cr= 2.34g Cr

How many moles of calcium are there in 525g calcium?

mole= mass/molar mass

525g/40.08g/mol=

13.1mol Ca

Complete practice problems p.316

(11 and 12).

Conversion from mass to atoms:

How many atoms of gold (Au) are in a pure gold nugget having a mass of 25.0g?

First, find the number of moles.

mol= mass/molar mass

mol= 25.0g/196.97g/mol=

0.127mol Au

Next, find the number of particles in 0.127

mol of Au using Avogadro’s #

0.127 x 6.02 x 10²³

= 7.65 x 10²² atoms Au

complete practice problems p. 318 (13).

Converting atoms to mass:

A party balloon contains 5.50 x 10²² atoms of helium gas. What is the mass in grams of the helium?

First, find the number of moles:

Mol=5.5 x 10²²/6.02 x 10²³=0.0914 mol He

Next, find the mass in grams:

mass=mol x molar mass

0.0914mol x 4.00g/mol= 0.366g He

Complete practice problems p. 318 (14).

Draw the diagram of conversions on p. 319

Sec 11.3Moles of

Compounds

Freon CCl2F2

Freon contains one atom of carbon, two atoms of chlorine and two atoms of fluorine.

The ratio is 1:2:2

Suppose you have one mole of freon:

The you would have Avogadro’s number of freon molecules.

One mole of freon would have one mole of carbon atoms, two moles of chlorine atoms and two moles of fluorine atoms.

Determine the moles of aluminum ions

(Al+3)in 1.25 moles of Al2O3

1 mole Al2O3 contains 2 mol Al+3 ions, therefore 1.25 mol contains

1.25 x 2 = 2.5 mol Al+3

Complete practice

problems p. 321.

The Molar Mass of Compounds:

The sum of the masses of every particle that makes up the compound.

Determine the molar mass of K2CrO4

(2 x 39.10) + 52.00 + (4 x 16.00) = 194.20g/mol

Complete practice problems p. 322

Converting moles of a compound to mass:

First, find the molar mass.

Next, multiply molar mass x moles to get the mass.

What is the mass of 2.5 moles of allyl sulfide (C3H5)2S?

Molar mass = (6x12.01) + (10x1.008) + (1x32.07)= 114.21g/mol

Mass = 2.5 x 114.21= 286g

Complete practice problems p. 323.

Converting the mass of a compound to moles:First find the molar mass.

Next, find the mol by dividing mass/molar mass

Calculate the number of moles in calcium hydroxide Ca(OH)2 in 325g

Molar mass= (1x40.08) + (2x16.00) + (2x1.008)= 74.096g/mol

mol = 325/74.096= 4.39mol

Complete practice problems p. 324 (30).

Converting the mass of a compound to number of

particles:

First find the molar mass.

Next, divide mass by molar mass to find mol.

Then, multiply by Avogadro’s number to find the number of particles.

Use the ratio of each particle from the formula to find the number of each particle.

A sample of aluminum chloride AlCl3 has a mass of 35.6g.

a. How many Al+3 ?b. How many Cl1-?c. What is the mass in

grams of one formula unit of aluminum chloride?

Molar mass = (1x26.98) + (3x35.45) = 133.33g/mol

Mass/molar mass=mol

35.6/133.33= 0.267mol

Mol x Avogadro’s #= number of particles.

0.267 x 6.02 x 10²³= 1.61 x 10²³ formula units AlCl3

Number of Al+3= 1 x 1.61 x 10²³= 1.61 x 10²³ Al+3

Number of Cl1-= 3 x 1.61 x 10²³ = 4.83 x 10²³ Cl1-

To calculate the mass of one formula unit, divide the molar mass by Avogadro’s #

133.33/6.02 x 10²³ = 2.21 x 10 ²²־ g AlCl3/formula unit

Complete practice problems p. 326

Copy diagram on p. 327.

Sec 11.4

Empirical and Molecular Formulas

Percent Composition:

The percent by mass of each element in a compound.

If a 100g sample of a compound contains 55g of element x and 45g of element y,

Use the equation:

percent by mass= mass of element/mass of compound x100

For element x:55g/100gx100=55%

For element y:45g/100gx100=45%

Percent composition can be determined from the chemical formula.

H2OFirst find the molar mass of water:

(2x1.01)+(1x16)=18.02g/mol

Use the molar mass as the mass of the compound water.

Next, divide the molar mass of hydrogen (element) over the molar mass of water (compound) x100

2.02g/18.02gx100==11.2% HDo the same with oxygen:

16g/18.02gx100=88.80% O

Percent compositions add up to 100

Complete practice problems p. 331.

Empirical Formula for a Compound:

The formula with the smallest whole- number mole ratio of the elements.

The empirical formula for hydrogen peroxide is HO

The molecular formula is H2O2

In both formulas, the ratio of O to H is 1:1

Finding empirical formula:

First, find the number of moles for each element: mol=mass/molar mass

Then change ratios to whole numbers by dividing by the lowest number of moles.

The percent composition of an oxide of sulfur is 40.05% S and 59.95% O

100g of the oxide contains 40.05 g S and 59.95g O

Find the mol for each:

40.05g/32.07g/mol=

1.249 mol S59.95g/16g/mol=3.747 mol O

Change to whole numbers by dividing both by 1.249

1.249/1.249=1 mol3.747/1.249=3 molThe simplest whole number ratio of S to O is 1:3, the empirical formula is SO3

Complete practice problems p. 333

Molecular formula:The actual number of atoms of each element in one molecule or formula unit of a substance.

To find the molecular formula:

First, find the number of moles for each element.

Next, find the simplest ratio by dividing number of moles of each element by smallest #

The simplest ratio is the empirical formula.

Find the molar mass of the empirical formula.

Divide the experimentally determined molar mass by the molar mass of the empirical formula. This is n

Multiply the subscripts in the empirical formula by n to determine the molecular formula.

Molecular formula =

(empirical formula) n

Complete practice problems p. 335 and 337

Write the diagram on p. 337