1. Basic chemical calculations

2. When solving numerical problems, always ask yourself whether your answer makes sense !!!

3. How many grams of NaCl were excreted in urine in 24 hours? 20 ml of an average sample was titrated by direct argentometric method and the consumption of the standard solution of AgNO 3 (c = 100 mmol/L) was 34.2 mL. Volume of urine = 1 500 mL/day, M(NaCl) = 58.5 g/mol ANSWERS IN THE TESTS: 1 500 kg i.e. 1.5 ton !!! Correct answer: 15 g

4. SI prefixes (metric prefixes) 10 -1 decid 10 1 decada 10 -2 centic10 2 hectoh 10 -3 milim10 3 kilok 10 -6 micro  10 6 megaM 10 -9 nanon10 9 gigaG 10 -12 picop10 12 teraT 10 -15 femtof10 15 petaP 10 -18 attoa10 18 exaE 10 -21 zeptoz10 21 zettaZ 10 -24 yoctoy10 24 yottaY

5. Converting units 5 mL = 0.005 L 0.750 L = 750 mL 1 L = 1 dm 3 1 mL = 1 cm 3 1  L  = 1 mm 3 0.1 mol/L = 100 mmol/L 50 mg = 0.050 g

6. Converting units Erythrocyte volume: 85 fL ( = 85  m 3 ) 85 fL = 85 x 10 -15 L = 85 x 10 -15 dm 3 = = 85 x 10 -18 m 3 = 85  m 3 Erythrocyte number: 5 x 10 6 / mm 3 = 5 x 10 6 /  L = 5 x 10 12 / L

7. Rounding off the results Numbers obtained by measurement are always INEXACT ! "exact" numbers - in mathematics: 10 = 10.000….. "measured values" e.g. by measuring the volume 10 mL - it is NOT exactly 10.00000… mL Uncertainties ("errors") always exist in measured quantities!

8. Significant figures Is there any difference between 4.0 g and 4.00 g ? Answer: YES ! 4.0 g2 significant figures 4.00 g 3 significant figures  4.00 g is "more precise" than 4.0 g E.g.4.003 4 significant figures 6.023 x 10 23 4 significant figures 0.0012 2 significant figures 5000 ? 1,2,3 or 4 significant figures

9. When rounding off the results, you have to consider significant figures of given numbers ! The precision of the result is limited by the precision of the measurements ! Rules 1) Multiplication and division result – must be with the same number of significant figures as the measurement with the FEWEST signif. fig. e.g. 6.221 cm x 5.2 cm = 32.3492 cm 2 --> round off to 32 cm 2

10. 2) Addition and subtraction result – cannot have more digits to the right of the decimal point than any of the original numbers e.g. 20.41 decimal place 1.3223 decimal places 83 ZERO decimal places 104.722 round off to 105 Rounding off: digits 5,6,7,8,9 --> round up digits 0,1,2,3,4 --> round down

11. CONCLUSION: Your calculator can give the result like this: 100 / 7 = 14.28571429 !!! DON‘T give as a result of the calculation a number with 10 digits, which shows your calculator, round it off to the "reasonable number" of decimal places calculation with more steps – round off only THE FINAL RESULT

12. You have to be familiar with your calculator ! E.g.1) 10 3 = 1000 !!! 10 EXP 3 = 10 x 10 3 = 10000 !!! 2) 50 2 x 5 50 : 2 x 5 = 125 !!! = 5 !!!

13. Uncertainties of quantitative methods 1. PRECISION = how closely individual measurements agree with one another 2. ACCURACY = how closely measurements agree with the correct ("true") value "true" value measured values

14. Precision and accuracy - shooting on the target a) good precision good accuracy b) good precision poor accuracy c) poor precision, but in average good accuracy d) poor precision poor accuracy

15. Density (  - relation between mass and volume - the amount of mass in a unit volume of substance  = m / V Mind the units ! SI unit: kg/m 3 other units: g/cm 3 kg/dm 3 note: cm 3 = mLdm 3 = L

16. Examples water 1 000 kg/m 3 1 g/cm 3 1 kg/dm 3 Au19 300 kg/m 3 19.3 g/cm 3 19.3 kg/dm 3

17. Calculate the density of 90% H 2 SO 4, if the mass of 200 mL of the solution is 363 g.  m / V  / 200 = 1.815 g/cm 3

18. Often we know the volume and we need to calculate the mass and vice versa! m = V x  V = m / 

19. What is the mass of the solution of KOH, if the volume is 2.5 L and density 1.29 g/cm 3 ?  m  V x  m  x 1.29 = 3 225 g units ! 2.5 L = 2 500 mL ( cm 3 )

20. What is the volume of the solution of HNO 3 if the mass is 150 g and the density 1.46 g/cm 3 ?  V  m /  V  / 1.46 = 102.7 cm 3 ( mL )

21. Amount of substance ( n ) - base SI quantity -is in a close relation to the NUMBER of ELEMENTARY ENTITIES (atoms, molecules, electrons, …) unit: MOLE ( 1 mol ) 1 mol = as many objects as the number of atoms in 12 g of the carbon isotope 12 6 C

22. 1 mol = 6.023 x 10 23 elementary entities = AVOGADRO’s number N A N A = 6.0221367 x 10 23 /mol number of entities = n x N A analogy: „counting units“ 1 pair = 2 1 dozen = 12 1 gross = 144 1 mol = 6.023 x 10 23

23. Calculate the number of elementary units present in 2.5 mol cations Ca 2+.  number of Ca 2+  n x N A  number of Ca 2+  x  x     x  

24. Calculate the number of protons released during complete dissociation of 2  mol H 3 PO 4. H 3 PO 4 --> 3 H + + PO 4 3- n (H+) = 3 x n (H3PO4) n (H+) = 6  mol  number of H +  n (H+) x N A  number of H +  x    x  x    x  

25. Calculate the number of C atoms in 0.350 mol of glucose. C 6 H 12 O 6 n (C) = 6 x n glukosa n (C) = 2.1 mol number of C  n (C) x N A  number of C  x  x    x  

26. Calculate the amount of substance: 2.71 x 10 24 molecules of NaCl  n = number of NaCl / N A  n =  x    x      mol

27. Molar mass ( M ) - the mass of 1 mol of a substance - unit: g / mol - can be calculated with the use of relative atomic masses (PERIODIC TABLE) Relative atomic mass A r Relative molecular mass M r expressed in atomic mass units note: atomic mass unit = 1/12 of the mass of atom 12 6 C u = 1.6605 x 10 -24 g A r = m atom / u M r = m molecule / u A r ( 12 6 C ) = 12.00A r ( H ) = 1.008 relative molecular mass - no real unit in biochemistry: Dalton ( Da ) e.g. protein 55 kDa

28. What is a molar mass of glucose ? C 6 H 12 O 6 A r (C) = 12.0 A r (H) = 1.0 A r (O) = 16.0  M = 6 x 12 + 12 x 1 + 6 x 16 = 180 g/mol

29. Often we know the mass and need to calculate the amount of substance and vice versa! n = m / M  m = n x M

30. How many moles of NaCl are present in 100 g of this substance ? M = 58.5 g/mol  n = m / M  n =    mol

31. Calculate the mass of 0.433 mol of calcium nitrate. M = 164.2 g/mol  m = n x M  m = 0.433 x 164.2 = 71.1 g

32. How many molecules of glucose are in 5.23 g C 6 H 12 O 6 ? M = 180 g/mol n = m / M n = 5.23 / 180 = 0.029056 mol number of molecules = n x N A number of molecules = 0.029056 x 6.023 x 10 23 = 1.75 x 10 22

33. Solution composition – "Concentration" solute = the substance which dissolves solvent = the liquid which does the dissolving  A solution is prepared by dissolving a solute in a solvent.

34. Solution composition – "Concentration" to designate amount of solute disolved in a solution "number of different ways to express concentration" Molar concentration (molarity) c mol/L Mass concentration  g/L Mass fraction w Mass percentage % % w/w Volume fraction Volume percentage % v/v

35. Molar concentration ( c ) (substance concentration, MOLARITY) - the number of moles of substance in 1 L of solution c = n / V unit: mol / L Mind the units ! volume must be in LITRES

36. Calculate molar concentration of NaCl solution, if 250 mL contain 0.1 mol NaCl.  c  n / V  c   mol/L units ! 250 mL = 0.250 L

37. What is the substance concentration of a solution, if it contains 15 g NaOH in 600 mL of solution. ( M (NaOH) = 40.0 g/mol ) n = m / M (NaOH)  c  n / V c  m / ( M (NaOH) x V ) c = 15 / ( 40 x 0.6 ) = 0.625 mol/L

38. How many moles of H + are present in 2 L of H 2 SO 4 solution, if the concentration is 0.1 mol/L ? H 2 SO 4 --> 2 H + + SO 4 2- n(H 2 SO 4 ) = c x V n(H 2 SO 4 ) = 0.1 x 2 = 0.2 mol n(H + ) = 2 x n(H 2 SO 4 ) n(H + ) = 2 x 0.2 = 0.4 mol

39. Mass concentration (  ) mass of substance in 1 L of solution  = m solute / V unit: g / l

40. What is the mass concentration of a solution, which contains 7.0 g of KCl in 750 ml ?  m KCl / V  g/l

41. How many grams of AgNO 3 do we need to prepare 7 L of solution of mass concentration 0.5 g/L ? m AgNO3 =  x V m AgNO3  x  g

42.  = c x M  c =  / M Why ?  = m solute / V m solute = n x M  = n x M V c = n / V Interconverting substance concentration ( c ) and mass concentration ( m )

43. What is the mass concentration of the NaOH solution, if the substance concentration is 0.5 mol/L ? M = 40 g/mol  c x M  x  g/L

44. Calculate the substance concentration of the NaCl solution, if the mass concentration is 10 g/L ? M = 58.5 g/mol c  M c  mol/L

45. m  c x V x M Why? m = n x M n = c x V Calculation of the mass necessary for making a solution of given substance concentration

46. How many grams of Na 2 SO 4 (M = 142 g/mol) are necessary for 1 500 mL of solution (c= 0.1 mol/L) ? m  c x V x M m  x  x  21.3 g

47. Mass fraction ( w ) ratio between a mass of the dissolved substance (m solute ) and the total solution mass (m solution ) w = m solute / m solution unit: - Mass percentage: mass fraction x 100 % ( i.e. grams of substance in 100 g of solution ) e.g. w = 0.15 15 % solution

48. Volume fraction (  ) analogy of mass fraction  = V solute / V solution unit: - Volume percentage: volume fraction x 100 % The use: ETHANOL in alcoholic drinks !!! e.g. alc. 11.5 % vol.

49. "Percent concentration" - summary "concentration 10 %" can be confusing ! it‘s better to specify it: % w/w percent by mass (mass percentage) % v/v percent by volume (volume percentage) % w/v percent by mass over volume (mass-volume percentage) w... weight v... volume

50. Calculate %(w/w) concentration of a solution prepared from 15 g NaOH and 80 mL of water. w = m NaOH / m solution w   ( i.e. 15.8 % ) note: density of water 1.0 g/cm 3

51. How many grams of AgNO 3 are necessary to prepare 700 g of 2 % (w/w) solution ? w = 0.02 w = m / m solution m = w x m solution m  x  g

52.  x w M x c M  we need to know the density of the solution! density must be in g/dm 3 w = c = Interconverting substance concentration (c) and mass fraction (w)

53. What is the substance concentration of 10 %(w/w) solution of Na 2 CO 3 ? Density of this solution is 1.1 g/cm 3. M = 106 g/mol  x w M 1 100 x 0.1 106 c = = 1.04 mol/L

54. What is the substance concentration of 10 %(w/w) solution of Na 2 CO 3 ? Density of this solution is 1.1 g/cm 3. M = 106 g/mol other way of calculation: 10 %(w/w) --> 10 g Na 2 CO 3 in 100 g of solution 10 g Na 2 CO 3 is 10 / 106 = 0.09434 mol the volume of 100 g of solution is 100 / 1.1 = 90.91 mL substance conc.: c = n / V c = 0.09434 / 0.09091 =1.04 mol/L

55. Mixing of solutions ( dilution = particular case of mixing ) These rules must be applied: 1) the mass is the sum of masses of the components: m 1 + m 2 = m ( conservation of the mass, NOT THE VOLUME !!! ) 2) the mass of the solute present in the new solution formed by mixing is the sum of masses of the solute dissolved in the components: m 1 w 1 + m 2 w 2 = mw Equation: m 1 w 1 + m 2 w 2 = (m 1 + m 2 ) w

56. Calculate the concentration of a solution %(w/w) prepared by mixing 300 g 70 % and 500 g 20 % H 2 SO 4 m 1 w 1 + m 2 w 2 = (m 1 + m 2 ) w 300 x 0.7 + 500 x 0.2 = ( 300 + 500 ) w 310 = 800 w w  that is 38.75 % )

57. "Cross rule" = an easy way to do such calculations a %( c - b ) mass portions c % b % ( a - c ) mass portions a, b … original concentrations c … new concentration

58. How many g of 60 %(w/w) HNO 3 do you need to prepare 1 200 g of 10 %(w/w) solution ? dilution: 60 % HNO 3 + water ( 0 % ) 60 % 10 - 0 = 10 portions 10 % the sum is 60 port. 0 % 60 - 10 = 50 portions 1 200 g ….. 60 portions --> 1 portion: 1 200 / 60 = 20 g We need: 10 portions of 60 % HNO 3 …. i.e. 10 x 20 = 200 g of 60 % HNO 3 50 portions of water …. i.e. 50 x 20 = 1000 g of water

59. Chemical equations a A + b B --> c C + d D Dalton’s law: ratio of amounts of substance of reactants can be expressed in small whole numbers n (A)a n (B)b a, b … stoichiometric coeficients = = stoichiometric factor

60. a) H 2 SO 4 + 2 NaOH --> Na 2 SO 4 + 2 H 2 O n(H 2 SO 4 ) / n(NaOH) = 1 / 2 b) 2 AgNO 3 + K 2 CrO 4 --> 2 KNO 3 + Ag 2 CrO 4 n(AgNO 3 ) / n(K 2 CrO 4 ) = 2 / 1 c) 2 KMnO 4 + 5 (COOH) 2 + 3 H 2 SO 4 --> K 2 SO 4 + 2 MnSO 4 + 10 CO 2 + 8 H 2 O n(KMnO 4 ) / n( (COOH) 2 ) = 2 / 5 MANGANOMETRY !

61. How many moles of NaOH are necessary for a full neutralization of 1.5 mol of oxalic acid? 2 NaOH + (COOH) 2 --> (COONa) 2 + 2 H 2 O n(NaOH) / n( (COOH) 2 ) = 2 / 1 n(NaOH) = 2 x n( (COOH) 2 ) n(NaOH) = 2 x 1.5 = 3 mol

62. How many g of HCl are necessary for a full neutralization of 10 g Na 2 CO 3 ? M(HCl) = 36.5 g/mol M(Na 2 CO 3 ) = 106 g/mol 2 HCl + Na 2 CO 3 --> 2 NaCl + CO 2 + H 2 O n(HCl) / n(Na 2 CO 3 ) = 2 / 1 [ m(HCl) / M(HCl) ] / [ m(Na 2 CO 3 ) / M(Na 2 CO 3 ) ] = 2 m(HCl) = 2 x [ m(Na 2 CO 3 ) / M(Na 2 CO 3 ) ] x M(HCl) m(HCl) = 2 x ( 10 / 106 ) x 36.5 = 6.89 g

63. How many g of HCl are necessary for a full neutralization of 10 g Na 2 CO 3 ? M(HCl) = 36.5 g/mol M(Na 2 CO 3 ) = 106 g/mol 2 HCl + Na 2 CO 3 --> 2 NaCl + CO 2 + H 2 O 2 x 36.5 g HCl ………………… 106 g Na 2 CO 3 ? g HCl ………………… 10 g Na 2 CO 3 ? 10 2 x 36.5 106 ? = 6.89 g =

64. How many grams of gold are in a 160 g piece marked 18 carat ? w = 18 / 24 = 0.750 w = m pure Au / m alloy m pure Au = m alloy x w m pure Au = 160 x 0.750 = 120 g note: "purity" of gold: CARAT ( pure gold: 24 carat ) THOUSANDS ( pure gold: 1000 / 1000 ) 18-carat gold = 18 / 24 = 750 / 1000 = 0.750 i.e. 75% gold