practise ch3 chemical equation n formulae

8/3/2019 Practise Ch3 Chemical Equation n Formulae

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Chapter 3: Chemical Formulae and Equations

Subtitle 3.1: Relative Atomic Mass and Relative Molecular

Mass

Concept:We can determine the mass of an atom relative to a standard atom

A Helium atom is 4 times heavier compare to a hydrogen atom.Helium is said to have relative atomic mass of 4

Important !!!Define:

Define

Important!!!!Relative mass does not have any unit.

Numerical problemsA. About Relative Atomic Mass

How to measuremass of an atom?

Standard atom1. hydrogen2. oxygen

3. carbon-12 Solid & easy to

handle Also used as

standard for massspectrometer

Check to:page 176 of text book

Look at Arof allelements listed in

periodic table

from periodic table:

Ar of Nitrogen atomis 14.

The average mass of anitrogen atom is 14times larger than 1/12of a carbon-12 atom.

Mr of WaterMolecule is 18The averagemass of one watermolecule is 18 timeslarger than 1/12of a carbon-12 atom

Get Arvalue fromperiodic table

Hydrogen asstandard atom

not use any morebecause gasseous formare difficult to handle

Relative atomic mass, A r- of an element is the average mass of one atom of the element whencompared with 1/12 of the mass ofan atom of carbon-12

1/12 of oneatomcarbon-12

How many heliumatoms arehere?????

Relative atomic mass of an element= The average mass of one atom of an element

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

Relative molecular mass, Mr-of a molecule is the average mass of the molecule when compared with1/12 of themass ofan atom of carbon-12

Relative molecular mass of an element= The average mass of one molecule

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

helium atom


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1. How many times is copper atom heavierthan two helium atom?Solution:

Mass of a copper atom = Ar of copperMass of 2 helium atom 2 x Ar of helium

= 642 x 4

= 8 times

2. How many magnesium atom have the same mass as two silver atoms ?Solution:Lets the number of magnesium atoms = nMass of n magnesium atoms = mass of 2 silver atomsSo, n x Ar of magnesium = 2 x __________

n x 24 =

=

Do It Yourself

1. How many times is one atom of silicon heavier than one atom of lithium

2. Calculate the number of atoms of lithium that have the same mass as twoatoms of nitrogen

3. The mass of one atom Y is A times larger than the mass of one nitrogen.Calculate the relative atomic mass of Y.

Get Ar value from

periodic table

Form 4 text bookQuick reviewpage 30


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B. About Relative Molecular Mass

To determine Relative Molecular Mass, Mr

Molecularsubstance

Relative Molecular Mass

Carbon dioxide,CO2

Ar of C + ( 2 x Ar of O) = 12 + (2 x 16 ) = 44

Nitrogen gas, N2 2 x Ar of N = 2 x 14 = 28

Relative formula mass is used to replace Mr forionic substances

Ionic substance Relative formula mass

Sodium Hydroxide, NaOH Ar of Na + Ar of O + Ar of H= 23 + 16 + 1 = 40

Aluminium sulphate, Al2(SO4)3

2 x Ar of Al +3 ( Ar of S + 4 x Ar of

O )=

Hydrated copper(II) sulphate, CuSO4.5H2O

Ar of Cu + Ar of S + 4 x Ar of O+ 5 ( Mr of H2O)

=

Do it yourself

1. Calculate the relative molecular mass of

a) Bromine, Br2

c) Ammonia, NH3

b) Methane, CH4 d) Glucose, C6H12O6


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2. Calculate relative formula mass of

a) Zinc oxide, ZnO c) Copper(II) hydroxide, Cu(OH)2

b) Magnesium nitrate, Mg(NO)3 d) Hydrated sodium carbonate,

Na2CO3.10H2O

B. The Mole and the Number of Particles

Definition of mole

The word pair and dozen represent a fixed number of objects.

In chemistry, we use the unit mole to measure the amount of substance. The symbol of mole is mol.

1 mol of substance = the number of particles in 12 g of carbon-12.= 6.02 x 1023 particles.

The value of 6.02 x 1023 is called as the Avogadro constant (NA). To determine the number of moles or the number of particles:

Form 4 practicalbook

Try this 3.1page 17

Number of particles = Number of moles 6.02 x 1023

2310x6.02

particlesofNumber=molesofNumber

Practical BookActivity 3.2, page 17


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Example 1:

i. 1 mol of iron atom = 6.02 x 1023 iron atoms

ii. 1 mol of hydrogen molecule = 6.02 x 1023 hydrogen molecules

iii. 1 mol of sodium chloride = 6.02 x 1023 formula units of sodium chloride

Example 2:A closed glass bottle contains 0.5 mol of oxygen gas, O2.

i. How many oxygen molecules, O2 are there in the bottle?ii. How many oxygen atoms are there in the bottle?

Solution:

i. Number of oxygen molecules = 0.5 x 6.02 x 1023= 3.01 x 1023

ii. 1 oxygen molecule, O2 has 2 oxygen atoms.Therefore, number of oxygen atoms

= number of oxygen molecules 2

= 3.01 1023 2

= 6.02 1023

Example 3:Find the number of moles of molecules in a sample containing 9.03 1023 molecules of carbon dioxide,CO2.

Solution:

Number of moles =23

23

10026

10029

.

.

= 1.5 mol.

Do it yourself

[Avogadro constant = mol-1]

1 Define a mole?

A mole is the amount of substance which has the same number of particles as there in 12 g carbon-12.

2 Calculate the number of atoms in 2 mol carbon.

Number of atoms = 2 6.02 x 1023= 1.2 1024 atoms.

3 How many ions are there in 1.5 mol sodium chloride, NaCl?

1 formula unit sodium chloride, NaCl has 2 ions which are 1 sodium ion and 1 chloride ion.Thus, number of ions = number of formula units x 2

= 1.5 6.02 x 1023 2


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= 1.806 1024 ions.

4 Calculate the number of moles of bromine molecules which consists 1.5 1022 of brominemolecules.

Number of moles = 1.5 10226.02 1023

= 0.025 mol.

5 How many atoms are there in 2 mol of ammonia, NH3?

1 ammonia molecule, NH3 has 4 atoms which are 1 nitrogen atom and 3 hydrogen atoms.Thus, number of atoms = number of molecules x 4

= 2 6.02 x 1023 4= 4.8 1024 atoms.

C. The mole and the mass of substances

Molar mass is

Unit of molar mass is g mol-1 or grams per mole.

The molar mass of a substance = the mass of 1 mol of the substance= the mass of NA number of particles= the mass of 6.02 x 1023 particles

Example:

Element/ Compound Relative mass Mass of 1mol

Molar mass

Lithium, Li 7 7g 7g mol-1

Iron, Fe 56 56g

Magnesium oxide, MgO 24+16=40 40g mol-1

Carbon dioxide, CO2 12+16x2=44

*1 : The value of molar mass of an element is equal to its relative atomic mass*2 : The value of molar mass of a compound is equal to its relative molecular or formulamass

Formula: Number of moles = mass

Relative atomic mass

(or relative molecular mass or relative formula mass)

*122

*2


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Example:

1. Calculate the number of moles found in 20g of magnesium oxide, MgO.(Relative atomic mass: Mg, 24; O, 16)

Solution:

Number of moles = massRelative formula mass

= 2024 + 16

= 0.5 mol

2. Calculate the mass in gram found in 0.2 mol of magnesium oxide, MgO.(Relative atomic mass: Mg, 24; O, 16)

Solution:

Number of moles = mass

Relative formula mass

Mass = number of moles x relative formula mass= 0.2 x (24 + 16)g= 8g

3. How many magnesium ions are there in 30g of magnesium oxide, MgO.(Relative atomic mass: Mg, 24; O, 16. Avogradro constant: 6.02 x 1023)

Solution:

The relative formula mass of magnesium oxide, MgO = 24 + 16= 40

Therefore, the molar mass of magnesium oxide, MgO = 40g mol-1

Number of moles of 30g magnesium oxide, MgO =mass of MgO

Relative formula mass ofMgO

= 30g40 g mol-1

= 0.75 mol

The number of formula units of MgO = 0.75x 6.02 x 1023 = 4.515 x 1023

Each formula units of MgO has 1 magnesium ions.

Therefore, the number of magnesium ions = the number of formula units of MgO x 1= 4.515 x 1023 x 1= 4.515 x 1023


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4. Calculate the mass in gram of 3 x 1022 units of magnesium oxide, MgO.(Relative atomic mass: Mg, 24; O, 16. Avogradro constant: 6.02 x 1023)

Solution:

Number of moles = number of particlesNA

Mass = number of particlesRelative formula mass NA

Mass = number of particles x relative formula massNA

Mass of 3x1022 units of magnesium oxide, MgO= 3 x 1022 x (24+16)6 X 1023

= 0.05 X 40= 2 g

Do It Yourself

1. Calculate the number of moles found in 9.5g of magnesium chloride, MgCl2.

(Relative atomic mass: Mg, 24; Cl, 35.5)

2. Calculate the mass in gram found in 0.3 mol of magnesium chloride, MgCl2.(Relative atomic mass: Mg, 24; Cl, 35.5)

3. How many chloride ions are there in 19g of magnesium chloride, MgCl2.(Relative atomic mass: Mg, 24; Cl, 35.5. Avogradro constant: 6.02 x 1023)

4. Calculate the mass in gram of 3 x 1022 units of magnesium chloride, MgCl2 .(Relative atomic mass: Mg, 24; Cl, 35.5. Avogradro constant: 6.02 x 1023)

Form 4 TextBookWork This Out 3 Page 35

Quick Review C

Pa e 35


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E. Chemical Formulae

A chemical formula is a representation of a chemical substance using letters for atoms and subscriptnumbers to show the numbers of each type of atoms that are present in the substance.

Examples : (a) Glucose

C6 H12O6

(b) Sodium hydroxide

Mg (OH)2

(1) Empirical Formulae

(i) The empirical formula of a compound gives the simplest whole number ratio ofatoms ofeach element present in the compound.

(ii) Steps in determining the empirical formula of a compound.

i. find the mass of each element in the compound

ii. convert the masses to the numbers of moles of atoms

iii. find the simplest ratio of moles of the elements

Example : 2.24 g of iron combines chemically with 0.96g of oxygen to form an oxide. What is theempirical formula of the oxide ?

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

Element Iron, Fe Oxygen, O

Mass (g) 2.24 0.96

Number of moles of atoms 2.24 = 0.0456

0.96 = 0.0616

Ratio of moles 0.04 =10.04

0.06 =1.50.04

Simplest ratio of moles 1 2 = 2 1.5 2 = 3

The empirical formula of the oxide is Fe2O3.

Show the symbols forcarbon, hydrogen and

oxygen

Show the numbers ofcarbon, hydrogen and

Show the symbols formagnesium, oxygen

and hydrogen.

Show the numbers of

magnesium, oxygen

and hydrogen.


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Do it Yourself

1. The table below shows the relative atomic mass and the mass of elements V and O in an oxide.

Element V O

Relative Atomic Mass 56 16

Mass(g) 5.6 2.4

What is the empirical formula of this compound ?

element V Oxygen, O

Mass (g) 5.6 2.4

Number of moles of atoms

Ratio of moles

Simplest ratio of moles

The empirical formula of the oxide is

2. Copper (II) iodide constains 20.13% of copper by mass. Find its empirical formula. [ Relativeatomic mass : Cu,64 ; I, 127 ]

Based on its percentage composition, 100g of copper(II) iodine contains 20.13g of copper. So,taking 100g of the compound.

element K Cl

Mass (g)

Number of moles of atoms

Ratio of moles

Simplest ratio of moles

The empirical formula of the oxide is .

3. A potassium compound has a percentage composition as the followingK, 31.84% ; Cl, 28.8% ; O, 39.18%

What is the empirical formula of the potassium compound ? [ Relative atomic mass : O, 16;Cl,35.5; K, 39 ]

Based on its percentage composition, 100g of compound contains 31.84g of potassium, 28.98g ofchlorine and 39.18g of oxygen. So, by taking 100g of the compound:

element K Cl O

Mass (g)

Number of moles ofatoms

Ratio of moles

Simplest ratio ofmoles

1 mole of potassium atoms combines with 1 mole of chlorine atoms and 3 moles of oxygenatoms.Therefore, the empirical formula of the potassium compound is KClO3. Form 4 TextBook

Work this out 3.7Page 42


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(2) Molecular Formulae

(i) The molecular formula of a compound gives the actual number of atoms of each elementpresent in a molecule of the compound.

(ii) The molecular formula of a compound is a multiple of its empirical formula.

(iii) Relating empirical formula to molecular formula

Compound Empirical formula Molecular formula n

Water H2O H2O = (H2O)1 1

Ethene CH2 C2H4 = (CH2)2 2

Ethane CH3 C2H6 = (CH3)2 2

propane CH2 C3H9 = (CH3)3 3

glucose CH2O C6H12O6 = (CH2O)2 6

(iv) Calculation involving molecular formulae

Example :The empirical formula of a compound is CH. Its relative molecular mass is 78. Find itsmolecular formula. [ Relative atomic mass : H, 1; C, 12 ]

Let the molecular formula be (CH)n.

The relative molecular mass = n[ 12 + 1 ]= 13n

However, its molar mass is 78.Therefore, 13n = 78

n = 78/13

= 6

Hence, the molecular formula of the compound is (CH)6 or C6H6.

Do it yourself

1. A carbon compound has an empirical formula of CH2 and a relativemolecular mass of 70. Find the molecular formula of the compound. [ Relative atomic mass :

H, 1; C, 12 ]

Molecular formula = ( Empirical formula )n


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Hence, the molecular formula of the compound is (CH2)5 or C5H10.

2. 2.07 g of element Z reacts with bromine to form 3.67g of a compound withan empirical formula of ZBr2. Find the relative atomic mass of element Z. [ Relative atomicmass: Br, 80 ]

element Z Br Mass (g)

Number of moles ofatoms

Simplest ratio of moles(from the emp for given)

Based on the empirical formula ZBr2 , the ratio ofatoms of Z : Br is 1 : 2

herefore,

2.07 : 0.02 = 1 : 2z

2.07/0.02z =

z = 207

The atomic mass of the element Z is 207.

(3) Ionic Formulae

(i) Ionic compounds are compounds consisting of anions and cations.

(ii) The formulae of some common cations

Cation ( positive ion ) Formula of cation Charge of cation

Sodium ion Na+ +1

Potassium ion K+ +1

Silver ion Ag+ +1

Hydrogen ion H+ +1

Ammonium ion NH4+ +1

Copper (II) ion Cu2+ +2

Calcium ion Ca2+ +2

Magnesium ion Mg2+ +2

Zinc ion Zn2+ +2

Barium ion Ba

2+

+2Iron (II) ion Fe2+ +2

Copper (I) ion Cu+ +1

Tin (II) ion Sn2+ +2

Lead (II) ion Pb2+ +2

Aluminium ion Al3+ +3

Iron (III) ion Fe3+ +3

Chromium (III) ion Cr 3+ +3

FORM 4 TextbookWork this out 3.8Page 44


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(iii) The formulae of some common anions

Anion ( negative ion ) Formula of anion Charge of anion

Fluoride ion F- -1

Chloride ion Cl- -1

Bromide ion Br - -1

Iodide ion I- -1

Hydroxide ion OH- -1

Nitrate ion NO3- -1

Nitrite ion NO2- -1

Hydride ion H- -1

Oxide ion O2- -2

Phosphate ion PO43- -3

Carbonate ion CO32- -2

Sulphate ion SO42- -2

Chromate (VI) ion Cr 2O72- -2

(iv) The chemical formulae of ionic compounds are electrically neutral because the total ofpositive charges are equal to the total of negative charges

(v) The chemical formula of an ionic compound can be constructed as the following :i. identify and write down the formula of its cation and anionii. determine the number of cations and anions by balancing the positive and

negative charges.iii. Write the formula of the compoundiv. The number of cations and anions are written as subscript numbers.

MgCl2

Magnesium chloride

Magnesium ion, Mg2+ Chloride ion, Cl-

1 magnesium ion, Mg2+

Total of positive charges=1 (+2)=+2

2 chloride ions, Cl-

Total of negative charges= 2 (-1)= -2


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Do it yourself

1. magnesium chloride

2. aluminium oxide

3. aluminiuim hydroxide

4. sodium sulphate

(4) Naming of chemical compounds

1. Chemical compounds are named systematically according to the guidelines given by theInternational Union of Pure and Applied Chemistry (IUPAC).

2. For ionic compounds, the name of the cation comes first, followed by the name of anion.

cation anion Name of ionic compound

Sodium ion Chloride ion Sodium chloride

Magnesium ion Oxide ion Megnesium oxide

Aluminium ion Oxide ion Aluminium oxide

Zinc ion Sulphate ion Zinc sulphate

3. Transition metals can form more than one ions, Roman numerals ( such as I, II, III ) are used todifferentiate the ions.

Fe2+ - iron (II) ionFe3+ - iron (III) ion

4. For simple molecular compounds, the name of the first element is maintained. However, the

name of the second element is added with an ide .Examples : HCl hydrogen chloride

HF - hydrogen flouride

5. Greek prefixes are used to show the number of atoms of each element in a compound.Examples : CO carbon monoxide

CO2 carbon dioxideCCl4 carbon tetrachloride / tetrachloromethaneSO3 sulphurtrioxide

6. Table below shows the meaning of the prefixes.

prefix meaning prefix meaning

Mono- 1 Hexa- 6di- 2 Hepta- 7

Tri- 3 Octa- 8

Tetra- 4 Nona- 9

Penta- 5 Deca- 10

Form 4 TextbookWork This Out 3.9Page 46



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F. CHEMICAL EQUATION

A) Qualitative aspect of chemical equation

A chemical equation is a shorthand description of a chemical reaction.

The starting substances are called reactants.

The new substances formed are called products.

The reactants are written at the left-hand side of the equation.

The products are written at the right-hand side of the equation.

A chemical equation also shows the states of each substance.

Symbol Physical states of substances

s Solid

Liquid

g Gas

aq Aqueous solution

Example :

Do It Yourself 3f

Identify the reactants, products and the state of each substance. Present your answer in the form of a

table.

Solution :

Reactants Products

1

23

Reactants Products

C (s) + O2 (g) CO2 (g)

Zn (s) + Cl2 (g) ZnCl2 (s)

1. HCl (aq) + NaOH (aq) NaCl (aq) + H2O (i)

2. CuCO3(s) CuO (s) + CO2 (g)

3. HCl (g) + NH3 (g) NH4Cl (s)



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B) Writing chemical equation

A chemical equation must be balanced. There must always be the same number of atom of each elementon each side of the equation.

Example :

Magnesium reacts with dilute hydrochloric acid, HCl to produce magnesium chloride, MgCl2 and hydrogengas, H2. Write an equation to represent the reaction.

Solution :

Do It Yourself 3.f B

Write a chemical equation for each of the following reactions.

1. A solution of silver nitrate is added to a solution of sodium chloride. A precipitate of silver chloride and asolution of sodium nitrate are produced.

2. Nitrogen gas reacts with hydrogen gas to produce ammonia gas.

3. When solid lead (II) carbonate is heated strongly, it decomposes into solid lead (II)oxide and carbon dioxide gas is released.

STEP 1Write the equation in words. The reactants are written on the left whereas theproducts are written on the right.

STEP 2 Write the correct chemical formula for each reactants and products.

STEP 3Balance the equation. You just need to adjust the coefficients in front of the chemicalformulae and not the subscripts in the formulae.

STEP 4 Put the state symbols in the equation.


STEP 1 Magnesium + hydrochloric acid Magnesium chloride + hydrogen gas

Reactants Products

STEP 2 Mg + HCl MgCl2 + H2

STEP 3 Mg + 2HCl MgCl2 + H2

STEP 4 Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)


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C) Quantitative aspect of chemical equation

The coefficients in a balanced equation tell us the exact proportions of reactants and products in achemical reaction.

Example :

The equation tell us that 2 moles of hydrogen reacts with 1 mole of oxygen to produce 2 moles of water.

Or

The equation tell us that 2 molecules of hydrogen reacts with 1 molecule of oxygen to produce 2molecules of water.

D) Numerical problems involving chemical equation

Stoichiometry is a study of quantitative composition of substances involved in chemical reactions.

We can always make use of the stoichiometric coefficients in a chemical equation to solve variousnumerical problems.

Generally the steps involved in stoichiometric calculations are as follows.

Example :

Copper (II) oxide, CuO reacts with aluminium according to the following equation.

Calculate the mass of aluminium required to react completely with 12 g of copper (II) oxide, CuO.[Relative atomic mass : O, 16 ; Al, 27 ; Cu, 64]

Solution :

The number of moles of 12g of Copper (II) oxide, CuO = 12 g(64 + 16) g mol-1

= 12 g = 0.15 mol80 g mol-1

2H2(g) + O2 (g) 2H2O (l)

STEP 1 Write the balanced equation of the reaction.

STEP 2 Compare the mole ratio.

STEP 3 Identify the information given and you want to find.

STEP 4 Calculate the number of moles.

3CuO (s) + 2Al (s) Al2O3 (s) + 3Cu (s)

3CuO (s) + 2Al (s) Al2O3 (s) + 3Cu (s)

3 mole 2 mole


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Based on the chemical equation, 3 mole of Copper (II) oxide, CuO requires 2 mole of aluminium.Therefore, the number of aluminium required by 0.15 mole of Copper (II) oxide, CuO

= 0.15 mole x 2 mole = 2 mole3 mole

Thus the mass of aluminium required

= 0.1 mol x 29 g mol-1 = 2.7 g

Do It Yourself 3f (D)

How many moles of potassium are needed to reacts with 0.5 mole of bromine gas ?

Solution :

Information : ? mole 0.5 mole

Based on the equation, 1 mole of bromine gas reacts with 2 moles of potassium.

Therefore, 0.5 mole of bromine gas will react with

2 x 0.5 = 1 mole of potassium.

2. 1.35 g of aluminium reacts with excessive copper (II) oxide powder to produce aluminium oxide powderand copper. Find the number of copper atoms produced.[Relative atomic mass : Al, 27 ; Avogadro constant : 6.02 x 1023 mol-1]

1. 2K (s) + Br2 (g) 2KBr(s)

2K (s) + Br2 (g) 2KBr(s)

2 mole 1 mole


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3.

What is the mass of zinc needed to produce 2.4 dm3 of hydrogen gas at room conditions ?[Relative atomic mass : Zn, 65 ; Molar volume 24 dm3 mol-1 at room conditions]

More Exercises:

1. CuCO3 CuO + CO2

In this reaction, 3.1 g of copper(II) carbonate are heated in a laboratory. Find :

(a) the mass of copper (II) oxide that being produced.(b) the volume of carbon dioxide gas produced at s.t.p

2. CaCO3 CaO + CO2

In this reaction, 300 cm3 gas carbon dioxide are produced at room temperature, when calciumcarbonate are heated. Find:

(a) the mass of calcium carbonate used.(b) mass of calcium oxide produced.

3. 2Na + 2H2O 2NaOH + H2

When 0.23 g of sodium is added to water, the metal will react vigorously at the surface of the water, find(a) the mass sodium hydroxide produced.(b) volume of hydrogen gasses being produced at temperature room.

4. 2Mg + O2 2MgO

A strip of magnesium has a weight of 1.2 g are being burn with sufficient oxygen to producedmagnesium oxide. Find:

(a) the mass magnesium oxide being produced.

(b) the mass of oxygen that needed for this reaction.

5. C3H8 + 5O2 3CO2 + 4H2O

Propane gas was burned in oxygen follow as equation above. If 3.36 dm3 of carbon dioxide gas areproduced in this reaction at s.t.p, find

(a) the mass of propane burned(b) volume of oxygen gas that reacted

Zn (s) + 2HNO3 (aq) Zn(NO3)2 (aq) + H2 (g)


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6. 2Al + 3CuO Al2O3 + 3Cu

1.35g of aluminium powder and copper (II) oxide was heated strongly in laboratory to producedaluminium oxide and copper. Find

(a) the mass of copper (II) oxide reacted(b) the mass of aluminium oxide produced.(c) the mass of copper produced

practise ch3 chemical equation n formulae

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