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# Calculations for IB Chemistry

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A collection of all the calculations needed to master IB Chemistry including the human biochemistry optional unit.

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### Calculations for IB Chemistry

1. 1. Mr Field
2. 2. INTRODUCTION ο§ The purpose of this presentation, is to collect all the calculations you need for IB Chemistry in one place. ο§ This will help you to memorise them. ο§ It will not help you to master themβ¦the only shortcut to mastery is many hours of dedicated practice. ο§ Good luck!
3. 3. PURPOSE: CALCULATING A QUANTITY IN MOLES FROM A NUMBER OF PARTICLES The Calculation Notes π π= πΏ n = the quantity in moles N = the number of particles L = Avogadroβs constant, 6.02x1023 ο§ When calculating numbers of atoms within molecules, multiply the number of particles (N) by the number of atoms in the formula ο§ To find out numbers of particles, rearrange to: N = n.L STOICHIOMETRY
4. 4. PURPOSE: DETERMINE RELATIVE MOLECULAR OR FORMULA MASS, MR The Calculation ππ = (ππ’ππππ ππ ππ‘πππ . ππ‘ππππ πππ π ) Notes ο§ Mr has no unit as it is a relative value ο§ To calculate molar mass, Mm, just stick a βgβ for grams on the end STOICHIOMETRY
5. 5. PURPOSE: CALCULATE A QUANTITY IN MOLES, FROM A MASS OF A SUBSTANCE The Calculation Notes π π= ππ n = the quantity in moles m = the mass of substance you are given in grams Mm = the molar mass of the substance ο§ To determine the mass of a given number of moles of a substance use: ο§ π= π π. π ο§ To determine the molar mass of a given mass of a substance: π ο§ ππ= π STOICHIOMETRY
6. 6. PURPOSE: DETERMINE EMPIRICAL FORMULA FROM % COMPOSITION BY MASS The Calculation 1. Divide each % by the atomic mass 2. Divide each answer to Step 1 by the smallest answer to Step 1 3. Notes ο§ You can follow the same method if you Multiply all answers to Step 1 to remove any obvious fractions have plain composition by mass rather than % composition by mass. a. If there is a β.5β multiply everything by 2 b. If there is a β.33β multiply everything by 3 etc STOICHIOMETRY
7. 7. PURPOSE: DETERMINE MOLECULAR FORMULA FROM EMPIRICAL FORMULA The Calculation ππ πΉπ = . πΉ π(πΉπ ) π Notes ο§ This and the previous calculation are often combined together in exam questions Fm = molecular formula Mr = relative molecular mass Fe = empirical formula m(Fe) = empirical formula mass STOICHIOMETRY
8. 8. PURPOSE: USE MOLE RATIOS TO DETERMINE THE NUMBER OF MOLES OF βBβ THAT CAN BE MADE FROM βAβ The Calculation ππ’ππππ ππ π΅ ππ πππ’ππ‘πππ π π΅ = π π΄ . ππ’ππππ ππ π΄ ππ πππ’ππ‘πππ Notes ο§ The second term in this equation is the mole ratio ο§ You must use a fully balanced equation ο§ This is the central step in many stoichiometry calculations STOICHIOMETRY
9. 9. PURPOSE: CALCULATE THEORETICAL YIELD The Calculation Notes ο§ Use mole ratios to determine the ο§ n/a expected quantity of product in moles ο§ Use π = π π . π to determine the expected mass. STOICHIOMETRY
10. 10. PURPOSE: DETERMINE LIMITING AND EXCESS REACTANTS The Calculation Notes ο§ Divide moles of each reactant by their ο§ You would often then need to use mole coefficient in the balanced equation ο§ Smallest value ο  limiting ratios to determine a quantity of product (in moles). ο§ Largest value ο  excess STOICHIOMETRY
11. 11. PURPOSE: CALCULATING PERCENTAGE YIELD. The Calculation πππ‘π’ππ π¦ππππ % πππππ = . 100 π‘βπππππ‘ππππ π¦ππππ Notes ο§ Actual and theoretical yield must have the same units. ο§ You might sometimes be required to rearrange this equation, or use it to work backwards from this to find the amount of reactant you started with. STOICHIOMETRY
12. 12. PURPOSE: APPLY AVOGADROβS LAW TO CALCULATE REACTING VOLUMES OF GASES The Calculation Notes π1 π2 = π1 π2 V1 = the initial volume of gas n1 = the initial quantity of gas in moles V2 = the final volume of gas n2 = the final volume of gas in moles ο§ Only applies when temperature and pressure remain constant. ο§ Units of V do not matter. But must be the same. ο§ This is really a special case of the Ideal Gas Law where the pressure, temperature and gas constant terms cancel each other out. STOICHIOMETRY
13. 13. PURPOSE: CALCULATE MOLAR QUANTITIES OF GASES AT STANDARD TEMPERATURE AND PRESSURE The Calculation Notes π π= 22.4 ο§ Only applies at standard conditions: ο§ 273 K (0oC) ο§ 101,000 Pa (1.00 atm) n = quantity in moles V = the volume of gas in dm3 22.4 is the molar volume of an ideal gas at 273K and 101,000 Pa ο§ If volume of gas is given in m3, use 2.24x10-5 as your molar volume. ο§ Molar volumes are given in the data booklet and do not need memorising. STOICHIOMETRY
14. 14. PURPOSE: THE IDEAL GAS EQUATION The Calculation ππ = πππ Notes ο§ In practice, you can often use: ο§ V in units of dm3 P = pressure in Pa V = volume of gas in m3 n = quantity of gas in moles R = the gas constant, 8.31 T = temperature in Kelvin (OC + 273) ο§ Pa in units of kPa ο§ You will need to be comfortable rearranging this equation to change the subject. ο§ This takes time to use, so only use it in non-standard conditions, or when the laws in Calculation 13 would not be quicker. STOICHIOMETRY
15. 15. PURPOSE: RELATIONSHIP BETWEEN TEMPERATURE, PRESSURE AND VOLUME The Calculation Charlesβ Law, at fixed pressure: Notes π1 π1 π2 π2 = Gay-Lussacβs Law, at fixed volume: π1 π1 = ο§ These only work where the quantity in π2 π2 Boyleβs Law, at fixed temperature: π1 π1 = π2 π2 moles remains fixed. ο§ All of these are just special cases of the ideal gas law, where the remaining terms just cancel each other out. V1 and V2 are initial and final volume P1 and P2 are initial and final pressure T1 and T2 are initial and final temperature STOICHIOMETRY
16. 16. PURPOSE: MOLAR CONCENTRATION The Calculation Notes π π= π c = concentration in mol ο§ Units are pronounced βmoles per decimetre cubedβ dm-3 ο§ You need to be able to use any rearrangement of this equation n = the quantity in moles V = the volume in dm3 (litres) STOICHIOMETRY
17. 17. PURPOSE: CONCENTRATION BY MASS The Calculation Notes π π= π c = concentration in g dm-3 m = the quantity in grams V = the volume in dm3 (litres) ο§ Units are pronounced βgrams per decimetre cubedβ ο§ You need to be able to use any rearrangement of this equation ο§ Generally, if you have a concentration like this, you should convert it into a molar concentration before proceeding. STOICHIOMETRY
18. 18. PURPOSE: DETERMINE RELATIVE ATOMIC MASS FROM % ABUNDANCE DATA The Calculation π΄π = % πππ’ππππππ (ππ‘ππππ πππ π . ) 100 Notes ο§ % abundance may be given in the question, or you may need to read it from a mass spectrum ο§ If you convert the percentages to Ar = relative atomic mass decimals (i.e. 0.8 for 80%, 0.25 for 25%), there is no need to divide by 100. ATOMIC STRUCTURE
19. 19. PURPOSE: DETERMINE % ABUNDANCE FROM RELATIVE ATOMIC MASS The Calculation Notes If there are two isotopes, label one of them βaβ and one βbβ. Now: π΄ π = ο§ Ar, Ia and Ib will be provided in the question, so you can plug the numbers in, and then rearrange to find x. π₯πΌ π +π¦πΌ π π₯+π¦ However, since x+y = 100%, y = 100-x so: π΄π = π₯πΌ π +(100βπ₯)πΌ π 100 Ar = relative atomic mass Ia = the mass of isotope a Ib = the abundance of isotope b x and y is the abundance of each isotope ο§ To find y, simply do y=100-x ο§ If you have three isotopes, you must know the abundance of at least one in order to find the other two. You would also need to subtract the abundance of this one from the 100, before doing the rest of the sum. ATOMIC STRUCTURE
20. 20. PURPOSE: CALCULATING THE HEAT CHANGE OF A PURE SUBSTANCE The Calculation π = ππβπ Notes ο§ Be careful of the units of massβ¦you may need to convert kg into g q = the heat change in Joules m = the mass of substance in grams ο§ Be careful of the units for specific heat capacity, if it is J K-1 kg-1 you will need to convert your mass into kg. c = specific heat capacity in J K-1 g-1 βT = temperature rise in K or OC ENERGETICS
21. 21. PURPOSE: CALCULATING AN ENTHALPY CHANGE FROM EXPERIMENTAL DATA The Calculation βπ» = βππβπ βH = the enthalpy change in Joules m = the mass of solution in grams c = specific heat capacity of water: 4.18 J K-1 g-1 βT = temperature rise in K or OC Notes ο§ The minus sign is needed to ensure that an exothermic reaction has a negative enthalpy change. ο§ Units are J or kJ mol-1 ο§ The mass of solution is assumed to be the same as its volume in cm3. ο§ The specific heat capacity of the reactants is ignored. ENERGETICS
22. 22. PURPOSE: CALCULATING βHR USING A HESS CYCLE The Calculation Notes Once you have produced you Hess cycle: ο§ See the βEnergeticsβ PowerPoint for 1. Write the relevant βH onto each arrow 2. Multiply each βH in accordance with the stoichiometry 3. advice on constructing Hess cycles. To do your sum, add when you go with an arrow, and subtract when you go against one. ENERGETICS
23. 23. PURPOSE: CALCULATING βHR FROM AVERAGE BOND ENTHALPIES The Calculation Notes Method 1: Make a Hess cycle, then do as in Calculation 20. ο§ It is more reliable to use Hess cycles Method 2: βπ» π ο§ Average bond enthalpies can be found = πππππ‘πππ‘ πππππ  β and you can easily forget whether it is reactants β products or vice versa. in Table 10 of the data booklet. (πππππ’ππ‘ πππππ ) ο§ You only need to worry about the bonds that broken and made. If a bond, for example a C-H is present at the start and finish, you can ignore itβ¦.this can save time in exams. ENERGETICS
24. 24. PURPOSE: CALCULATING βHR FROM ENTHALPIES OF FORMATION The Calculation Notes Method 1: Make a Hess cycle, then do as in Calculation 20. ο§ It is more reliable to use Hess cycles Method 2: βπ» π ο§ βHof for elements in their standard = βπ» ππ (πππππ’ππ‘π ) β and you can easily forget whether it is products β reactants or vice versa. states is zero. βπ» ππ (πππππ‘πππ‘π ) βHr = enthalpy change of reaction βHof = enthalpy change of formation ο§ βHof values for many compounds can be found in Table 11 of the data booklet. ο§ In some questions, you may also need to take a state change into account, if standard states are not used. ENERGETICS
25. 25. PURPOSE: CALCULATING βHR FROM ENTHALPIES OF COMBUSTION The Calculation Notes Method 1: Make a Hess cycle, then do as in Calculation 20. ο§ It is more reliable to use Hess cycles Method 2: βπ» π ο§ βHoc for CO2 and H2O is zero. = βπ» ππ (πππππ‘πππ‘π ) β βπ» ππ (πππππ’ππ‘π ) and you can easily forget whether it is reactants β products or vice versa. ο§ βHoc values for many compounds can be found in Table 12 of the data booklet. βHr = enthalpy change of reaction βHoc = enthalpy change of combustion ENERGETICS
26. 26. PURPOSE: CALCULATING LATTICE ENTHALPY The Calculation Notes You need to build a Born-Haber cycleβ¦.see the Energetics PowerPoint for help. ο§ n/a ENERGETICS
27. 27. PURPOSE: CALCULATING βS FROM STANDARD ENTROPY VALUES The Calculation βπ π = π π (πππππ’ππ‘π ) β Notes π π (πππππ’ππ‘π ) ο§ Units are J K-1 mol-1 ο§ So values can be found in Table 11 of the data booklet βSo = standard entropy change of reaction ο§ You cannot assume that So of an element is zero. It is not. So = standard entropy of each substance ENERGETICS
28. 28. PURPOSE: CALCULATING βGR STANDARD GIBBβS FREE ENERGY OF FORMATION VALUES The Calculation Notes Method 1: Make a Hess cycle, then do similar to Calculation 20. ο§ Units are J or kJ mol-1 ο§ It is more reliable to use Hess cycles and you can easily forget whether it is products β reactants or vice versa. Method 2: βπΊ π = βπΊ ππ (πππππ’ππ‘π ) β βπΊ ππ (πππππ‘πππ‘π ) βGr = Gibbβs free energy of reaction βGof = Gibbβs free energy of formation ο§ βGof for elements in their standard states is zero. ο§ βGof values for many compounds can be found in Table 11 of the data booklet. ENERGETICS
29. 29. PURPOSE: CALCULATING βGR FROM EXPERIMENTAL DATA The Calculation βπΊ = βπ» β πβπ βG = Gibbβs free energy βH = Enthalpy change Notes ο§ If βH is in kJ mol-1, you will need to divide βS by 1000 to convert it to units of kJ K-1 mol-1 ο§ You may first need to calculate βH and βS using Calculations 23 and 24. T = Temperature in Kelvin βS = Entropy change ENERGETICS
30. 30. PURPOSE: CALCULATE THE RATE OF A REACTION The Calculation ββ[π] β[π] πππ‘π = = βπ‘ βπ‘ β[R] = change in reactant concentration Notes ο§ Units are mol dm-3 s-1 ο§ The minus sign in front of β[R] is because the concentration of reactants decreases β[P] = change in product concentration βt = change in time KINETICS
31. 31. PURPOSE: CALCULATE THE GRADIENT OF A SLOPE The Calculation πβππππ ππ π¦ πΊπππππππ‘ = πβππππ ππ π₯ Notes ο§ Used for calculating rate from graph of concentration (y-axis) over time (xaxis) KINETICS
32. 32. PURPOSE: DETERMINE THE ORDER OF REACTION WITH RESPECT TO A REACTANT, X. The Calculation 1. 2. Identify two experiments, where the concentration of βxβ has changed, but all others have remained the same. Compare the change in [x] to the change in rate: a. If doubling [x] has no effect on rate, Notes ο§ Sometimes, you canβt find two cases where only [x] has changed, in which case you may need to take into account the order of reaction with respect to other reactants. then 0th order. b. If doubling [x] doubles rate, then 1st order. 3. If doubling [x] quadruples rate, then 2nd order. KINETICS
33. 33. PURPOSE: DEDUCE A RATE EXPRESSION The Calculation πππ‘π = π[π΄] π₯ [π΅] π¦ [πΆ] π§ Notes ο§ Reactants with a reaction order of zero can be omitted from the rate equation k = rate constant (see below) [A/B/C] = concentration of each reactant x/y/z = order of reaction with respect to each reactant ο§ Given suitable information, you may need to calculate the value of the rate constant if given rates, concentrations and reaction order, or the expected rate given the other information. KINETICS
34. 34. PURPOSE: DETERMINING THE UNITS FOR THE RATE CONSTANT The Calculation πππ ππβ3 π  β1 ππππ‘π  = πππ π₯ (ππβ3 ) π₯ ο§ mol and dm-3 terms on the top and the bottom should be cancelled out ο§ Remaining mol and dm-3 terms on the bottom should then be brought to the top by inverting their indices. Notes ο§ If you canβt understand this, try to memorise: ο§ 0th order: mol dm-3 s-1 ο§ 1st order: s-1 ο§ 2nd order: mol-1 dm3 s-1 ο§ 3rd order: mol-2 dm6 s-1 x = the overall order of the reaction KINETICS
35. 35. PURPOSE: DETERMINING ACTIVATION ENERGY Notes The Calculation ο§ The Arrhenius equation: π = π΄π ο§ This rearranges to: ln π = βπΈ π 1 .π π βπΈ π ππ + ln π΄ ο§ Which is basically the equation for a straight line in the form βy=mx+cβ where: βyβ is ln k; βxβ is 1/T; βmβ is βEa/R; βcβ is ln A ο§ So, if we do a reaction at a range of temperatures and calculate ln k, then: ο§ Where: ο§ ο§ ο§ ο§ ο§ ο§ k = rate constant A = Arrhenius constant e = exponential constant Ea = activation energy R = gas constant, 8.31 T = temperature in Kelvin 1. Draw a graph with 1/T along the x-axis, and ln k on the y-axis ο§ Units of Ea are kJ mol-1 2. Draw a straight line of best fit. ο§ From the graph, you might also be asked 3. Determine the gradient of the line 4. Then: πΈ π = βπΊπππππππ‘ π to calculate A: ο§ βln Aβ, is the y-intercept of the graph, so simply raise βeβ to the power of this intercept. KINETICS
36. 36. PURPOSE: DETERMINING THE EQUILIBRIUM CONSTANT EXPRESSION The Calculation For the reaction: wA + zB ο  yC + zD Notes ο§ Only applies to reactants for which there is a concentration: ο§ Aqueous substances are included ο§ Solids and pure liquids are omitted as they do not have a concentration per se [πΆ] π¦ [π·] π§ πΎπ = [π΄] π€ [π΅] π§ ο§ For reactions involving gases, use partial pressures instead of concentrations. ο§ At SL, you only need be able to construct the expression. Kc = equilibrium constant ο§ At HL, you may need to calculate Kc from the expression and suitable data, or to determine reactant equilibrium concentrations of reactants, given Kc and suitable information. EQUILIBRIUM
37. 37. PURPOSE: DETERMINING CHANGES IN [H+] GIVEN CHANGES IN PH The Calculation Notes ο§ For each increase of β1β on the pH scale, ο§ At SL, you do not need to be able to divide [H+] by 10. ο§ For each decrease of β1β on the pH scale, calculate pH, just to understand it in relative terms. multiply [H+] by 10 ACIDS AND BASES
38. 38. PURPOSE: DEDUCE [H+] AND [OH-] GIVEN KW The Calculation Notes πΎ π€ = π» + . [ππ»β ] ο§ At standard conditions, Kw = 1.00x10-14 ο§ Kw varies with temperature, so it is So: π»+ πΎπ€ = [ππ»β ] And: ππ»β important to know how to do this calculation. πΎπ€ = [π» + ] ACIDS AND BASES
39. 39. PURPOSE: CALCULATING PH FROM [H+] AND VICE VERSA The Calculation Notes ππ» = βπππ10 [π» + ] [π»+ ] = 10βππ» ο§ With strong acids, you can assume [H+] is the same as the concentration of the acid (adjusted for the stoichiometry) ο§ With weak acids, you will need to calculate [H+] using Ka or pKa. ACIDS AND BASES
40. 40. PURPOSE: CALCULATING PH FROM [OH-] AND VICE VERSA The Calculation Notes πππ» = βπππ10 [ππ»β ] [ππ» β ] = 10βπππ» ο§ With strong bases, you can assume [OH- ] is the same as the concentration of the acid (adjusted for the stoichiometry) ο§ With weak acids, you will need to calculate [OH-] using Kb or pKb. ACIDS AND BASES
41. 41. PURPOSE: DETERMINING KA AND KB OF ACIDS/BASES AND THEIR CONJUGATE BASES/ACIDS The Calculation Notes πΎ π€ = πΎ π. πΎ π So: And: πΎπ€ πΎπ = πΎπ ο§ This is useful when trying of determine the strength the conjugate base of a weak acid, and the conjugate acid of a weak base. πΎπ€ πΎπ = πΎπ ACIDS AND BASES
42. 42. PURPOSE: DETERMINING PKA AND PKB OF ACIDS/BASES AND THEIR CONJUGATE BASES/ACIDS The Calculation ππΎ π€ = ππΎ π + ππΎ π So: ππΎ π = ππΎ π€ β ππΎ π Notes ο§ This is useful when trying of determine the strength the conjugate base of a weak acid, and the conjugate acid of a weak base. And: ππΎ π = ππΎ π€ β ππΎ π ACIDS AND BASES
43. 43. PURPOSE: DETERMINING PH FROM POH AND VICE VERSA The Calculation ππΎ π€ = ππΎ π + ππΎ π Notes ο§ This is useful to quickly and easily calculate one of pH/pOH from the other (or [H+]/[OH-]). So: ππ» = 14 β πππ» And: πππ» = 14 β ππ» ACIDS AND BASES
44. 44. PURPOSE: CALCULATING PH OF A SOLUTION OF A WEAK ACID Notes The Calculation π»+ [π΄β ] πΎπ = [π»π΄] ο§ We assume that [HA] is equal to the concentration stated in the question, as only a very small amount has dissociated. ο§ You may need to work backwards from pH to Since [H+] = [A-]: πΎπ = π»+ 2 work out Ka: [π» + ] = 10βππ» [π»π΄] π»+ 2 So: [π» + ] = πΎ π . [π»π΄] ο§ Then: πΎ π = [π»π΄] ο§ You will not need to work out problems for Then: ππ» = βπππ10 [π» + ] polyprotic weak acids that would require quadratics. ACIDS AND BASES
45. 45. PURPOSE: CALCULATING POH OF A SOLUTION OF A WEAK BASE Notes The Calculation π΅ + [ππ» β ] πΎπ = [π΅ππ»] Since [B+] = [OH-]: ππ» β 2 πΎπ = [π΅ππ»] ο§ We assume that [BOH] is equal to the concentration stated in the question, as only a very small amount has dissociated. ο§ You may need to work backwards from pOH to work out Kb: [ππ» β ] = 10βπππ» So: [ππ» β ] = Then: πΎ π . [π΅ππ»] πππ» = βπππ10 [ππ» β ] ο§ Then: ππ» β 2 πΎπ = [π΅ππ»] ACIDS AND BASES
46. 46. PURPOSE: DETERMINING PH OF ACIDIC BUFFER SOLUTIONS (AND ALKALI BUFFERS) Notes The Calculation π»+ [π΄β ] πΎπ = [π»π΄] Now [H+] is not equal to [A-]: So: [π»+ ] πΎ π . [π΄β ] = [π»π΄] ο§ You may need to use your stoichiometry to calculate the [HA] and [A-] in the buffer solution first. ο§ Since additional A- is added, we now need to use the concentration of A- from the buffer as our [A-]. ο§ [HA] should either be that stated in the question, or that calculated via stoichiometry, depending on the context. ο§ For alkali buffers, do the same process Then: ππ» = βπππ10 [π»+ ] but with OH-, Kb etc. ACIDS AND BASES
47. 47. PURPOSE: CALCULATE EOCELL The Calculation π πΈ ππππ = πΈ π πππ‘βπππ β πΈ π (anode) Eocell = cell potential Eo = standard electrode potential Notes ο§ The value should always be positive, so if you get it the wrong way round, just take off the minus sign. ο§ Eo can be found in the data booklet. ο§ You do not need to take the stoichiometry of the reaction into account, just use the Eoas they are in the data booklet. OXIDATION AND REDUCTION
48. 48. PURPOSE: CALCULATING RELATIVE UNCERTAINTIES (IN %) The Calculation πππππ‘ππ£π π’πππππ‘ππππ‘π¦ = Notes πππ πππ’π‘π π’πππππ‘ππππ‘π¦ .100 π ππ§π ππ π£πππ’π ο§ Large measurements have lower relative uncertainties MEASUREMENT AND PROCESSING
49. 49. PURPOSE: CALCULATING ABSOLUTE UNCERTAINTY WHEN ADDITION/SUBTRACTION IS INVOLVED The Calculation Notes πππ‘ππ π’πππππ‘ππππ‘π¦ = ο§ This only works when adding and (πππ πππ’π‘π π’πππππ‘ππππ‘πππ ) subtracting values with the same units MEASUREMENT AND PROCESSING
50. 50. PURPOSE: CALCULATING RELATIVE UNCERTAINTY WHEN MULTIPLICATION/DIVISION IS INVOLVED The Calculation Notes πππ‘ππ π’πππππ‘ππππ‘π¦ = ο§ For use when you are (πππππ‘ππ£π π’πππππ‘ππππ‘πππ ) multiplying/dividing values with different units MEASUREMENT AND PROCESSING
51. 51. PURPOSE: CALCULATING THE ENERGY VALUE OF FOOD FROM COMBUSTION DATA The Calculation Notes Use: ο§ You may need to convert the energy π = ππβπ value into a value per mole by dividing q by the number of moles of substance burnt. q = the heat change in Joules m = the mass of substance in grams c = specific heat capacity in J K-1 g-1 βT = temperature rise in K or OC HUMAN BIOCHEMISTRY
52. 52. PURPOSE: CALCULATING IODINE NUMBERS The Calculation Notes 100 ο§ π(πΌ2 ) = π πΌ2 . π(πππππ) ο§ ο§ N(I2) = the iodine number ο§ m(I2) = the mass of iodine reacting in g ο§ All units should be grams ο§ You may need to convert from data involving bromine to an βiodine equivalentββ¦just use your stoichiometry ο§ m(lipid) = the mass of lipid involved in g HUMAN BIOCHEMISTRY
53. 53. PURPOSE: CALCULATE THE NUMBER OF DOUBLE BONDS IN A LIPID USING IODINE NUMBER. The Calculation Notes Calculate the quantity in moles of I2 corresponding to the iodine number: π(πΌ2 ) π(πΌ2 ) = π π (πΌ2 ) N(I2) = the iodine number Calculate the quantity in moles of lipid in 100g: 100 π(πππππ) = π π (πππππ) Then the number of double bonds is: π(πΌ2 ) π·ππ’πππ πππππ  = π(πππππ) n = quantities in moles Mm = molar masses ο§ This works because each mole of double bonds reacts with one mole of iodine. HUMAN BIOCHEMISTRY