Download topic: chemical formula, chemical equations and stoichiometry

Lampiran O
Bahan Pengajaran
TOPIC:
CHEMICAL FORMULA, CHEMICAL EQUATIONS AND
STOICHIOMETRY PROBLEM
(FORMULA KIMIA, PERSAMAAN KIMIA DAN MASALAH STOIKIOMETRI)
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Task
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Sub-topic
Learning outcomes
Activity 1.1
Able to
Relative Atomic Mass 1. understand the concept of Relative Atomic Mass, RAM and
and Relative Molecular
Relative Molecular Mass, RMM.
Mass
2. Able to calculate the Relative Molecular Mass.
Activity 1.2
Able to
The Mole and Number of 1. define a mole.
Particles
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.
Activity 1.3
Able to
Mole and Mass of 1. state the meaning of molar mass.
Substances.
2. relate the molar mass to Avogadro Constant.
3. convert mass of a substance to mole.
4. Solve numerical problem
Activity 1.4
Able to
Number moles of Gas and 1. state the meaning of Molar Volume.
Molar Volume.
2. solve numerical problem.
Activity 1.5
Able to
Empirical Formula and 1. state the meaning of empirical formula and molecular
Molecular Formula.
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.
Experiment 1.6
Able to
Empirical formula of 1. carry out the experiment 3.5
Magnesium oxide.
2. determine the empirical formula of magnesium oxide.
Activity 1.7
Able to
Chemical Formula of 1. write ionic formulae of ions.
ionic compound
2. construct chemical compound using IUPAC
nomenclature.
3. naming chemical formula
Activity 1.8
Able to
Chemical Equation
1. write chemical equation from word equation to symbol.
2. balancing chemical equation
3. use the chemical equations in stoichiometry problem.
Activity 1.9 Balancing Able to
Chemical Equation
1. write and balancing chemical equation
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.
2.
To understand the concept of Relative Atomic Mass, RAM and
Relative Molecular Mass, RMM.
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.
2.
3.
4.
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.
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.
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.
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.
2.
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
What are
= 24x1 + 32x1+16x4+ 1x14+ 16x7 =141
the steps
How do you think to solve the problem?
Step 1. Write the chemical formula
385
3.
4.
Step 2. Find the relative atom mass(RAM) of each elements in a compound.
Step 3. Relationship;
Relative molecular mass of Molar Mass is equal total mass of all elements by multiply RAM
of every element with number of its atoms.
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]
How many atoms of nitrogen have the same masses of one atom of iron?
[ N,14; Fe,56]
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]
The element chlorine forms a compound with oxygen, with molecule formula is Cl 2On. The
relative molecule mass of the compound is 83. What is the value of n? [Cl.35.5; O,16]
3.
4.
5.
Activity 1.2 The Mole and Number of Particles
Learning Outcomes: You should be able to
1.
2.
3.
Define a mole
State the meaning of Avogadro Constant, N A
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 2:
Model particle
Declarative
Knowledge Figure 3: Model of particles
of Sodium
chloride
of Calcium atom
1. Definition
i.
ii.
iii.
Figure 4 ; Model of
water molecule
A mole is an amount of substance that contain as many particles as the number of atom in
exactly 12 g of carbon -12.
Avogadro constant, NA, is defined as the number of particles in one mole of particles or one
mole of substance contains 6.02 x1023 particles
There are three type of particles
a. Atom
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2.
3.
b. Molecule
c. Ion
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 23……atoms Na
b. One mole of molecular substances contains 6.02 x10 23 molecules
Example : 1 mole of water, H2O contains…6.02 x 1023…water molecules .
c. One mole of ionic substances contains 6.02 x1023 formula units
Example: 1 mole of sodium chloride, NaCl contains 6.02 x 1023Na+ ion and 6.02 x 1023
chloride ion,Cl-.
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.
2.
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
Analogy : 1 dozen contain 12 item.
1 mole contains 6.02x 1023 particles
Exercise 1.2
[Avogadro constant NA= 6.02 X 1023 mol-1]
How to do?
1. Calculate the number of particles
a. 3.5 moles lead, Pb.
Problem solving skills.
b. 0.5 moles lead (II) ion, Pb2+.
1. State the problem
c. 0.25 moles carbon dioxide, CO2
2. Identify information given.
2. Find the number of moles of
24
3. Formula
a. 1.204 x 10 molecules of chlorine gas, Cl2
23
4. Solution
b. 12.04 x10 atom of ammonia, NH3
5. Check
theofanswer
3. Calculate the number of particles and state the type of particle
in each
the following,
(a) 1/4 mol of copper, Cu
(unit and value of the quantity)
(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
Learning outcomes
Students be able to
1.
2.
3.
4.
State the meaning of molar mass
Relate molar mass to Avogadro constant.
Relate molar mass of a substance to its relative atomic mass or relative
molecular mass.
Solve numerical problem to convert number of mole of a given substances
to its mass and vice-versa.
387
Declarative knowledga:
molar mass, Avogadro constant, NA, Relative atomic Mass,RAM, Relative Molecular Mass, RMM.
Figure 5: solid of
Sulphur,S
1.
2.
3.
Figure 6; Solid of
Aluminium, Al
Figure 7; Crystal of
Copper(II) sulphate, CuSO4
Definition
The mass of one mole of any substances, in gram is called Molar Mass.
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.
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]
How to do?
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)
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)
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 x1023]
Problem solving
1. Problem: Convert gram to mole
2. Formula; Mole = mass / molar mass
3.Solution:
388
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 1023
=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 1023
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 1023
Mass of 0.26 moles Ag = 0.26 x 108 = 28.08 g
Exercise 1.3
1.
2.
3.
Find the number of particles of 60g of oxygen gas, O 2?
Find (a) number of moles (b) mass from the number of 1.2 x10 23 atom of ozone, O3.
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=6.023x1023]
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.
Problem solving skills
Strategy:
1. State the problem
1. Students remember the definition of Molar Volume
2. Identify information given
and the relations between molar mass, molar volume
3. State the formula or relationship.
and Avogadro constant, NA
4. Use information given in
2. Analysis the numerical problem.
formula to solve problem.
5. Checking the answer
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DECLARATIVE KNOWLEDGE
Figure 8: Bromine gas ,Br2
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 dm3.
3.
At standard temperature and pressure or s.t.p condition or at temperature 0 oC and pressure 1
atmosphere, 1 mole of gas or molar volume occupies a volume of 22.4 dm3.
4.
Therefore at standard temperature and pressure (s.t.p) 1 mole of gas occupies 22.4 dm 3, examples;
a.
1 mole of hydrogen gas, H2 (= 2g) occupies .22.4… dm3 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 ……22.4…dm3 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 1023 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 contains…6.02x1023….molecules.
c.
At s.t.p …22.4…dm3 (1 mole) of neon gas, Ne, contains…6.02x1023….. atom of Ne.
PROCEDURAL KNOWLEDGE
Calculating Involving Gas
Example 1
Calculate the volume occupied by the following gases at s.t.p.
CO2
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 x1023 , 1 mole gas occupies 22.4 dm3 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, NH 3 occupies 3.5 x 22.4 dm3 =78.4 dm3.
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 dm3 at stp
c. Formula : mole = number particles
6.02 x1023
, volume gas = mole x molar volume
Solution:
1.5 x1023 molecules = 1.5 x1023 mole ÷ 6x1023 =0.25 mole
Volume of oxygen gas, O2 = 0.25 mole x 22.4 dm3 = 5.6 dm3
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 25 oC
a. 1.2 dm3 of methane , CH4
b. 6 dm3 of ethane, C2H6
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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 25oC equal to 24 dm3,
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
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.
DECLARATIVE KNOWLEDGE
1. Define chemical formula
Remember
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.
Molecule formula = (Empirical Formula)n
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.
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PROCEDURAL KNOWLEDGE
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
Water
Carbon dioxide
Magnesium oxide
Aluminium chloride
Ammonia
Glucose
Ethene
Propene
Butene
Ethane
Benzene
Molecular
formula
H2O
CO2
MgO
AlCl3
NH3
C6H12O6
C2H4
C3H6
C4H8
C2H6
C6H6
Number of atoms
Actual
Simplest ratio
H:O = 2:1
H:1 = 2:1
C:O = 1 :2
Mg:O =
Al:Cl=
N:H =
Empirical
Formula
H2O
CO2
Number of n
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)
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.
1
2
3
4
5
Element
Mass(g)
RAM
Number of moles
Ratio of mole
divided by smallest
mole
Empirical Formula
Na
2.3
23
2.3
23
= 0.1
0.1
0.05
=2
Na2O
O
0.8
16
0.8
16
0.05
0.05
0.05
=1
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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 C 2H5. 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 P 2O. 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.
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
Result repeated until a constant mass is obtained
Magnesium ribbon
Heat
Figure: Set- up of apparatus
Crucible and lid
393
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
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
DECLARATIVE KNOWLEDGE
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.
394
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+.
PROCEDURAL KNOWLEDGE
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: O21 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: OH1 NH4+ cation and 1 OH- anion combine to give a compound with zero net charge,
Formula of ammonium hydroxide is NH4OH
+1 + -1 =0
B. Charge on Cation DOES NOT EQUAL Charge on Anion
a. Aluminium chloride




CATION: aluminium (Group III or 13), charge is +3: Al 3+
ANION: chloride (Group VII or 17), charge is -1: Cl1 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
395
b. Sodium oxide




CATION: sodium (Group I), charge is +1: Na +
ANION: oxide (Group VI), charge is -2: O22 Na+ cations and 1 O2- anion combine to give a compound with zero net charge:
(2 x +1) + -2 = 0 (Na+, Na+, O2-)
Formula of sodium oxide is Na2O
c. Ammonium sulphate



CATION: ammonium, charge is +1: NH4+
ANION: sulfate, charge is -2: SO422 NH4+ cations and 1 SO42- 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: Ba 2+
ANION: hydroxide, charge is -1: OH1 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 O2Step 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=O2Step 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
Number of atom
mono
1
di
2
tri
3
tetra
4
penta
5
heksa
6
hepta
7
oct
8
Example :
Molecular Formula
CO
CO2
SO3
CCl4
Name
Carbon monoxide
Carbon dioxide
Sulphur trioxide
Carbon tetrachloride
396
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 ………………..….
(b) MgO ……………….…….
(c) LiCl ……………….………
(d) NaOH …………………..…
(e) ZnSO4 ………………………
3 Name the following molecule
(f) NaHCO3 ……………………….…
(g) KBr …………………………….
(h) Ca(OH)2 ………………..………
(i) BaSO4 ………………..……….
(j) AgCl ……………………………
(a) NO ………………..……….
(b) NO2 ……………..…………
(c) SO2 ……….………………
(d) SO3 ……….………………
(e) CS2 …………….………….
4. Give the molecule formula of these compounds
(f) BF3
……………….…………
(g) N2O4 ………………………….…
(h) H2O ……………………………..
(i)NH3 ………………………………
(j) CO …………………………….
(a) Ammonia
………………
(b) Sulfur dioxide
.........................
(c) Nitrogen monoxide ………………
(d) Nitrogen dioxide ………….……..
(e) Water …………………………..
(f)Carbon dioxide ……………………..
(g) Carbon monoxide…………………
(h) Sulphur trioxide ………………….
(i) Carbon tetrachloride ……………….
(j) Boron triflouride …………………..
ACTIVITY 1.8:
CHEMICAL EQUATIONS
Write a balanced equation to represent each of the reaction:
Reactant
2.
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)
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
1.
Product
Mg = 1
Mg = 1
O =2
Reactant
O=1
Product
Mg = 2
Mg = 2
O =2
O=2
………………………………………………………………………
397
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)
.......................................................................................................................................................
398
8. 2NH3(aq) + H2SO4 → (NH4)2SO4(aq)
.......................................................................................................................................................
ACTIVITY 1.10 :CHEMICAL EQUATIONS AND STOICHIOMETRY PROBLEM
DECLARATIVE KNOWLEDGE
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)
1.
a)
b)
3.
a)
b)
4.
5.
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 dm3 at s.t.p, Avogadro constant:
6.02 x 1023 particles]
When sodium nitrate is heated, it decomposes into sodium nitrite and oxygen gas.
2NaNO3 (p) → 2NaNO2 (p) + O2 (g)
Find the mass of sodium nitrite produced if 34 g of sodium nitrate is heated.
(27.6g)
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]
Reaction of calcium carbonate with excessive dilute hydochloric acid will produce calcium
chloride, water and carbon dioxide gas.
Find the mass of calcium carbonate needed to react completely with 0.5 mole of dilute hydochloric
acid?
(25g)
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)
3
[Relative atomic mass: Ca, 40; C, 12; O, 16: Molar volume: 24 dm at room temperature]
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]
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)
23
[Relative atomic mass; Al, 27; Avogadro constant: 6.02 x 10 particles]
FORMULAE OF MOLES
The steps involved in stoichiometric
calculations are as follows:
Mass (g)
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
Mole =
Ar or Mr
1.
1. 2.
No. of particles
Mole =
e
3.
Mole
ele
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
.
23
6.02 X 10
=
at s.t.p
Volume of gas
22, 400
or
Mole
=
Volume of gas
24, 000
at room temperature
s.t.p
399
ACTIVITY 1.1 RELATIVE ATOMIC MASS AND RELATIVE MOLECULAR
MASS
Answer Relative Molecular Mass (Exercise 1.1)
No
1
Answer
(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
No
2
Answer
2x1 +16n=36
16n= 36-2=34
n =34/16= 2.125=2
3
56/16= 4
4
9x10=90
5
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.
2.
(a) Aim of Problem: Calculate the number of particles of 3.5 mole lead, Pb
Given: NA = 6.023 x1023 particles mole -1
Relation of concept: mole → particle
Number of particles = mole x NA
Solution:
Number of particle = 3.5mole x 6.02 x10 23 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, Pb 2+ ion
Given: NA = 6.02 x1023 particles mole -1
Relation of concept: mole → particle
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 x1023 particles mole -1
Solution
Number of particles= 0.25 mol molecule CO2 x 6.023 x1023molscules mole-1
=1.505x 1023 molecules
(a)Aim of Problem: Find the number of moles of chlorine gas, Cl2
Given :NA = 6.023 x1023 particles mole-1
Relation of concept: particles→ mole
Number of mole = Number of particle/NA
Solution
Number of moles =1.204 x1024 molecule gas chlorine =0.199x1024-23 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
400
3.
4.
Relation of concept: particles→ mole
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 1023
(a) Aim of Problem : Calculate the number of particles ¼ mol of copper atom
Given: NA = 6.023 x1023 particles mole-1
Relation of concept: mole→ particles
Number of particle = number of mole x N A
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
Relation of concept: mole→ particles
Number of particle =Number of mole x N A
Solution
Number of ions for 0.3 mole of calcium ions, Ca 2+
= 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
Relation of concept: mole→ particles
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
(a) Aim of Problem : Calculate the number of mole
Given:NA = 6.023 x1023 particles mole-1
Relation of concept: particles → mole
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
Relation of concept: particles → mole
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
401
5.
(c) Aim of Problem : Calculate the number of mole
Given:NA = 6.023 x1023 particles mole-1
Relation of concept: particles → mole
Number of mole= number of particle
NA
Solution:
Number mole of 3x1021 sulphur trioxide molecules, SO3
= 3 x1021 molecules SO3
6.023 x1023
=0.498 x 10-2 mole
=4.98 x10-3 mole
(a) Aim of problem: How many atoms
Given:NA = 6.023 x1023 particles mole-1
Relation of concept: mole → particle
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, O 2.?
Information given: RAM oxygen =16, NA=6.023x1023
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 x10 23 atom of ozone, O3
Given: NA=6.023x1023
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, O 3
Information given: RAM oxygen =16, number of 1.2 x10 23 atom
Relation of concept; Mole → mass(g)
Solution:
1.2 x1023 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
402
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 x1023 = 1.50 x1023 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 1023x 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 dm3.
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
6x1023
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 x1023mole 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.
c.
Number of mole = 240cm3 x 6x1023 =0.06 x1023
1000x24
0.48 mole
1000x24
Number of mole = 0.48x 6x1023
=1.2x1019
1000x24
403
ACTIVITY 1.5(a) CHEMICAL FORMULA
Exercise 1.5 Answer
Substances
Water
Carbon dioxide
Magnesium oxide
Aluminium chloride
Ammonia
Glucose
Ethene
Propene
Butene
Ethane
Benzene
Molecular
formula
H2O
CO2
MgO
AlCl3
NH3
C6H12O6
C2H4
C3H6
C4H8
C2H6
C6H6
Number of atoms
Actual
Simplest ratio
H:O = 2:1
H:1 = 2:1
C:O = 1 :2
C:O = 1:2
Mg:O = 1 : 1
Mg: O= 1:1
Al:Cl= 1:3
Al: Cl=1:3
N:H = 1:3
N:H= 1:3
C:H:O =6:12:6
C:H:O=1:2:1
C: H = 2:4
C:H =1:2
C:H =3:6
C:H=1:2
C:H=4:8
C:H=1:2
C:H=2:6
C:H=1:3
C:H=6:6
C:H=1:1
Empirical
Formula
H2O
CO2
MgO
AlCl3
NH3
CH2O
CH2
CH2
CH2
CH3
CH
Number of n
1
1
1
1
1
6
2
3
4
2
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 with…1 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 carbonate……K2CO3…….
(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 oxide………Al2O3…….
(q) Magnesium oxide……….MgO….…
2. Name the ionic compound with each of the following formulae
(a) Ba(NO3)2 ………Barium nitrate….
(f) NaHCO3…Sodium hydrogen carbonate..
(b) MgO
(g) KBr …...Potassium bromide…………
(c) LiCl
………...Magnesium oxide…
………….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…………….
404
(b) NO2
(c) SO2
(d) SO3
(e) CS2
……………nitrogen dioxide………
……………sulphur dioxide………
……………sulphur trioxide………
……………carbon disulphide…….
(g) N2O4 …..…Dinitrogen tetraoxide………
(h) H2O ………Dihydrogen oxide = water…
(i)NH3 …….….Nitogen trihydrogen = ammonia.
(j) CO ……. …carbon monoxide…………….
6. Give the molecule formula of these compounds
(a) Ammonia
…….NH3…………
(b) Sulfur dioxide
.........SO2................
(c) Nitrogen monoxide ……NO………….
(d) Nitrogen dioxide ……. ..NO2…………
(e) Water …………………H2O………….
N
ACTIVITY 1.8:
(f)Carbon dioxide ………CO2……………..
(g) Carbon monoxide……CO………………
(h) Sulphur trioxide ..……SO3…………….
(i) Carbon tetrachloride …CCl4…………….
(j) Boron triflouride …..…BF3…………….
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)...........................
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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.
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