1. Find the relative Atomic Mass (Ar) for the following elements.

- O - 16Ar
- C - 12Ar
- S - 32Ar
- P -31Ar
- Mg - 24Ar
- Fe - 56Ar

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

H = 1Ar O = 16Ar

2H = 2 x 1 = 2

1O = 1 x 16 = 16

2 + 16 = 18

H2O = 18 Mr

H = 1Ar

S = 32 Ar

O = 16 Ar

2H = 2 x 1 = 2

1S = 1 x 32 = 32

4O = 4 x 16 = 64

2 + 32 + 64 = 98

H2SO4 = 98Mr

Ca = 40Ar

C = 12 Ar

O = 16 Ar

1 Ca = 1 x 40 = 40

1 C = 1 x 12

3 O = 3 x 16 = 48

40 + 12 + 48 = 100

CaCO3 = 100 Mr

Mg = 24 Ar

O = 16 Ar

1 Mg = 1 x 24 = 24

1 O = 1 x 16

24 + 16 = 40

MgO = 40 Mr

Na = 23 Ar

C = 12 Ar

O = 16 Ar

2Na = 2 x 23 = 46

1 C = 1 x 12 = 12

3 O = 3 x 16 = 48

46 + 12 + 48 = 106

Na2CO3 = 106 Mr

Mg = 24 Ar

N = 14 Ar

O = 16 Ar

1 Mg = 1 x 24 = 24

2 N = 2 x 14 = 28

6 O = 6 x 16 = 96

24 + 28 + 96 = 148

Mg(NO3)2 = 148 Mr

Fe = 56 Ar

Cl = 35.5 Ar

1 Fe = 1 x 56 = 56

3 Cl = 3 x 35.5 = 111.5

56 + 111.5 = 167.5

FeCl3 = 167.5 Mr

C = 12 Ar

H = 1 Ar

O = 16 Ar

2 C = 2 x 12 = 24

4 H = 4 x 1 = 4

2 O = 2 x 16 = 32

24 + 4 + 32 = 60

CH3COOH = 60 Mr

H = 1 Ar

P = 31 Ar

O = 16 Ar

3 H = 3 x 1 = 3

1 P = 1 x 31 = 31

4 O = 4 x 16 = 64

3 + 31 + 64 = 98

H3PO4 = 98 Mr

C = 12 Ar

H = 1 Ar

O = 16 Ar

6 C = 6 x 12 = 72

8 H = 8 x 1 = 8

7 O = 7 x 16 = 112

72 + 8 + 112 = 192

C6H8O7 = 192 Mr

** **K
= 39 Ar

N = 14 Ar

O = 16 Ar

1 K = 1 x 39 = 39

1 N = 1 x 14 = 14

3 O = 3 x 16 = 48

39 + 14 + 48 = 101

KNO3 = 101 Mr

**2. Find the percentage of oxygen in the following compounds. Which compound has the greatest percentage
of oxygen? **

First we need to find the Relative Atomic Mass of each element

Ca = 40 Ar

O = 16 Ar

Then we need to find the Relative Formula mass of the compound

1 Ca = 1 x 40 = 40

1 O = 1 x 16 = 16

40 + 16 = 56

CaO = 56 Mr

Then we need to divide the Relative Atomic Mass of Oxygen
by the Formula Mass of CaO and multiply by 100.

16 / 56 = .29 x 100 = 29 %

The percentage of oxygen in CaO is 29%

First we need to find the relative atomic masses of the
elements.

H = 1 Ar

N = 14 Ar

O = 16 Ar

Then we need to find the mass of oxygen in the formula

3 O = 3 x 16 = 48

Then we need to find the relative formula mass
of the compound

1 H = 1 x 1 = 1

1 N = 1 x 14 = 14

3 O = 3 x 16 = 48

1 + 14 + 48 = 63

HNO3 = 63 Mr

Then we need to divide the mass of oxygen in the
formula by the relative formula mass of HNO3 and multiply by 100.

48/63 = 0.68 x 100 = 68% oxygen by
mass in HNO3

First we need to find the relative atomic masses for the elements in the compound

C = 12 Ar

H = 1Ar

O = 16 Ar

Then we need to find the mass of oxygen in the compound

1 O = 1 x 16 = 16

Then we need to find the relative formula mass of the compound.

2 C = 2 x 12 = 24

6 H = 6 x 1 = 6

1 O = 1 x 16 = 16

24 + 16 + 6 = 46

C2H5OH = 46 Mr

Then we need to divide the mass of oxygen in the compound by the
relative formula mass of the compound and multiply by 100.

16/46 = 0.35 x 100 = 35% oxygen by mass in C2H5OH

** **First we need to find the relative atomic masses
for the elements in the compound.

C = 12 Ar

H = 1 Ar

O = 16 Ar

Then we need to find the mass of oxygen in the compound.

1 O = 1 x 16 = 16

Then we need to find the relative formula mass of the compound

6 C = 6 x 12 = 72

12 H = 12 x 1 = 12

1 O = 1 x 16 = 16

72 + 12 + 16 = 100

C6H12O = 100 Mr

Then we need to divide the mass of oxygen by the mass of the
compound and multiply by 100.

16/100 = 0.16 x 100 = 16% oxygen by mass in C6H12O

The compound that has the greatest percentage of oxygen by mass is HNO3.

**3. Find the percentage of carbon in the following compounds. Which compound has the least percentage
of carbon?**

First we need to find the relative atomic masses of the elements in the compound.

C = 12 Ar

O = 16 Ar

Then we need to find the mass of carbon in the compound.

1 C = 1 x 12 = 12

Then we need to find the relative formula mass of the compound.

1 C = 12

2 O = 2 x 16 = 32

12 + 32 = 44

CO2 = 44 Mr

Then we need to divide the mass of carbon by the mass of the compound
and multiply by 100.

12/44= 0.27 x 100 = 27 % carbon by mass in CO2

First we need to find the relative atomic mass of the elements in the
compound

C = 12 Ar

H = 1 Ar

Then we need to find the mass of carbon in the compound.

2 C = 2 x 12 = 24

Then we need to find the relative formula mass of the compound.

2 C = 2 x 12 = 24

2 H = 2 x 1 = 2

24 + 2 = 26

C2H2 = 26 Mr

Then we need to divide the mass of carbon by the mass of the compound.

24/26 = 0.92 x 100 = 92% carbon by mass in C2H2

First we need to find the relative atomic mass of the elements in
the compound

Mg = 24 Ar

C = 12 Ar

O = 16 Ar

Then we need to find the mass of carbon in the compound.

1 C = 1 x 12 = 12

Then we need to find the relative formula mass of the compound.

1 Mg = 1 x 24 = 24

1 C = 1 x 12

3 O = 3 x 16 = 48

24 + 12 + 48 = 84

MgCO3 = 84 Mr

Then we need to divide the mass of carbon by the mass of the compound
and multiply by 100.

12 / 84 = 0.14 x 100 = 14% carbon by mass in MgCO3

MgCO3 has the least percentage of carbon by mass

**4. Sebastian needs to make 20 g of ZnCl2. He knows he can make ZnCl2 by reacting Zn with HCl acid.**

** Zn + 2HCl
--> ZnCl2 + H2**

**Determine the relative atomic masses and the relative formula masses for the reactants and the products.**

Relative Atomic Masses

Zn = 65 Ar

H = 1 Ar

Cl = 35.5 Ar

Relative Formula Masses

HCl

1 H = 1 x 1 = 1

1 Cl = 1 x 35.5 = 35.5

1 + 35.5 = 36.5

HCl = 36.5 Mr

ZnCl2

1 Zn = 1 x 56 = 56

2 Cl = 2 x 35.5 = 71

56 + 71 = 127

ZnCl2 = 127 Mr

H2

2 H = 2 x 1 = 2

H2 = 2 Mr

check that you calculations are correct by substituting
the atomic and formula masses in your equation.

Zn + 2HCl --> ZnCl2 + H2

56 Ar + 2 x 36.5 Mr = 127 Mr
+ 2 Ar

56 + 73 = 127 + 2

129 = 129

The relative formula and
atomic masses are correct because mass cannot be created or destroyed - therefore the mass of the reactants must equal the
mass of the products.

**Determine the ratio of Zn to ZnCl2**

** **According to our relative formula and atomic masses
we need 56 g of Zn to form 127 g of ZnCl2. Therefore our ratio is

56 g Zn = 127 g ZnCl2

**Determine the ratio of HCl to ZnCl2**

** **According to our relative formula masses ( and our equation)
that 73 g of HCl are needed to make 127 g of ZnCl2. Therefore our ratio is

73
g HCl = 127 g ZnCl2

**Determine how much Zn and HCl you need to form 20 g of ZnCl2.**

** **To determine how much Zn we need to start with we need
to cross multiply.

First we need to set up our ratio

56 g Zn = 127 g ZnCl2

X g Zn = 20 g ZnCl2

Then we need to cross multiply

(56 g Zn x 20 g ZnCl2) / 127 g ZnCl2 = 8.8 g of Zn

If you look at the equation, the ZnCl2 cross each other out to leave you with Zn.

Therefore we need 8.8 g of Zn to form 20 g of ZnCl2.

To determine how much HCl we need to form 20 g of ZnCl2 we need to cross multiply.

First we have to set up our ratio.

73 g HCl = 127 g ZnCl2

X
g HCl = 20 g ZnCl2

Then we need to cross multiply

(73
g HCl x 20 g ZnCl2) / 127 g ZnCl2 = 11.5 g HCl

11.5 g of HCl are needed to make 20 g of ZnCl2

**5. Jane wants to make 100 g of Hg. To do this she is going to heat HgO to form Hg and O2.**

**What type of reaction is this?**

** **This is a decomposition reaction.

** 2
HgO --> 2 Hg + O2**

**Determine the relative formula and relative atomic masses of the reactants and of the products.**

** **Relative Atomic Masses

Hg = 201 Ar

O = 16 Ar

Relative Formula Masses

HgO

1 Hg = 1 x 201 = 201

1 O = 1 x 16 = 16

201 + 16 = 217

HgO = 127 Mr

O2

2 O = 2 x 16 = 32

to make sure that your relative atomic and
formula masses are correct substitute them for the formulas in your equation.

2HgO --> 2Hg + O2

2 x 217 --> 2 x 201 + 32

434 --> 402 + 32

434 --> 434

The relative formula and atomic masses are correct because mass cannot be created or destroyed
and the mass of the reactants must equal the mass of the products.

**Determine the ratio of HgO to Hg**

** **According to our relative atomic and formula masses and
our equation, 434 g of HgO are needed to make 402 g of Hg. Therefore our ratio is

434 g HgO = 402 g Hg

**Determine how much HgO Jane needs to heat to form 100g of Hg.**

** **To determine how much HgO Jane needs to start with to
form 100g of Hg we need to cross multiply. First we have to set up a ratio.

434 g HgO = 402 g Hg

X g HgO = 100 g Hg

Then we need to cross multiply

(434 g HgO x 100 g Hg) / 402 g Hg = 108 g HgO

Jane needs to heat 108 g HgO to form 100 g Hg.