1.2 Formulas 3 hours
Assessment statement
1.2.1 Define the terms relative atomic mass (A r
) and relative molecular mass (M r
).
1.2.2 Calculate the mass of one mole of a species from its formula.
1.2.3 Solve problems involving the relationship between the amount of substance in moles, mass and molar mass.
1.2.4 Distinguish between the terms empirical formula and molecular formula.
1.2.5 Determine the empirical formula from the percentage composition or from other experimental data.
1.2.6 Determine the molecular formula when given both the empirical formula and experimental data.
Obj Teacher’s notes
1
2
3
2
3
3
The term molar mass (in g mol –1 ) will be used.
Aim 7: Virtual experiments can be used to demonstrate this.
1

• Chemical formulas show the ratio of the number of atoms present in a compound or ion.
• Magnesium chloride MgCl
2
– 2 chloride ions for every 1 magnesium ion (Ionic compound)
• Acetic Acid CH
3
COOH
– 2 atoms carbon, 2 atoms oxygen, 4 atoms hydrogen in one molecule of acetic acid (molecular compound)
• Ionic compound consist of a metal (positive ion) and nonmetal (negative ion).
• Molecular (covalent) compounds are made of non-metals only.
2

r
• The mass of an atom is compared with that of an atom of carbon-12. (Carbon-12 standard)
• The relative atomic mass of carbon-12 is taken to be 12.
• Relative masses are based on elements having isotopes
(same identity, different “stuff” within core)
• For example, naturally occurring chlorine consists of atoms of
relative isotopic masses 35 (75%) and 37 (25%).
– Its relative atomic mass (A r
) is 35.5
.
3

r
• The MASS of a MOLECULE
• The mass of a molecule is compared with that of an atom of carbon-12. (Carbon-12 standard)
• A relative molecular mass can be calculated easily by adding together the relative atomic masses of the constituent atoms that have units g mol -1
• e. g., ethanol: CH
3
CH
2
(C
2
+ H
6
OH, has a M r of 46.08 g mol -1
+ O = 24.02 + 6.06 + 16.00 = 46.08)
4

• Find the molecular mass of Glucose C
6
H
12
O
6.
• Solving Process:
– Add the atomic masses of all the atoms in the C
6
H
12
O
6
• 6 C atoms 6 x 12.0 = 72.0 g mol -1
• 12 H atoms 12 x 1.0 = 12.0 g mol -1
• 6 O atoms 6 x 16.0 = 96.0 g mol -1
Formula mass of glucose = 180.0 g mol -1
– This mass represents one mole of glucose.
5

• Find the formula mass of Copper (II) phosphate,
Cu
3
(PO
4
)
2
.
• Solving Process:
– Add the atomic masses of all the atoms in the Cu
3
(PO
4
)
2 formula.
Remember the subscript applies the entire polyatomic ion.
• 3 Cu atoms
• 2 P atoms
• 8 O atoms
3 x 63.5 = 190.5 g mol -1
2 x 30.9 = 61.8 g mol -1
8 x 16.0 = 128.0 g mol -1
Formula mass of Cu
3
(PO
4
)
2
= 380.3 g mol -1
– This mass represents one mole of Copper (II) phosphate.
6

1.1 The mole concept and Avogadro’s constant
2 hours
Assessment statement
1.1.1 Apply the mole concept to substances.
Obj Teacher’s notes
2 The mole concept applies to all kinds of particles: atoms, molecules, ions, electrons, formula units, and so on. The amount of substance is measured in moles (mol). The approximate value of Avogadro’s constant (L),
6.02 × 10 23 mol –1 , should be known.
1.1.2 Determine the number of particles and the amount of substance (in moles).
TOK: Chemistry deals with enormous differences in scale. The magnitude of Avogadro’s constant is beyond the scale of our everyday experience.
3 Convert between the amount of substance (in moles) and the number of atoms, molecules, ions, electrons and formula units.
7

• Chemists do not deal with amounts of substance by counting out individual atoms or molecules.
• Atomic Mass Units (amu) as a Very Small Number
• It is important to obtain a relationship between mass and the number of particles in a compound.
• Molecular masses are in atomic mass units, which is a very small unit.
• An atomic mass unit is only 1.66 x 10 -24 grams.
• We need to use a larger unit for everyday use such as the gram for laboratory use.
• We mass quantities of substances on balances.
8

• Chemists have found that
6.022 x 10 23 atoms of
Hydrogen have a mass of
1.0079 g.
• This number 6.022 x 10 23 is called Avogadro’s number.
• Commonly referred to as the
Mole.
9

• Avogadro’s number is an accepted SI standard.
• The symbol used to represent Avogadro's number is N
A
.
• This quantity can be expressed as 6.02205 x 10 23 to be more precise.
• This number of whatever we are questioning is called one mole (mol) of things.
• Dozen: 12 of something. Whether feathers or dump trucks.
• Same number, different masses
• Mole: just a name representing a number
(6.022 x10 23 ) (just like dozen)
10

• A mole of feathers and a mole of trucks are the same number
– Are they the same mass?
• One mole of particles (atoms, ions, molecules, electrons, formula units) has a mass (in grams)
– Happens to be equivalent in # to that of one particle in atomic mass units.
• If a mole of any particle has a mass of 4.02 g mol -1 , then a single particle has a mass of 4.02 amu (Atomic
Mass Units)
• If you have a single particle with a mass of 53.03 amu, then a mole of the same particles will have a mass of
53.03 g.
11

g mole
6.022 x 10 23 mole
mole g
22.4 L mole mole
6.022 x 10 23
.
Stuff:
• Atoms
• Molecules
• Particles
• Electrons
• Ions
• Formula Units mole
.
22.4 L
For Gases only:
@ STP
(Standard Temp & Pressure)
(0 o C (273 K) & 1 atm)
12

• It is important to note that one mole of atoms contains
6.022 x 10 23 atoms.
• One mole of molecules contains 6.022 x 10 23 molecules.
• One mole of ions contains 6.022 x 10 23 ions.
• N
A
, therefore can have any of these units (but the SAME number).
13

g mole
6.022 x 10 23 mole
mole g
22.4 L mole mole
6.022 x 10 23
.
Stuff:
• Atoms
• Molecules
• Particles
• Electrons
• Ions
• Formula Units mole
.
22.4 L
For Gases only:
@ STP
(Standard Temp & Pressure)
(0 o C (273 K) & 1 atm)
14

• 1.20 x 10 25 molecules of NH
3
• Solving Process: will be how many moles?
– We have molecules and wish to covert to moles.
– One mole equals 6.022 x 10 23 molecules.
– 1 mol/6.022 x 10 23 molecules
– 1.20 x 10 25 molecules NH
3
1 mol NH
3
.
6.022 x 10 23 molecules NH
3
= 19.9 mol NH
3
15

• Convert to the indicated units for each question.
1. 9.03 x 10 23 atoms bromine to moles of bromine.
2. 3.5 mol of sodium to atoms of sodium
3. 2 mol of Calcium chloride (CaCl
2
) to: a) Formula Units of calcium chloride b) Number of calcium ions c) Number of chloride ions
4. 4.75 mol of electrons to # of electrons.
5. 5.08 x 10 23 molecules of oxygen to moles of oxygen molecules. Moles of oxygen to atoms.*
*note: oxygen is a diatomic molecule, O
2
.
16

17

• From the reaction involving nitrogen with hydrogen to form ammonia
N
2
+ 3H
2

2NH
3
• The most important part of this BALANCED equation is that 1 mole of nitrogen reacts with 3 moles of hydrogen to form 2 moles of Ammonia.
• Based on this information, we can write ratios that relate moles of reactants to each other and to moles of products.
18

• How many moles of ammonia are produced when
0.60 mol of nitrogen reacts with hydrogen?
N
2
+ 3H
2

2NH
3
• The mole ratio of nitrogen to ammonia is 1:2 meaning for every 1 mole of Nitrogen, 2 moles of ammonia are produced.
19

VOLUME
 22.4 L x 22.4 L x molar mass
MOLES
MASS
 molar mass
Mole ratio
Mole ratio
VOLUME
MASS
 22.4 L x 22.4 L x molar mass
 molar mass
MOLES
 6.022 x 10 23 x 6.022 x 10 23
 6.022 x 10 23 x 6.022 x 10 23
MOLECULES MOLECULES
20

• Calculate the number of grams of NH
3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen.
• Solution:
– Balance the equation (if needed)
– Grams of H
2 grams of NH
3 to moles of H
2 to moles of NH
3 to
21

1. How many grams of acetylene are produced by adding water to 5.00 g of CaC
2
?
Acetylene gas (C
2 calcium carbide.
H
2
) is produced by adding water to
CaC
2
+ 2H
2
O

C
2
H
2
+ Ca(OH)
2
2. Using the same equation, determine how many moles of CaC water.
2 are needed to react completely with 49.0g of
22

• How many molecules of oxygen are produced when
29.2 g of water is decomposed by electrolysis according to this balanced equation?
2H
2
O (l) → 2H
2
(g) + O
2
(g)
23

1. How many molecules of oxygen are produced by the decomposition of 6.54 g of potassium chlorate
(KClO
3
)?
2KClO
3
(s) → 2KCl (s) + 3O
2
(g)
2. The last step in the production of nitric acid is the reaction of nitrogen dioxide with water. How many grams of nitrogen dioxide must react with water to produce 5.00 x 10 22 molecules of nitrogen monoxide?
3NO
2
(g) + H
2
O (l) → 2HNO
3
(aq) + NO (g)
24

• How many moles are represented by 21 g of ammonia,
NH
3
?
• Solving process:
– We have grams and want to convert to moles.
– Using the atomic mass from the periodic table, we can calculate the atomic mass of NH
3
(17 g mol -1 )
– 1 mole of NH
3 has a mass of 17 g.
– We have a sample of 21 g of NH
3
– Divide our sample by the mass of 1 mole of NH
3
– We end up with 1.24 mol of Ammonia
25

• What mass is 20 mol of NH
3
?
• Solving Process:
– Using your periodic table, determine the molecular mass of NH
3
(17.0 g mol -1 )
– 17.0 g for 1 mol of NH
3
– 20 mol NH
3 x 17.0 g NH
3
1 mol NH
3
= 340 g of NH
3 in 20 mol NH
3
26

• How many atoms are in a 10.0 g sample of calcium metal?
• Solving Process:
– 1 formula mass of calcium metal is 40.1 g mol -1 .
– 10 g of calcium is 0.249 mol calcium
(10 g/40.1 g per mol)
– There are 6.02 x 10 23 (N
A
) atoms in one mole
– N
A
(0.249 mol calcium) = # atoms in 10 g of calcium metal
= 1.50 x 10 23 atoms in 10 g of calcium
27

Make the following conversions:
1. 7.92 formula units of Water (H
2
O) to moles.
2. 2.65 mol of methane (CH
4
) to molecules.
3. 207 g of sodium nitrite (NaNO
3
) to moles.
4. 85.7 g of gold (III) sulfate (AuSO
4
) to formula units.
5. 5.97 x 10 23 formula units of magnesium bicarbonate
(Mg(HCO
3
)
2
)to grams
6. 0.458 moles of sodium thiosulfate (Na
2
S
2
O
3
) to grams
7. 1.99 x 10 24 formula units of ammonium phosphate
((NH
4
)
3
PO
4
)to moles.
8. 55 g of Mercury (Hg) to moles.
28