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Chemistry:- Three Main Core Subjects: 1. Inorganic Chemistry 2. Organic Chemistry

Chemistry:- Three Main Core Subjects: 1. Inorganic Chemistry 2. Organic Chemistry 3. Physical Chemistry Also 4. Analytical Chemistry Foundation Chemistry will introduce the essential “foundations” from each of these core areas - upon which further study in Chemistry

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Chemistry:- Three Main Core Subjects: 1. Inorganic Chemistry 2. Organic Chemistry

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  1. Chemistry:- Three Main Core Subjects: 1. Inorganic Chemistry 2. Organic Chemistry 3. Physical Chemistry Also 4. Analytical Chemistry Foundation Chemistry will introduce the essential “foundations” from each of these core areas - upon which further study in Chemistry will build and advance in the future. Many areas of chemistry involve calculations - so a basic working knowledge of mathematics is essential. The course will start off with a summary of some essential maths and then move on to appropriate areas from I O P and A [in no particular order] to address the SYLLABUS and LEARNING OUTCOMES - as provided at the start of the handout. Dr. David Johnson: CH0004 Lecture Notes - Term 1

  2. ABSOLUTELY ESSENTIAL!!!!! Know How to Use You Calculator Casio fx-83 or fx-85 series Dr. David Johnson: CH0004 Lecture Notes - Term 1

  3. Essential Mathematics: (or arithmetic or ability to do calculations!!) Significant Figures & Rounding Up: Often when doing a calculation using a calculator the display often shows an answer containing many more digits than are required or necessary. The usual procedure is then to round up the number and display the value to an appropriate number of significant figures - dropping the irrelevant insignificant figures For instance - if you have done a titration calculation involving a titration result of 11.2 cm3 - and the calculated answer comes to 0.1258469723 mol dm-3, you should quote a result of 0.126 mol dm-3 (which has been rounded up to 3SF) 14.87 4SF 0.0001487 4SF 0.00014870 5SF 1487001 7SF 1487000 7 SF 1.487 X 106 4SF (standard form - see later) Dr. David Johnson: CH0004 Lecture Notes - Term 1

  4. Significant Figures & Rounding Up: to 3 SF 0.1246 becomes 0.125 0.1254 becomes 0.125 0.1256 becomes 0.126 When doing a calculation involving several stages - give each answer one more SF than necessary and then round up in the final stage. SF is not the same as DECIMAL PLACES e.g. (17.1 x 0.103) / 21.4 = 0.0823037 = 0.0823 to 3 SF = 0.082 to 3 DP quote your answers in the required or most appropriate form (e.g. 0.0823) - usually based around the least reliable value e.g. 2.6387 g 121.2 g 5.23 g 129.0687 g 129.1 g (to 1 DP and 4 SF) Dr. David Johnson: CH0004 Lecture Notes - Term 1

  5. Standard Form: (or Scientific Notation) 21568 = 2.1568 x 104 (10 x 10 x 10 x 10) 0.00001254 = 1.254 x 10-5 1 / (10 x 10 x 10 x 10 x 10) 10 is the base (see later) and 4 or -5 is the exponent When you multiply numbers in standard form - the exponents are added (the principal of logarithms - later) e.g. (2 x 104) x (3.1 x 102) = (2 x 3.1) x (104 x 102) = 6.2 x 106 Using a Calculator: 1.345 x 107 1.345 EXP 7 0.000125 1.25 EXP - 4 108 1 EXP 8 Dr. David Johnson: CH0004 Lecture Notes - Term 1

  6. Logarithms: The logarithm (or “log”) of a number N is the power to which 10 must be raised to give N. If N = 1 since 100 =1 log N = 0 If N = 10,000 since 105 = 10,000 log N = 5 If N = 0.01 since 10-2 = 0.01 log N = -2 If N = 432 since 102.635 = 432 log N = 2.635 Natural Logarithms: The natural logarithm (or “ln”) of a number N is the power to which the special value “e” must be raised to give N. “e” is a constant and is 2.7183 If N = 1 since e0 =1 ln N = 0 If N = 77.2 since e4.346 = 77.2 ln N = 4.346 If N = 0.02 since e-3.912 = 0.02 ln N = -3.912 Relationship: ln N = 2.303 log N Calculator: log 2.315 type it as its written (log key) Dr. David Johnson: CH0004 Lecture Notes - Term 1

  7. Use of Logarithms: Multiplication: logs are added log(A x B) = logA + logB Division: logs are subtracted: log(A/B) = logA - logB Powers: same as multiplication or division log A2 = 2 logA log A-5 = -5 logA log A = 0.5 log A Antilogarithms: If you want to convert logN to N - you use the “antilog” function (or inverse log on calculators). Dr. David Johnson: CH0004 Lecture Notes - Term 1

  8. SI Units (Systeme Internationale): Property Unit Abbreviation length meter m mass kilogram kg time second s temperature kelvin K amount mole mol current ampere A luminosity candela cd called SI BASE Units Base units may be changed by orders of magnitude (i.e. multiples of x 10) by use of PREFIXES: e.g. milligram (mg) All other units are derived from Base Units Confusion: many branches of science (including chemistry) still use OLD UNITS and SYSTEMS (e.g. mile, inch, pounds, ounces, pints, Torr, eV, degrees fahrenheit etc) Dr. David Johnson: CH0004 Lecture Notes - Term 1

  9. Unit Prefixes: PrefixAbbreviationMeaning giga- G x 109 mega- M x 106 kilo- k x 103 deci- d x 10-1 centi- c x 10-2 milli- m x 10-3 micro-  x 10-6 nano- n x 10-9 pico- px 10-12 femto- f x 10-15 e.g. kilometre (km) nanogram (ng) gigabyte (Gb) millimole (mmole) microamp (A) decimetre cubed (dm3) - Litre (a “derived unit”) Dr. David Johnson: CH0004 Lecture Notes - Term 1

  10. Derived Units: All physical quantities can be expressed as combinations of the Base Units. Area: SI unit: m2 rectangle A = l x w circle A = r2 Volume: SI unit: m3 cube V = l3 sphere V = 4/3 r3 Density: = mass / volume SI unit: kg m-3 Force: = mass x acceleration = (mass x distance) / time 2 SI unit: kg m / s2 (“Newton”) Energy = force x distance = (mass x distance2) / time 2 SI unit kg m2 / s2 Nt m (“Joule”) Dr. David Johnson: CH0004 Lecture Notes - Term 1

  11. Converting SI Units: just move the decimal place - or adjust the exponential in standard form 15,256 g = 15.256 kg (kilograms) 85.62 m = 8562 cm (centimetres) = 85620 mm (millimetres) 25 cm3 = 0.025 dm3 (or Litres, L) = 0.025 x 10-3 m3 (metres cubed) = 2.5 x 10-5 m3 256.3 nm = 0.2563 m (microns) 570.5 mmoles = 0.5705 moles Dr. David Johnson: CH0004 Lecture Notes - Term 1

  12. Converting to Other Units: need a CONVERSION FACTOR e.g. 1 inch is 2.54 cm how many inches in 57.27 cm = 57.27 / 2.54 = 22.55 inches how many cm in 11.61 inches = 11.61 x 2.54 = 29.49 cm e.g. 1 eV (electron Volt) is 1.602 x 10-19 Joules how many Joules in 1270.6 eV = 1270.6 x 1.602 x 1019= 2.036 x 10-16 J how many eV in 47.8 kJ = (47.8 x 1000) / 1.602 x 1019 = 22.984 x 10-23 eV Dr. David Johnson: CH0004 Lecture Notes - Term 1

  13. Converting to Other Units: e.g. Degrees Centigade to Kelvins T (in K) = T (0C) + 273.15 what is 25.1 0C in Kelvins = 25.1 + 273.15 = 298.3 K Dr. David Johnson: CH0004 Lecture Notes - Term 1

  14. Rules for Rounding Off: • if the digit to be dropped (i.e. first insignificant figure) > 5 - ROUND UP • if the digit to be dropped < 5 - ROUND DOWN • When do you Round Off? • At the very end of the series of calculations Dr. David Johnson: CH0004 Lecture Notes - Term 1

  15. Rules for Significant Figures: • all nonzero digits are significant • 275.8564 (7 SF) • interior zeros are significant • 12.0024 (6 SF) • trailing zeros (after decimal point) are significant • 2.6500 (5 SF) • leading zeros are not significant • 0.00021 (2 SF) Dr. David Johnson: CH0004 Lecture Notes - Term 1

  16. Rules for Significant Figures: Addition & Subtraction: Rule: round off to the least uncertain decimal place 17.85 + 547.1 + 2.5478 = 567.5 Multiplication & Division: Rule: the answer cannot have more SF’s than the fewest number present in the original numbers 17.589 x 5.23 = 92.0 Dr. David Johnson: CH0004 Lecture Notes - Term 1

  17. Numbers & Equations: Having now discussed some of the basic mathematics that is required to perform simple chemical calculations - let us now turn out attention to a couple of examples of where this may be applied. Certain aspects of chemistry, particularly physical and analytical chemistry, involve calculations based on key equations. The understanding of the equations will be essential to that particular subject - but there is no point in understanding the chemistry unless it can be applied numerically. Example 1 - the Ideal Gas Law: PV = nRT P = pressure (in Nm-2) V = volume in M3 R is a constant (8.314 J K-1 mol-1) T is the temperature (in K) Dr. David Johnson: CH0004 Lecture Notes - Term 1

  18. Example Calculation 1: Given the equation PV = nRT (the ideal gas equation), calculate the volume of a gas (V, in m3) where the pressure (P) is 2.457 x 105 Nm-2, the number of moles (n) is 0.721 and the temperature (T) is 315 K. R is a constant and has a value of 8.314 J K-1 mol-1. V = nRT / P = (0.721 x 8.314 x 315) / 2.457 x 105 = 7.69 x 10-3 m3= 7.69 dm3 (Litres) Example 2 - the Nernst Equation: E = E0 - 0.0592 log (X/Y) E = cell potential (in Volts) E0 = a constant (X/Y) is a ratio of the concentration of 2 chemicals Dr. David Johnson: CH0004 Lecture Notes - Term 1

  19. Example Calculation 2: Given the equation: E = E0 - 0.0592 log (X / Y) the Nernst equation - calculate the cell potential (E, in Volts) where the ratio [X/Y] is 0.161 and the constant E0 has a value of 1.635 Volts. E = 1.635 - (0.0592 x log [0.161]) = 1.682 Volts Dr. David Johnson: CH0004 Lecture Notes - Term 1

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