3.1 Relative Atomic MassRelative Atomic Mass1. The mass of an atom when compared to another is known as the relative atomic mass (RAM).2. The relative atomic mass (RAM) of an element is the average mass of one atom of the element when compared with 1/12 of the mass of an atom of carbon-12, which taken as 12 units.3. 1/12 of the mass of an atom of carbon-12 is named as 1 atomic mass unit (amu).4. The mass of one carbon atom is 12 amu.

Example 1The mass of a sodium atom is 23 times greater than 1/12 of the mass of carbon-12 atom. What is the relative atomic mass of sodium?

Answer:23

Example 2:The mass of element A is twice of the mass of carbon, therefore its relative atomic mass is __________. (Relative atomic mass of carbon = 12)

Answer:Relative Atomic Mass of Element A = 2 x 12 = 24

Example 3:An atom of element X is 13 times heavier than one atom of helium. Calculate the relative atomic mass of X.( Ar: He = 4 )

Answer:Relative Atomic Mass of X = 13 x 4 = 52

Example 4:How many times that the mass of 2 bromine atoms are greater than 4 neon atoms? (Ar: Ne = 20; Br = 80 )

Answer:2(80)4(20)=2

The mass of 2 bromine atoms is 2 time greater than the mass of 4 neon atoms.

Example 54 atoms of element L have same mass as 1 tellurium atom. Find the relative atomic mass of L. (Ar: Te = 128 )

Answer:4L=1(128)L=1284=32

The relative atomic mass of L = 32

Relative Molecular MassThe relative molecular mass (Mr) of an element is the average mass of one molecule of the element/compound when compared with of the mass of an atom of carbon-12, which taken as 12 units.

The relative molecular mass of a molecule is equal to the sum of the relative atomic mass of all the atoms in the molecule.ExampleFind the relative molecular mass of carbon dioxide.[ JAR: C = 12; O = 16 ]

Answer:

The formula of carbon dioxide = CO2

Relative molecular mass of CO2= 1 x 12 + 2 x 16 = 44

ExampleWhat is the relative molecular mass of aluminium sulphate [ Al2(SO4)3]?( Ar: O = 16, S = 32; Al = 27 )Answer:

Relative molecular mass ofAl2(SO4)3= 2 x 27 + 3( 32 + 4 x 16) = 342Relative Molecular Mass - Example 1Question:Given that the formula of a compound is KXO3and its relative molecular mass is 167. Find the relative atomic mass of element X?(Ar: O = 16; K = 39 )

AnswerLet's say the relative atomic mass of element X = m

The relative moleculaar mass of KXO3= 39 + m + 3(16) = 167m = 167 - 39 - 3(16) = 80

relative atomic mass of element X = 80Relative Molecular Mass - Example 2Question:The general formula of a hydrocarbon is CnH2nand the relative molecular mass of the hydrocarbon is 84. Find the value of n.

Answer:

Relative molecular mass ofCnH2n= n(12) + 2n(1) = 8414n = 84n = 84/14 = 6Concept of Mole1. A mole is defined as the amount of substance which contains the Avogadro Number of particles.2. The Avogadro Number (or Avogadro Constant) is defined as the number of atoms in 12 g of the carbon-12 isotope, which is equal to 6.02 x 1023.1 mole = 6.02 x 1023

Example:1 mol of atoms =6.02 x 1023atoms mol of atoms =3.01 x 1023atoms mol of atoms =1.505 x 1023atoms.

Note:The particles in a material can be atoms, molecules or ions.Therefore1 mol potassium atom =6.02 x 1023potassium atom1 mol carbon dioxide molecules =6.02 x 1023of carbon dioxide molecules.1 mol sulphate ions =6.02 x 1023sulphate ions.

Example:Find the number of atoms in:a. 2 mol ferumb. 3.6 mol zinkc. 2.8 mol zinkd. mol ferumAnswer:a. 2 mol iron = 2 x6.02 x 1023= 1.204x 1024iron atomsb. 3.6 mol zink = 3.6 x6.02 x 1023= 2.167x 1024zink atomsc. 2.8 mol zink = 2.8 x6.02 x 1023= 1.686x 1024zink atomsd. mol iron = x6.02 x 1023= 1.505x 1023iron atoms.

Mole and mol1. Mole is the unit of amount of substance.2. mol is the symbol of mole.Number of Mole and Number of Particles1. We have just learn that, mole is aquantity, and it is equal to 6.02 x 1023. The number6.02 x 1023is called the Avogadro constant. 2. Therefore, if we are given the number of mole of substance, and asked to find the number of particles (atoms, molecules or ions) in it, we multiply the number of mole by the Avogadro constant.3. Likewise, if we are given the number of particles, and asked to find the number of mole of the particles, we divide the number of particles by the Avogadro constant.

ExampleWhich contains more atoms, 1 mol of helium or 1 mol of uranium? Which has a greater mass? [ RAM: He=4; U=238 ]

Answer:1 mol of helium and 1 mol of uranium has equal number of atoms.

The mass of one Uranium atom ss greater than the mass of one helium atom.

ExampleFind the number of atoms in 2.5 mol of gold.

Answer:Number of atoms= Number of mole x Avogadro constant= 2.5 x6.02 x 1023=1.505 x 1024

ExampleHow many moles of magnesium that contain 2.76x 1023of magnesium atom?

Answer:Number of mole= Number of atomsAvogadro constant=2.76x 10236.02 x 1023= 0.46 molNumber of Mole of Atoms and Number of Mole of Molecules1. All molecules contain more than one atom.2. For example, in a carbon dioxide molecule (CO2), there are 3 atoms - 1 carbon atom and 2 oxygen atoms.3. Therefore, in 5carbon dioxide molecules, there will be 15 atoms- 5 carbon atom and 10 oxygen atoms.4. Similarly,in 1 molecarbon dioxide molecules, there will be 3 mole atoms- 1 mole carbon atoms and 2 mole oxygen atoms, and in 3molecarbon dioxide molecules, there will be 9 mole atoms- 3 mole carbon atoms and 6 mole oxygen atoms, so on and so forth.

Example:a. How many hydrogen molecules can be made with 20 mol of hydrogen atoms?b. How many hydrogen atoms are there in 20 mol of hydrogen molecules?Answer:a.Each hydrogen molecules (H2) consist of 2 hydrogen atoms.Therefore,20 mol of hydrogen atoms can make 10 mole of hydrogen molecules.

b. Number of mole of hydrogen atoms in 20 mole of hydrogen molecules= 2 x 20 mole = 40 mole.

Example:a. Find the number of carbon atom in 0.75 mol of carbon dioxide.b. Find the number of oxygen atom in 0.75 mol of carbon dioxide.Answer:a.Each carbon dioxide molecules(CO2)consist of 1 carbon atom and 2 oxygen molecules.Therefore, 0.75 mole of carbon dioxide contain 0.75 mole carbon atoms.

b.Number of mole of oxygen atoms in 0.75 mole of carbon dioxide= 2 x 0.75 mole = 1.50 moles.Number of Mole and Mass of a SubstanceMolar Mass

1. The mass of one mole of atoms is its relative atomic mass in grams, and is called a molar mass.2. A Molar mass is the mass of a substance containing the Avogadro Constant of particles.

ElementRelative atomic massMass of 1 mol of particleMolar Mass

Hydrogen11 g1 g mol-1

Carbon1212 g12 g mol-1

Oxygen1616 g16 g mol-1

Copper6427 g27 g mol-1

Iron5656 g56 g mol-1

Example

1. The relative atomic mass of copper is 64, therefore the mass of 1 mole copper is64g2. The RAM of sodium is 23, therefore the mass of 2 mole sodium is46g3. The RAM of nitrogen is 14, therefore the mass of 2.5 mole l nitrogen is35g4. The RAM of calcium is 20, therefore the mass of 0.3 mole calcium is6gConclusion

Note:The relationship between the mass of a substance and the number of mole of the particles in the substance can be summarised by using the following equation:

n=

If you are given the mass of substance and asked to find the number of mole of the substance (or vice versa), the problem can be solved by using this equation.

Example:Find the number of mol of atoms in 4.6g sodium [Relative atomic mass: Na=23]

Answer:Number of mole,n= =0.2mol

ExampleHow many moles of each substance are there in 191 g NaOH [Relative atomic mass: Na=23, O=16, H=1]

Answer:Relative Formula Mass of NaOH = 23 + 16 + 1 = 40Number of mole of 191g NaOH,n= =4.775mol

Example 1 - Number of Mole and Mass of a SubstanceQuestionWhat is the mass of7.12 mol NaI.[Relative atomic gas: Iodine = 131; Sodium = 23]

Answer:The relative formula mass of NaI = 23 + 131 = 154.

The mass of 7.12 mol NaI= Number of mole x Relative Foemula Mass of NaI= 7.12 x 154= 1096.48gMass of Substance and Number of Particles

1. Sometime, you may be given the mass of a substance, and asked to find the number of particles (or vice versa).2. To solve the problem, we must find the number of mole of the substance then only we can find the number of particles (or mass of the substance).

Example:

How many molecules are there in 16 g of oxygen (O2)?[ Ar: O =.16, Avogadro Number = 6 x 1023]

Answer:The relative molecular mass of O2= 2(16) = 32Numberofmoleofoxygenmolecules===0.5molExample 1 - Mass of Substance and Number of ParticlesQuestion:Find the number of atoms in 34g of ammonia gas (NH3). [Relative atomic mass: Nitrogen: 14; Hydrogen: 1]

Answer:Relative molecular mass ofNH3= 14 + 3(1)= 17

Number of mole of ammonia=3417=2mole

Number of NH3molecules = 2 x 6.02 x 1023= 1.204 x 1024

EachNH3molecule contain 4 atoms (1 nitrogen atom and 3 hydrogen atoms).

Therefore, number of atoms= 4 x1.204 x 1024=4.816 x 1024Example 2 - Mass of Substance and Number of ParticlesFind the mass of methane (CH4) that contain 1.806 x 1023of methane molecules. [Relative atomic mass: Carbon: 12; Hydrogen: 1]

Answer:The relative molecular mass ofCH4= 12 + 4(1)= 16

NumberofmoleofCH4=1.80610236.021023=0.3mole

Mass ofCH4= Number of mole x molar mass= 0.3 x 16 = 4.8gNumber of Mole and Volume of GasMolar Volume of Gas1. Equal amount of gas occupies same volume of space.2. Molar volume of a gas is defined as the volume occupied by one mol the gas at certain condition3. At room temperature(25C) and pressure (1 atm), one mole of any gas occupies a volume of 24 dm (24 000 cm )4. At standard temperature(0C) and pressure (1 atm) one mole of any gas occupies a volume of 22.4 dm(22400cm).

Number of mole of gassesVolume at s.t.p. (dm3)

1 mol of O2gas1 x 22.4 = 22.4

1 mol of H2gas1 x 22.4 = 22.4

2 mol of He gas2 x 22.4 = 44.8

3.5 mol of N2gas3.5 x 22.4 = 78.4

Number of mole of gassesVolume at r.t.p. (dm3)

1 mol ofO2gas1 x 24 = 24

1 mol ofH2gas1 x 24 = 24

2 mol of He gas2 x 24 = 48

3.5 mol ofN2gas3.5 x 24 = 84

Example:What volume (at room temperature and pressure r.t.p.) would2 moles of oxygen gasoccupy? (Molar Gas Volume at r.t.p. = 24 dm)

Answer:Volume of gas = 2 x 24 = 48 dm

Example:A sample of ozone gas has a volume of 960cm at room temperature and pressure. Find the number of mole of the ozone. [Molar volume at r.t.p. = 24.0dm]

Answer:Numberofmole= =0.04molVolume of Gas, Mass of Substance and Number of Particles

Figure below shows the relationship between the number of mole of particles in a substance with the mass, number of particles and volume of gas.

Example

Find the number of particle in 1200cm of nitrogen gas at room temperature. (Molar Gas Volume at r.t.p. = 24 dm)

Answer:

Numberofmoleofnitrogengas=1200cm324000cm3=0.05mol

Numberofnitrogenmolecules=0.05mol6.021023=3.011022Example 1 - Volume of Gas, Mass of Substance and Number of ParticlesQuestion:A container contains 3.913 x 1023methane molecules. What is the volume of the methane gas at r.t.p.? [Avogrado constant = 6.02 x 1023; Molar volume of gasses at r.t.p. = 24,000 cm3/mol]

Answer:

Numberofmoleofmethanegas,n=3.91310236.021023=0.65molVolumeofgas=0.6524000cm3=15600cm3

Example 2 - Volume of Gas, Mass of Substance and Number of ParticlesQuestion:A gas jar containing 200cm of oxygen gas at s.t.p. Find the mass of the oxygen gas. [Relative atomic mass: Oxygen = 16; Molar volume of gas at s.t.p. = 22.4 dm]

Answer:

Numberofmoleofoxygen gas,n=200cm322400cm3=1112molRelativemolecularmassofoxygen=216=32Massofoxygen=(1112)32gmol1=0.286gChemical Formulae1. Each element is represented by its own chemical symbol.2. In an element, the chemical formula represents the atoms in the molecule.3. The chemical formula tells us:a. the types of atoms or ions in the compound,b. the number of atoms or ions in the compound,4. For example, the chemical formula for ethene is C2H4. This shows that ethene is the result of the combination of the elements carbon and hydrogen, and there are 2 carbon atoms and 4 hydrogen atoms in each molecule of ethene.5. Generally, chemical formula can bedividedintoa. Empirical formulab. Molecular formula6. Theempirical formulaof a substance is the chemical formula that gives the simplest whole-number ratio of atoms of each element in the substance.7. Themolecular formulaof a substance is the chemical formula that gives the actual number of atoms of each element in the substance.Empirical Formula1. The empirical formula of a substance is the chemical formula that gives the simplest whole-number ratio of atoms of each element in the substance.2. Empirical = information gained by means of observation, experience, or experiment.Example:Chemical SubstancesMolecular FormulaEmpirical Formula

GlucoseC6H12O6CH2O

WaterH2OH2O

Carbon DioxideCO2CO2

BenzeneC6H6CH

ButaneC4H8CH2

Finding Empirical FormulaSteps to determine the empirical formula of a compoundSTEP 1: Find the massSTEP 2: Find the moleSTEP 3: Find the simplest ratio

Example:In a chemical reaction, 4.23g of iron reacts completely with 1.80g of oxygen gas, producing iron oxide. Calculate the empirical formula of iron oxide. [Relative atomic mass: Iron = 56; Oxygen = 16]

Answer:ElementFeO

Mass4.23g1.80g

Number of mole =0.0755 =0.1125

Simple ratio =1 =1.5

Ratio in round number23

The empirical formula of iron oxide = Fe2O3

Example 1 - Empirical Formula

Example 2 - Empirical Formula

Molecular Formula1. Themolecular formulaof a substance is the chemical formula that gives the actual number of atoms of each element in the substance.2. A molecular formula is the same as or a multiple of the empirical formula.3. For example, the empirical of carbon dioxide is CO2and the molecular formula is alsoCO2.4. Whereas, the empirical formula of ethane is CH3while the molecular formula of ethane isC2H6.

Finding Molecular FormulaExampleGiven that the empirical formula of benzene is CH and its relative molecular mass is 78. Find the molecular formula of benzene. [Relative Atomic Mass: Carbon: 12; Hydrogen: 1]

Answer:Let's say the molecular formula of benzene is CnHn.

The relative molecular mass ofCnHn= n(12) + n(1)= 13n

13n = 78n = 78/13 = 6

Therefore, the molecular formula of benzene=C6H6Example 1 - Molecular Formula

Percentage of Composition of a Compound

Formula of Ions1. Ionic compounds made up of positive ions and negative ions.2. To write the formula of an ionic compound, we need to know the symbol and charge of the ions in the compound.Positive ions that you need to know in SPM form 4 chemistry:IonSymbolIonSymbolIonSymbol

PotassiumK+CalciumCa2+AluminiumAl3+

SodiumNa+MagnesiumMg2+Iron (III)Fe3+

LithiumLi+ZincZn2+

HydrogenH+BariumBa2+

Argentums(I)Ag+Iron (II)Fe2+

Mercury(I)Hg+Tin (II)Sn2+

AmmoniumNH4+Lead(II)Pb2+

Copper(II)Cu2+

Manganese(II)Mn2+

Negative ions that you need to know in SPM form 4 ChemistrySulphateCarbonateNitrateHydroxide

SO42-CO32-NO3-OH-

FluorideChlorideBromideIodideOxide

F-Cl-Br-I-O2-

Negative ions that you need to know in SPM form 5 ChemistryPhosphateEthanoatManganate(VII)Dichromate(VI)Thiosulphate

PO43-CH3COO-MnO4-Cr2O72-S2O32-

Chemical Formula of Ionic Compounds2 Requirements to form the formula of an ionic compound

1. Have at least 2 types of ions that contain opposite charge.2. The amount of positive charge/charges must be equal to the amount of negative charge/charges in the compound.Example 1 If the Charge of the Positive Ions = Charge of Negative IonsWrite the formula of each of the following compound

a. Potassium bromideb. Sodium chlorideAnswer:a.Potassium bromide

b.Sodium chloride

Example 2 If the Charge of the Positive Ions Charge of Negative IonsWrite the formula of each of the following compounda. Calcium iodideb. Sodium oxideAnswer:a.Calcium iodide

b.Sodium oxide

Example 3 If there is more than 1 element in the ionWrite the formula of each of the compounda. Ammonium sulphateb. Zinc nitrateAnswer:a.Ammonium sulphate

b.Zinc nitrate

Example 4 Formulae of AcidWrite the formula of the following acida. Hydrochloric acidb. Sulphuric acidc. Nitric acidd. Ethanoic acide. Phosphoric acidf. Carbonic acidAnswer:a.Hydrochloric acid

b.Sulphuric acid

c.Nitric acid

d.Ethanoic acid

e.Phosphoric acid

f.Carbonic acid

Formula of MoleculeIn SPM, you need to know the formulae of the following molecules

FluorineF2Hydrogen chlorideHCl

ChlorineCl2Hydrogen bromideHBr

BromineBr2Hydrogen iodideHI

IodineI2Nitric oxideNO

Carbon dioxideCO2Nitrogen dioxideNO2

waterH2ONitrous oxideN2O

AmmoniaNH3TetrachloromethaneCCl4

Sulphur trioxideSO3Hydrogen sulphideH2F

Sulphur dioxideSO2GlucoseC6H12O62

Chemical EquationWriting Chemical EquationsA Balance EquationAccording to the Law of Conservation of Mass, atoms cannot be created or eliminated in the process of chemistry reaction. Therefore, the amount of atom of each element before reaction must be equal after reaction.Example:Balance the following chemical equations:

a. Mg + O2 MgOb. CuO + H2 Cu + H2Oc. Na + H2O NaOH + H2d. Li + Cl2 LiClAnswer:a.2Mg + O2 2MgO

b.CuO + H2 Cu + H2O

c.2Na + 2H2O 2NaOH + H2d.2Li + Cl2 2LiCl

Writing a Balance EquationExample:Lead(II) nitrate aqueous react with potassium iodide aqueous to produce lead(II) iodide precipitate and potassium nitrate aqueous.

Answer:

Mind Map - Chemical Formulae and Equations