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Chemical calculations

Chemical calculations. When solving numerical problems, always ask yourself whether your answer makes sense!. How much of NaCl was excreted to urine within 24 hours ?

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Chemical calculations

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  1. Chemicalcalculations

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

  3. How much ofNaClwasexcreted to urine within 24 hours? 20 ml ofanaverage sample wastitrated by directargentometricmethodandtheconsumptionofthe standard solutionof 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 of our students: 1 500 kg i.e. 1.5 ton !!!

  4. SI base units Système International d'Unités All other units can be derived from SEVEN base units.

  5. SI prefixes (metricprefixes) for making larger for making smaller 10-1 deci d 101 deca da 10-2 centi c 102 hecto h 10-3 mili m 103 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

  6. Convertingunits How it works? 0.750 L = mL 0.750 L = 750 mL 5 mL = 0.005 L 5 mL = L 20 mL = mL 20 mL = 0.02 mL

  7. Unitsofvolume 1 L = 1 dm3 1 mL = 1 cm3 1 mL = 1 mm3

  8. Converting units Erythrocytevolume: 85 fL ( = 85 mm3 ) 85 fL = 85 x 10-15 L = 85 x 10-15 dm3 = = 85 x 10-18 m3 = 85 mm3 Erythrocytenumber: 5 x 106 / mm3 = 5 x 106 / mL = 5 x 1012 / L

  9. Rounding off the results Numbersobtained by measurement are alwaysINEXACT ! "exact" numbers - in mathematics: 10 = 10.000….. "measuredvalues" e.g. by measuringthevolume 10 mL - itis NOT exactly 10.00000… mL Uncertainties ("errors") alwaysexist in measuredquantities!

  10. Significant figures Answer: YES ! 4.0 g 2 significantfigures 4.00 g 3 significantfigures  4.00 g is "more precise" than 4.0 g E.g. 4.003 4 significantfigures 6.023 x 1023 4 significantfigures 0.0012 2 significantfigures 5000 ? 1,2,3 or 4 significantfigures Is there any difference between 4.0 g and 4.00 g ?

  11. Whenroundingofftheresults, youhave to consider significantfiguresofgivennumbers ! Theprecisionoftheresultis limited by theprecision ofthemeasurements ! Rules 1)Multiplication and division result – mustbewiththesamenumberofsignificantfigures as themeasurementwiththe FEWEST signif. fig. e.g. 6.221 cm x5.2 cm = 32.3492 cm2 --> roundoff to 32 cm2

  12. 2)Addition and subtraction result – cannothave more digits to therightofthedecimal point thananyoftheoriginalnumbers e.g. 20.4 1 decimal place 1.322 3 decimalplaces 83 ZEROdecimalplaces 104.722 roundoff to 105 Roundingoff:digits 5,6,7,8,9 --> round up digits 0,1,2,3,4 --> rounddown

  13. CONCLUSION: Yourcalculatorcangivetheresultlikethis: 100 / 7 = 14.28571429 !!! DON‘T give as a resultofthecalculation a number with 10 digits, whichshowsyourcalculator, round itoff to the "reasonablenumber" ofdecimalplaces calculationwith more steps – roundoff THE FINAL RESULT

  14. Youhave to befamiliarwithyourcalculator ! E.g. 1) 103 = 1000 !!! 10 EXP 3 = 10 x 103 = 10000 !!! 2) 50 2 x 5 50 : 2 x 5 = 125 !!! = 5 !!!

  15. Uncertainties of quantitative methods PRECISION= howcloselyindividualmeasurements agreewithoneanother ACCURACY= howcloselymeasurementsagree withthecorrect ("true") value "true" valuemeasuredvalues

  16. Precision and accuracy - shooting on the target a) goodprecision goodaccuracy b) goodprecision pooraccuracy c) poorprecision, but in average goodaccuracy d) poorprecision pooraccuracy

  17. Amountof substance (n) - base SI quantity • is in a closerelation to the NUMBER of ELEMENTARY ENTITIES (atoms, molecules, electrons, …) unit: MOLE ( 1 mol ) 1 mol = as many objects as thenumberofatoms in 12 g ofthecarbon isotope 126C

  18. 1 mol = 6.023 x 1023elementaryentities = AVOGADRO’snumber NA NA = 6.0221367 x 1023 /mol number of entities = n x NA analogy: "countingunits" 1 pair = 2 1 dozen = 121 gross = 144 1 mol = 6.023 x 1023

  19. Molarmass(M) - themassof 1 mol of a substance - unit: g / mol - canbecalculatedwiththe use ofrelativeatomicmasses(PERIODIC TABLE) Relativeatomicmass Ar Relativemolecularmass Mr atomicmass unit = 1/12 ofthemassof atom 126C u = 1.6605 x 10-24 g Ar = matom / u Ar ( 126C ) = 12.00 Ar ( H ) = 1.008 Mr = mmolecule / u relativemolecularmass - no real unit in biochemistry: Dalton ( Da ) e.g. protein 55 kDa

  20. Density (r) - relationbetweenmass and volume - theamountofmass in a unit volumeof substance r = m / V Mind theunits ! SI unit: kg/m3 otherunits: g/cm3 kg/dm3 note: cm3 = mL dm3 = L

  21. Solutions solute = the substance whichdissolves solvent =theliquidwhichdoesthedissolving  A solutionisprepared by dissolving a solute in a solvent.

  22. Solutioncomposition – "Concentration" "numberofdifferentwayshow to express concentration" Molarconcentration(molarity) c mol/L Massconcentrationm g/L Massfraction w Masspercentage % % w/w Volumefraction Volumepercentage% v/v

  23. 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

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

  25. Mass fraction ( w ) ratio between a massofthedissolved substance (msolute) and thetotalsolutionmass (msolution) w = msolute / msolution unit: - Masspercentage:massfraction x 100 % ( i.e. gramsof substance in 100 g ofsolution ) e.g. w = 0.15 15 % solution

  26. 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.

  27. "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

  28. Mixing of solutions ( dilution = particular case of mixing ) These rules must be applied: 1) the mass is the sum of masses of the components: m1 + m2 = 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: m1w1 + m2w2 = mw Equation: m1w1 + m2w2 = (m1 + m2) w

  29. "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

  30. Exercises

  31. 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

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

  33. 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

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

  35. 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 )

  36. 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 )

  37. 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

  38. 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

  39. 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

  40. 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

  41. 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

  42. 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

  43. 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

  44. 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

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

  46. 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

  47. 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

  48. 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

  49. 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.

  50. 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

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