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ATOMS, MOLECULES & STOICHIOMETRY

ATOMS, MOLECULES & STOICHIOMETRY. I. INTRODUCTION. II. MOLE CONCEPT. III. CALCULATIONS ON REACTIONS. VI. MASS SPECTRO-METRY. V. REDOX REACTIONS. IV. VOLUMETRIC ANALYSIS. A 6-Course Menu. 12 C. I INTRODUCTION. Relative Atomic Mass Scale: Carbon-12 Scale

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ATOMS, MOLECULES & STOICHIOMETRY

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  1. ATOMS, MOLECULES & STOICHIOMETRY

  2. I.INTRODUCTION II. MOLE CONCEPT III. CALCULATIONS ON REACTIONS VI. MASS SPECTRO-METRY V. REDOX REACTIONS IV. VOLUMETRIC ANALYSIS A 6-Course Menu

  3. 12C I INTRODUCTION • 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) An atom of 12C is assigned a mass of 12 units, => meaning 1/12 of the mass of a 12C atom is 1 unit.

  4. Why 12C as the Standard Atom? • 12C is a readily available common element. • (ii) Most abundant isotope of carbon.

  5. 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. 35Cl & 37Cl, 79Br & 81Br (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.

  6. Relative abundance 9 mass spectrum of neon gas 1 m/e 19 20 21 22 23 (c) Relative abundances of isotopes are determined from the mass spectrum of the element. e.g. rel. abundance of 20Ne and 22Ne is 90% and 10% respectively.

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

  8. Example:

  9. B. Relative Atomic Mass (Ar) (Term used for ATOMS) 1. Definition: Ratio of the average mass of one atom of an element compared to 1/12of the mass of a 12C atom. (No unit) 2. Ar is calculated as the weighted mean of the relative isotopic masses of the isotopes according to their abundance.

  10. Rel. abundance 9 mass spectrum of neon gas 1 m/e 20 21 22 Eg 1.1 There are 90% 20Ne and 10% 22Ne naturally occurring, thus Ar of neon = 20 x 90/100 + 22 x 10/100 = 20.2

  11. C. Relative Molecular Mass (Mr) • (Term used for MOLECULES) • Definition: • Ratio of the mass of one molecule compared to 1/12of the mass of a 12C atom. (No unit) • 2. Mr iscalculated as the sum of all relative atomic masses from the formula of the molecule. Eg 1.2 Mr of chlorine gas, Cl2 = Mr of octane, C8H18 = 35.5  2 = 71.0 8  12.0 + 18  1.0 = 114

  12. D. Relative Formula Mass (Mr) (Term used for IONIC COMPOUNDS) For ionic compounds like NaCl, it is NOT a molecule. • Definition: • Ratio of the mass of one formula unit of an ionic compound compared to 1/12of the mass of a 12C atom. (No unit)

  13. 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

  14. What is a MOLE? II MOLE CONCEPT A. Mole & Avogadro Constant (L) 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 6.02 x 1023

  15. 1. Definition: A mole of substance is the amount of that substance which contains Avogadro’s numberof particles. 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) ..named in honour of Italian Chemist, Amedeo Avogadro, but he did not come up with the number!! So who did?

  16. No. of particles ______________ No. of moles of particles = 6.02 x 1023 0.0204 N2 molecules 4.04 x 1024 He 2. Since 1 mole of particles = 6.02 x 1023particles, Eg. 2.1 1.81 x 1023

  17. Why the concept of a MOLE? - used to represent a large amount of substances.. e.g. Mass of ONE atomof Carbon-12 = 12 / 6.02 x 1023 g = 2  10-23 g or 2  10-26 kg (0. 0000000000 0000000000 002 g )

  18. Mass of A = no. of moles of A  Molar mass of A Mass of A __________ No. of moles of A = Molar mass of A 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 g 1H atoms

  19. 120.4 g mol-1 -- MgSO4 120.4 ammonia 17.0 -- 17.0 g mol-1 ammonium carbonate -- 96.0 g mol-1 96.0 2. Molar mass is the mass of the substance per unit mole. Unit: g mol-1 Eg.

  20. Fluorine 1.05  1023 38.0 6.65 Hydrogen chloride 6.33  10-3 36.5 0.231 SO2 64.1 0.115 1.08  1021 2.78  10-3 1.67  1021 C6H12O6 Eg.

  21. 2 1 2 x 6.02 x 1023 6.02 x 1023 2 1 C. Stoichiometry & Related Calculations • 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.  One mole of AmBn contains m moles of A & n moles of B. AmBn  mA  nB no. of moles of A = m  no. of moles of AmBn no. of moles of B = n  no. of moles of AmBn

  22. 2 mol 1 mol 5 mol 10 mol 0.15 mol 0.15 mol Eg. 2.4

  23. 2. C4 H8 C atoms H atoms (a) 0.1 mol 0.8 mol 0.4 mol (b) 8 mol 2 mol 16 mol (c) 0.032 mol 0.004 mol 0.016 mol Eg. 2.4 Can you calculate the % by mass of Ca in CaBr2 ? [Answer : 20.1 %]

  24. mass of A in AmBn mass of AmBn % by mass of A in AmBn = -----------------------  100% = -------------------------------------------------  100% = ---------------------------  100% no. of moles of A in AmBn Molar mass of A Molar Mass of AmBn m Molar mass of A Molar Mass of AmBn

  25. Eg. 2.5 • Calculate the % by mass of O in aluminium oxide. Al2O3 (Ar: Al 27.0; O 16.0) % by mass of O = -----------------------  100% = 3  16.0 2  27.0 + 3  16.0 47.1%

  26. (b) Calculate the % by mass of of H2O in CuSO4.5H2O. Ar: Cu 63.5; S 32.1; O 16.0; H 1.0 Mr(H2O) = 1.0  2 + 16.0 = 18.0 % by mass of H2O = ----------------------------------------  100% = 5  18.0 63.5 + 32.1 + 4  16.0 + 5  18.0 36.1%

  27. 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

  28. O Mn mass in 100 g 72.0 % 28.0 % 72.0/54.9 = 1.31 28.0/16.0 = 1.75 no. of mol 1.31/1.31=1 3 mole ratio 1.75/1.31=1.34 4 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? Ans: Mn3O4

  29. N C H Why total not 100% exactly? mass in 100 g 63.2 % 12.3 % 24.6 % 12.3/1.0 = 12.3 24.6/14.0 = 1.76 63.2/12.0 = 5.27 no. of mol 2.99  3 6.99  7 1 mole ratio (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? Empirical formula isC3H7N.

  30. 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.

  31. Same no. of moles of gas at same temp. & pressure V dm3 V dm3 r.t.p.: Room temperature and pressure 1 atmosphere 25C volume of 1 mole of any gas = 24.0 dm3 vol. of gas (dm3) _______________ No. of moles of gas at r.t.p. = 24.0 D. Calculations involving Gases 1 mol of any gas (i.e. 6.02  1023 gas particles) occupies the same volume under the same temperature and pressure.

  32. s.t.p.: standard temperature and pressure 1 atmosphere 0C volume of 1 mole of any gas = 22.4 dm3 vol. of gas (dm3) _______________ No. of moles of gas at s.t.p. = 22.4

  33. Eg 2.7 For each of the following pair, which container has a greater volume? Y Y Y X

  34. 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 contains 6 g of H2 at r.t.p. 96 dm3 of CO2 at r.t.p.

  35. Container X contains Container Y contains 7 g of CO at s.t.p. 0.70 mol of Cl2 at 0oC and 1 atm 0.70 mol of Cl2 at 0oC & 1 atm = 0.70 mol of Cl2 at s.t.p. 7 g of CO = 7 / 28.0 = 0.25 mol of CO (at s.t.p.)

  36. Container X contains Container Y contains 0.17 g of NH3 & 0.4 mol of N2 at 500oC and 1 atm 3.01  1022 CH4 molecules at 500oC & 1 atm (0.17/17.0) = 0.01 mol of NH3 0.01 mol NH3 + 0.4 mol N2 = 0.41 mol of gas 3.01  1022 CH4 = 3.01  1022/ 6.02  1023 = 0.05 mol of CH4

  37. no. of moles of A (mol) volume (dm3) mass of A (g) / molar massof A volume (dm3) Conc. of A in g dm-3 Molar massof A  No. of moles of A = conc. of A volume (mol dm-3) (dm3) E. Calculations involving Solutions Concentration 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= —————–———— = ——–————————------------- = —————————

  38. Eg 2.8 5.30 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. Recall vol. is in dm3. (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

  39. 2.02 / 98.0 0.0206 0.0206 x 2 0.0412 0.0206 x 1 0.0206 0.17/17.0 0.0100 0.0100 x 39.1 0.391 0.0100 x 56.1 0.561 Eg 2.9 (a) Given: 2.02 g dm-3 H2SO4, [H2SO4(aq)] = = mol dm-3 [H+(aq)] = = mol dm-3 [SO42-(aq)] = = mol dm-3 (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

  40. At r.t.p. (25C & 1 atm) no. of moles of GAS = no. of moles of particles = At s.t.p. (0oC and 1atm) no. of moles of GAS = no. of mole of A Mole = AmBn m A n B In aqueous solution no. of moles of A= no. of moles of A = m x no. of moles of AmBn no. of particles 6.02 x 1023 vol. of gas(dm3) 24 (dm3 ) mass of A Molar Mass of A vol. of gas(dm3) 22.4 (dm3) Vol(dm3 )xConc(mol/dm3 )

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