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Chapter 6 Review Answers
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Chapter 6 Review Answers

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

Fe = 56 Ar

 

b.      Cobalt

Co = 59 Ar

 

c.       Radium

Ra = 226 Ar

 

d.      Sulphur

S = 32 Ar

 

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

Ca = 40 Ar x 1 = 40

C = 12 Ar x 1 = 12

O = 16 Ar x 3 = 48

 

40 + 12 + 48 = 100 Mr

The relative formula mass (Mr) of calcium carbonate is 100 Mr

 

b.      NaCl

Na = 23

Cl = 35.5

23 + 35.5 = 58.5 Mr

The relative formula mass (Mr) of sodium chloride is 58.5 Mr

 

c.       Al2O3

Al = 27 Ar x 2 = 54

O = 16 Ar x 3 = 48

54 + 48 = 102 Mr

The relative formula mass (Mr) of aluminum oxide is 102 Mr

 

d.      Fe2O3

Fe = 56 Ar x 2 = 112

O = 16 Ar x 3 = 48

112 + 48 = 160 Mr

The relative formula mass (Mr) of Iron(III)oxide is 160 Mr

 

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

H = 1 Ar x 2 = 2

S = 32 Ar x 1 = 32

O = 16 Ar x 4 = 64

2 + 32 + 64 = 98 Mr

The relative formula mass of sulphuric acid is 98 Mr

 

Sulphur has a mass of 32 in this compound.

 

Percentage by mass of sulphur = 32/98 x 100 = 33%

 

b.      MgSO4

Mg = 24 Ar x 1 = 24

S = 32 Ar x 1 = 32

O = 16 Ar x 4 = 64

24 + 32 + 64 = 118 Mr

The relative formula mass of magnesium sulphate is 118 Mr

 

Sulphur has a mass of 32 in this compound

 

Percentage by mass of sulphur = 32 / 118 x 100 = 27 %

 

c.       SO2

S = 32 Ar x 1 = 32

O = 16 Ar x 2 = 32

32 + 32 = 64 Mr

The relative formula mass of sulphur dioxide is 64 Mr

 

Sulphur has a mass of 32 in this compound

 

Percentage by mass of sulphur = 32 / 64 x 100 = 50%

 

d.      ZnSO4

Zn = 65 Ar x 1 = 65

S = 32 Ar x 1 = 32

O = 16 Ar x 4 = 64

65 + 32 + 64 = 161 Mr

The relative formula mass of zinc sulphate is 161 Mr

 

Sulphur has a mass of 32 in this compound

 

Percentage by mass of sulphur = 32 / 161 x 100 = 20%

 

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 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?

2 Na + Cl2 2 NaCl

 

Na = 23 Ar

2 Na = 23 x 2 = 46 g

 

Na = 23 Ar x 1 = 23

Cl = 35.5 Ar x 1 = 35.5

23 + 35.5 = 58.5 Mr

NaCl = 58.5 Mr

2 NaCl = 58.5 x 2 = 117 g

 

46 g of Na = 117 g of NaCl

X g of Na = 20 g of NaCl

 

X g of Na = 20 g x 46 g

                          117 g

 

X g of Na = 7.9 g

 

To make 20 g of NaCl you need to react 7.9 g of Na.

 

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?

FeSO4 + 2NaOH Fe(OH)2 + Na2SO4

 

Fe = 56 Ar x 1 = 56

S = 32 Ar x 1 = 32

O = 16 Ar x 4 = 64

56 + 32 + 64 = 152 Mr

FeSO4 = 152 Mr

 

Fe = 56 Ar x 1 = 56

O = 16 Ar x 2 = 32

H = 1 Ar x 2 = 2

56 + 32 + 2 = 90 Mr

Fe(OH)2 = 90 Mr

 

152 g of FeSO4 = 90 g of Fe(OH)2

100 g of FeSO4 = X g of Fe(OH)2

 

X g of Fe(OH)2 = 100 x 90

                                 152

 

X g of Fe(OH)2 = 59 g

 

100 g of FeSO4 reacts to form 59 g of Fe(OH)2

 

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?

CaCO3 + 2HCl CaCl2 + CO2 + H2O

 

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 Mr

 

C = 12 Ar x 1 = 12

O = 16 Ar x 2 = 32

12 + 32 = 44 Mr

CO2 = 44 Mr

 

100 g CaCO3 = 44 g CO2

X g CaCO3 = 40 g CO2

 

X g CaCO3 = 40 x 100

                              44

 

X g CaCO3 = 91 g

 

91 g of CaCO3 reacts with HCl to form 40 g of CO2

 

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

2Mg + O2 2MgO

 

Mg = 24 Ar

2 Mg = 2 x 24 = 48 g

 

Mg = 24 Ar x 1 = 24

O = 16 Ar x 1 = 16

24 + 16 = 40 Mr

MgO = 40 Mr

2 MgO = 40 x 2 = 80 g

 

48 g Mg = 80 g MgO

56 g Mg = X g MgO

 

X g MgO = 56 x 80

                       48

 

X g MgO = 93 g

 

56 g Mg burns in oxygen to form 93 g of MgO

 

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

Li = 7 Ar

1 mole Li = 7 g

 

b.      Carbon

C = 12 Ar

1 mole C = 12 g

 

c.       Oxygen

O = 16 Ar

1 mole O = 16 g

 

d.      Aluminum

Al = 27 Ar

1 mole Al = 27 g

 

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

O2

O = 16 Ar x 2 = 32 Mr

1 mole O2 = 32 g

 

b.      Chlorine

Cl2

Cl = 35.5 Ar x 2 = 71 Mr

1 mole Cl2 = 71 g

 

c.       Bromine

Br2

Br = 80 Ar x 2 = 160 Mr

1 mole Br2 = 160 g

 

d.      Fluorine

Fl2

Fl = 19 Ar x 2 = 38 Mr

1 mole Fl2 = 38 g

 

 

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

MgO

 

Mg = 24 Ar x 1 = 24

O = 16 Ar x 1 = 16

24 + 16 = 40 Mr

1 mole MgO = 40 g

 

b.      Sodium hydroxide

NaOH

 

Na = 23 Ar x 1 = 23

O = 16 Ar x 1 = 16

H = 1 Ar x 1 = 1

23 + 16 + 1 = 40 Mr

1 mole NaOH = 40 g

 

c.       Potassium bromide

KBr

 

K = 39 Ar x 1 = 39

Br = 80 Ar x 1 = 80

39 + 80 = 119 Mr

1 mole KBr = 119 g

 

d.      Sodium chloride

NaCl

 

Na = 23 Ar x 1 = 23

Cl = 35.5 Ar x 1 = 35.5

23 + 35.5 = 58.5 Mr

1 mole NaCl = 58.5 g

 

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

MgSO4

 

Mg = 24 Ar x 1 = 24

S = 32 Ar x 1 = 32

O = 16 Ar x 4 = 64

24 + 32 + 64 = 120 Mr

1 mole MgSO4 = 120 g

 

120 g x 2 moles = 240 g

mole

 

2 moles MgSO4 = 240 g

 

b.      4 moles of potassium nitrate

KNO3

 

K = 39 Ar x 1 = 39

N = 14 Ar x 1 = 14

O = 16 Ar x 3 = 48

39 + 14 + 48 = 101 Mr

1 mole KNO3 = 101 g

 

101 g x 4 moles = 404 g

mole

 

4 moles KNO3 = 404 g

 

c.       0.5 moles of calcium oxide

CaO

 

Ca = 40 Ar x 1 = 40

O = 16 Ar x 1 = 16

40 + 16 = 56 Mr

1 mole CaO = 56 g

 

56 g x 0.5 moles = 28 g

mole

 

0.5 moles CaO = 28 g

 

d.      3.4 moles of iron(II)carbonate

FeCO3

 

Fe = 56 Ar x 1 = 56

C = 12 Ar x 1 = 12

O = 16 Ar x 3 = 48

56 + 12 + 48 = 116 Mr

1 mole FeCO3 = 116 g

 

116 g x 3.4 moles = 394 g

mole

 

3.4 moles FeCO3 = 394 g

 

 

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

1 mole x 6.02 x 1023 atoms = 6.02 x 1023 atoms

                         mole

 

b.      4 moles of hydrogen gas

4 moles x 6.02 x 1023 molecules = 24 x 1023 molecules

                              mole

 

c.       5.6 moles of nitric acid

5.6 moles x 6.02 x 1023 molecules = 34 x 1023 molecules

                              mole

d.      0.02 moles of ethane

0.02 moles x 6.02 x 1023 molecules = .12 x 1023 molecules

                              mole

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

K2O

K = 2 moles

O = 1 mole

 

b.      Lithium carbonate

Li2CO3

Li = 2 moles

C = 1 mole

O = 3 moles

 

c.       Calcium sulphate

CaSO4

Ca = 1 mole

S = 1 mole

O = 4 moles

 

d.      Iron(III)nitrate

Fe(NO3)3

Fe = 1 mole

N = 3 moles

O = 9 moles

 

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