Chemical calculations

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

i.e. 1.5 ton !!! Answers of our students: 1 500 kg How much of NaCl was excreted to urine within 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 of our students: 1 500 kg i.e. 1.5 ton !!!

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

SI prefixes (metric prefixes) for making smaller for making larger 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

Converting units 0.750 L = mL 0.750 L = 750 mL 5 mL = L 5 mL = 0.005 L How it works? 0.750 L = mL 0.750 L = 750 mL 5 mL = L 5 mL = 0.005 L 20 mL = mL 20 mL = 0.02 mL

Units of volume 1 L = 1 dm3 1 mL = 1 cm3 1 mL = 1 mm3

Converting units Erythrocyte volume: 85 fL ( = 85 mm3 ) 24.8.2018 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

Rounding off the results 24.8.2018 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!

Significant figures Is there any difference between 4.0 g and 4.00 g ? 24.8.2018 Significant figures Is there any difference between 4.0 g and 4.00 g ? 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

Rules e.g. 6.221 cm x 5.2 cm = 32.3492 cm2 --> round off to 32 cm2 24.8.2018 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 cm2 --> round off to 32 cm2

result – cannot have more digits to the right of the decimal point 24.8.2018 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

Your calculator can give the result like this: 24.8.2018 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 THE FINAL RESULT

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 !!!

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

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

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

1 mol = 6.023 x 1023 elementary entities = AVOGADRO’s number NA NA = 6.0221367 x 1023 /mol number of entities = n x NA analogy: "counting units" 1 pair = 2 1 dozen = 12 1 gross = 144 1 mol = 6.023 x 1023

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 Mr atomic mass unit = 1/12 of the mass of atom 126C u = 1.6605 x 10-24 g Ar = matom / u Ar ( 126C ) = 12.00 Ar ( H ) = 1.008 Mr = mmolecule / u relative molecular mass - no real unit in biochemistry: Dalton ( Da ) e.g. protein 55 kDa

Density (r) r = m / V Mind the units ! SI unit: kg/m3 - 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

Solutions solute = the substance which dissolves solvent = the liquid which does the dissolving  A solution is prepared by dissolving a solute in a solvent.

Solution composition – "Concentration" "number of different ways how 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

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

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

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

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.

"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

( dilution = particular case of mixing ) 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

"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

Exercises

Density ( r ) r = m / V Mind the units ! SI unit: kg/m3 - 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

24.8.2018 Examples water 1 000 kg/m3 1 g/cm3 1 kg/dm3 Au 19 300 kg/m3 19.3 g/cm3 19.3 kg/dm3

24.8.2018 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

Often we know the volume and we need to calculate the mass and vice versa! m = V x r V = m / r

What is the mass of the solution of KOH, 24.8.2018 What is the mass of the solution of KOH, if the volume is 2.5 L and density 1.29 g/cm3 ? m = V x r m = 2500 x 1.29 = 3 225 g units ! 2.5 L = 2 500 mL ( cm3 )

What is the volume of the solution of HNO3 if 24.8.2018 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 )

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

1 mol = 6.023 x 1023 elementary entities = AVOGADRO’s number NA NA = 6.0221367 x 1023 /mol number of entities = n x NA analogy: "counting units" 1 pair = 2 1 dozen = 12 1 gross = 144 1 mol = 6.023 x 1023

24.8.2018 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

Calculate the number of protons released 24.8.2018 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 x 10-6 x 6.023 x 1023 = 3.61 x 1018

Calculate the number of C atoms in 0.350 mol of glucose. 24.8.2018 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

24.8.2018 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

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 Mr expressed 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

What is a molar mass of glucose ? 24.8.2018 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

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

24.8.2018 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

24.8.2018 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

How many molecules of glucose are in 5.23 g C6H12O6 ? M = 180 g/mol 24.8.2018 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

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.

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

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

24.8.2018 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

24.8.2018 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

24.8.2018 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

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

24.8.2018 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

24.8.2018 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

Interconverting substance concentration ( c ) and mass concentration ( m ) m = c x M c = m / M Why ? m = msolute / V msolute = n x M m = n x M V c = n / V

24.8.2018 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

24.8.2018 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

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 m = c x V x M Why? m = n x M n = c x V

24.8.2018 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

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

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.

"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

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

24.8.2018 How many grams of AgNO3 are necessary to prepare 700 g of 2 % (w/w) solution ? w = 0.02 w = m / msolution m = w x msolution m = 0.02 x 700 = 14 g

Interconverting substance concentration (c) and mass fraction (w) r x w M x c M r we need to know the density of the solution! density must be in g/dm3 c = w =

What is the substance concentration of 10 %(w/w) solution of Na2CO3 What is the substance concentration of 10 %(w/w) solution of Na2CO3 ? Density of this solution is 1.1 g/cm3. M = 106 g/mol r x w M 1 100 x 0.1 106 c = c = = 1.04 mol/L

other way of calculation: What is the substance concentration of 10 %(w/w) solution of Na2CO3 ? Density of this solution is 1.1 g/cm3. M = 106 g/mol other way of calculation: 10 %(w/w) --> 10 g Na2CO3 in 100 g of solution 10 g Na2CO3 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

( dilution = particular case of mixing ) 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

Calculate the concentration of a solution %(w/w) 24.8.2018 Calculate the concentration of a solution %(w/w) prepared by mixing 300 g 70 % and 500 g 20 % H2SO4 m1w1 + m2w2 = (m1 + m2) w 300 x 0.7 + 500 x 0.2 = ( 300 + 500 ) w 310 = 800 w w = 310 / 800 = 0.3875 ( that is 38.75 % )

"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

dilution: 60 % HNO3 + water ( 0 % ) 60 % 10 - 0 = 10 portions How many g of 60 %(w/w) HNO3 do you need to prepare 1 200 g of 10 %(w/w) solution ? dilution: 60 % HNO3 + 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 % HNO3 …. i.e. 10 x 20 = 200 g of 60 % HNO3 50 portions of water …. i.e. 50 x 20 = 1000 g of water

= stoichiometric factor 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

a) H2SO4 + 2 NaOH --> Na2SO4 + 2 H2O n(H2SO4) / n(NaOH) = 1 / 2 b) 2 AgNO3 + K2CrO4 --> 2 KNO3 + Ag2CrO4 n(AgNO3) / n(K2CrO4) = 2 / 1 c) 2 KMnO4 + 5 (COOH)2 + 3 H2SO4 --> K2SO4 + 2 MnSO4 + 10 CO2 + 8 H2O n(KMnO4) / n( (COOH)2 ) = 2 / 5 MANGANOMETRY !

How many moles of NaOH are necessary for 24.8.2018 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 H2O n(NaOH) / n( (COOH)2 ) = 2 / 1 n(NaOH) = 2 x n( (COOH)2 ) n(NaOH) = 2 x 1.5 = 3 mol

2 HCl + Na2CO3 --> 2 NaCl + CO2 + H2O 24.8.2018 How many g of HCl are necessary for a full neutralization of 10 g Na2CO3 ? M(HCl) = 36.5 g/mol M(Na2CO3) = 106 g/mol 2 HCl + Na2CO3 --> 2 NaCl + CO2 + H2O n(HCl) / n(Na2CO3) = 2 / 1 [ m(HCl) / M(HCl) ] / [ m(Na2CO3) / M(Na2CO3) ] = 2 m(HCl) = 2 x [ m(Na2CO3) / M(Na2CO3) ] x M(HCl) m(HCl) = 2 x ( 10 / 106 ) x 36.5 = 6.89 g

How many g of HCl are necessary for a full neutralization of 24.8.2018 How many g of HCl are necessary for a full neutralization of 10 g Na2CO3 ? M(HCl) = 36.5 g/mol M(Na2CO3) = 106 g/mol 2 HCl + Na2CO3 --> 2 NaCl + CO2 + H2O 2 x 36.5 g HCl ………………… 106 g Na2CO3 ? g HCl ………………… 10 g Na2CO3 ? 10 2 x 36.5 106 ? = 6.89 g =