I. INTRODUCTION II. MOLE CONCEPT

III. CALCULATIONS ON REACTIONS

IV. VOLUMETRIC ANALYSIS

V. REDOX REACTIONS

VI. MASS SPECTRO-METRY

Relative Atomic Mass Scale: Carbon-12 Scale-Atoms are too small to be weighed Thus, compare the masses of atoms with the mass of a standard atom:

CARBON-12 ATOM (IUPAC Agreement 1960)

I INTRODUCTION

An atom of 12C is assigned a mass of 12 units, => meaning 1/12 of the mass of a 12C atom is 1 unit.

12C

Why 12C as the Standard Atom?

(i) 12C is a readily available common element.

(ii) Most abundant isotope of carbon.

(b) The relative abundance of isotopes refers to the percentage of the isotopes as they are found in the naturally occurring element.

e.g. Relative abundance of 35Cl & 37Cl is 75% and 25% respectively.

A. Relative Isotopic Mass1(a) Isotopes are atoms of the same element having

the same number of protons but different number of neutrons.

e.g. 35Cl & 37Cl, 79Br & 81Br

(c) Relative abundances of isotopes are determined from the mass spectrum of the element.

m/e

Relative abundance

9

1

19 20 21 22 23

mass spectrum of neon gas

e.g. rel. abundance of 20Ne and 22Ne is 90% and 10% respectively.

(b) Relative isotopic mass =

Mass of one atom of the isotope1/12 the mass of one atom of 12C

2. Definition of Relative Isotopic Mass: (Term used for ISOTOPES)

(a) Ratio of the mass of one atom of the isotope compared to 1/12 of the mass of a 12C atom.

(No unit)

Example:

Isotope Rel. isotopic mass Rel. abundance1H 1.0078 99.9852H 2.0140 0.015

16O 15.9949 99.76017O 16.9991 0.04018O 17.9991 0.20035Cl 34.9689 75.77037Cl 36.9659 24.230

(Term used for ATOMS)

1. Definition: Ratio of the average mass of one atom of an

element compared to 1/12 of the mass of a 12C atom. (No unit)

B. Relative Atomic Mass (Ar)

2. Ar is calculated as the weighted mean of the relative isotopic masses of the isotopes according to their abundance.

= 20 x 90/100 + 22 x 10/100 = 20.2

Eg 1.1 There are 90% 20Ne and 10% 22Ne naturally occurring, thus Ar of neon

m/e

Rel. abundance

9

1

20 21 22

mass spectrum of neon gas

(Term used for MOLECULES)

1. Definition: Ratio of the mass of one molecule compared to

1/12 of the mass of a 12C atom. (No unit)

2. Mr is calculated as the sum of all relative atomic masses from the formula of the molecule.

C. Relative Molecular Mass (Mr)

Eg 1.2 Mr of chlorine gas, Cl2 =

Mr of octane, C8H18 =

35.5 2 = 71.0

8 12.0 + 18 1.0 = 114

1. Definition: Ratio of the mass of one formula unit of an ionic

compound compared to 1/12 of the mass of a 12C atom. (No unit)

(Term used for IONIC COMPOUNDS)

For ionic compounds like NaCl, it is NOT a molecule.

D. Relative Formula Mass (Mr)

2. Rel. formula mass is calculated as the sum of all relative atomic masses from the formula of the species (which can be atoms, molecules, ions, ionic compounds etc.).

Eg 1.3 Mr of NaCl = 23.0 + 35.5 = 58.5

What is a MOLE?

II MOLE CONCEPT

1 dozen of eggs = 12 eggs

1 ream of papers = 500 pieces of papers

1 3.5” diskette = 1.44 MB of memory

1 mole of particles = _________ particles

A. Mole & Avogadro Constant (L)

6.02 x 1023

The Avogadro constant L is defined as 6.02 x 1023 mol-1, the number of carbon atoms present in 12.0 g of carbon-12. (note the unit of L)

1. Definition:

A mole of substance is the amount of that substance which contains Avogadro’s number of particles.

..named in honour of Italian Chemist, Amedeo Avogadro, but he did not come up with the number!! So who did?

2. Since 1 mole of particles = 6.02 x 1023 particles,

No. of moles of particles =No. of particles6.02 x 1023

______________

Eg. 2.1 no. of moles no. of particles

(a) 0.300 mol

(b)1.23 x 1022 N2

molecules (c) 6.71 mol He

1.81 x 1023

0.0204 N2 molecules

4.04 x 1024 He

Why the concept of a MOLE?- used to represent a large amount of substances..

e.g. Mass of ONE atom of Carbon-12

= 2 10-23 g or 2 10-26 kg = 12 / 6.02 x 1023 g

(0. 0000000000 0000000000 002 g )

B. Molar Mass & Related Calculations

1. Mass of one mole (6.02 x 1023) of 12C atoms = 12.0 g (i.e. Ar (C) g)

Mass of one mole of chlorine atoms =

and mass of one mole of carbon dioxide molecules =

Hence, mass of one mole of _________ = 1.0 g

35.5 g

44.0 g1H atoms

No. of moles of A =Mass of AMolar mass of A

__________

Mass of A = no. of moles of A Molar mass of A

Name FormulaRel.

molecular mass

Rel. formula

massMolar mass

magnesium sulphate

NH3

(NH4)2CO3

2. Molar mass is the mass of the substance per unit mole. Unit: g mol-1

Eg.

120.4MgSO4120.4

g mol-1--

--ammonia 17.0 17.0 g mol-1

--ammonium carbonate 96.0 96.0 g mol-1

Name Formula Mr No. of moles

No. of particles

Mass /g

F2 0.175

HCl 3.81 1021

Sulphur dioxide

1.80 10-

3

Glucose 180 0.500

SO2 64.1 1.08 1021 0.115

Eg.

Fluorine 38.0 6.65 1.05 1023

Hydrogen chloride

36.5 6.33 10-3 0.231

C6H12O6 2.78 10-3 1.67 1021

C. Stoichiometry & Related Calculations

1. One H2O molecule contains __ atoms of H & __ atom of O.

6.02 x 1023 H2O molecules contain _____________ atoms

of H & _________ atoms of O. One mole of H2O contains __ moles of H atoms & __ mole

of O atoms.

AmBn m A n B

no. of moles of A = m no. of moles of AmBn

no. of moles of B = n no. of moles of AmBn

One mole of AmBn contains m moles of A & n moles of B.

2 1

2 1

2 x 6.02 x 1023

6.02 x 1023

1. CaCl2 Ca2+ ions Cl- ions

(a) 1 mol(b) 5 mol(c) 0.3 mol

Eg. 2.4

1 mol 2 mol

5 mol 10 mol0.15 mol 0.15 mol

0.032 mol(c)8 mol(b)

0.1 mol(a)

H atomsC atomsC4 H82.

Eg. 2.4

0.4 mol 0.8 mol

2 mol 16 mol0.004 mol 0.016 mol

Can you calculate the % by mass of Ca in CaBr2 ?

[Answer : 20.1 %]

% by mass of A in AmBn = ----------------------- 100%

= ------------------------------------------------- 100%

= --------------------------- 100%

mass of A in AmBn

mass of AmBn

no. of moles of A in AmBn Molar mass of A

Molar Mass of AmBn

m Molar mass of A

Molar Mass of AmBn

Al2O3 (Ar: Al 27.0; O 16.0)

% by mass of O = ----------------------- 100%

=

Eg. 2.5(a) Calculate the % by mass of O in aluminium oxide.

3 16.0 2 27.0 + 3 16.0

47.1%

Mr(H2O) = 1.0 2 + 16.0 = 18.0

% by mass of H2O

= ---------------------------------------- 100%

=

Ar: Cu 63.5; S 32.1; O 16.0; H 1.0

(b) Calculate the % by mass of of H2O in CuSO4.5H2O.

5 18.0 63.5 + 32.1 + 4 16.0 + 5 18.0

36.1%

3. Empirical formula is the formula that shows the simplest whole-number ratio for number of atoms of different elements present in the substance.

Molecular formula is the formula that shows the actual number of atoms of different elements in one molecule of the compound. It is a simple multiple of the empirical formula.

For example,

Butane: E.F. = C2H5, M.F. = C4H10

Glucose: E.F. = CH2O, M.F. = C6H12O6

Eg 2.6(a) % by mass of elements of a compound X are Mn

72.0%, O 28.0%. Calculate the empirical formula of compound X.

How are % by mass of elements in a cpd det’d exptmentally?

Mn O

Ans: Mn3O4

72.0 % 28.0 %mass in 100 g

no. of mol 72.0/54.9= 1.31

28.0/16.0= 1.75

mole ratio 1.31/1.31=1 3

1.75/1.31=1.344

C H N

(b) % by mass of elements of a compound Y are C 63.2%, H 12.3%, N 24.6%. The Mr of Y is determined as 114. Calculate the empirical and molecular formulae of compound Y.

How is Mr det’d exptmentally?

63.2 % 12.3 %mass in 100 g 24.6 %

no. of mol 63.2/12.0= 5.27

12.3/1.0= 12.3

24.6/14.0= 1.76

mole ratio 16.99 72.99 3

Empirical formula is C3H7N.

Why total not 100% exactly?

The molecular formula is C3nH7nNn, where n is an integer. Given the Mr = 114,

3n(12.0) + 7n(1.0) + n(14.0) = 114

57n = 114

n = 2

Therefore, the molecular formula is C6H14N2.

25C 1 atmosphere

r.t.p.: Room temperature and pressure

D. Calculations involving Gases1 mol of any gas (i.e. 6.02 1023 gas particles) occupies the same volume under the same temperature and pressure.

volume of 1 mole of any gas = 24.0 dm3

No. of moles of gas at r.t.p. =vol. of gas (dm3)

24.0_______________

V dm3 V dm3Same no. of moles of gas at same temp. & pressure

0C 1 atmosphere

s.t.p.: standard temperature and pressure

volume of 1 mole of any gas = 22.4 dm3

No. of moles of gas at s.t.p. =vol. of gas (dm3)

22.4_______________

For each of the following pair, which container has a greater volume?

Eg 2.7

Container X contains Container Y contains Ans(a) 6 g of H2 at r.t.p. 96 dm3 of CO2 at r.t.p.

(b) 7 g of CO at s.t.p.0.70 mol of Cl2 at 0oC and

1 atm

(c)3.01 1022 CH4

molecules at 500oC and 1 atm

0.17 g of NH3 & 0.4 mol of N2 at 500oC and 1 atm

(d) 2 mol of Ne at 50oC and 1 atm

2 mol of Ar at 25oC and 1 atm

Y

Y

Y

X

Container X contains Container Y contains6 g of H2 at r.t.p. 96 dm3 of CO2 at r.t.p.

6 g of H2 = 6 / 2.0 = 3.0 mol of H2

Vol. of 3 mol of H2 at r.t.p. = 3.0 x 24.0 = 72 dm3

Container X contains Container Y contains7 g of CO at s.t.p. 0.70 mol of Cl2 at 0oC

and 1 atm

7 g of CO = 7 / 28.0 = 0.25 mol of CO (at s.t.p.)

0.70 mol of Cl2 at 0oC & 1 atm = 0.70 mol of Cl2 at s.t.p.

Container X contains Container Y contains3.01 1022 CH4

molecules at 500oC & 1 atm

0.17 g of NH3 & 0.4 mol of N2 at 500oC and

1 atm

3.01 1022 CH4 = 3.01 1022 / 6.02 1023

= 0.05 mol of CH4

(0.17/17.0) = 0.01 mol of NH3 0.01 mol NH3 + 0.4 mol N2 = 0.41 mol of gas

E. Calculations involving SolutionsConcentration of solution can be expressed in:(a) mol dm-3 (no. of mole of solute in 1 dm3 of solution)(b) g dm-3 (mass of solute in 1 dm3 of solution)

Conc. of A in mol dm-3 = —————–————

= ——–————————-------------

= —————————

no. of moles of A (mol)volume (dm3)

mass of A (g) / molar mass of A volume (dm3)

Conc. of A in g dm-3

Molar mass of A

No. of moles of A = conc. of A volume (mol dm -3) (dm3)

Eg 2.85.30 g anhydrous sodium carbonate were dissolved and volume of the resulting solution is 250 cm3. Calculate the concentration of (a) sodium carbonate in g dm-3,(b) sodium carbonate in mol dm-3,(c) Na+ ion in g dm-3.

(a) Concentration of Na2CO3 in g dm-3 = 5.30/0.250 = 21.2

(b) Concentration of Na2CO3 in mol dm-3 = 21.2/106 = 0.200

(c) Concentration of Na+ in g dm-3 = 0.200 x 2 x 23 = 9.20

Recall vol. is in

dm3.

Eg 2.9(a) Given: 2.02 g dm-3 H2SO4,

[H2SO4(aq)] = = mol dm-3

[H+(aq)] = = mol dm-3

[SO4

2-(aq)] = = mol dm-3

2.02 / 98.0 0.0206

0.04120.0206 x 2

0.02060.0206 x 1

(b) Given 0.17 g dm-3 OH- in KOH(aq),

[OH-(aq)] = = mol dm-3

[K+(aq)] = = g dm-3

[KOH(aq)] = = g dm-3

0.17/17.0 0.0100

0.0100 x 39.1 0.391

0.0100 x 56.1 0.561

no. of particles 6.02 x 1023

vol. of gas(dm3) 24 (dm3 )

vol. of gas(dm3) 22.4 (dm3)

Vol(dm3 )xConc(mol/dm3 )

Mole

no. of moles of A = m x no. of moles of AmBn

no. of molesof particles

=

no. of mole of A

=

AmBn m A n B

At r.t.p. (25C & 1 atm)

no. of moles of GAS

=

At s.t.p. (0oC and 1atm)

no. of moles of GAS

=

In aqueous solution

no. of moles of A=

mass of AMolar Mass of A