Chapter 6 Review

Quantitative Chemistry

Relative Atomic Mass (Ar)

The relative atomic mass of an element is the average
mass of the naturally occurring atoms of the element, using a scale where an atom of carbon-12 has a mass of exactly 12.

Isotopes are naturally occurring atoms of the same
element that have different masses. Isotopes have the same number of protons
but a different number of neutrons.

Example: Find the relative atomic mass of Lithium.

Look on the periodic table and find Lithium. In the periodic table the top number is the relative atomic mass.
In this case the relative atomic mass of Li is 7 Ar. If you are using
a different periodic table, the relative atomic mass is always bigger than the atomic number.

1. Find the relative atomic mass of the following atoms.

a.
Iron

b. Cobalt

c.
Radium

d. Sulphur

Relative Formula Mass

The relative formula mass (Mr) of a substance is the
sum of the relative atomic masses of the elements present in the formula unit.

If the substance is only made of simple molecules,
this mass may also be called the relative molecular mass (Mr)

Example : Find the relative formula mass (Mr) of Sulphuric Acid.

Sulphuric acid has the formula H2SO4

Step 1: Determine the relative atomic mass for all the atoms in the formula

H
= 1 Ar

S
= 32 Ar

O
= 16 Ar

Step 2: Multiply
the relative atomic masses of the atoms by the number of atoms in the formula

H = 1 Ar x 2 = 2

S
= 32 Ar x 1 = 32

O = 16 Ar x 4 = 64

Step 3: Add the totals for each individual element to determine the relative formula mass

2
+ 32 + 64 = 98 Mr

The relative formula mass of
Sulphuric acid is 98 Mr

2.
Find the relative formula mass (Mr) of the following compounds:

a.
CaCO3

b.
NaCl

c.
Al2O3

d.
Fe2O3

Percentage by mass

Percentage by mass is determined
by first finding the relative formula mass (Mr) then dividing the total mass of one element by the relative formula mass.

Example: Determine the percentage by mass of Iron in Iron(II) oxide.

Step one: Determine the relative formula mass of iron(II) oxide.

The formula for iron(II)oxide is FeO

Fe = 56

O = 16

56 + 16 = 72 Mr

Step two : Determine the mass of iron in the compound

Fe = 56 Ar

There is one atom of Fe in FeO

Fe = 56 x 1 = 56

Step three : Divide the mass of iron by the relative formula mass of iron(II)oxide

Fe/FeO = 56/72 = 0.78

Step four: Multiply result by 100

0.78 x 100 = 78 %

The percentage by mass of iron in iron oxide is 78%

3.
Determine the percentage by mass of sulphur in the following compounds:

a.
H2SO4

b.
MgSO4

c.
SO2

d.
ZnSO4

Compound formation and chemical formulas

A particular compound always contains the same elements.

These elements are always present in the same proportions by mass.

It does not matter where the compound is found or how it is made.

These proportions cannot be changed.

We can determine how much of
a compound will be made or how much reactant we need to start with by using ratios.

Example: Calcium carbonate when heated forms calcium oxide and carbon dioxide.
If 10 grams of calcium carbonate are heated how much calcium oxide is formed?

Step one : Write the balanced formula

CaCO3 à CaO + CO2

Step two : Determine the relative formula mass (Mr) of calcium carbonate.

Ca = 40 Ar x 1 = 40

C = 12 Ar x 1 = 12

O = 16 Ar x 3 = 48

40 + 12 + 48 = 100 Mr

CaCO3 = 100 g

Step three : Determine the relative formula mass (Mr ) of calcium oxide

Ca = 40 Ar x 1 = 40

O = 16 Ar x 1 = 16

40 + 16 = 56 Mr

CaO = 56 g

Step four : Set up your ratio

100 g of CaCO3 produces 56 g CaO

10 g of CaCO3 produces X g CaO

cross multiply

X = __10 g x 56 g__

100g

X = 5.6 g

So 10 g of CaCO3 when heated will produce 5.6 g of CaO

4.
Sodium reacts with chlorine gas to form sodium chloride. If you want to produce 20 g of sodium
chloride, how much sodium do you need to start with?

5.
Iron(II)sulphate reacts with sodium hydroxide to form a light
green precipitate of iron(II)hydroxide and sodium sulphate. If 100 g of iron(II)sulphate is reacted with excess sodium hydroxide, how much precipitate is produced?

6.
Calcium carbonate reacts with hydrochloric acid to form calcium
chloride, carbon dioxide and water. How much calcium carbonate must you start
with to form 40 g of carbon dioxide?

7.
Magnesium burns in air to form Magnesium oxide. How much magnesium oxide will be produced if 56 g of magnesium is burned?

The mole

One mole of a substance has a mass equal to its relative formula mass in grams.

One mole of a substance contains
6.02 x 1023 per mole (Avogadro constant) of atoms, molecules, or formula units, depending on the substance considered.

Example : Determine the mass of one mole of potassium atoms.

Step one : Determine the relative atomic mass of potassium

K = 39 Ar

1 mole K = 39 g

8.
Determine the mass of one mole of the following atoms:

a.
Lithium

b.
Carbon

c.
Oxygen

d.
Aluminum

Example : Determine the mass
of one mole of hydrogen molecules.

Step one : Determine the formula of hydrogen molecules

H2

Step two : Determine the relative formula mass

H = 1 Ar x 2 = 2 Mr

1 mole H2 = 2 g

9.
Determine the mass of one mole of the following molecules

a.
Oxygen

b.
Chlorine

c.
Bromine

d.
Fluorine

Example : Determine the mass
of one mole of water

Step one : Determine the formula

H2O

Step two : Determine the relative formula mass

H = 1 Ar x 2 = 2

O = 16 Ar x 1 = 16

2 + 16 = 18 Mr

1 mole H2O = 18 g

10. Determine the mass of one mole of the following compounds:

a.
Magnesium oxide

b.
Sodium hydroxide

c.
Potassium bromide

d.
Sodium chloride

Example : Determine the mass
of two moles of zinc chloride.

Step one : Determine the formula

ZnCl2

Step two : Determine the relative formula mass

Zn = 65 Ar x 1 = 65

Cl = 53.5 Ar x 2 = 71

65 + 71 = 136 Mr

ZnCl2 = 136 Mr

1 mole ZnCl2 = 136 g

Step three : Multiply the mass of one mole by the number of moles

__136 g__
x 2 moles = 272 g

mole

2 moles ZnCl2 = 272 g

11. Determine the mass of the following compounds:

a.
2 moles of magnesium sulphate

b.
4 moles of potassium nitrate

c.
0.5 moles of calcium oxide

d.
3.4 moles of iron(II)carbonate

Example : How many molecules are present in 0.5 moles of water?

Step one: Multiply the number of moles by the avogadro constant.

0.5 moles x __6.02 x 1023 molecules__ = 3.01 x 1023 molecules

mole

12. Determine the number of molecules or atoms in:

a.
1 mole of calcium atoms

b.
4 moles of hydrogen gas

c.
5.6 moles of nitric acid

d.
0.02 moles of ethane

Working out chemical formulas using
moles

Compounds always forms in fixed ratios.

Example : Determine the number
of moles of copper and the number of moles of oxygen in copper(II)oxide.

Step one : Determine the formula of copper(II)oxide

The formula of Copper(II)oxide
is CuO, as determined by the charges on the ions.

Step two : Using the small numbers
after the symbol for each atom to determine the number of moles. Remember, if
there is no number then there is one mole.

CuO

Cu = 1 mole

O = 1 mole

13. Determine the number of moles of each element in the following compounds:

a.
Potassium oxide

b.
Lithium carbonate

c.
Calcium sulphate

d.
Iron(III)nitrate