Basic chemical calculations (textbook: 35 - 42) 5.10.2017 Basic chemical calculations (textbook: 35 - 42)

When solving numerical problems, 5.10.2017 When solving numerical problems, always ask yourself whether your answer makes sense !!!

5.10.2017 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 !!!

Base physical quantities (SI) 5.10.2017 Base physical quantities (SI) Quantity Symbol Unit Symbol 1) MASS m kilogram kg 2) LENGTH l metre m 3) TIME t second s 4) EL. CURRENT I amper A 5) TEMPERATURE T kelvin K 6) LUMINOUS INTENSITY Iv candela cd 7) AMOUNT of SUBSTANCE n mole mol

SI prefixes (metric prefixes) 5.10.2017 SI prefixes (metric prefixes) 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 5 mL = 0.005 L 0.750 L = 750 mL 5.10.2017 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

Converting units Erythrocyte volume: 85 fL ( = 85 mm3 ) 5.10.2017 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 5.10.2017 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!

Your calculator can give the result: 100 / 7 = 14.28571429 !!! 5.10.2017 CONCLUSION: Your calculator can give the result: 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

You have to be familiar with your calculator ! 5.10.2017 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 5.10.2017 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

Precision and accuracy - shooting on the target 5.10.2017 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

Density ( r ) r = m / V Mind the units ! SI unit: kg/m3 5.10.2017 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

5.10.2017 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

5.10.2017 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

5.10.2017 Often we know the volume and need to calculate the mass and vice versa! m = V x r V = m / r

What is the mass of the solution of KOH, 5.10.2017 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 5.10.2017 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 ) 5.10.2017 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 5.10.2017 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

5.10.2017 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 5.10.2017 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. 5.10.2017 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

5.10.2017 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 5.10.2017 Molar mass ( M ) - the mass of 1 mol of a substance - unit: g / mol can be calculated with the use of relative atomic masses 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 „true“ unit in biochemistry: Dalton ( Da ) e.g. protein 55 kDa

What is a molar mass of glucose ? 5.10.2017 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

5.10.2017 Often we know the mass and need to calculate the amount of substance and vice versa! n = m / M m = n x M

5.10.2017 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

5.10.2017 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 5.10.2017 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“ 5.10.2017 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“ 5.10.2017 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) 5.10.2017 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

5.10.2017 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

5.10.2017 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

5.10.2017 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 ) 5.10.2017 Mass concentration ( m ) mass of substance in 1 L of solution m = msolute / V unit: g / l

5.10.2017 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

Interconverting substance concentration ( c ) 5.10.2017 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

5.10.2017 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

m = c x V x M Why? m = n x M n = c x V 5.10.2017 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

5.10.2017 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 5.10.2017 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 5.10.2017 „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

5.10.2017 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

Interconverting substance concentration (c) and mass fraction (w) 5.10.2017 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! Why? c = n / V w = msolute / msolvent n = msolute / M r = msolvent / V density must be in g/dm3 c = w =

5.10.2017 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

( dilution = particular case of mixing ) 5.10.2017 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) 5.10.2017 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 % )

= stoichiometric factor 5.10.2017 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 5.10.2017 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 5.10.2017 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 5.10.2017 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 grams of gold are in a 160 g piece marked 18 carat ? 5.10.2017 How many grams of gold are in a 160 g piece marked 18 carat ? w = 18 / 24 = 0.750 w = mpure Au / malloy mpure Au = malloy x w mpure 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