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Basic chemical calculations

Basic chemical calculations. When solving numerical problems, always ask yourself whether your answer makes sense !!!. How many grams of NaCl were excreted in urine in 24 hours?

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Basic chemical calculations

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  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 AgNO3 (c = 100 mmol/L) was 34.2 mL. Volume of urine = 1 500 mL/day, M(NaCl) = 58.5 g/mol Correct answer: 15 g ANSWERS IN THE TESTS: 1 500 kg i.e. 1.5 ton !!!

  4. SI prefixes (metric prefixes) 10-1 deci d 101 deca da 10-2 centi c 102 hecto h 10-3 mili m103 kilo k 10-6 micro m 106 mega M 10-9 nano n 109 giga G 10-12 pico p 1012 tera T 10-15 femto f 1015 peta P 10-18 atto a 1018 exa E 10-21 zepto z 1021 zetta Z 10-24 yocto y 1024 yotta Y

  5. Converting units 5 mL = 0.005 L 0.750 L = 750 mL 1 L = 1 dm3 1 mL = 1 cm3 1 mL= 1 mm3 0.1 mol/L = 100 mmol/L 50 mg = 0.050 g

  6. Converting units Erythrocyte volume: 85 fL ( = 85 mm3 ) 85 fL = 85 x 10-15 L = 85 x 10-15 dm3 = = 85 x 10-18 m3 = 85 mm3 Erythrocyte number: 5 x 106 / mm3 = 5 x 106 / mL = 5 x 1012 / 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 Answer: YES ! 4.0 g 2 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 1023 4 significant figures 0.0012 2 significant figures 5000 ? 1,2,3 or 4 significant figures Is there any difference between 4.0 g and 4.00 g ?

  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 x5.2 cm = 32.3492 cm2 --> round off to 32 cm2

  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. 20.4 1 decimal place 1.322 3 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) 103 = 1000 !!! 10 EXP 3 = 10 x 103 = 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 ( r ) - relation between mass and volume - the amount of mass in a unit volume of substance r = m / V Mind the units ! SI unit: kg/m3 other units: g/cm3 kg/dm3 note: cm3 = mL dm3 = L

  16. Exampleswater 1 000 kg/m3 1 g/cm31 kg/dm3Au 19 300 kg/m3 19.3 g/cm319.3 kg/dm3

  17. Calculate the density of 90% H2SO4, if the mass of 200 mL of the solution is 363 g. r = m / V r = 363 / 200 = 1.815 g/cm3

  18. Often we know the volume and we need to calculate the massand vice versa!m = V xr V = m / r

  19. What is the mass of the solution of KOH, if the volume is 2.5 L and density 1.29 g/cm3 ? m = V xr m = 2500x 1.29 = 3 225 g units ! 2.5 L = 2 500 mL ( cm3 )

  20. What is the volume of the solution of HNO3 if the mass is 150 g and the density 1.46 g/cm3 ? V = m / r V = 150 / 1.46 = 102.7 cm3 ( 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 126C

  22. 1 mol = 6.023 x 1023 elementary entities = AVOGADRO’snumber NA NA = 6.0221367 x 1023 /mol number of entities = n x NA analogy: „counting units“ 1 pair = 2 1 dozen = 121 gross = 144 1 mol = 6.023 x 1023

  23. Calculate the number of elementary units present in 2.5 mol cations Ca2+. number of Ca2+ = n x NA number of Ca2+ = 2.5 x 6.023 x 1023 = 1.506 x 1024

  24. Calculate the number of protons released during complete dissociation of 2 mmol H3PO4. H3PO4 --> 3 H+ + PO4 3- n(H+) = 3 x n(H3PO4) n(H+) = 6 mmol number of H+ = n(H+)x NA number of H+ = 6 x10-6x 6.023 x 1023 = 3.61 x 1018

  25. Calculate the number of C atoms in 0.350 mol of glucose. C6H12O6 n(C) = 6 x nglukosa n(C) = 2.1 mol number of C = n(C)x NA number of C = 2.1 x 6.023 x 1023 = 1.26 x 1024

  26. Calculate the amount of substance: 2.71 x 1024 molecules of NaCl n = number of NaCl / NA n = 2.71 x 1024 / ( 6.023 x 1023 )= 4.5 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 Ar Relative molecular mass Mrexpressed in atomic mass units note: atomic mass unit = 1/12 of the mass of atom 126C u = 1.6605 x 10-24 g Ar = matom / u Mr = mmolecule / u Ar ( 126C ) = 12.00 Ar ( 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 ? C6H12O6 Ar (C) = 12.0 Ar (H) = 1.0 Ar (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 / Mm = 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 = 100 / 58.5 = 1.71 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 C6H12O6 ? M = 180 g/mol n = m / M n = 5.23 / 180 = 0.029056 mol number of molecules = n x NA number of molecules = 0.029056 x 6.023 x 1023 = 1.75 x 1022

  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 m 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= 0.1 / 0.250 = 0.4 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 H2SO4 solution, if the concentration is 0.1 mol/L ? H2SO4 --> 2 H+ + SO4 2- n(H2SO4) = c x V n(H2SO4) = 0.1 x 2 = 0.2 mol n(H+) = 2 x n(H2SO4) n(H+) = 2 x 0.2 = 0.4 mol

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

  40. What is the mass concentration of a solution, which contains 7.0 g of KCl in 750 ml ? m = mKCl / V m = 7 / 0.75 = 9.33 g/l

  41. How many grams of AgNO3 do we need to prepare 7 L of solution of mass concentration 0.5 g/L ? mAgNO3 = m x V mAgNO3 = 7 x 0.5 = 3.5 g

  42. Interconverting substance concentration ( c ) and mass concentration ( m ) m = c x Mc = m / MWhy ? m = msolute / V msolute = n x Mm = n x M V c = n / V

  43. What is the mass concentration of the NaOH solution, if the substance concentration is 0.5 mol/L ? M = 40 g/mol m = c x M m = 0.5 x 40 = 20 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 / M c = 10 / 58.5 = 0.17 mol/L

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

  46. How many grams of Na2SO4 (M = 142 g/mol) are necessary for 1 500 mL of solution (c= 0.1 mol/L) ? m = c x V x M m = 0.1 x 1.5 x 142 = 21.3 g

  47. Mass fraction ( w ) ratio between a mass of the dissolved substance (msolute) and the total solution mass (msolution) w = msolute / msolution 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 ( j ) analogy of mass fraction j = Vsolute / Vsolution 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 = mNaOH / msolution w = 15 / ( 80 +15 ) = 0.158 ( i.e. 15.8 % ) note: density of water 1.0 g/cm3

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