Chapter 11.1 Atoms and Molecules

At the end of this topic, students should be able to:Identify and describe proton, electron and neutron.Define proton no.,Z, nucleon no., A and isotope. Write isotope notation.Define relative atomic mass and relative molecular mass based on the C-12 scale.Sketch and explain the function of the following main components of a simple mass spectrum:Analyze mass spectrum of an element.Name cations, anions and salt according to the IUPAC.

1.1 Atoms and MoleculesWhat is matter?

Matter is anything that has mass and occupies space.All matters consist of tiny particles called atoms.3 states of matter are solid, liquid and gas.For examples are metal, plastics, gas, etc.

1.1.1 Atoms

Protons and neutrons are found in the nucleus of the atom, while electrons surround the nucleus to form electron cloud.

The properties of these particles are summarized in the table.

Nucleonthe particles that are found in the nucleus.

consist proton and neutron.

By proton we can identified the element.

PROTON NUMBER AND NUCLEON NUMBERProton number (atomic number), Z.The proton number = the number of protons in the nucleus of an atom.

Nucleon number (mass number), A.The nucleon number = the total number of protons and neutrons in the nucleus of an atom.

1.1.2 Isotopic notation Nucleon numberProton numberAtom or ion

For neutral atom, number of protons = number of electrons.

For +ve ion, number of electron is less than number of protons.

For ve ion, number of electrons is more than number of protons.

Number of protons = proton number= 13

Number of neutrons = nucleon number proton number= 27 - 13= 14

Number of electron= proton number charge carried by species= 13 ( +3)= 10

Example:

Exercise:Determine the number of proton, neutron and electron in the following species.

a) b)

Exercise:The atomic number of lead (Pb) is 82 and the mass number is 207.

Write out the atomic notation for lead.Give the number of protons, neutrons and electrons.

1.1.3 ISOTOPES Two or more atoms of the same element having same proton number but different nucleon number.

Eg : 1) hydrogen isotope

2) carbon isotope 12 14 6 611H21H31HCC;

Isotopes of an element have the same:

number of protons (proton number)

charge of nucleus of the atoms (ionization energy; electron affinity; size of the atom; electronegativity are the same)

number of electrons in a neutral atom

electronic configuration (the number of valence electrons)

chemical properties

Isotopes of an element have different:

number of neutrons (nucleon number) in the nucleus of the atoms

relative isotopic mass

physical properties (e.g boiling point / melting point, density, effusion rate,)

1.1.4 Molecule

A molecule consists of a small number of atoms joined together by covalent bond.

Diatomic molecule: contains two atoms (example: H2, Cl2, HCl, CO)

Polyatomic molecule: contains more than two atoms (example: H2O, NH3)

1.1.5 Ion

An ion is a charged species formed from a neutral atom or molecule when electrons are gained or lost as the result of a chemical reaction.

Cation: a positively charged ion (number e < number p) (example: Mg2+, Al3+)

Anion: a negatively charged ion (number e > number p) (example: Cl, OH)

Monatomic ion: ion contains only one nucleus (example: Fe3+, S2)

Polyatomic ion: ion contains more than one nucleus (example: H3O+, CN)

1.1.6 RELATIVE MASSIsotopes carbon -12 as a reference or standard for comparing the masses of other atoms.

RELATIVE ATOMIC MASS, Ar

Average mass of one atom of the element relative to 1/12 times the mass of one atom of carbon-12.

The mass of C-12 is defined exactly 12.00 a.m.u.

1 atomic mass unit : a mass unit equal to 1/12 the mass of a C-12 atomAr = average mass of one atom of the element 1/12 x mass of one atom of C-12

Example:Oxygen consists of three isotopes 16O ;17O and 18O in the ratio of 99.76 : 0.04 : 0.20. Calculate the relative atomic mass (to 4 decimal point) of oxygen.

ANSWER :

Relative Atomic Mass

= (16 x 99.76) + (17 x 0.04) + (18 x 0.20) (99.76 + 0.04 + 0.20)

= 16.0044

Exercise:Determine the Ar for Ne consists of three isotopes 20Ne ; 21Ne and 22Ne in the ratio of 90.92 ; 0.26 and 8.82.

Answer:20.18

RELATIVE MOLECULAR MASS, MrThe mass of one molecule of the substance relative to 1/12 times the mass of one atom of carbon-12.

= sum of the relative atomic masses of all the atoms shown in the molecular formula.

Mr = average mass of one molecule of the substance1/12 x mass of one atom of C-12

Example:The relative molecular mass of carbon dioxide, CO2.

Answer:

Mr CO2 = Ar C + 2 Ar O = 12 + 2(16) = 44

Exercise:The relative molecular mass of ethanol, C2H5OH

Answer:46.00

1.1.7 Mass SpectrometerMass spectrometer is used to determine:

Relative atomic mass of an element

Relative molecular mass of a compound

Types of isotopes that are found in the naturally occurring element including the abundance of the isotopes and its relative isotopic mass.

Recognize the structure of the compound in an unknown sample

There are five main stages:

(a) Vaporisation Chamber ~Sample of the element is vaporised into gaseous atom

(b) Ionisation Chamber ~A gaseous sample (atom or molecule) is bombarded by a stream of high-energy electrons that are emitted from a hot filament. Collisions between the electrons and the gaseous atom (or molecule) produce positive ions by dislodging an electron from each atom or molecule.

M sample + e filament M+ sample + e sample + e filament M M+

(c) Acceleration Chamber (Electric field)

The positive ions are accelerated by an electric field towards the two oppositely charged plates. The electric field is produced by a high voltage between the two plates. The emerging ions are of high and constant velocity.

Magnetic field

The positive ions are separated and deflected into a circular path by a magnet according to its mass/charge (m/e) ratio.

Positive ions with small m/e ratio are deflected most and appear near A. Ions with large m/e ratio are deflected least and appear near B (Figure 1.1.2).

(e) Ion detector

The numbers of ions and types of isotopes are recorded as a mass spectrum.

In practice, the ion detector is kept in a fixed position. The magnetic field is varied so that the positive ions of different masses arrive at the detector at different times.

Mass spectrumthe horizontal axis the m/e ratio nucleon number isotopic mass relative atomic mass of the ions entering the detector.

The vertical axis - the abundance or detector current or relative abundance or ion intensity or percentage abundance of the ions.

Information from a mass spectrum of an elementthe isotopes which are present in the elementthe relative isotopic mass of each isotopethe abundance of each isotope

Thus, the relative atomic mass of the element can be determined

Relative abundancem/e2425260638.19.1The mass spectrum of magnesium shows that naturally occurring magnesium consists of three isotopes: 24Mg, 25Mg and 26Mg.

The height of each line is proportional to the abundance of each isotope.

In this example, 24Mg is the most abundance of the three isotopes.

Ar of Mg: = 24.33

where Q = the abundance of an isotope of the element = the percentage of the isotope found in the naturally occurring element

m = the relative isotopic mass of the element Average atomic mass =

Notes :The height of each peak measures the relative abundance of the ion which gives rise to that peak.

The total number of peaks in the mass spectrum of an element shows the types of naturally occurring isotopes.

3)The ratio of mass/charge for each species is found from the value of the accelerating voltage associated with a particular peak. Many ions have a charge of +1 elementary charge unit, and the ratio m/e is numerically equal to m, the mass of the ion. (1 elementary charge unit = 1.60x10 19 C)

4)The ion with the highest value of m/e is the molecular ion, and its mass gives the molecular mass of the compound

1.1.8 IUPAC Nomenclature for Cations, Anions and Salts Cations A positive ion is formed by the loss of a certain number of electrons from the given atom

For example:

Mg 2e- + Mg2+ (2e- were lost by Mg atom)

K1e- + K+ (1e- were lost by K atom)

Anions A negative ion is formed by gain of one or more electrons by the given atom

For example:

S + 2e-S2- (2e- were gain by sulphur atom)

Cl + 1e-Cl- (1e- were gain by chlorine atom)

Salts-Combination of cation and anion to form ionic compound

For example:

Na+ + Cl-NaCl

Mg2+ + 2Br-MgBr2

a) Naming monoatomic ionsMonoatomic ions:ions consisting of single atomic nucleus

Cation: named by just adding ion Ex:K+ : potassium ionCa2+ : calcium ion

Anion : named by adding suffix ideEx:Cl-: chloride ionS2-: sulphide ionO2-: Oxide ion

b) Naming ionic compoundsWhen naming an ionic compound made up of two element (binary): the name of the cation (metal ion) appear first, followed by the name of the anion(non-metal ion)

compoundMetal ionNon-metal ionNameNaFNa+, SodiumF-, FlourideSodium FlourideMgBrMg2+, MagnesiumBr-, BromideMagnesium Bromide

System for naming cation of transition metals Stock system/systematic system:- a Roman numeral that match the ionic charge is placed in bracket immediately after the element name of metal

- for example: Fe 2+ iron(II) ion

Common nomenclature system- suffix-ous indicates the lower ionic charge and the suffix-ic indicates the higher ionic charges- for example: Fe2+ ferrous ionFe3+ferric ion

Systematic (stock) and common names for ion and copper ions

formulachargeSystematic NameCommon NameFeCl22+Iron(II) chlorideFerrous chloride

FeCl33+Iron(III) chlorideFerric chloride

Cu2O1+Copper(I) oxideCupprous oxideCuO2+Copper(II) oxideCupric oxide

Rules for writing for an ionic compoundThe positive ion is always written first.

The ratio of positive ions to negative ions must be such that the total number of +ve charge equal the total number of -ve charge; the formula unit must be electrically neutral

The smallest set of subscripts that give electrical neutrality is always chosen. We always write empirical formulas for ionic compounds

c)Naming polyatomic ionsPolyatomic ions are composed of 2 or more atoms bonded togetherMost polyatomic ion consist to non-metal such as P, S, C, or N bonded to oxygen atoms-ate: most common polyatomic ion -ite : used for names of related ion that have one less oxygen atom

Chapter 11.2 Mole Concept

At the end of this class, students should be able to:Define mole in term of carbon-12 and Avogadro constant, NA.Interconvert between moles, mass, number of particles, molar volume of gas at s.t.p and room temperature.Determine empirical and molecular formulae from mass composition or combustion data.

N + 3H NH3

1 MOLEATOM N3 MOLEATOM H1 MOLEMOLECULE NH3

MOLE CONCEPTMole: a substance quantity that contain a number of particles such as atom, ion, or molecule that have equal number of atom in 12.00 grams of C-12.

Therefore, one mole of particles contains 6.023 x 1023 particles. This number is known as Avogadro Number (symbol: L @ NA)

NA = 6.02 x 1023 mol-1

Examples:

1 mole of Cu contain 6.023 x 1023 atoms.1 mole of CO2 gas contain 6.023 x 1023 molecules.1 mole of Cl- ion contain 6.023 x 1023 ions.1 mole of electron contain 6.023 x 1023 electrons. No. of moles = no. of particles NA

MOLAR MASSThe mass of one mole of an element is the relative atomic mass expressed in grams.The mass of one mole of a compound is the relative molecular mass of the compound expressed in grams.Unit for molar mass = g/mole or gmole-1

No. of moles = mass of substances molar mass

EXAMPLE:1.0 mol of chlorine atoms = 6.023x1023 chlorine atoms= 35.5 g Cl

1.0 mol of chlorine molecules = 6.023x1023 chlorine molecules= 2(35.5) g = 71.0 g Cl2= 6.023x1023 x 2 chlorine atoms

Examples:Mass of 1 mole of Mg = 24.0 gMass of 1 mole of chlorine gas ( Cl2) = 71.0 gMass of 1 mole of KNO3 = 101.0 g

Tips:

/ molar mass x NAg mol no. of atom

Example 1:Calculate the number of mole in 6.4 g Cu2.5 g H2O

Answer:

Mole Cu = 6.4 g = 0.1 mole 64 g/moleb)Mole H2O = 2.5 g = 0.139 mole 18 g/mole

Example 2:How many atoms H in 1 mole CH4?

Answer:

1 mole CH4 consist 4 moles atom Hno. of atoms H in 1 mole CH4= 4 x 6.023 x 1023= 2.4092 x 1024 atoms

Example 3:The mass of NH3 = 0.25 g. Calculate : number of mole of NH3number of molecule of NH3 , number of N atom and H atom.[ Ar : N = 14 ; H = 1 ]

Answer :

a) No. of mole = 0.25 g = 0.015 mole 17 gmole-1

b) No. of molecule = 0.015 mole x 6.023 x 1023 molecule/mole= 9.035 x 1021 molecules of NH3

No. of N atom = 0.015 mole x 1 x 6.023 x 1023 atom/mole= 9.035 x 1021 atoms of N

No. of H atom = 0.015 mole x 3 x 6.023 x 1023 atom/mole= 2.71 x 1022 atoms of H

Exercise 1:Calculate the number of ion Ca2+ and ion Cl- in 1 mole CaCl2.

Answer:

Ion Ca2+ = 6.023 x 1023Ion Cl- = 2 x 6.023 x 1023 = 1.2046 x 1024

Exercise 2:A sample consist of 64.0 g oxygen gas. Calculate :a) No. of mole of the oxygen gasNo. of oxygen atom in the sample.

Answer:2 mole2.41 x 1024 atoms

MOLE CONCEPT OF GASESMolar volume of any gases at STP, Vm = 22.4 dm3mole-1STP = Standard Temperature and PressureT = 273.15 KP = 1 atm @ 760 mmHg

Mole of gases at STP,

n = Volume of gas (L)Vm (22.4 dm3mole-1)

Note: Do not use 22.4 L as the molar volume at temperatures and pressures other than 273 K and 1 atm.

E.g :

How many moles are there in 6.5 L oxygen at STP ?Answer:

Mole of O2 = 6.5 L = 0.29 mole 22.4 L mole-1

Empirical FormulaThe simplest formula for a compound.

It does not express the real composition of molecule.

For example: methane, CH4

Example: A sample of a brown-colored gas that is a major air pollutant is found to contain 2.34 g of N and 5.34 g of O. What is the simplest formula of the compound?(N=14.0 g mol-1 ,O =16.0 g mol-1)

Solution:

Elements present NOMass of compound2.34 g5.34 gNumber of mole, n2.34 g14 g mol-1= 0.167 mol5.34 g16 g mol-1= 0.334 mol Simplest mole ratio0.167 mol0.167 mol=10.334 mol0.167 mol= 2Empirical FormulaNO2

2. What is the empirical formula for the compound composed of 43.7% P and 56.3% O by weight? ( P = 31.0 , O= 16.0 )

Elements present POMass of compound43.7 g56.3 gNumber of mole, n43.7 g31 g mol-1= 1.41 mol56.3 g16 g mol-1= 3.52 mol Simplest mole ratio1.41 mol1.41 mol=13.52 mol1.41 mol= 2.5Empirical Formula(ratio x 2)P2O5

Molecular formulaShows the actual number of the atom in one molecule of the compound

Where n = 1,2,3,4..Molecular formula = n x Empirical formula

The Mr for the hydrocarbon, CH is 78. Determine its molecular formula. ( C= 12 , H= 1)

Example:

Let the molecular formula be (CH)n

Mr of compound = 78 (CH)n= ( 12 + 1)n= 13n

13n= 78 n= 78 13= 6Molecular formula = n x Empirical formula:. Molecular formula= C6H6

Concentration UnitsThe concentration of solutions is the quantity of dissolved substance per unit quantity of solvent in a solution.

It can be expressed in1. Molarity, M2. Molality, m (molal concentration)3. Mole fraction, X4. Percent by mass, % w/w5. Percent by mass/volume, % v/v

1. Molarity, MNo. of moles of solute dissolved in 1L (1 dm3 @ 1000 cm3) of solution.

Unit : mole L-1 @ mole dm-3 @ M.Molarity, M = no.of mole solute no.of L solution* 1 L = 1 dm3 1 mL = 1 cm3 1 dm3 = 1000 cm3

Example:A solution prepared by dissolving 3.42 g of sucrose ,C12H22O11 in 500 cm3 of water. Calculate its molarity.

Answer:M = 3.42 g C12H22O11 x 1 mole C12H22O11 500 cm3 342 g C12H22O11 x 1000 cm3 1 L = 0.02 mol / L

Exercise: How many grams of solute are present in 783 ml of an aqueous solution of 0.35 M KOH?

Answer:

15.621 g

The number of mole of solute per unit mass of solvent in kg.

(unit: mol kg1 @ molal, m)2) Molal Concentration @ molality, mMolality = moles of soluteMass of solvent in kg1 kg= 1 dm3 = 1 L

Example:What is the molality when 0.75 mol is dissolved in 2.50 dm3 of solvent?

Answer:Molality, m= moles of soluteMass of solvent in kg= 0.75 mol 2.50 kg = 0.30 mol kg-1

1 dm3 = 1 kg

3. Mole fraction, X

The number of moles of one component in a mixture divide by the total number of moles of all substances present in the mixture.

Note that the total mole fraction in a mixture is equal to one.

XA = nA ntotal

ntotal = nA + nB + ..

Example:A solution is prepared by dissolving 32.0 g of methanol (CH3OH) in 72.0 g of water. Calculate the mole fraction of methanol in the solution.Answer:No. of moles CH3OH = 32.0 = 1.0 mole 32.0 No. of moles H2O = 72.0 = 4.0 moles 18.0X CH3OH = 1 = 0.2 1 + 4

4. Percent by mass, % w/w

The percentage of the mass of solute per mass of solution.

Note:Mass of solution = mass of solute + mass of solvent% w/w = mass solute x 100% mass solution

Example:8g of sugar is dissolved in 42g of water. What is the concentration (% w/w) of the solution formed?

Answer:% w/w = mass solute x 100 mass solution = 8 g x 100 (42 + 8)g = 16 %

5. Percent by volume/volume, % v/vPercentage by mass/volume is defined as the percentage of volume of solute in mL per volume of solution in mL.

% v/v = volume of solute, mL x 100% volume of solution, mL5% v/v of KCl 5 mL of KCl dissolved in 100 mL of solutionNote : mass of solution = volume of solution x density of solution

Example:A 20 mL sample of plant food solution contains 1.4 mL of glucose. What is the concentration (in % v/v) of glucose in the plant food?

Answer:% v/v = volume of solute (mL) x 100 % volume of solution (mL) = 1.4 mL x 100 % 20 mL = 7 %

Oxidation NumberOxidation numbers are used to keep track of how many electrons are lost or gained by each atom.

the change in the oxidation state of a species shows either it has undergo oxidation or reduction.

General rules:

The oxidation state for any element in its free state is zero, 0.Na, Mg, H2, O2, Be, P4

Any ions composed of only one atom (monatomic ions) has an oxidation number equal to its ionic charge.

All group I elements form +1 ionsAll group II elements form +2 ions

The oxidation number of oxygen is always -2, except in peroxides ion (O22-) and hydrogen peroxide (H2O2) where it is -1.

The oxidation number of hydrogen is always +1, except when it is bonded to metals in binary compound (LiH, NaH, CaH2) where it is -1.

Fluorine has an oxidation number -1 in all its compounds.

Other halogens (Cl, Br, and I) have ve oxidation numbers when they occur as halide ions in their compounds. When combined with oxygen (in oxoacids and oxoanions) they have +ve oxidation numbers.

The oxidation number of Cl is -1 in HCl, but the oxidation number of Cl is +1 in HOCl.

The sum of the oxidation numbers of all of the atoms in a neutral compound is 0.CO2, FeSO4, NaCl, FeCl3

The sum of the oxidation numbers of all the atoms shown in the formula for a polyatomic ion or complex ion equals the electrical charge on the ion.Cr2O7 2- , NO3 -

Example

What are the oxidation number of Al2O3

Step 1: The oxidation no of oxygen is -2Step 2: The total oxidation no of Al2O3 is zero Assume oxidation no of Al = x

2(x) + 3(-2) = 0 2x 6 = 02x =+6 x = +3

b) Cr2072-

Step 1: The oxidation no of oxygen is -2Step 2: The total oxidation no of Cr2072- is -2 Assume oxidation no of Cr = x 2x + 7(-2) = -2 2x 14 = -2 2x = +12 x = +6

Exercises :

What are the oxidation number of

a) Mn atom in MnO4 - =

b) Br - =

c) Cl atom in ClO3 - =

d) S atom in K2S =

e) N atom in NH3 =

REDOX REACTIONReaction that involves transfer of electrons between elements in the reactants and products.

Oxidation ~ loss of electrons ( oxidising agent must gain electrons)

Reduction ~ gain of electrons ( reducing agent must lose of electrons)

EXAMPLEZn (s) + Cu2+ (aq) Zn2+ (aq) + Cu(s)

OXIDISING AGENTREDUCING AGENTOXIDATION REACTIONREDUCTION REACTION

CHEMICAL REACTIONa process in which one set of substances called reactants is converted to a new set of substances called products.

CHEMICAL EQUATION

Shorthand expression for a chemical reaction. aA + bB cC + dD (reactants) (products)

= irreversible reaction = reversible reaction

When reactions involve different phase it is usual to put the phase in brackets after the symbol (s=solid; l=liquid; g=gas; aq=aqueous).The numbers a, b, c and d showing the relative numbers of molecules reacting, are called the stoichiometric coefficients.Eg:Given :CaCO3 (s) + 2HCl (aq) CaCl2 (aq) + CO2 (g) + H2O (l)

Reactants : CaCO3 and HCl

Products : CaCl2 , CO2 and H2O

BALANCING A CHEMICAL EQUATION2 methods: 1) by inspection

2) by Ion Electron

HCl + Ca CaCl2 + H2Step 1:Count the no of moles of atoms of eachelement on both product and reactant sides

By Inspection Method

ReactantsProducts1 mol H atom2 mol H atoms1 mol Cl atoms2 mol Cl atoms1 mol Ca atoms1 mol Ca atoms

Step 2:Determine which elements are not balanced The no of moles of H and Cl are not balanced

Balance one element at a time using coefficients

~ Insertion of a 2 before HCl on the reactant side should be balance the equation:

2HCl + Ca CaCl2 + H2Step 3:

Step 4:Check the mass balanceThe equation is balanced

ReactantsProducts2 mol H atoms2 mol H atoms2 mol Cl atoms2 mol Cl atoms1 mol Ca atoms1 mol Ca atoms

Balancing a Chemical Equation The Ion Electron Method (or Half-Reaction Method)

is to split an unbalanced redox equation into two half-reactions, oxidation and reduction half-reactions

Both half-reactions are then balanced in according to the charge of electrons.

It is applied the best for redox reactions with ions and in aqueous solution.

Step-by-step in balancing redox reactions using this method : Separate into half-reactions.

Balance the oxidizing and reducing elements in each half-reaction.

Balance atoms other than O and H.

In acidic solution, balance oxygen with H2O and hydrogen with H+. In basic solution, balance oxygen with H2O, hydrogen with H+ and OH-.

Balance each half-reaction for charge with e-.

Multiply each half-reaction by a coefficient so the numbers of electrons are the same in each and add both half-equations together.

Cancel common reactants and products.

Eliminate fractions by multiplying a common factor if needed.

Complete the equation by final inspection.

STOICHIOMETRY

Quantitative study of reactants and products in a chemical reaction.

The coefficients in a balanced equation tell us the ratio of amounts (in moles @ mass) of reactants and products.

Stoichiometry Steps1. Write a balanced equation.2. Identify known & unknown.3. Line up conversion factors.Mole ratio - moles moles

Molar mass -moles gramsMolarity - moles liters solnMolar volume -moles liters gasCore step in all stoichiometry problems!!Mole ratio - moles moles4. Check answer.

Given :CaCO3 (s) + 2HCl (aq) CaCl2 (aq) + CO2 (g) + H2O (l)Information from the chemical equation :

1 mole of CaCO3 react with 2 moles of HCl to yield 1 mole of CaCl2, 1 mole of CO2 and 1 mole of H2O.Example:

1 molecule CaCO3 + 2 molecules HCl 1 molecule CaCl2 + 1 molecule CO2 + 1 molecule H2O

In terms of equivalent :

i. Mole equivalent

1 mole of CaCO3 2 moles of HCl 1 mole of CaCl2 1 mole of CO2 1 mole of H2O.

ii. Mass equivalent

1 mole x 100 g/mole of CaCO3 2 moles x

36.5 g/mole of HCl 1 mole x 111 g/mole of

CaCl2 1 mole x 44 g/mole of CO2 1 mole

x 18 g/mole of H2O.

100 g of CaCO3 73 g of HCl 111 g of CaCl2 44 g of CO2 18 g of H2O.

Example:How many moles of hydrochloric acid do we need to react with 0.45 mole of zinc to form zinc chloride and hydrogen gas?

Answer: Zn + 2HCl ZnCl2 + H2

Moles of HCl = 0.45 mol Zn x 2 mol HCl 1 mol Zn = 0.90 mol HCl

Example 2: How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?

Answer:Moles of KClO3 = 9 mol O2 x 2 mol KClO3 3 mol O2 = 6 mol KClO3

2KClO3 2KCl + 3O2

Exercise:1.Balance the following chemical equations by inspection and write its mole equivalent and mass equivalent.NH3 + CuO Cu + N2 + H2OAnswer:2NH3 + 3CuO 3Cu + N2 + 3H2O2 mole of NH3 3 mole of CuO 3 mole of Cu 1 mole of N2 3 mole of H2O

34 g of NH3 238.8 g of CuO 190.8 g of Cu 28 g of N2 54 g of H2O

2. Ammonia gas, NH3 is prepared by reaction between N2 gas and H2 gas.a)Write the balance equation

b)Write the mole equivalent and mass equivalent

c)How many moles of ammonia yield from 0.481 mole of H2 ?

ANSWER : N2 + 3 H2 2 NH3

b) Mole equivalent1 mole of N2 3 mole of H2 2 mole of NH3

Mass equivalent28 g of N2 6 g of H2 34 g of NH3

c) 0.321 mole

LIMITING REACTANT

Limiting reactant / reagent : The reactant which is completely consumed before the others ( finished first )

Excess reactant : A reactant that remains after a reaction is over.

Limiting reactant will determine the mole/mass of products.

Limiting reactant can also be identify by the smallest ratio of

the number of moles of reactant the stoichiometric coefficient

Limiting Reagents1.2

1. Write a balanced equation.

2. For each reactant, calculate the amount of product formed.

3. Smaller answer indicates:limiting reactantamount of product

Example :Urea, (NH2)2CO is prepared by reacting NH3 with CO2.2 NH3 + CO2 (NH2)2CO + H2OIf 637.2 g of NH3 are allowed to react with 1142 g of CO2 : Which of the two reactants is the limiting reagent ?

b) Calculate the mass of (NH2)2CO formed.

c) How much of the excess reagent (in grams) is left at the end of reaction ?

Answer :a) Moles of NH3 = 637.2 g / 17.0 gmol-1 = 37.48 mol

Mole of CO2 = 1142 g / 44.0 gmol-1 = 25.95 molNH3 : 37.48 / 2= 18.74 mol

CO2 : 25.95 / 1= 25.95 mol

Since the NH3 has the smallest ratio, NH3 is the limiting reagent.

From the balanced equation :

2 mole NH3 1 mole (NH2)2CO

34 g NH3 60 g (NH2)2CO

637.2 g NH3 60 g (NH2)2CO x 637.2 g NH3 34 g NH3

= 1124.5 g (NH2)2CO

OR 2 mole NH3 1 mole (NH2)2CO

37.48 mol NH3 1 mol (NH2)2CO x 37.48 mol NH3 2 mol NH3

= 18.74 mol (NH2)2CO

Mass of (NH2)2CO = 18.74 mol x 60 g mole-1

= 1124.4 g

34 g NH3 44 g CO2

637.2 g NH3 44 g CO2 x 637.2 g NH3 34 g NH3

= 824.6 g (used)

Excess reactant ( CO2 ) = (1142 - 824.6) g = 317.4 g

Exercise:How many moles of Fe3O4 can be obtained by reacting 16.8 g of Fe with 10.0 g of H2O?

3Fe (s) + 4H2O (g) Fe3O4 (s) + 4H2 (g)

Answer:

0.10 mol Fe3O4

REACTION YIELD & PERCENTAGEYIELDTo determine how efficient a given reaction by using percentage yield% yield = actual yield x 100% theoretical yieldTheoretical yield the maximum obtainable yield predicted by the balanced (stoichiometric) equation.

Actual yield the amount of product actually obtained from a reaction.

Percent Yield

Example:A student completely reacts 5.00 g of magnesium with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the student's percentage yield?

Answer:

2Mg(s) + O2(g) ----> 2MgO(s)

% yield = 97.2

Exercise:For the balanced equation shown below, if the reaction of 85.8 grams of Fe produces 95.8 grams of FeCl3, what is the percent yield? 2Fe + 3Cl2 2FeCl3

Answer:

38.48%

Exercise:For the balanced equation shown below, if the reaction of 50.7 grams of C3H6S produces a 39.9% yield, how many grams of CO2 would be produced ? C3H6S + 6O2 3CO2 + 3H2O + SO3 Answer:36.08 g CO2

Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.

MiVi = MfVfMi = 4.00Mf = 0.200Vf = 0.06 LVi = ? L= 0.003 L = 3 mL3 mL of acid+ 57 mL of water= 60 mL of solutionExample

Exercise 1:Calculate the new concentration (molarity) if enough water is added to 100mL of 0.25M sodium chloride to make up 1.5L.

Answer:

0.017 M

Exercise 2:Calculate the volume to which 500mL of 0.02M coppper sulfate solution must be diluted to make a new concentration of 0.001M.

Answer: 10.00L

TitrationTitrationis a common laboratory method ofquantitativechemical analysisthat is used to determine the unknownconcentration of a knownreactant.

Type of titration:Acid-base titrationRedox titrationBack titration

1) Acid base TitrationAn acid-base titration is a procedure which is used to determine the concentrationof anacidorbase.

A measuredvolumeof anacidorbaseof known concentrationis reacted with a sample to the equivalence point.

Acid base titration

Steps in Calculations

Write the balanced chemical equation for the reaction

Extract all the relevant information from the question

Check that data for consistency, for example, concentrations are usually given in M or mol L-1but volumes are often given in mL. You will need to convert the mL to L for consistency. The easiest way to do this is to multiply the volume in mL x 10-3

Calculate the moles of reactant (n) for which you have both the volume(V) and concentration(M) : n = M x V

From the balanced chemical equation find the mole ratio known reactant : unknown reactant

Use the mole ratio to calculate the moles of the unknown reactant

From the volume(V) of unknown reactant and its previously calculated moles(n), calculate its concentration(M): M = n V

Example 1 :30 mL of 0.10M NaOH neutralised 25.0mL of hydrochloric acid. Determine the concentration of the acid.Answer:NaOH + HCl NaCl + H2OMa Va = aMb Vb b Ma (0.025 L) = 1(0.10 M) (0.030 L) 1Ma = 0.12 M or mol / L

Example 2:In an acid-base titration, 17.45 mL of 0.180 M nitric acid, HNO3, were completely neutralized by 14.76 mL of aluminium hydroxide, Al(OH)3. Calculate the concentration of the aluminium hydroxide.

Answer: 0.0711 M

Redox titrationRedox titration(also calledoxidation-reduction titration) is a type oftitrationbased on aredox reactionbetween theanalyteandtitrant.

Redox titration may involve the use of aredox indicatorand/or apotentiometer.

Example:How many millilitres of a 0.206 M HI solution are needed to reduce 22.5 mL of a 0.374 M KMnO4 solution according to the following equation:

10HI + 2KMnO4 + 3H2SO4 5I + 2MnSO4 + K2SO4 + 8H2O

Answer:Ma Va = aMb Vb b

(0.206 M) Va = 10(0.374 M) (22.5 mL) 2

Va = 204 mL

Exercises:1)How many mL of 0.20 M HBr is needed to neutralize 23.45 mL of 0.11 M KOH? What is the balanced equation for the reaction between the two solutions?

Answer:HBr(aq)+ KOH(aq)----->KBr(aq)+ H2O(l)

X = 0.043 mL

2)Lab technician wants to determine the molarity of an unknown acid. What is the molarity of the solution if 13.61 mL of the acid is neutralized by 34.69 mL of 0.025 M KOH?

Answer:

X = 0.064 M

Prepared by:Chem_mkpm@tie

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