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

Why 12 C as the Standard Atom? (i) 12 C 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 35 Cl & 37 Cl is 75% and 25% respectively. A.Relative Isotopic Mass 1(a) Isotopes are atoms of the same element having the same number of protons but different number of neutrons. e.g. 35 Cl & 37 Cl, 79 Br & 81 Br

(c) Relative abundances of isotopes are determined from the mass spectrum of the element. m/e Relative abundance mass spectrum of neon gas e.g. rel. abundance of 20 Ne and 22 Ne is 90% and 10% respectively.

(b) Relative isotopic mass = Mass of one atom of the isotope 1 / 12  the mass of one atom of 12 C 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 12 C atom. (No unit)

Example: IsotopeRel. isotopic massRel. abundance 1H1H H2H O O O Cl Cl

(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 12 C atom. (No unit) B.Relative Atomic Mass (A r ) 2. A r is calculated as the weighted mean of the relative isotopic masses of the isotopes according to their abundance.

= 20 x 90/ x 10/100 = 20.2 Eg 1.1 There are 90% 20 Ne and 10% 22 Ne naturally occurring, thus A r of neon m/e Rel. abundance 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 12 C atom. (No unit) 2. M r is calculated as the sum of all relative atomic masses from the formula of the molecule. C.Relative Molecular Mass (M r ) Eg 1.2 M r of chlorine gas, Cl 2 = M r of octane, C 8 H 18 = 35.5  2 =   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 12 C atom. (No unit) (Term used for IONIC COMPOUNDS) For ionic compounds like NaCl, it is NOT a molecule. D.Relative Formula Mass (M r )

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 M r of NaCl = = 58.5

What is a MOLE? IIMOLE 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 10 23

The Avogadro constant L is defined as 6.02 x 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 particles, No. of moles of particles = No. of particles 6.02 x ______________ Eg. 2.1 no. of molesno. of particles (a)0.300 mol (b) 1.23 x N 2 molecules (c)6.71 mol He 1.81 x N 2 molecules 4.04 x 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  g or 2  kg = 12 / 6.02 x g ( g )

B.Molar Mass & Related Calculations 1. Mass of one mole (6.02 x ) of 12 C atoms = 12.0 g (i.e. A r (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 g 1 H atoms No. of moles of A = Mass of A Molar mass of A __________ Mass of A = no. of moles of A  Molar mass of A

NameFormula Rel. molecular mass Rel. formula mass Molar mass magnesium sulphate NH 3 (NH 4 ) 2 CO 3 2. Molar mass is the mass of the substance per unit mole. Unit: g mol -1 Eg MgSO g mol ammonia g mol ammonium carbonate g mol -1

NameFormulaMrMr No. of moles No. of particles Mass /g F2F HCl 3.81  Sulphur dioxide 1.80  Glucose SO  Eg. Fluorine  Hydrogen chloride  C 6 H 12 O   10 21

C.Stoichiometry & Related Calculations 1.One H 2 O molecule contains __ atoms of H & __ atom of O x H 2 O molecules contain _____________ atoms of H & _________ atoms of O. One mole of H 2 O contains __ moles of H atoms & __ mole of O atoms. A m B n  m A  n B no. of moles of A = m  no. of moles of A m B n no. of moles of B = n  no. of moles of A m B n  One mole of A m B n contains m moles of A & n moles of B x 6.02 x x 10 23

1.CaCl 2 Ca 2+ ionsCl - ions (a)1 mol (b)5 mol (c)0.3 mol Eg mol 2 mol 5 mol10 mol 0.15 mol

0.032 mol(c) 8 mol(b) 0.1 mol(a) H atomsC atomsC 4 H 8 2. Eg mol 0.8 mol 2 mol16 mol mol0.016 mol Can you calculate the % by mass of Ca in CaBr 2 ? [Answer : 20.1 %]

% by mass of A in A m B n =  100% =  100% =  100% mass of A in A m B n mass of A m B n no. of moles of A in A m B n  Molar mass of A Molar Mass of A m B n m  Molar mass of A Molar Mass of A m B n

Al 2 O 3 (A r : 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    %

M r (H 2 O) = 1.0  = 18.0 % by mass of H 2 O =  100% = A r : Cu 63.5; S 32.1; O 16.0; H 1.0 (b) Calculate the % by mass of of H 2 O in CuSO 4.5H 2 O. 5    %

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. = C 2 H 5, M.F. = C 4 H 10 Glucose:E.F. = CH 2 O, M.F. = C 6 H 12 O 6

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: Mn 3 O %28.0 %mass in 100 g no. of mol 72.0/54.9 = /16.0 = 1.75 mole ratio 1.31/1.31= /1.31=1.34 4

CH N (b) % by mass of elements of a compound Y are C 63.2%, H 12.3%, N 24.6%. The M r of Y is determined as 114. Calculate the empirical and molecular formulae of compound Y. How is M r det’d exptmentally? 63.2 %12.3 %mass in 100 g24.6 % no. of mol 63.2/12.0 = /1.0 = /14.0 = 1.76 mole ratio   3 Empirical formula is C 3 H 7 N. Why total not 100% exactly?

The molecular formula is C 3n H 7n N n, where n is an integer. Given the M r = 114, 3n(12.0) + 7n(1.0) + n(14.0) = n = 114 n = 2 Therefore, the molecular formula is C 6 H 14 N 2.

25  C 1 atmosphere r.t.p.: Room temperature and pressure D.Calculations involving Gases 1 mol of any gas (i.e  gas particles) occupies the same volume under the same temperature and pressure. volume of 1 mole of any gas =24.0 dm 3 No. of moles of gas at r.t.p. = vol. of gas (dm 3 ) 24.0 _______________ V dm 3 Same no. of moles of gas at same temp. & pressure

0C0C 1 atmosphere s.t.p.: standard temperature and pressure volume of 1 mole of any gas =22.4 dm 3 No. of moles of gas at s.t.p. = vol. of gas (dm 3 ) 22.4 _______________

For each of the following pair, which container has a greater volume? Eg 2.7 Container X containsContainer Y containsAns (a) 6 g of H 2 at r.t.p.96 dm 3 of CO 2 at r.t.p. (b) 7 g of CO at s.t.p mol of Cl 2 at 0 o C and 1 atm (c) 3.01  CH 4 molecules at 500 o C and 1 atm 0.17 g of NH 3 & 0.4 mol of N 2 at 500 o C and 1 atm (d) 2 mol of Ne at 50 o C and 1 atm 2 mol of Ar at 25 o C and 1 atm Y Y Y X

Container X containsContainer Y contains 6 g of H 2 at r.t.p.96 dm 3 of CO 2 at r.t.p. 6 g of H 2 = 6 / 2.0 = 3.0 mol of H 2 Vol. of 3 mol of H 2 at r.t.p. = 3.0 x 24.0 = 72 dm 3

Container X containsContainer Y contains 7 g of CO at s.t.p.0.70 mol of Cl 2 at 0 o C and 1 atm 7 g of CO = 7 / 28.0 = 0.25 mol of CO (at s.t.p.) 0.70 mol of Cl 2 at 0 o C & 1 atm = 0.70 mol of Cl 2 at s.t.p.

Container X containsContainer Y contains 3.01  CH 4 molecules at 500 o C & 1 atm 0.17 g of NH 3 & 0.4 mol of N 2 at 500 o C and 1 atm 3.01  CH 4 = 3.01  / 6.02  = 0.05 mol of CH 4 (0.17/17.0) = 0.01 mol of NH mol NH mol N 2 = 0.41 mol of gas

E.Calculations involving Solutions Concentration of solution can be expressed in: (a) mol dm -3 (no. of mole of solute in 1 dm 3 of solution) (b) g dm -3 (mass of solute in 1 dm 3 of solution) Conc. of A in mol dm -3 = —————–———— = ——–———————— = ————————— no. of moles of A (mol) volume (dm 3 ) mass of A (g) / molar mass of A volume (dm 3 ) Conc. of A in g dm -3 Molar mass of A  No. of moles of A = conc. of A  volume (mol dm -3) (dm 3 )

Eg g anhydrous sodium carbonate were dissolved and volume of the resulting solution is 250 cm 3. 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 Na 2 CO 3 in g dm -3 = 5.30/0.250 = 21.2 (b) Concentration of Na 2 CO 3 in mol dm -3 = 21.2/106 = (c) Concentration of Na + in g dm -3 = x 2 x 23 = 9.20 Recall vol. is in dm 3.

Eg 2.9 (a) Given: 2.02 g dm -3 H 2 SO 4, [H 2 SO 4 (aq)] = =mol dm -3 [H + (aq)]==mol dm -3 [SO 4 2- (aq)]==mol dm / x 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 / x x

no. of particles 6.02 x vol. of gas(dm 3 ) 24 (dm 3 ) vol. of gas(dm 3 ) 22.4 (dm 3) Vol (dm 3 ) xConc (mol/dm 3 ) Mole no. of moles of A = m x no. of moles of A m B n no. of moles of particles = no. of mole of A = A m B n  m A  n B At r.t.p. (25  C & 1 atm) no. of moles of GAS = At s.t.p. (0 o C and 1atm) no. of moles of GAS = In aqueous solution no. of moles of A= mass of A Molar Mass of A