topic: chemical formula, chemical equations...

383

Lampiran O

Bahan Pengajaran

TOPIC:

CHEMICAL FORMULA, CHEMICAL EQUATIONS AND

STOICHIOMETRY PROBLEM

(FORMULA KIMIA, PERSAMAAN KIMIA DAN MASALAH STOIKIOMETRI)

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Task Sub-topic Learning outcomes

1 Activity 1.1

Relative Atomic Mass

and Relative Molecular

Mass

Able to

1. understand the concept of Relative Atomic Mass, RAM and Relative Molecular Mass, RMM.

2. Able to calculate the Relative Molecular Mass.

2 Activity 1.2

The Mole and Number of

Particles

Able to

1. define a mole. 2. state the meaning of molar mass or Relative Molecular Mass. 3. relate the number of particles in one moles to substances

using Avogadro Constants.

4. Solve the numerical problem to convert the number of mole to the number of particles.

3 Activity 1.3

Mole and Mass of

Substances.

Able to

1. state the meaning of molar mass. 2. relate the molar mass to Avogadro Constant. 3. convert mass of a substance to mole. 4. Solve numerical problem

4 Activity 1.4

Number moles of Gas and

Molar Volume.

Able to

1. state the meaning of Molar Volume. 2. solve numerical problem.

5 Activity 1.5

Empirical Formula and

Molecular Formula.

Able to

1. state the meaning of empirical formula and molecular formula.

2. determine the empirical formula and molecular formula. 3. compare and contrast the empirical formula and the

molecular formula.

4. solve numerical problem involving the empirical formula and the molecular formula.

6 Experiment 1.6

Empirical formula of

Magnesium oxide.

Able to

1. carry out the experiment 3.5 2. determine the empirical formula of magnesium oxide.

7 Activity 1.7

Chemical Formula of

ionic compound

Able to

1. write ionic formulae of ions.

2. construct chemical compound using IUPAC

nomenclature.

3. naming chemical formula

8 Activity 1.8

Chemical Equation

Able to

1. write chemical equation from word equation to symbol. 2. balancing chemical equation 3. use the chemical equations in stoichiometry problem.

10 Activity 1.9 Balancing

Chemical Equation

Able to

1. write and balancing chemical equation

11 Activity 1.10

Chemical equation and

Stoichiometry Problem

Able to

1. understang meaning the chemical equation. 2. solve Stoichiometry Problem

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Activity 1.1 RELATIVE ATOMIC MASS AND RELATIVE MOLECULAR MASS

Goal : Students be able

1. To understand the concept of Relative Atomic Mass, RAM and

Relative Molecular Mass, RMM.

2. To calculate the Relative Molecular Mass of given chemical formulae.

Introduction( Declarative Knowledge)

1. The mass of an atom is very light, we cannot measure a single atom directly on a balance, but

we can determine the mass of one atom relative to another.

Relative Atomic Mass, RAM

1. We need to compare the atomic mass Of particular element with that of a standard atom. The

atom mass obtained in such a way is called relative atomic mass RAM.

2. During the 19th century, hydrogen atom was chosen as the standard atom. It is used in the

comparison of the atomic mass of the other elements as it was the lightest atom.

3. It 1961 , Carbon atom as chosen as the standard element to be compare with. The relative

atomic mass of carbon was fixed at 12.00 for carbon-12.

4. Therefore, relative atomic mass was define as; the number of times the mass of one atom of

an element is heavier than 1/12 times of one atom carbon-12 atom. The relative atomic

mass C-12 is 12. So

The relative atomic mass of an element = Mass of 1 atom of the element

1/12 x mass of 1 atom of carbon-12

The relative molecular mass, RMM, of a compound is defined as,

The number of times the mass of one molecule of a compound is heavier than 1/12 times

the mass of one carbon-12 atom. The relative atomic mass of C-12 is 12.

The relative Molecular Mass = Mass of 1 molecule of the compound

1/12 x mass of 1 atom of C-12

Calculating Relative Molecular Mass (RMM) or Molar Mass

(Procedural Knowedge)

1. The relative molecule Mass of a molecule can be calculate by adding up all the mass of all

atoms,

Example, Calculate the relative molecular mass of

a. ammonia (NH3)

b. hydrated magnesium sulphate

[Relative Atomic Mass; H,7; N,14;Mg, 24; S, 32; O,16]

a. Relative molecule Mass of Ammonia, NH3

=Mass of one nitrogen atom + mass of three hydrogen atoms

=RAM of nitrogen x number of nitrogen atom + RAM of hydrogen x number of hydrogen

atom

Molar Mass of ammonia, NH3 = 14 x1 + 1x3 = 17

b. The relative molecule mass, MgSO4. 7H2O

Relative molecular mass

=Total mass of magnesium atoms + mass of sulphur atom+ mass of oxygen atom + mass of 7

water molecules

=RAM of Mg x number of Mg atom + RAM of sulphur x number of Sulphur atoms + RAM of

hydrogen x number of Hydrogen atom + RAM Oxygen x number of Oxygen atoms.

Molar mass of hydrated magnesium sulphate MgSO4. 7H2O

= 24x1 + 32x1+16x4+ 1x14+ 16x7 =141

2. How do you think to solve the problem?

Step 1. Write the chemical formula

What are

the steps

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Step 2. Find the relative atom mass(RAM) of each elements in a compound.

Step 3. Relationship;

3. Relative molecular mass of Molar Mass is equal total mass of all elements by multiply RAM

of every element with number of its atoms.

4. Step 4. Checking the answer.

Exercise 1.1

1. `Calculate the relative molecular mass and relative formula mass of

a. HCl k. CaCO3

b. Na2CO3 l. H3PO4

c. Al(NO3)3 m. Cu(NO3)2

d. CuSO4.5H2O n. (NH4)2SO4

e. I2 o. CO2

f. SF6 p.FeSO4

g. NO2 q. Mg(OH)2

h. CCl4 r. MgSO4.7H2O

i. FeCl2 s. FeSO4

j. NaCl t. Al2(SO4)3

[H,1;Cl,35.5; N,14 ;Na,23; C,12; O,16; Al,27; Cu,64; S,32; I,127; F,18; Fe, 56; Ca,40; P,31]

2. The relative molecule mass of H2On is 36. Calculate the value of n.

[ Relative atomic mass; H,1 ; O,16]

3. How many atoms of nitrogen have the same masses of one atom of iron? [ N,14; Fe,56]

4. The mass of an atom of an element Y is ten times than the mass of an atom of beryllium. What is

the relative atom of mass Y? [Relative atom mass Be, 9]

5. The element chlorine forms a compound with oxygen, with molecule formula is Cl2On. The

relative molecule mass of the compound is 83. What is the value of n? [Cl.35.5; O,16]

Activity 1.2 The Mole and Number of Particles

Declarative Knowledge

1. Definition i. A mole is an amount of substance that contain as many particles as the number of atom in

exactly 12 g of carbon -12.

ii. Avogadro constant, NA, is defined as the number of particles in one mole of particles or one mole of substance contains 6.02 x10

23 particles

iii. There are three type of particles a. Atom

Learning Outcomes: You should be able to

1. Define a mole

2. State the meaning of Avogadro Constant, NA

3. Relate the number of particles in one mole of substances using Avogadro

Constant.

4. Solve numerical problem to convert the number of mole to number of

particles and vice-versa.

Figure 3: Model of particles

of Calcium atom

Figure 4 ; Model of

water molecule

Figure 2: Model particle

of Sodium chloride

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b. Molecule c. Ion

2. Relation between number of particles in one mole of substances using Avogadro Constant., Example

a. One mole of atomic substances contains 6.02 x1023 atoms. Example : 1 mole of sodium atom, Na, contain 6.02 x 10

23atoms Na

b. One mole of molecular substances contains 6.02 x1023 molecules Example : 1 mole of water, H2O contains6.02 x 10

23water molecules .

c. One mole of ionic substances contains 6.02 x1023 formula units Example: 1 mole of sodium chloride, NaCl contains 6.02 x 10

23Na

+ ion and 6.02 x 10

23

chloride ion,Cl-.

3. Numerical Problem of converting mole particles.

PROCEDURAL KNOWLEDGE

Calculating the number of moles and number of particle

Strategy 1: Remember :

[ 1 mole of a substances contains 6.02 x 1023

molecules or atoms or ions]

Strategy 2:

1. Remember the formula or relationship between number of particles, number of mole and Avogadro number or constant.

Number of particles = Number of mole x NA

Number of moles = Number of particles NA

2. Analogy : 1 dozen contain 12 item. 1 mole contains 6.02x 10

23 particles

Exercise 1.2

[Avogadro constant NA= 6.02 X 1023 mol-1]

1. Calculate the number of particles

a. 3.5 moles lead, Pb. b. 0.5 moles lead (II) ion, Pb2+. c. 0.25 moles carbon dioxide, CO2

2. Find the number of moles of

a. 1.204 x 1024 molecules of chlorine gas, Cl2 b. 12.04 x1023 atom of ammonia, NH3

3. Calculate the number of particles and state the type of particle in each of the following, (a) 1/4 mol of copper, Cu (b) 0.3 mole of calcium ions, Ca2+ (c) 1.2 moles of chlorine, Cl2

4. Find the number of moles in the following substances: (a) 4.5 x 1022 atoms of magnesium, Mg (b) 7.2 x 1024 bromide ions, Br- (c) 3x1021 sulphur trioxide molecules, SO3,

5. How many atoms are there in; (a) 1/8 mole tetrachloromethane, CCl4 (b) 0.075 mole sulphur trioxide, SO3.

Activity 1.3 Mole and Mass of substances

How to do?

Problem solving skills.

1. State the problem

2. Identify information given.

3. Formula

4. Solution

5. Check the answer

(unit and value of the quantity)

Learning outcomes

Students be able to

1. State the meaning of molar mass 2. Relate molar mass to Avogadro constant. 3. Relate molar mass of a substance to its relative atomic mass or relative

molecular mass.

4. Solve numerical problem to convert number of mole of a given substances to its mass and vice-versa.

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Declarative knowledga:

molar mass, Avogadro constant, NA, Relative atomic Mass,RAM, Relative Molecular Mass, RMM.

Definition

1. The mass of one mole of any substances, in gram is called Molar Mass. 2. The molar mass of any substances is numerically equal to its relative atomic mass, RAM or

Relative molecular Mass,RMM or relative formula mass, RFM in grams.

3. The molar mass of a substances = The mass of 1 mole of the substances = The mass of NA number of particles.

= The mass of 6.02 x 1023

particles.

PPROCEDURAL KNOWLEDGE

Calculation involving number of moles, mass and number of particles and Avogadro constant.

(Procedural metacognition)

Example: 1[ Convert mole to mass]

Calculate the mass of the following:

(a) 0.3 mole of chlorine, Cl2 (b) 1/8 mole of CO2 (c) 3.2 moles ammonia gas, NH3

Problem solving:

1.Problem:Convert 0.3 mole chlorine, Cl2 to mass

2.Formula: Mass = Number of mole x molar mass (RMM/RAM)

3. Solution

a. 0.3 mole Cl2= 0.3 x (35.5x2) =21.3 g

b. 1/8 mole CO2 = 1/8x 44g = 5.5g

c. 3.2 mole NH3 = 3.2x (14+3x1)g= 54.4g

Example 2 ( convert gram to number mole)

Find the number of moles in

(a) 31.5g copper, Cu (b) 45g ethane gas,C2H6

[Relative atomic mass; H,1: C,12; Cu,64]

Problem solving

1. Problem: Convert gram to mole

2. Formula; Mole = mass / molar mass

3.Solution:

a. Number of moles 31.5 g Cu = 31.5=0.49 mole 64

b. Number of moles 45 g C2H6 = 45 = 1.5 mole 30

Example 3 (convert mass to number of mole)

Calculate the number of moles in

(a) 20g methane, CH4 (b) 372.6 g lead, Pb (c) 4.68 g water, H2O [Relative atomic mass; H,1; C,12; O,16; Pb,207; NA=6.023 x10

23]

Problem solving

1. Problem: Convert gram to mole

2. Formula; Mole = mass / molar mass

3.Solution:

Figure 6; Solid of

Aluminium, Al

Figure 7; Crystal of

Copper(II) sulphate, CuSO4

Figure 5: solid of

Sulphur,S

Problem solving skills.


2. Identify information given.

3. Formula

4. Solution

5. Check the answer

(unit and value of the quantity)

How to do?

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a. 20g methane CH4 = 20 = 20 =1.25 mole 12+4 16

b. 372.6g lead, Pb = 372.6 = 1.8 mole 207

c. 4.66g water,H2O = 4.66= 0.26 mole 18

Example 4 [Convert particles mole mass(g)]

Find the mass of the following;

(a) 2 x1023 atoms of oxygen, O (b) 1.2 x 1023 molecule of nitrogen dioxide, NO2 (c) 1.56 x 1023 atoms of silver, Ag

[Relative atomic mass; O,16; N,14; Ag,108; NA=6.023 x 1023

]

Problem solving

1. Problem: Convert particle to mole to mass(g)

2. Formula; Mole = Number of particles

NA

Mass(g) = mole x molar mass(RMM/RAM)

3.Solution

(a) Number of moles in 2 x1023 atoms of oxygen, O = 2 x1023 6.02 x 10

23

=3.3 moles

Mass of 3.3 moles= 3.3 x 16= 52.8 g

(b) Number of moles in 1.2 x 1023 =0.02 moles NO2 6.02 x 10

23

Mass of 0.02 moles = 0.02 x (14+ 2x16) =0.02 x 46=0.92 g

(c) Number of moles of 1.56 x 1023 = 0.26 moles Ag 6.023 x 10

23

Mass of 0.26 moles Ag = 0.26 x 108 = 28.08 g

Exercise 1.3

1. Find the number of particles of 60g of oxygen gas, O2? 2. Find (a) number of moles (b) mass from the number of 1.2 x1023 atom of ozone, O3. 3. Calculate, (a) number of molecule and number of atoms (b) mass of 0.8 moles molecule of

benzene, C6H6.

4. Calculate 5 g of Neon, Ne (a) Number of moles (5) Number of particles (6) Calculate 1.2 moles atoms of hydrogen peroxide, H2O2 of

(a) Mass (b) Number of molecules and number of atoms.

(7) Calculate 1x1023 molecules of ethanol, C2H5OH of (a) Mass (b) Number of moles

(8) Calculate the number of moles 7.4 g calcium hydroxide, Ca(OH)2 [Relative atomic mass; H,1; O,16; Ca,40, Ne=20, NA=6.023x10

23]

Activity 1.4 Number Moles of Gas and Its Molar Volume

Goal : Students be able to 1) understand the concept of Molar Volume

2) calculate the numerical problem.

Strategy:

1. Students remember the definition of Molar Volume

and the relations between molar mass, molar volume

and Avogadro constant, NA

2. Analysis the numerical problem.

Problem solving skills


2. Identify information given

3. State the formula or relationship.

4. Use information given in

formula to solve problem.

5. Checking the answer

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DECLARATIVE KNOWLEDGE

Introduction

All gases compound such as hydrogen gas, H2, nitrogen gas, N2, Oxygen gas,O2, Flourine gas, F2,

Bromine gas, Br2, chlorine gas, Cl2, carbon dioxide, CO2, Carbon monoxide,CO, Ammonia gas,

NH3,Tetrachloride methane, CCl 4 are involving molar volume.

1. The molar volume of a gas is define as the volume occupied by 1 mole of a gas at particular temperature and pressure.

2. At room condition ( temperature is 25oC and pressure at 1 atmosphere), 1 mole of gas or molar volume occupies a volume of 24 dm

3.

3. At standard temperature and pressure or s.t.p condition or at temperature 0oC and pressure 1 atmosphere, 1 mole of gas or molar volume occupies a volume of 22.4 dm

3.

4. Therefore at standard temperature and pressure (s.t.p) 1 mole of gas occupies 22.4 dm3, examples; a. 1 mole of hydrogen gas, H2 (= 2g) occupies .22.4 dm

3 at s.t.p.

b. 1 mole of chlorine, Cl2 (=71g) occupies 22.4.. dm3 at s.t.p.

c. 1 mole of neon, Ne (=20g) occupies 22.4..dm3 at s.t.p. d. 1 mole of carbon dioxide, CO2 (=44g) occupies 22.4dm

3 at s.t.p.

5. 1 mole of any gas at the same temperature and pressure contains the same number of particles, that is 6.02 x 10

23 particles (atoms, molecules or ions) or Avogadro constant.

a. At s.t.p 22.4 dm3 (1 mole) of gas hydrogen, H2, contains 6.02 x1023

molecules H2.

b. At s.t.p 22.4 dm3 (1 mole) of carbon dioxide, CO2 contains6.02x1023

.molecules.

c. At s.t.p 22.4dm3 (1 mole) of neon gas, Ne, contains6.02x1023.. atom of Ne.


Calculating Involving Gas

Example 1

Calculate the volume occupied by the following gases at s.t.p.

a. 3.5 moles of ammonia gas, NH3 b. 2.8g carbon dioxide,CO2 c. 1.5 x1023 molecules of oxygen, O2 [Na =6.02 x10

23 , 1 mole gas occupies 22.4 dm

3 at s.t.p]

Problem solving:

1.Problem:Convert mole to volume of gas

2. Information given: 1 mole gas occupies 22.4 dm3 at s.t.p

3.Formula: Mass = Number of mole x molar gas

4. Solution

a. 1 mole of ammonia gas, NH3 occupies 22.4 dm3 at s.t.p.

3.5 mole of ammonia gas, NH3 occupies 3.5 x 22.4 dm3 =78.4 dm

3.

b. Formula ; mole = mass , Volume gas = mole x molar volume

molar mass

So0lution: 2.8g carbon monoxide, CO = 2.8/28 = 0.1 mole ,[RMM CO =12+16=28]

Volume of carbon monoxide gas, CO = 0.1mole x 22.4 dm3 = 2.24 dm

3 at stp

c. Formula : mole = number particles , volume gas = mole x molar volume

6.02 x1023

Solution:

1.5 x1023 molecules = 1.5 x1023 mole 6x1023 =0.25 mole Volume of oxygen gas, O2 = 0.25 mole x 22.4 dm

3 = 5.6 dm

3

Strategy : We use the formula ,

i. number of mole = volume of gas occupies

Molar volume

i. Number of molecule = number of molex NA ii. Mass = Number of moles x RMM

Exercise 1.4

1. Calculate the volume occupied by the following gases at s.t.p a. 0.5 mole of neon gas, Ne b. 1.5 moles of carbon dioxide, CO2 c. 4.8 x 1023 molecules chlorine, Cl2 d. 0.5 g of neon gas, Ne

2. Find the number of molecule of the following gases at 25oC a. 1.2 dm3 of methane , CH4 b. 6 dm3 of ethane, C2H6

CO2

Figure 8: Bromine gas ,Br2

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c. 180 cm3 of hydrogen , H2 3. Calculate the mass of the following gases at room conditions.

a. 3.6 dm3 of ammonia, NH3 b. 240 cm3 of dinitrogen oxide, N2O c. 0.48 cm3 of carbon monoxide , CO

[ Molar volume at s.t.p = 22.4 dm3 , Molar volume at 25

oC equal to 24 dm

3,

NA = 6 x 1023

, H,1; C,12 ; O,16 ; N. 14; Cl, 35.5; Ne,20]

Activity 1.5 Chemical Formula

1.5.1 Empirical Formula and Molecular Formula


1. Define chemical formula

Chemical formula is a representation of a chemical substances using letters for atoms and subscript

number to show the numbers of each type of atoms that are present in the substances.

Example; Chemical for water: H2O, (H represent for hydrogen atom, O represent for oxygen atom and

subscript 2 represent number of hydrogen atom and number of oxygen atom is one).

2. There is two type of chemical formula to represent of substances.

a. Empirical Formula

b. Molecular Formula. C. Ionic chemical formula

3. Empirical Formula of a compound is formula to show the simplest whole number ratio of atoms of

each element in the compound.

a.Example : molecular formula of glucose is C6H12O6

The ratio of carbon atom to hydrogen atom to oxygen atom in glucose molecule is

C : H : O= 6 : 12 : 6 =1 : 2 : 1 .

Empirical formula of glucose is CH2O.

b. The empirical formula can be determined if we follow the following data

i. mass of element ( % mass of element)

ii. Relative atomic mass for each elements

iii. find simplest ratio of atom for each elements

iv. write the empirical formula for letter of each symbol elements and subscript number of each

elements.

4 . Molecular Formula of a compound is formula to show the actual number of atoms of each element

that are present in a molecule of a compound.

Molecular formula is n times of empirical formula. Whereby n is subscript positive integer.

5. The molecular of a compound can be determined if we know the following data

a. The empirical formula

b. Its relative molecular mass or molar mass.

Learning outcomes

You should be able to;

1. state the meaning of chemical formula

2. state the meaning of empirical formula

3. state the meaning of molecular formula.

4. determine the empirical formula and molecular formula of substances.

5. compare and contrast empirical formula and molecular formula.

6. solve numerical problems involving empirical and molecular formula.

Molecule formula = (Empirical Formula)n

Remember

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Example

A compound has an empirical formula of CH2 and relative molecular mass is 70. Find the molecular

formula of the compound.

[Relative atomic mass: H,1; C,12]

Solution

Relative molecular mass, (C1H2)n =70

(1x12 + 2x1)n =70

14n =70

n = 5

So, the molecular formula of the compound is = (CH2)5 which is C5H10

EXERCISE 1.5 (a)

Substances Molecular

formula

Number of atoms Empirical

Formula

Number of n

Actual Simplest ratio

Water H2O H:O = 2:1 H:1 = 2:1 H2O

Carbon dioxide CO2 C:O = 1 :2 CO2

Magnesium oxide MgO Mg:O =

Aluminium chloride AlCl3 Al:Cl=

Ammonia NH3 N:H =

Glucose C6H12O6

Ethene C2H4

Propene C3H6

Butene C4H8

Ethane C2H6

Benzene C6H6

Example

2.3 g of sodium react with 0.8 g oxygen to produce an oxide.

Find the empirical formula of the oxide.

[Relative atomic mass: Na, 23; O, 16]

(Use table to solve the problem)

Element Na O

1 Mass(g) 2.3 0.8

2 RAM 23 16

3 Number of moles 2.3

23

= 0.1

0.8

16

0.05

4 Ratio of mole

divided by smallest

mole

0.1

0.05

=2

0.05

0.05

=1

5 Empirical Formula Na2O

Problem solving skill

1. Aim: to find the empirical formula

2. Information given: RAM Na,23; O,16; 2.3 g Na.

3.Relationship of concept or formula:

Mole = mass (g)

RAM

Empirical formula as ratio simplest mole atom of each elements

4. Solution:a. Mass (g) (or % mass) convert to mole atom

(mole atom=mass/RAM)

b. Divide with the smallest mole to get the whole

number of simplest ratio for each elements,

c. Write empirical formula; write symbol of

element and its number of elements.

5. Checking: Checking the procedure and the answer once or twice.

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Exercise 1.5(b)

1. 1.69 g of iron react with 0.72 g of oxygen. Find the empirical formula of the iron oxide.

[Relative atomic mass; Fe, 56: O, 16].

(Fe2O3).

2. 2.7 g of X metal react with chlorine to form 13.35 g of X chloride. What is the empirical formula of

this compound? [Relative atomic mass: X, 27: Cl, 35.5]

(XCl3).

3. 1.72 g of metal X oxide contain 0.8 g of oxygen. Find the empirical formula of oxide.

[Relative atomic mass: O,16; X,46]

(X2O5).

4. Y oxide contain 0.5 mol of element Y atom that react with 4 g of oxygen. Find the empirical formula of the oxide. [Relative atomic mass: O, 16]

(Y2O).

5. A compound that make up of carbon, hydrogen and oxygen contain 40% of carbon, 6.65% of

hydrogen and X % of oxygen. What is the empirical of this compound?

[Relative atomic mass: C,12; H,1: O,16]

(CH2O).

6. An organic compound has an empirical formula of C2H5. If it has a relative molecular mass of 58,

what is the molecular formula of this compound?

[Relative atomic mass: C,12; H,1]

(C4H10).

7. 3.9 g of element Y react with 0.8 g of element Z to produce a compound with a molecular formula

Y2Z. Find the relative atomic mass of element Z?

[Relative atomic mass: Y, 39]

(16)

8. 4.6 g of element P react with 1.6 g of oxygen to form P oxide with a formula of P2O. Find the relative

atomic mass of element P?

[Relative atomic mass: O, 16]

(23)

9 An element E formed EH4 hydride that contains 87.5% of E by mass. Find the relative atomic mass of

element E? [Relative atomic mass: H,1]

(28)

10 What is the mass of metal T that combine with 4.8 g of oxygen to form metal oxide with an empirical

formula of T2O3? [Relative atomic mass: T, 56; O,16] ( 11.2g)

Experiment 1.6 Empirical Formula of Magnesium Oxide

Aim: To determine the empirical formula of magnesium oxide

Material: Magnesium ribbon, sand paper

Apparatus: Crucible with lid, bunsen burner, tong, fireclay triangle, chemical balance, tripod stand.

Figure: Set- up of apparatus

Result

Magnesium ribbon

Crucible and lid

Heat

Procedure

1. A crucible and its lid are weighted. 2. 10 cm of magnesium ribbon is cleaned with a sand paper. 3. The magnesium ribbon is coiled loosely and place in the

crucible together with the lid

4. The crucible together with the lid and magnesium ribbon are weight again.

5. The crucible is heated strongly without its lid. When the magnesium starts to burn, the crucible is covered with its lid.

6. The lid of the crucible is lifted from time to time using a pair of tongs.

7. When the magnesium ribbon stops burning, the lid is removed and the crucial is heated.

8. The crucible with lid is allowed to cool down to room temperature.

9. The crucible with its lid and its content are weighing are repeated until a constant mass is obtained

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Result,

Mass of crucible + lid =...g.

Mass of crucible + lid + magnesium =g

Mass of crucible + lid + magnesium oxide from the first heating =..g

Mass of crucible + lid + magnesium oxide from the second heating =.g

Precaution

1. Sand paper is used to removed oxide layer on the surface of the magnesium ribbon. 2. The crucible lid is lifted to allow oxygen from their air to react with magnesium. 3. The crucible lid is then closed quickly to prevent fumes of magnesium oxide from escaping. 4. The crucible of heating, cooling and weighing are repeated until constant mass is obtained to

ensure magnesium react completely with oxygen to form magnesium oxide.

Calculation

Elements Magnseium, Mg Oxygen, O

Mass (g)

Number of mole

Simplest ratio

Empirical formula of

magnesium oxide

Analysis

1. How many mol of magnesium react with 1 mol of oxygen atom? 2. Write the chemical equation of the reaction. 3. Why the magnesium ribbon is cleaned with sand paper? 4. Why the lid is closed quickly while burning the magnesium ribbon? 5. Why the lid is lifted while burning the magnesium ribbon? 6. Why the product is heated, cooled and weighted repeatedly? 7. What is the empirical formula of magnesium oxide?

Activity 1.7 Chemical Formula of Ionic Compound


1. Define Ionic Formula

a. Chemical formula of ionic compound is called ionic formula.

b. Ionic compound are made up of positively charge ions, called cation and negatively charge ions

called anions.

c. Number of positively charge ions are same with number of negatively charge ions in a ionic

compound. So charge of the ionic compound is neutral.

d. Particle in ionic compound in aqueous solution or liquid is ions.

In solid state, ion positive and ion negative are hold by strong electrostatic forces.

Learning Outcome

You should be able to:

1. write ionic formulae of ions.

2. construct chemical formula of ionic compounds

3. state names of chemical compounds using IUPAC nomenclature.

4. use symbols and chemical formula for easy and systematic communication in the field of

chemistry.

Sodium chloride, NaCl Sodium oxide, Na2O

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2. Positive charge ions are from metal in Group 1, 2, 3 and transition metal. Number of charge depends

on position of metal in Group.

a. Group 1 positive one charge (+)

b. Group 2 positive two charge (+2)

c. Group 3 positive three charge (+3)

d. Transition metal various positive charge.

3. Negatively charge ion are from non metal,

a. Group 15 negatively three charge (-3)

b. Group 16 negatively two charge (-2)

c. Group 17 negatively one charge(-1)

Naming the Ionic compound.

1. Chemistry have devised clear and systematic ways of naming chemical compound. It is based on the

recommendation of the International Union of Pure and Applied Chemistry (IUPAC)

2. Name of negative charge ion is added ide to the name of element.

Example : Oxygen become oxide

Chlorine become chloride

The end of the name ate means contain oxygen

Negatively charge ion consists of two elements called polyatomic ions.

3. a. Name of positively charge is same as name of their metal for Group 1, 2, 3 and 4.

b. For transition metal, roman numeral are used in their name to distinguish the different type of ion.

Example: Ion iron(II), formula ion Fe2+

and Ion iron(III), formula ion is Fe3+

.


Writing chemical ionic Formula

Steps to write chemical ionic formula

Step one - Write the symbols for the ions in the compound.

Step two - Look up the oxidation numbers of the ions involved and write them as superscripts to the right

of the elemental symbols.

Step three - Use the correct combination of ions to produce a compound with a net charge of zero.

Multiple Ion are indicated with subscripts. Multiple polyatomic ions must be enclosed in parenthesis.

Example;

A. Charge in cation EQUAL charge in anion

a. Magnesium oxide

CATION: magnesium (Group II) charge is +2: Mg2+

ANION: oxide ion (Group VI or16) charge is -2: O2-

1 Mg2+ cation and 1 O2- anion combine to give a compound with zero net charge, +2 + -2 = 0

Formula of magnesium oxide is MgO

b. Ammonium hydroxide

CATION: ammonium ion, charge is +1: NH4+

ANION: hydroxide ion, charge is -1: OH-

1 NH4+ cation and 1 OH

- anion combine to give a compound with zero net charge, +1 + -1 =0

Formula of ammonium hydroxide is NH4OH

B. Charge on Cation DOES NOT EQUAL Charge on Anion

a. Aluminium chloride

CATION: aluminium (Group III or 13), charge is +3: Al3+

ANION: chloride (Group VII or 17), charge is -1: Cl-

1 Al3+ cation and 3 Cl- anions combine to give a compound with zero net charge: ( +3) + (3 x -1) = 0

Formula of aluminium chloride is AlCl3

396

b. Sodium oxide

CATION: sodium (Group I), charge is +1: Na+

ANION: oxide (Group VI), charge is -2: O2-

2 Na+ cations and 1 O2- anion combine to give a compound with zero net charge: (2 x +1) + -2 = 0 (Na

+, Na

+, O

2-)

Formula of sodium oxide is Na2O

c. Ammonium sulphate

CATION: ammonium, charge is +1: NH4+

ANION: sulfate, charge is -2: SO42-

2 NH4+ cations and 1 SO4

2- anion combine to give a compound with zero net charge:

(2 x +1) + -2 = 0

Formula of ammonium sufate is (NH4)2SO4

c. Barium hydroxide

CATION: barium (Group II or 2), charge is +2: Ba2+

ANION: hydroxide, charge is -1: OH-

1 Ba2+ cation and 2 OH- anions combine to give a compound with zero net charge: (+2) + (2 x -1) = 0

Formula of barium hydroxide is Ba(OH)2

MOLECULAR FORMULA

1. Molecular compound are held together by covalent bond, or share pair of electrons between non-metal

element and non-metal element .

2.Writing Molecular Formula

Example1 Write the correct formula for nitrogen (IV) oxide

Step 1- write the symbols for the element involve Nitrogen , N Oxide, O

Step 2- Use the roman numeral as the apparent charge of the first element. Find the apparent chart of the

second element by looking table 1

Nitrogen = N4+

Oxide O2-

Step 3- Determine the ratio by which the elements will bond to show the net charge is zero. Use subscript

to indicate the number of atoms of each element present.

Example: ( +4) + 2(-2) = 0

So the molecular formula is NO2 Example 2 Write the correct formula for nitrogen(III) oxide

Step 1: Write symbol of elements : Nitrogen = N, Oxide =O

Step 2: Use the roman as the apparent charge of the first element. Find the apparent charge of the second

Elements. Nitrogen =N3+

Oxide=O2-

Step 3: Determine the ration by which the element will bond to show a net charge of zero. Use subscript

to indicate the number of atoms of each present.

In this case, 2(+3) +3(-2) = 0

So molecular formula is N2O3

3. Naming molecular Formula

Naming molecular formula use Greek prefix to show the number of atom of each element. The Greek

prefix use as below,

The Greek Prefix mono di tri tetra penta heksa hepta oct

Number of atom 1 2 3 4 5 6 7 8

Example :

Molecular Formula Name

CO Carbon monoxide

CO2 Carbon dioxide

SO3 Sulphur trioxide

CCl4 Carbon tetrachloride

397

CHEMICAL FORMULAE OF IONIC COMPOUNDS and MOLECULE FORMULA

Name: Form: ..Date:

Exercise 1.7

1. Construct a chemical formula for each of the following ionic compounds

(a) Magnesium chloride .. (j) lead (II) sulphite

(b)Potassium carbonate.. (k) Iron(II) hydroxide .

(c)Calcium carbonate .. . (l) Potassium chlorate(V) .

(d) Calcium sulphate .. (m) Ammonium phosphate.

(e) Copper (I) oxide . (n) Sodium bromate(I) ..

(f)Silver iodide . (o) Copper(II) sulphate

(g) Zinc nitrate . (p)Lead(II) sulphate

(i) Aluminium oxide. (q) Magnesium oxide

2. Name the ionic compound with each of the following formulae

(a) Ba(NO3)2 ... (f) NaHCO3 .

(b) MgO .. (g) KBr .

(c) LiCl . (h) Ca(OH)2 ..

(d) NaOH .. (i) BaSO4 ...

(e) ZnSO4 (j) AgCl

3 Name the following molecule

(a) NO ... (f) BF3 .

(b) NO2 .. (g) N2O4 .

(c) SO2 . (h) H2O ..

(d) SO3 . (i)NH3

(e) CS2 .. (j) CO .

4. Give the molecule formula of these compounds

(a) Ammonia (f)Carbon dioxide ..

(b) Sulfur dioxide ......................... (g) Carbon monoxide

(c) Nitrogen monoxide (h) Sulphur trioxide .

(d) Nitrogen dioxide ... (i) Carbon tetrachloride .

(e) Water .. (j) Boron triflouride ..

ACTIVITY 1.8: CHEMICAL EQUATIONS

Write a balanced equation to represent each of the reaction:

1. Magnesium + oxygen magnesium oxide Mg + O2 MgO

................................................................................................

(Identify the formulae of the reactants and products not balance)

2Mg + O2 2MgO

...............................................................................................

(Balance the equation by adjusting the coefficients)

2. Zinc + hydrochloric acid zink chloride + hydrogen

.................................................................................................

3. Sodium hydroxide + sulphuric acid sodium sulphate + water

4. Lead (II) nitrate + potassium iodide lead(II) iodide + potassium nitrate

Reactant Product

Mg = 1 Mg = 1

O = 2 O = 1 Reactant Product

Mg = 2 Mg = 2

O = 2 O = 2

398

5. Copper (II) carbonate is added to hydrochloric acid solution. Copper (II) chloride solution, water and carbon dioxide gas are produced.

ACTIVITY 1.9 BALANCING CHEMICAL EQUATION

Learning outcomes

You should be able to

1. write chemical equation from statement of chemical reactions

2. balancing chemical equation

3. understand the chemical equations

Exercise

A. Write a balanced chemical equations to represent each of the reaction:

1. When copper (II) carbonate is heated, copper(II) oxide solid, carbon dioxide gas are produce.

.......................................................................................................................................................

2. When calcium carbonate is heated, calcium oxide solid, carbon dioxide gas are produce.

.....................................................................................................................................................

3. When lead(II) nitrate is heated, lead(II) oxide solid, nitrogen dioxide gas are produce.

.................................................................................................................................................

4.When zinc nitrate solid is heated, zinc oxide solid, and nitrogen dioxide gas are produce.

............................................................................................................................................. ..

5. Sodium hydroxide solution react with nitric acid solution produce sodium nitrate solution and water

..................................................................................................................................................... ..

6. Potassium hydroxide solution reacts with hydrochloric acid solution produce potassium chloride

solution and water.

...................................................................................................................................... .................

7. When magnesium metal reacts with sulphuric acid solution, produce magnesium sulphate solution and

hydrogen gas.

......................................................................................................................................................

8.When sodium metal reacts with water, sodium hydroxide solution and hydrogen gas are produce.

.................................................................................................................................. .....................

9. When barium nitrate solutiom reacts with potassium chromate(VI) solution, barium chromate(VI)

precipate and potassium nitrate solution is produce.

.......................................................................................................................................................

10. When aluminium carbonate solution reacts with hydrochloric acid solution, aluminium chloride

solution, carbon dioxide gas and water are produce.

.......................................................................................................................................................

B. Write the meaning of chemical quationS below

1. 2 Na(s) + 2H2O(l) 2 NaOH(aq) + H2(g)

2 mole of sodium metal react two mole of water produce 2 mole of sodium hydroxide solution and

one mole hydrogen gas..

2. MgCO3(s) MgO(s) + CO2(g)

.......................................................................................................................................................

3. LiOH(aq) + HCl(aq) LiCl(aq) + H2O(l)

.......................................................................................................................................................

4. ZnCO3(aq) + H2SO4(aq) ZnSO4(aq) + CO2(g) + H2O(l)

.......................................................................................................................................................

5. Mg(s) + CuSO4(aq) MgSO4(aq) + Cu(s)

.......................................................................................................................................................

6. AgNO3(aq) + KCl(aq) AgCl(s) + KNO3(aq)

............................................................................................................................. ..........................

7. NaNO3(s) NaNO2(s) +O2(g)

.......................................................................................................................................................

399

8. 2NH3(aq) + H2SO4 (NH4)2SO4(aq)

.......................................................................................................................................................

ACTIVITY 1.10 :CHEMICAL EQUATIONS AND STOICHIOMETRY PROBLEM


Aim: Quantitative and Qualitative Chemical equation

Solving problems using chemical equations (Stoichiometry)

Exercise 1.10

1.Methane gas (CH4) burns in excess oxygen to produce carbon dioxide gas and water according to the

following equation:

CH4 (g) + 2O2(g) CO2(g) + 2H2O(g)

a) If 4.0 g of methane reacts with excessive oxygen gas, find

i) the mass of water produced? (9g)

ii) the volume of carbon dioxide gas released at s.t.p (5600 cm3)

iii) the number of particles (molecules) of carbon dioxide gas released? (1.505 x 1023

)

b) Calculate the mass of methane burned, if 14.4 g of water produced. (6.4g) [Relative atomic mass: ; C, 12; H, 1; O, 16;Molar volume: 22.4 dm

3 at s.t.p, Avogadro constant:

6.02 x 1023

particles]

1. When sodium nitrate is heated, it decomposes into sodium nitrite and oxygen gas. 2NaNO3 (p) 2NaNO2 (p) + O2 (g)

a) Find the mass of sodium nitrite produced if 34 g of sodium nitrate is heated. (27.6g)

b) Find the mass of sodium nitrate needed to produce 600 cm3 oxygen gas at room condition.

(4.25g)

[Relative atomic mass: Na, 23; N, 14; O, 16; Molar volume: 24 dm3 at room temperature]

3. Reaction of calcium carbonate with excessive dilute hydochloric acid will produce calcium

chloride, water and carbon dioxide gas.

a) Find the mass of calcium carbonate needed to react completely with 0.5 mole of dilute hydochloric

acid? (25g)

b) Find the volume of carbon dioxide gas released if 50 g of calcium carbonate reacts completely with

excessive dilute hydochloric acid at room conditions. (1 200 cm3)

[Relative atomic mass: Ca, 40; C, 12; O, 16: Molar volume: 24 dm3 at room temperature]

4. If 448 cm3 hydrogen gas at s.t.p burns completely with excessive oxygen gas, what is the mass of

water produced? (0.36g)

[Relative atomic mass: ; H, 1; O, 16;Molar volume: 22.4 dm3 at s.t.p]

5. 1.35 g of aluminium reacts with excessive copper (II) oxide to produce aluminium oxide powder and

copper. Find the number of copper atoms produced.

(4.515 x 1022

)

[Relative atomic mass; Al, 27; Avogadro constant: 6.02 x 1023

particles]

FORMULAE OF MOLES

1.

2.

Mole

e

No. of particles

6.02 X 1023 =

Mole

Mass (g)

Ar or Mr =

1.

Mole

ele

Volume of gas

22, 400 =

at s.t.p

3. Mole Volume of gas

24, 000 =

at room temperature

s.t.p

or

The steps involved in stoichiometric

calculations are as follows:

1. Write the balanced equation of the reaction. 2. Gather all information from the question:

Convert the given information to the number

of moles if necessary

3. Based on the equation, compare the ratio of moles of the related substances

4. Calculate the answer proportionately. Then, convert the answers to the required units .

400

ACTIVITY 1.1 RELATIVE ATOMIC MASS AND RELATIVE MOLECULAR

MASS

Answer Relative Molecular Mass (Exercise 1.1)

No Answer No Answer

1

(a) HCl = 36.5

(b) Na2CO3 =106 (c) Al(NO3)2 =151 (d) CuSO4.5H2O =250 (e) I2 =254 (f) SF6 =140 (g) NO2 =46 (h) CCl4 =154 (i) FeCl2 =127 (j) NaCl =58.5 (k) CaCO3=100 (l) H3PO4 =98 (m) Cu(NO3)2 =188 (n) (NH4)2SO4 =132 (o) CO2 =44 (p) FeSO4 =152 (q) Mg(OH)2 =58 (r) MgSO4.7H2O =246 (s) FeSO4 =152 (t) Al2(SO4)3 =342

2

3

4

5

2x1 +16n=36

16n= 36-2=34

n =34/16= 2.125=2

56/16= 4

9x10=90

35.5x 2+ 16n= 83

16n =83-71=12

n =12/16

=0.75=1

ACTIVITY 1.2 THE MOLE AND NUMBER OF PARTICLES

(Execise 1.2 Answer)

1. (a) Aim of Problem: Calculate the number of particles of 3.5 mole lead, Pb Given: NA = 6.023 x10

23 particles mole

-1

Relation of concept: mole particle

Number of particles = mole x NA

Solution:

Number of particle = 3.5mole x 6.02 x1023

atom Pb mole-1

=21.08x1023

=2.11x10x1023

atom

=2.11x 1024

atom Pb

(b). Aim :Calculate the number of particles 0.5 mols lead, Pb2+

ion

Given: NA = 6.02 x1023

particles mole -1


Number of particles = mole x NA

Solution

Number of particles= 0.5 mole ions Pb2+

x 6.02 x1023

ions mole-1

=3.01 x1023

ion Pb2+

(c) Aim of Problem: Calculate the number of particles of0.25 moles of carbon dioxide Given: NA = 6.023 x10

23 particles mole

-1

Solution

Number of particles= 0.25 mol molecule CO2 x 6.023 x1023

molscules mole-1

=1.505x 1023

molecules

2. (a)Aim of Problem: Find the number of moles of chlorine gas, Cl2 Given :NA = 6.023 x10

23 particles

mole

-1

Relation of concept: particles mole

Number of mole = Number of particle/NA

Solution Number of moles =1.204 x10

24 molecule gas chlorine =0.199x10

24-23 mole

6.023X1023

molecules

=0.2x10=2 mole

(b) Aim of Problem: Find the number of moles of 12.04x1023

of ammonia

Given, NA = 6.023 x1023

particles mole

-1

401


Number of mole = number of particles

NA Solution

Number of moles =12.04 x1023

atom of ammonia, NH3 = 1.99 =2 mole 6.023 x 10

23

3. (a) Aim of Problem : Calculate the number of particles mol of copper atom Given: NA = 6.023 x10

23 particles

mole

-1

Relation of concept: mole particles

Number of particle = number of mole x NA

Solution

Number of particles for 1/4 mol of copper, Cu

=1x 6.023 x1023

atom copper

4

= 1.505 x1023

atom Cu

(b) Aim of Problem : Calculate the number of particles

Given:NA = 6.023 x1023

particles mole

-1


Number of particle =Number of mole x NA

Solution

Number of ions for 0.3 mole of calcium ions, Ca2+

= 0.3 x 6.023 x1023

ion Calcium

= 1.81 x 1023

ions

(c) Aim of Problem : Calculate the number of particles

Given:NA = 6.023 x1023

particles mole

-1


Solution

Number of particles of 1.2 moles of chlorine, Cl2

= 1.2 x6.023 x1023

molecule chlorine

=7.23 x1023

molecule

= 1.2 x 6.023 x 1023

x 2 atom chlorine

=14.46 x1023

=1.4x1024

atom

4. (a) Aim of Problem : Calculate the number of mole Given:NA = 6.023 x10

23 particles

mole

-1


Number of mole = Number of particle

NA

Solution:

Number mole of 4.5 x 1022

atoms of magnesium, Mg

= 4.5 x1022

atom Mg

6.023 x1023

atom Mg

= 0.7 x 1022-23

= 0.7x10-1

= 0.07 mole

(b) Aim of Problem : Calculate the number of mole

Given:NA = 6.023 x1023

particles mole

-1


Number of mole = Number of particle NA

Solution:

Number mole of 7.2 x 1024

bromide ions, Br-

= 7.2 x1024

bromide ion

6.023 x 1023

bromide ion

=11.95 mole

=1.196x 101 mole

402

(c) Aim of Problem : Calculate the number of mole

Given:NA = 6.023 x1023

particles mole

-1


Number of mole= number of particle

NA

Solution:

Number mole of 3x1021

sulphur trioxide molecules, SO3 = 3 x10

21 molecules SO3

6.023 x1023

=0.498 x 10-2

mole

=4.98 x10-3

mole

5. (a) Aim of problem: How many atoms Given:NA = 6.023 x10

23 particles

mole

-1


Number particle = number of mole x NA x number of atom per molecule

Solution:

Number of atom for 1/8 mole tetrachloromethane, CCl4

=1 x 6.023 x1023

molekul CCl4

8

= 1 x6.023 x1023

x 5 atom = 3.76 x1023

atom

8

(b) Number of atom in SO3=0.075 x4 x 6.02x1023

atom =1.806x1023

atom

ACTIVITY 1.3 MOLE AND MASS OF SUBSTANCES

(Exercise 1.3 Answer)

1. Problem/aim: Find the number of particles of 60g of oxygen gas, O2.? Information given: RAM oxygen =16, NA=6.023x10

23

Relation of concept or formula : Mass mole particles

Solution;

Number of mole 60 g oxygen, O2 = 60 mole

2x16

Number of particles O2 = 60 x 6.023x1023

molecules= 1.12 x1024

molecules

2x16

Checking: unit is molecule

2. (a) Aim of problem : Find number of moles the number of 1.2 x1023 atom of ozone, O3 Given: NA=6.023x10

23

Relation of concept: Particles mole

Solution: Number of mole = 1.2 x1023

atom of ozone, O3 = 0.20 mole

6.023x1023

Checking: unit; mole (b) Aim of Problem : Find mass the number of of azone, O3

Information given: RAM oxygen =16, number of 1.2 x1023

atom

Relation of concept; Mole mass(g)

Solution: 1.2 x10

23 atom =0.20 mole

6.023x1023

Mass = 0.20 x RMM O3 = 0.20 x 16x3 g=9.6 g

Checking: unit is gram

3. (a) Aim of problem :Calculate number of molecule and number of atoms of 0.8 moles molecule benzene, C6H6.

Given : NA=6.023x1023

Relation of concept: mole molecule atom

Solution : 0.8 mole molecules benzene

= 0.8 x 6.023x1023

molecules

= 4.816 x 1023

molecule

= 0.8 x 6.023x1023

x 12 atom = 5.78 x1024

atom

Aim of problem :

Calculate mass of 0.8 moles molecule of benzene, C6H6.

Information given : RAM; C=12; H,1

Relation of concept : mole mass(g)

Solution: 0.8 mole x (12x6+1x6) g = 62.4 g

403

Checking : unit is gram.

4. Calculate 5 g of Neon, Ne (a) Number of moles

5 = 0.25 mole

20

(b) Number of particles 5x 6.03 x10

23 = 1.50 x10

23 atom

20

5.Calculate 1.2 moles atoms of hydrogen peroxide, H2O2 of

(a) Mass= 1.2 x (2+16x2) = 40.8 g (b)Number of molecules and number of atoms.

=1.2 x 6.03 x1023

= 7.23 x1023

mlecules

7.23x 1023

x 4 =2.9x 1024

atom

6.Calculate 1x1023

molecules of ethanol, C2H5OH of

(a)Mass

=1x1023

x (12x6+6+16) = 7.6 g

6.02 x1023

(b) Number of moles

= 1x1023

= 0.17 moles a. x1023

7.Calculate the number of moles 7.4 g calcium hydroxide, Ca(OH)2 = 7.4 = 7.4 =0.1 mole ( 40 +16x2+2) 74

ACTIVITY 1.4 NUMBER MOLES OF GAS AND MOLAR VOLUME

Exercise 1.4 Answer

1. a Volume of neon gas, Ne = 0.5 x 22.4 dm3 = 11.2 dm

3.

a. Volume 1.5 mole x 22.4 dm3 = 33.6 dm3 b. 4.8 x1023 mole

6 x1023

Volume = 4.8 x1023

x22.4 dm3 = 17.92dm3

6x10

23

d. Number of mole =0.5g/20 mole

Volume of neon gas at s.t.p = 0.5x22.4 dm= 0.56dm3

20

2.a. Number of mole 1.2 = 0.05 mole

24 dm3

Number of molecule = 1.2 x6x1023

= 3 mole x1023

Molecules

24

b. number of mole =6 = 0.25 mole

24

Number of molecule = 6 x6x1023

=15 x1023

mole

24

c.Number of mole = 180cm3

1000x 24dm3

Number of molecule = 180x6x1023

molecules =0.45 x1023

mole 4.5 x 1022

molecules

1000x24

3. a.Number of mole = 3.6 mole=0.15 mole

24

Mass = 3.6 xRMM NH3 = 3.6 x[14+3]/24 =0.025 g

b. Number of mole = 240cm3 x 6x10

23 =0.06 x10

23

1000x24

c. 0.48 mole

1000x24

Number of mole = 0.48x 6x1023

=1.2x1019

1000x24

404

ACTIVITY 1.5(a) CHEMICAL FORMULA

Exercise 1.5 Answer

Substances Molecular

formula

Number of atoms Empirical

Formula

Number of n

Actual Simplest ratio

Water H2O H:O = 2:1 H:1 = 2:1 H2O 1

Carbon dioxide CO2 C:O = 1 :2 C:O = 1:2 CO2 1

Magnesium oxide MgO Mg:O = 1 : 1 Mg: O= 1:1 MgO 1

Aluminium chloride AlCl3 Al:Cl= 1:3 Al: Cl=1:3 AlCl3 1

Ammonia NH3 N:H = 1:3 N:H= 1:3 NH3 1

Glucose C6H12O6 C:H:O =6:12:6 C:H:O=1:2:1 CH2O 6

Ethene C2H4 C: H = 2:4 C:H =1:2 CH2 2

Propene C3H6 C:H =3:6 C:H=1:2 CH2 3

Butene C4H8 C:H=4:8 C:H=1:2 CH2 4

Ethane C2H6 C:H=2:6 C:H=1:3 CH3 2

Benzene C6H6 C:H=6:6 C:H=1:1 CH 6

Experiment 1.6 Empirical Formula of Magnesium Oxide

Answer

Result

Mass of crucible + lid = 27.20 g

Mass of crucible + lid + magnesium = 27.30 g

Mass of crucible + lid + magnesium oxide from the first heating = 27.50 g

Mass of crucibke + lid + magnesium oxide from the second heating= 27.50g

Analysis

2. 1mol.. of magnesium atom has reacted with1 mol ..of oxygen atoms.

3. 2 .Mg(s) + O2(g) 2 MgO.. 4. To remove the layer of magnesium oxide on its surface. 5. To avoid white fume of magnesium oxide from escaping to the surrounding. 6. To allow oxygen from outside to diffuse into the crucible so that the burning process of magnesium

occurs continuously.

7. To ensure that all the magnesium react completely with oxygen. 8. Conclusion: Empirical formula of magnesium oxide is MgO

ACTIVITY 1.7(b) CHEMICAL FORMULA OF IONIC COMPOUND AND

MOLECULAR COMPOUND

Answer

1.(a) Magnesium chloride MgCl2 (j) lead (II) sulphite .PbSO3

(b)Potassium carbonateK2CO3. (k) Iron(II) hydroxide ..Fe(OH)2

(c)Calcium carbonate CaCO3 (l) Potassium chlorate(V).. KClO3

(d) Calcium sulphate CaSO4 (m) Ammonium phosphate..(NH4)3PO3

(e) Copper (I) oxide .Cu2O. (n) Sodium bromate(I) NaBrO..

(f)Silver iodide .AgI (o) Copper(II) sulphate CuSO4.

(g) Zinc nitrate .Zn(NO3)2 (p)Lead(II) sulphate PbSO4

(i) Aluminium oxideAl2O3. (q) Magnesium oxide.MgO.

2. Name the ionic compound with each of the following formulae

(a) Ba(NO3)2 Barium nitrate. (f) NaHCO3Sodium hydrogen carbonate..

(b) MgO ...Magnesium oxide (g) KBr ...Potassium bromide

(c) LiCl .Lithium chloride (h) Ca(OH)2 ..Calcium hydoxide

(d) NaOH Sodium hydroxide (i) BaSO4 Barium sulphate....

(e) ZnSO4 .Zinc sulphate (j) AgCl Silver chloride...

3.Name the following molecule

(a) NO nitrogen monoxide (f) BF3 ..Boron trifloride.

405

(b) NO2 nitrogen dioxide (g) N2O4 ..Dinitrogen tetraoxide

(c) SO2 sulphur dioxide (h) H2O Dihydrogen oxide = water

(d) SO3 sulphur trioxide (i)NH3 ..Nitogen trihydrogen = ammonia.

(e) CS2 carbon disulphide. (j) CO . carbon monoxide.

6. Give the molecule formula of these compounds

(a) Ammonia .NH3 (f)Carbon dioxide CO2..

(b) Sulfur dioxide .........SO2................ (g) Carbon monoxideCO

(c) Nitrogen monoxide NO. (h) Sulphur trioxide ..SO3.

(d) Nitrogen dioxide . ..NO2 (i) Carbon tetrachloride CCl4.

(e) Water H2O. (j) Boron triflouride ..BF3. N

ACTIVITY 1.8: CHEMICAL EQUATIONS

Answer

2. Zinc + hydrochloric acid zink chloride + hydrogen

.....Zn + HCl ZnCl2 + H2.............

.... .. Zn + 2HCl ZnCl2 + H2........

3. Sodium hydroxide + sulphuric acid sodium sulphate + water

NaOH + H2SO4 Na2SO4 + H2O

2NaOH + H2SO4 Na2SO4 + 2H2O

4.Lead (II) nitrate + potassium iodide lead(II) iodide + potassium nitrate

Pb(NO3)2 + KI PB(NO3)2 + KNO3

Pb(NO3)2 + 2KI PbI2 + 2KNO3

5. Copper (II) carbonate is added to hydrochloric acid solution. Copper (II) chloride solution, water and

carbon dioxide gas are produced.

CuCO3 + HCl CuCl2 + H2O + CO2

CuCO3 + 2HCl CuCl2 + H2O + CO2.

ACTIVITY 1.9 BALANCING CHEMICAL EQUATION

Exercise ( Answer)

A. Write a balanced chemical equations to represent each of the reaction:

1. When copper (II) carbonate is heated, copper(II) oxide solid, carbon dioxide gas are produce.

........................CuCO3(s) CuO(s) + CO2(g) ............................................................ 2. When calcium carbonate is heated, calcium oxide solid, carbon dioxide gas are produce.

.....................CaCO3 CaO + CO2(g)...................................................................... 3. When lead(II) nitrate is heated, lead(II) oxide solid, nitrogen dioxide gas are produce.

....................Pb(NO3)2 (S) PbO(s) + NO2(g) + O2(g)...................................................

........................ 2Pb(NO3)2 (S)2 PbO(s) + 4NO2(g) + O2(g)........................................ 4.When zinc nitrate solid is heated, zinc oxide solid, and nitrogen dioxide gas are produce.

.......................Zn(NO3)2. ZnO + NO2(g) + O2(g)........................................................

........................ 2Zn(NO3)2(s) 2 ZnO(s) +4 NO2(g) + O2(g)........................................ 5. Sodium hydroxide solution react with nitric acid solution produce sodium nitrate solution and water

...........................NaOH(aq) + HNO3(aq) NaNO3 + H2O ......................................... 6. Potassium hydroxide solution reacts with hydrochloric acid solution produce potassium chloride

solution and water.

....................................KOH(aq) + HCl(aq) KCl(aq) + H2O(l)...................................... 7. When magnesium metal reacts with sulphuric acid solution, produce magnesium sulphate solution

and hydrogen gas.

........................Mg(s) + H2SO4(aq) MgSO4(aq) + H2(g)............................................... 8.When sodium metal reacts with water, sodium hydroxide solution and hydrogen gas are produce.

...............................Na(s) + H2O(l) NaOH(aq) + H2(g) ................................................. 9. When barium nitrate solutiom reacts with potassium chromate(VI) solution, barium chromate(VI)

precipate and potassium nitrate solution is produce.

................Ba(NO3)2(aq) + K2CrO4(aq) BaCrO4(aq) + 2KNO3(aq).................................. 10. When aluminium carbonate solution reacts with hydrochloric acid solution, aluminium chloride

solution, carbon dioxide gas and water are produce.

......................Al2(CO3)3(aq).+ HCl(aq) AlCl3(aq) + CO2(g) + H2O(l)...........................

406

B. Write the meaning of chemical quation below

1. 2 Na(s) + 2H2O(l) 2 NaOH(aq) + H2(g)

2 mole of sodium metal react two mole of water produce 2 mole of sodium hydroxide solution and

one mole hydrogen gas..

2. MgCO3(s) MgO(s) + CO2(g)

...1 mole of magnesium carbonate solid is heated produce 1 mole of magnesium oxide solid and 1

mole of carbon dioxide gas...................................................

3. LiOH(aq) + HCl(aq) LiCl(aq) + H2O(l)

.......1 mole of lithium hydroxide solution reacts with 1 mole of hydrochloric acid solution produce1

mole of lithuim chloride solution aqueous and 1 mole of water...............................

4. ZnCO3(aq) + H2SO4(aq) ZnSO4(aq) + CO2(g) + H2O(l)

.....1 mole of zinc carbonte solution react with 1 mole of sulphuric solution. Produce 1 mole of zinc

sulphate and 1 mole of carbon dioxide gas nad 1 mole of water ...............................

5. Mg(s) + CuSO4(aq) MgSO4(aq) + Cu(s)

.......1 mole of magnesium metal react with 1 mole of copper(II) sulphate solution produce 1 mol of

magnesium sulphate solution and 1 mole of copper metal....................................

6. AgNO3(aq) + KCl(aq) AgCl(s) + KNO3(aq)

1 mole of silver nitrate solution react with 1 mol of potassium chlorida produce 1 mol of silver

chloride solution and 1 mol of potassium nitrate solution............................................

7. NaNO3(s) NaNO2(s) +O2(g)

......1 mole of sodium nitrate solidn is heated produce 1 mole of sodium nitrite solution and 1 mole of

oxygen gas.......................................................................................................

8. 2NH3(aq) + H2SO4 (NH4)2SO4(aq)

......2 mole of ammonia aqueous react with 1 mole of sulphuric acid produce 1 mole of ammoniu

sulphate solution...........................................................................

Disahkan oleh,

Pengesahan Pertama,

Dengan ini saya telah membaca dan menyemak Bahan Pengajaran bagi Tajuk Formula Kimia,

Persamaan kimia dan Stoikiometri seperti di atas untuk kajian ini.

(Pn Jamaliah binti Ghazali )

(Guru Penolong Kanan Kokurikulum dan mengajar Kimia)

Sekolah di mana Kajian dijalankan,

Pengesahan Kedua,

Dengan ini saya telah membaca dan menyemak Bahan Pengajaran bagi Tajuk Formula Kimia,

Persamaan kimia dan Stoikiometri seperti di atas untuk kajian ini.

(Pn Nur Azleena binti Hussen)

Guru Kimia

Sekolah di mana Kajian dijalankan.

topic: chemical formula, chemical equations...

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