The Mole

Molar mass, percent composition, empirical and molecular formulas

How many atoms are in here? What about weight?

.

Atomic Mass

A single atom has a very small mass. on the order of 10-23 grams per atom 1 atom has a mass of about

0.000 000 000 000 000 000 000 01g Because this mass is so small, we use a unit called

amu to describe the mass of a single atom. 1 atom of carbon has a mass of 12.011 amu 1 atom of hydrogen has a mass of 1.00794 amu 1 atom of gold has a mass of 196.967 amu

Formula Mass

Formula Mass – the mass, in amu, of one formula unit of a compound.

To calculate the formula mass of a compound; Count and record the number of each atom

in the formula. Multiply the number of atoms by the mass of

that atom from the periodic table. Add these products together.

Definitions:

Atom: The smallest particle of an element (ex: He).

Molecule: The smallest particle of a compound (ex: H2O)

Atomic Weight: The decimal number on the periodic chart (Units: amu)

Formula Mass: add up the total atomic weights of the elements in a compound (Units: amu).

Formula Mass

Calculate the formula mass of Ca

1. Ca = 1(Ca)2. Ca = 1(40.08) = 40.083. 40.08 amu4. Units = amu

Ca

Cl

Cl

Formula Mass

Calculate the formula mass of CaCl2

1. CaCl2 = 1(Ca) + 2(Cl)

2. CaCl2 = 1(40.08) + 2(35.45)= 110.981. 110.982. Units = amu

Ca

The Mole

The mass of a single atom is too small a quantity for use in lab.

We need to come up with a way to take atomic masses from the periodic table and turn them into a mass that is useable in lab.

In 1811, Amedeo Avogadro determined the number of atoms in one mole of a substance.

The Mole

The mole establishes a relationship between the atomic mass unit (on periodic table) and the gram (used in lab).

Mole (mol) - The amount of a pure substance that contains 6.02 x 1023 particles of that substance.

The mole is used to describe a huge amount of any extremely small particle. A mole of gold, a mole of salt and a mole of water each contain 6.02 x 1023 individual units.

The Mole

One mole of oxygen, one mole of salt and one mole of water.

Each sample contains one mole, 6.02 x 1023, particles.

Do you think that one mole of gold atoms is going to weigh the same as one mole of aluminum atoms? Why?

It doesn’t make much sense to weigh atoms in amu, because the units are so small……

• If you express atomic mass (amu) in GRAMS instead of amu (like we did yesterday), than you have one MOLE of atoms.

Molar Mass

A molar mass is calculated the same way that you calculated formula mass (HUP, TWO, THREE, FOUR).

The only difference is the unit that will be placed on the final answer because you are calculating the mass of one mole of a substance.

If the formula mass of CaCl2 is 110.98 amu, then the molar mass of CaCl2 is 110.98 grams.

Formula vs. Molecular Mass

The Mole

There are 6.02 x 1023 carbon atoms in 12 grams, one mole, of carbon.

There are 6.02 x 1023 gold atoms in 197 grams, one mole, of gold.

There are 6.02 x 1023 formula units of calcium chloride in 110.98 grams, one mole, of CaCl2.

So, how do we use the mole in Chemistry?

We use the mole to convert between the mass, volume and number of particles in a substance.

We can make conversions from:Moles to particles and particles to moles.Moles to grams and grams to moles.Moles to volume of a gas and volume of

a gas to moles.Any combination of the above categories!

Molar Conversions

number ofrepresentative

particles

mass

1 mol22.4L

22.4L1 mol

6.02 x 10 particles1 mol

23

6.02 x 10 particles231 mol

1 molmolar mass

molar mass1 mol

moles

volumeat STP

Converting from Moles to Particles.

How many particles of Sodium Chloride are there in 2.50 moles of Sodium Chloride (NaCl)?

2.50 mol NaCl x 6.02 x 1023part. = 1.51 x 1024

1 mole part. NaCl

Converting from Moles to Particles

How many particles of gold are there in 4.25 moles of Au?

4.25 mol Au x 6.02 x 1023 part.= 2.56x1024 part 1 mole Au

Converting from Particles to Moles.

How many moles of gold are there in 1.5 x 1024

atoms of gold?

1.5 x 1024 atoms Au x 1 mole = 2.5 moles Au

6.02 x 1023 atoms

Converting from Particles to Moles

How many moles of calcium chloride are there in 3.6 x 1024 particles of CaCl2?

3.6 x 1024 units x 1 mole CaCl2 = 6.0 mol CaCl26.02 x 1023 part.

Converting from Mass to Moles

How many moles of gold are there in 25.0 grams of gold?

25.0 g Au x 1 mole Au = 0.127 mol Au

196.97 g Au

Converting from Mass to Moles

How many moles of calcium chloride are there in 75.5 grams of CaCl2?

75.5 g CaCl2 x 1 mole CaCl2 = 0.680mol CaCl2

110.98 g CaCl2

Converting from Moles to Mass

How many grams of Carbon are there in 4.2 moles of Carbon?

4.2 mole C x 12.01 g C = 50. g C

1 mole C

Converting from Moles to Mass

How many grams of water are there in 2.75 moles of water?

The molar mass of water is: 16.00 + 1.01 + 1.01 = 18.02 g/mol

2.75 moles H2O x 18.02 g H2O = 49.6 g H2O

1 mole H2O

Volume of Gases

One mole of gas (6.02 x 1023 particles of gas) occupies a volume of 22.4 liters at standard temperature and pressure.

This is true if the gas is monatomic like Helium (He), diatomic like Oxygen (O2) or a compound like Carbon dioxide (CO2).

Converting from Volume to Moles of a gas

How many moles of oxygen are there in 10.0L of oxygen?

10.0L O2 x 1 mole O2 = 0.446 mole O2

22.4 L O2

Converting from Volume to Moles of a gas

How many moles of Carbon dioxide (CO2) are there in 2.50 L of CO2 gas?

2.50 L CO2 x 1 mole CO2 = 0.112 mole CO2

22.4 L CO2

Converting from Moles to Volume of a gas

How many liters of Helium are there in 3.6 moles of Helium?

3.6 moles He x 22.4 L He = 81 L He

1 mole He

Converting from Moles to Volume of a gas

What volume will 8.24 moles of CO2 occupy?

8.24 mol CO2 x 22.4 L CO2 = 185 L CO2

1 mole CO2

Multi-Step Conversions

If you need to convert from: Mass to particles or particles to mass, Mass to volume or volume to mass, Volume to particles or particles to volume,

then you need to perform a 2-step conversion.

Notice the mole is not present in the starting or ending quantities. You will ALWAYS use the mole when converting between these units!

Multi-Step Conversions

How many formula units of calcium chloride are there in 10.0 grams of CaCl2?

10.0 g CaCl2 x 1 mole CaCl2 x 6.02 x 1023 units =

110.98 g CaCl2 1 mole CaCl2

5.42 x 1022 formula units of CaCl2

Multi-Step Conversions

What volume would 2.25 grams of hydrogen gas (H2) occupy at standard temperature and pressure?

2.25 g H2 x 1 mole H2 x 22.4 L = 25.0 L H2

2.02 g H2 1 mole

Multi-Step Conversions

How many atoms of Argon gas are there in a typical light bulb (0.20 liters) at standard temperature and pressure?

0.20 L Ar x 1 mole x 6.02 x 1023 atoms = 5.4 x 1021

22.4 L 1 mole atoms Ar

Percent Composition

The Percent Composition – the percent by mass of each element in a compound.

To calculate %Comp:

mass of element x 100 = % by mass

mass of compound

Percent Composition

Determine the percent by mass of each element in calcium chloride (CaCl2).

Ca 1 x 40.08 = 40.08 40.08 x 100 = 36.11% Ca

110.98

Cl 2 x 35.45 = 70.90 70.90 x 100 = 63.89% Cl

110.98

40.08 + 70.90 =110.98

(molar mass)

Percent Composition

Calculate the percent by mass of each element in Calcium sulfate, CaSO4.

Ca 1 x 40.08 = 40.08 40.08 x 100 = 29.44% Ca 136.15

S 1 x 32.07 = 32.07 32.07 x 100 = 23.55% S 136.15

O 4 x 16.00 = 64.00 64.00 x 100 = 47.01% O

136.15

40.08 + 32.07 + 64.00 = 136.15 molar mass

Percent Composition

You can determine the %Comp of a substance experimentally.Measure the mass of a sampleDecompose the sample (usually by heating) to

separate the component substances.Measure the mass of the substance that

remains.Calculate %Comp as before

Empirical Formula

Empirical Formula – the smallest whole number mole ratio of elements in a compound.

The empirical formula for a class of molecules is often the same for each sample.Examples: C5H10O5 = CH2O

C6H12O6 = CH2O

C11H22O11 = CH2O

Calculating Empirical Formula

1) Assume you are working with a 100.0g sample so the mass of each element will be the same as the percent of that element (for simplicity).

2) Convert each mass in grams to moles using the molar mass of the element.

3) Find the whole number ratio of these calculated amounts by dividing each mole value by the smallest number of moles calculated.

4) Round to whole numbers and use the ratio to determine the empirical formula.

Calculating Empirical Formula

A compound was found to contain 29.6% Calcium, 23.7% Sulfur and 46.8% Oxygen. What is the empirical formula for the compound?

If you had a 100 gram sample, you would have 29.6g Ca, 23.7g S and 46.8g O.

Calculating Empirical Formula

Ca 29.6g x 1 mole = 0.739 mol Ca 40.08g

S 23.7g x 1 mole = 0.739 mole S 32.07g

O 46.8 x 1 mole = 2.93 mole O 16.00g

0.739 = 1 0.739 = 1 2.93 = 3.960.739 0.739 0.739

Round to the nearest whole numbers, 1 : 1 : 4 Therefore the empirical formula is CaSO4

Molecular Formulas

Molecular Formula – the actual number of atoms of each element in one molecule of a compound.

The ratio of molecular mass to empirical mass is the same as the ratio of molecular formula to empirical formula.

Calculating Molecular Formula

Example:Molecular formula Molecular Mass

C6H12O6 180.18

Empirical formula Empirical Mass

CH2O 30.03

Ratio C6H12O6 = 6 Ratio 180.18 = 6

CH2O 30.03

Calculating Molecular Formula

Ribose has a molar mass of 150g/mol and a chemical composition of 40.0%C, 6.67% hydrogen and 53.3% oxygen. What is the molecular formula for ribose?Begin by calculating the empirical

formula for Ribose.

Calculating the Molecular Formula for Ribose

40.0g x 1 mole = 3.33 mole C 12.01g

6.67g x 1 mole = 6.60 mole H 1.01g

53.3g x 1 mole = 3.33 mole O 16.00g

3.33 = 1 6.60 = 1.98 3.33 = 13.33 3.33 3.33

Empirical Formula for RiboseCH2O

Calculating Molecular Formula

Once you know the empirical formula, you can calculate the molecular formula.

Covalent bonds can form in many different ratios.

Use the empirical formula and the ratio of molar mass to empirical formula mass to determine the molecular formula.

Calculating the Molecular Formula for Ribose

The molecular mass of Ribose is 150 g/mol The empirical mass of Ribose is 30 g/mol

(CH2O = 12.01 + (2 x 1.01) + 16.00 = 30.03 g/mol)

150 = 5 Therefore, the Molecular

30 formula is 5 times the

Empirical Formula

C5H10O5

The Formula of a Hydrate

Hydrate – a compound that has a specific number of water molecules bound to its atoms.

Examples:

CaCl2 . 2H2O

FePO4 . 4H2O

MgSO4 . 7H2O

Determining the Formula for a Hydrate

To determine the formula of a hydrate, the sample must be heated to remove all of the water.

The mass of the anhydrous sample is subtracted from the mass of the hydrated sample to determine the mass of water removed.

The mass of the anhydrous sample and the mass of the water are used to determine the empirical formula and molecular formula as before.

Determining the Molecular Formula of a Hydrate

Cobalt (II) chloride is a hydrated salt. If 11.75g of this hydrate is heated, 9.25g

of anhydrous cobalt chloride remains. What is the formula for this hydrate?

11.75g hydrated salt- 9.25g anhydrous salt

2.50g water

Determining the Molecular Formula of a Hydrate

9.25g CoCl2 x 1 mole = 0.0712 mol CoCl2

129.83g

2.50g H2O x 1 mole = 0.139 mol H2O 18.02g

0.0712 = 1 0.139 = 1.95

0.0712 0.0712

1:2 ratio Therefore the formula for the hydrate is

CoCl2 . 2H2O