Matter1

Chapter 1 : Matter

1.1 Atoms and molecules

# Identify and describe protons, electrons and

neutrons as sub-atomic particles.

# Define proton number, nucleon number and isotopes

and write isotope denotation.

# Define relative atomic mass and relative molecular

mass based on the C-12 scale.

Objectives :

LECTURE 1

What is matter?

1. Matter is made up of elements, compounds and mixtures

2. The basic building block of matter is called atom.

3. An element consists of same kind of atoms e.g. Fe

Anything that occupies space and possesses mass is called matter.

Example: air, earth, animals, trees, atoms,…

The three states of matter are solid, liquid and gas

3. A compound is made up of 2 or more kinds of atoms chemically bonded in a definite proportion e.g. NaCl, CO2.

4. A compound may consist of molecules or ions. E.g. NH3, K+Br-.

5. A molecule is made up of 2 or more atoms of the same or different element/s bonded covalently to form a discrete unit.e.g. O3 , H2O.

6. An ion is an atom/group of atoms either gained electron/s (anion) or lost electron/s(kation). E.g. S2- , Cu+, SO4

2- .

Matter is made up of very tiny particles called atoms.

. Atom is spherical and consists of a nucleus surrounded with a cloud of moving –ve electrons (e)

THE FUNDAMENTAL PARTICLES OF ATOMS

nucleuselectron

cloud

. Nucleus of an atom is small, dense & consists of +ve proton/s (p) and neutral neutron/s(n)Atoms consist of protons, electrons and neutrons.

Name charge mass/amu

1. Proton, p 1+ 1.007277

2. Electron, e 1- 0.0005480

3. Neutron, n 0 1.0008605

• Nucleus = p + n.

Mass of proton mass of neutron

me : mp = 1: 1836

Mass of atom mass of nucleon (p + n).

Properties of subatomic particles

PROTON NUMBER AND NUCLEON NUMBER

Total number of protons is called proton number, Z ( or atomic number)

Total number of protons and neutrons is called nucleon number, A (or mass number)

Z determine the identity of an element.

Symbol notation of atom :

A X where A= nucleon/ mass no.

= Z + n

Z = atomic/ p no. = p

No. of e- = Z– (charge of atom/ion)

Z

charge

e.g. 23 Na+

11

Total charge on the ion = 1+

nucleon no., A = No. of (p+ n) = 23

proton no., Z = no. of p = 11

no. of n in nucleus of = A - Z = 23 – 11= 12

No. of e- = Z – charge = 11- (+1) = 10

Find: the no. of I) proton, ii) neutron, iii) electron?

Example 1.1.2: Give the number of protons, neutrons and electrons in each atom of the following species:(a) b)Hg200

80

2+ 2178 O

Symbol No. of charge

p n e atom

80 120 78 2 2+

8 9 10 1 2-

Hg20080

2178 O

2+

2

2

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.

ISOTOPES

<>Atoms with the same proton number but different in nucleon numbers. Hence same chemical properties but different physical properties.

E.g :12 14 13

6 6 6C

Ex. What is the no. of (I) p ii) n iii) e in a) 27Al3+ b) 31P3- ?

1315

CC

Ex. 2 Give the isotopes’ names of hydrogen and oxygen.and write their respective atomic denotation.

Ans: H: protium, deuterium and tritium

O : oxygen- 16, oxygen-17 and oxygen -18

1 2 3

H H H

1 1 1

16 17 18

O O O

8 8 8

No. Symbolnotation

Z A No. of p

No. of n

No. of e-

Atomic mass

a. 35 Cl-

17

17 35 17 18 18 35

b. 24 Mg2+

12

c. 16 *32 18 32

d. 56 26 23

e. 82 125 80

f. -----Ca ----

-----

18 40

Complete the following table:

No.

Symbolnotation

Z A No. of p

No. of n No. of e Atomic mass

a. 35 Cl-

17

17 35 17 18 18 35

b. 24 Mg2+

12

12 24 12 12 12 - 2= 10

24

c. 32 S 2-

16

16 *32 16 16 18 32

d. 56 Fe3+

26

26 56 26 30 23 56

e. 207 Pb2+

82

82 207 82 125 80 207

f. 40 2+-----Ca ----

20-----

20 40 20 20 18 40

RELATIVE MASS

<>Isotope Carbon-12 as a reference or standard for comparing the masses of other atoms.

Definition: Relative atomic mass => mass of 1atom of an element(a.m.u.)

1/12mass of 1 atom of C-12 isotope(a.m.u.)

e.g. Relative atomic mass of Na

= mass of 1atom Na

1/12 mass of 1 atom 12C isotope

= 23 a.m.u./1.0 a.m.u = 23

Relative molecular mass

Definition: Sum of all the relative atomic masses of all atoms present in a molecule.

e.g. : Molecular formula of sulphuric acid is H2SO4, determine its relative molecular mass. [ RAM: H=1.00; S=32.0; O=16.0]

Ans:RMM H2SO4 = 2X1.00 + 32.0 + 4X16.0 =98.0

1. For each of the substances:: i) Na2CO3 ii. MgO. [ Mg=24.0; O= 16.0; Na=23.0; C=12.0]

Find its relative molecular mass

Exercise:

.2. .Determine the relative atomic mass of an element X if the ratio of the atomic mass of X to carbon-12 is 0.75

3. The relative atomic mass of chlorine atom is 35.5 Explain why this no. is not a whole no..

4. Relative atomic mass of bromine atom is 80. What is the mass of Br compared to carbon-12 atom?

Ans: 9.00

Ans: 6.7

The mass spectrometer is used to :

i. determine the atomic mass and molecular mass of substances.

ii. reveal the presence of isotopes of elementsand its relative abundance.

MASS SPECTROMETRY

Who invented the mass spectrometer?

Vaporisation chamber

Acceleration chamber (Electric field)

Magnetic field

Ionization chamber

Ion detectorvacuum

Figure : Diagramatic representation of Mass Spectrometer

AB

Vaporisation Chamber

Sample of the element is vaporised into gaseous atom

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 + e M+ + e + e

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

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 spectrum :

The horizontal axis = m/e ratio or nucleon number or isotopic mass or 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.

The height is proportional to the amount of each isotope present.

Relative abundance

63

8.1 9.1

0 24 25 26 m/e

The 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 Mg

= (24 a.m.u. x 63) + (25 a.m.u. x 8.1) + (26 a.m.u x 9.1)

(63 + 8.1 + 9.1)

= 24.33

Ar =

ii

ii

i

Q

mQ

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

General formula for Ar

. The relative atomic mass of an element of 2 isotopes,= ( % abundance X1 x isotopic mass X1 +

% abundance X2 x isotopic mass X2 ) ÷sum of % abundance X1,X2

Mass spectrum of an atom shows the presence of isotopes

by representative lines with lengths proportional to

the abundance of their ions

Eg.: Boron comprises of 2 isotopes, i.e. 14.6 % 10B

dan 85.4 % 11B with their isotopic masses of 10.01294

dan 11.00931 respectively. Calculate the relative atomic

mass of B.

Ans:RAM Boron = % 10B x Ar + % 11B x Ar

100 % = 14.6 x 10.01294 + 85.4 x 11.00931 %

100%= 10.8

1. The atomic masses of 6Li and 7Li are 6.0151 amu and 7.0160 amu respectively. What is the relative abundance of each isotope if the relative atomic mass of lithium is 6.941 amu?

Ans:(92.95%, 7.05 %)

2. Naturally occurring iridium, Ir is composed of 2 isotopes and in the ratio of 5:8. The relative isotopic mass of and are 191.021 and 193.025 respectively. Calculate the relative atomic mass of iridium.

Ans:(192.254)

Exercise your mind:

Objectives:

#1 . Write chemical formulae of ionic compounds

#2 Define mole in term of mass

#2. Relate the no. of mole with Avogadro constant

#3. perform calculation involving converting :

i. Mole to mass

ii. mass to mole

#4. Relate the number of particles to mass through Avogadro ‘s constant, NA

LECTURE 2

Chemical formulae of ionic compounds:

Ionic compounds consists of :

Cation(+ve ion) and anion(-ve ion) chemically bonded together by ionic bond.

types : I. between monoatomic cation and anion

e.g K+Cl-

2. Monoatomic cation and polyatomic anion

e.g. Mg2+CO32-

3. Polyatomic cation and monoatomic anion

e.g. ( NH4+)

2S2-

4. Polyatomic cation and anion e.g. . NH4+NO3

-

5. Other combinations.

+ve Ion -ve Ion

K+ Potassium Br- Bromide

NH4+ Ammonium S2- Sulphide

Ba2+ Barium N3- Nitride

Al3+ Aluminium OH- Hydroxide

Fe2+ Iron(II) NO3- Nitrate

Fe3+ Iron(III) MnO4- Permanganate

Mn2+ Manganese (II) CO32- Carbonate

Mn4+ Manganese(IV) SO42- Sulphate

Cr3+ Chromium(III) CrO42- Chromate

IUPAC names of some common anions & caions

+ve Ion -ve Ion

H+ Hydrogen Cl- Chloride

Cu+ Copper (I) O2- peroxide

Ag+ Silver ClO3- Chlorate

Co2+ Cobalt (II) ClO2- Chlorite

Ba2+ Barium NO2- Nitrite

Zn2+ Zinc CN - Cyanide

Pb2+ Lead(II) O2- Oxide

Sn4+ Tin(IV) SO32- Sulphite

Pb4+ Lead(IV) PO43- Phosphate

IUPAC names of some common anions & caions

Writing chemical formulae of ionic compounds

e.g. Potassium sulphite:

. K+ SO32-

2- 1+ X2 =2

x2

K2SO3

Steps: : i. write symbol of cation followed by that of anion.

ii. Combine the 2 , make sure their total charge = 0

e.g. 2K+ + SO32- = (2+) + (2-) = 0

Hence the chemical formula for potassium sulphite is K2SO3

Exercise

1. Try write chemical formula for:

a. Calcium hydrogen carbonate

b. Aluminium oxide

c. Sodium nitride

d. Potasium dichromate

e Barium sulphide

f. Potassium permanganate

2. Name the following compounds using IUPAC Nomenclature:

i. K2CrO4

ii. Na2Cr2O7

iii. Ca(OH)2

iv. ZnSO4

v. Al2(CO3)3

vi. HNO3

vii. MgS

viii. MgSO4.7H2O

The amount of substance which contains equal amount of elementary entities ( atoms, molecules, ions etc.) as there are 12C atoms in 12.00g of 12C.

•In 12.00 g of 12C, there are 6.022045 x 1023 atoms which is 1 mole of the substance carbon-12

•Therefore 1 mole is just a unit to represent an amount equal to 6.022045 x 1023

•1 mole = 6.022045 x 1023 = Avogadro constant , NA

MOLE:

MOLE:

12.00g of carbon-12 will have the same number of atoms as that of 1H atoms in 1.00 g of hydrogen-1,

i.e. 1 mole of any substance consists of 6.022045 x 1023 = NA particles ( atoms, ions, molecules) of it.

Exercise 1. Which substance has more atoms?

i. 1 mole of Na or 1 mole of B?

ii. 0.11 mole of Mg or 0.72 mole of Cl?

Symbol used for mole is n

MOLAR MASS:

Is the mass of 1 mole of an element or compound.

E.g. 1 mole of copper,Cu, contains 6.022045 x 1023 atoms and weighs 63.5 g. What is the molar mass of Cu?

Ans: 63.5 g mol-1 Unit used = g mol-1

Mass of Cu atom = 63.5 amu (from mass spectrometry)

1 mole Cu atoms

≡ 6.022 x 1023 atoms mole-1Cu x 63.5 a.m..u/atom Cu

Since 1 a.m.u. = 1.6605 x 10-24 g

(from Std.atomic mass of 12C scale)

Hence molar mass of Cu, MCu =( 3.824 x 1025 a.m.u. mol-1 x 1.66505 x 10-24 g / 1 a.m.u.) = 63.5 g mol-1

molar mass is taken as the mass with value of the relative atomic mass ,Ar or molecular mass, Mr in unit g mole-1

MOLAR MASS OF ATOM

(a) 30.97 g mole-1

Exercise 2: (a) What is the molar mass of phosphorus with Its relative atomic mass of 30.97?

(b) 0.50 mole of K weighs 19.50 g. What is its molar mass?

(b) 39.0 g mole-1

MOLAR MASS OF a molecule/compound

Is the sum of atomic masses of the elements in 1 mole of the molecule or compound.

e.g. Find the molar mass of Cl2

Ans: Molar mass of Cl2 , MCl = 2 x MCl

= 2 x 35.5 g mole-1

= 70.0 g mole-1

2

MOLAR MASS OF a molecule/compound

Exercise 3: Find the molar mass of SO2.

[Given Ar of : S = 32.0, O = 16.0]

Ans : 64.0 g mole-1

Relate mass to mole of a substance

Relate mole to mass to become quantity we can measure, ie. That can be weighed.

Mass (g) = moles(n) x molar mass(g mole-1)

i.e.

m = n x M

e.g.1. . What is the mass of 2.50 moles of iron, Fe?

[ Fe = 56.0]

From m = n x M,

Hence mass of Fe, m = 2.50 moles x 56.0 g mole-1

= 140.0 g

e.g. 2 : How may mole of N2 contained in 1.40 g of N2 (g)?

.From m = n x M,

I.e. n = m / M

Ans: Mole of N2, n = 1.40 g / 28.0 g mole-1

= 0.050 mole

Exercise 4: [Given Ar : C = 12.00 ; O = 16.0, Ca = 40.0]

(a) Find the mass of 0.25 mole of CO.

(b) State the mass of 1.50 mole of Ca.

(c ) Find the amount(moles)

i) of element in 54.0g of Al.

ii) of ion in 4.8 g of Mg2+

Ans: (a) 7.00 g (b) 60.0 g( c) I) 0.2 mole

ii) 0.2 mole

no. of particles = n x NA

e.g. 1. Find the no. of ion in 0.80 mole of O2-

Ans: No. of ion O2- = 0.80 mole x 6.022 x 1023 ions mole-1

= 4.48 x 1024 ions

Relate mole to no. of elementary particles

e.g. Find the amount(moles) in 6.022 x 1024 atoms of Na?

Ans: n = 6.022 x 1024 atoms = 10.00 moles

6.022 x 1023 atoms mole-1

Exercise 5: Find the no. of

i) atoms in 1.20 mole of S.

ii) Molecules in 40.0 g of CO2i) 7.23 x 1023

ii) 5.47 x 1023

Exercise 6: How many moles are there in 2.50 x 1026of NO2 molecules? What is the mass of NO2?

[ N= 14.0; O = 16.0}

Ans: 4.15 x 102 mole ; 19090 g

Summary/review

#1 Mole is an amount of substance containing the same no. of elementary. particles as that of

12C atoms in 12.00g of 12C.

#2 molar mass of different substances contain the same no. of elementary particles.

e.g. the no. of : Na atoms in 23.0 g = CO2 molecules in 44.0g

#3 Molar mass Ξ 1 mole Ξ 6.022 x 1023 ( i.e.NA)

#4 Molar atomic mass obtained by expressing the Ar in g mole-1

#5 Molar molecular mass = sum of molar atomic masses of all elements in molecule.

Relate mole to mass and no. of particles.

Summary/review

# 1 Convert mole to mass: m = n x M

#2 Convert mass to mole: n = m / M

#3 Convert mole to no. of elementary particles:

no. of particles = m x NA

#4 Convert no of elementary particles to mole:

n = no. of particles / NA

#5 Relate mole to no. of elementary particles

No. of particles(atom/ ion/molecule) = n x NA

#6 Relate no. of elementary particles to mole

n = No. of particles / NA

THE MOLE CONCEPT

Objectives:

#1 . Perform calculation involving converting

i. mass to No. of elementary particles

ii. no. of particles to mass

#2. Relate mole with molar volume of gases

#4. Relate volume of a gas to quantity in moles at :

i. STP ii. room condition

LECTURE 3

Convert mass to No. of elementary particles

mass mole No. of particles

e.g. How many molecules are found in 34.0 g of NH3 ?

Ans: molar mass of NH3 = 14.0 + 3.00 = 17.0 g mole-1

moles of NH3 , n = m / M

= 34.0 g / 17.0 g mole-1

= 2.00 mole

No. of NH3 molecules

= 2.00 mole x 6.022 x 1023 molecules mole-1

= 1.20 x 1024molecules

i.e. No. of particles = m x NA

M

Convert no. of particles to mass

No. of particles mole mass

i.e. m = no. of particles x M NA

e.g. What is the mass of 1.807 x 1025 formula units of sodium hydroxide, NaOH ?

Ans: molar mass, MNaOH = 23.0+16.0+1.00 = 40.0 g mole-1

Hence mass of NAOH,

mNaOH = 1.807 x 1024 f-units x 40.0 g mole-1

6.022 x 1023 f-units mole-1

= 1200 g

Exercises

1. How many molecules are there in 3.65 g of hydrogen chloride, HCl ?

2. How may oxygen atoms in 20.0 g of SO3 molecules?

3. Find the mass of 1.25 x 1022 Ca2+ ions ?

4. Which is heavier, 6.022 x 1030 H2 molecules or 1.205 x 1021 atom Na?

5. What mass of sulphur dioxide, SO2 contains the same no. of molecules as are in 1.00 g of ammonia?

#1. 6.022 x 1022

#2. 3x1.506 x 1023

#3. 0.830g#4. H2

#5. 3.77 g

Relate mole with molar volume of gases

Molar volume of any gases at STP, Vm = 22.4 L mole-1

STP =standard temperature and pressure

Where T = 273.15 K P = 1 atm @ 760 mm Hg

Amount of gas at STP, n = Vgas (L) Vm (22.4 L mole-1 )

e.g. How many moles are there in 6.5 L oxygen at STP ?

n = 6.5 L O2 x 1 mole O2 = 0.29 mole

22.4 L O2

Relate volume with molar volume of gases




Amount of gas at STP, n = Vgas (L) Vm (22.4 L mole-1 )

e.g. How many moles are there in 6.5 L oxygen at STP ?

n = 6.5 L O2 x 1 mole O2 = 0.29 mole

22.4 L O2

Relate volume of a gas to quantity in moles




At STP Vgas (L) = n x Vm (22.4 L mol-1 )

e.g. Find volume of 0.25 mole of HCl gas at STP ?

ANS: VHCl = 0.25 mol HCl x 22.4 L mol-1

= 5.6 L

Exercise/e.g:

Calculate the mass of 4.80 L of hydrogen gas at room conditions.

Solution:

using Vgas (L) = n x Vm (24 L mol-1 )

n = ( 4.80 / 24 ) mol

mass of hydrogen = ( 4.80 / 24 ) mol x 2.00 g mol-1

= 0.40 g

Molar volume of any gases at room conditions, Vm = 24L mole-1

Room temperature & pressure

Where T = 250C@298 K P = 1 atm @ 760 mm Hg

mailto:250C@298

mailto:250C@298

mailto:250C@298

Exercise:1. A balloon filled with gas at STP occupies a volume of 2.24 L.

Calculate the amount of hydrogen gas. [Ans: 0.1 mole.]

2. Calculate the volume of 24 x 1023 molecules of gas at STP.[Ans: 89.6 L ]

3. A sample of CO2 has a volume of 56 mL at STP.Calculate: (a) the no. of moles of CO2 molecules:

(b) the no. of CO2 molecules ( c) the no. of oxygen atoms in the sample ( d) the mass of CO2

4. For a sample of chlorine gas with a mass of 71.0g,Find: a) the amount( mole) of its molecules b) its volume at STP c) its volume at room conditions.

Ans: 3 (a) 0.0025 b) 1.5 x 1021 c) 3.0 x 1021

Summary/review

#1 Relate mass to No. of elementary particles : No. of particles = m x NA

M

where m = mass of substance

M = its molar mass; NA = 6.022 x 1022

#2 Relate mole with molar volume of gasesAmount of gas at STP, n = Vgas (L)

Vm (22.4 L mol-1 )

Amount of gas at room conditions, n = Vgas (L) Vm (24 L mol-1 )

THE MOLE CONCEPT

Objectives:

#1 . Define

i. Empirical formula

ii. Molecular formula of a chemical compound.

#2. Determine the empirical formula and molecular formula of a compound from it’s

a) mass composition, b) combustion data.

#3 Determine the % composition by mass of a chemical compound.

Chemical Formula

Empirical Formula indicates which elements are present and the simplest whole-number ratio of their atoms in a molecule.

Molecular Formula shows the exact number of atoms of each element in the smallest unit of a substance.

molecular formula = (empirical formula)n

Molecules Empirical formula

Molecular Formula

n

Water H2O H2O 1

Hydrogen peroxide HO H2O2 2

Benzene CH C6H6 6

Ethyne CH C2H2 2

Which 2 compounds have same empirical formula and molecular formula ?

Give 2 molecules that might have different molecular formulae but the same empirical formula

1. Calculating the empirical formula from the masses of constituents

E.g. 18.3 g sample of hydrated compound contained 4.0 g of calcium, 7.1 g of chlorine and 7.2 g of water only. Calculate its empirical formula.

10.0

20.0

0.40

0.4

5.35

1.7

18

2.7

10.0

10.0

Constituent Ca Cl H2O

Mass/g 4.0 7.1 7.2

Amount/mol = 0.10 = 0.20 = 0.40

Simplest ratio of relative amount

= 1.0 = 2.0 = 4.0

1 2 4

10.0

20.0

10.0

40.0

Empirical formula = CaCl2.4H2O

Exercise

50.0 g of phenol. A general disinfectant, has 38.29g C, 3.21 g H ang 8.50 g O. Determine its empirical formula.

Ans: C6H6O

Calculating the empirical formula from percentage composition by mass

Example: Ascorbic acid (vitamin C) cures scurvy and may help prevent the common cold. It is composed of 40.92% carbon, 4.58% hydrogen and 54.50% oxygen by mass. The molar mass of ascorbic acid is 176 g mol1. Determine its empirical formula and molecular formula

12

92.40

16

50.5441.341.3

Element C H O

Mass/g 40.92 4.58 54.50

Amount/mol = 3.41 = 4.58 = 3.41

Simplest ratio

= 1 = 1.33

= 1

1x3 1.33x3 1x3

3 4 3

Empirical formula = C3H4O3

mass formula empirical

massmolar n =

= = 2 88

176

Molecular formula = (C3H4O3)n

= (C3H4O3)2

= C6H8O6

Exercise

1) Allicin is the compound responsible for the characteristic smell of garlic. An analysis of the compound gives the following percent composition by mass: C: 44.4%; H: 6.21%; S: 39.5%; O: 9.86%. Calculate its empirical formula. What is its molecular formula given that its molar mass is about 162 g mol1.

2) Determine the formula of a mineral with the following mass composition: Na=12.1%, Al=14.2%, Si=22.1%, O=42.1%, H2O=9.48%.

Calculating the empirical formula from elemental analysis data(Combustion data)

E.g 1.00 g sample of compound A was burnt in excess oxygen producing 2.52 g of CO2 and 0.443 g of H2O.

Determine the empirical formula of the compound

1 mol CO2 contains 1 mol C & 1 mol H2O 2 mol

H

22C CO g 44.0

C g 12.0 x CO g 2.52 m = 0.688

OH g 18.0

H g 2 x OH g 0.443

22H

m

the mass of oxygen = msample – (mC + mH)

= 1.00 g – (0.688 g + 0.0492 g) = 0.263 g O

0.12

688.0

00.1

0492.00.16

263.0

0164.0

0573.00164.0

0492.0

0164.0

0164.0

Element C H O

Mass/g 0.688 0.0492 0.263

Amount/mol = 0.0573

= 0.0492 = 0.0164

Simplest ratio of relative amount

= 3.49 = 3.00 = 1.00

3.49x2 3.00x2 1.00x2

7 6 2

Empirical formula = C7H6O2

Exercise

1. 0.535 g sample of thiophene, a carbon-hydrogen-sulphur compound used in manufacturing of pharmaceuticals, is burned completely in excess oxygen to yield 1.119 g CO2, 0.229 g H2O, and 0.407 g SO2.

Determine empirical formula of thiopene.

Percentage composition from formula

For an element in a compound = atoms of element x RAM x 100

Relative formula mass

e.g. Calculate the % composition of hydrogen in C12H22O11

Where RFM C12H22O11 = 342.0

ANS: % C = 22(1.0) X 100 = 6.4 %

342.0

Exercise: Calculate the % of nitrogen, by mass, in Ca(NO3)3

Ans: 17.1

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