1. Basic chemical calculations (textbook: 35 - 42)
Basic chemical calculations (textbook: )

2. When solving numerical problems,
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 = mL/day, M(NaCl) = 58.5 g/mol
Correct answer: 15 g
ANSWERS IN THE TESTS: kg
i.e. 1.5 ton !!!

4. Base physical quantities (SI)
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

5. SI prefixes (metric prefixes)
SI prefixes (metric prefixes)
deci d deca da
centi c hecto h
mili m kilo k
micro m mega M
nano n giga G
pico p tera T
femto f peta P
atto a exa E
zepto z zetta Z
yocto y yotta Y

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

7. Converting units Erythrocyte volume: 85 fL ( = 85 mm3 )
Converting units
Erythrocyte volume: 85 fL ( = 85 mm3 )
85 fL = 85 x L = 85 x dm3 =
= 85 x m3 = 85 mm3
Erythrocyte number: 5 x 106 / mm3 = 5 x 106 / mL
= 5 x 1012 / L

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

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

10. You have to be familiar with your calculator !
You have to be familiar with your calculator !
E.g. 1) = !!!
10 EXP 3 = 10 x 103 = !!! 2)
2 x 5
50 : 2 x 5 = !!!
= !!!

11. Uncertainties of quantitative methods
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

12. Precision and accuracy - shooting on the target
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

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

14.
Examples water kg/m g/cm kg/dm Au kg/m g/cm kg/dm3

15.
Calculate the density of 90% H2SO4, if the mass of 200 mL of the solution is 363 g. r = m / V r = 363 / 200 = g/cm3

16.
Often we know the volume and need to calculate the mass and vice versa! m = V x r V = m / r

17. What is the mass of the solution of KOH,
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 = g
units ! L = mL ( cm3 )

18. What is the volume of the solution of HNO3 if
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 = cm3 ( mL )

19. Amount of substance ( n )
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

20. 1 mol = 6.023 x 1023 elementary entities
1 mol = x 1023 elementary entities
= AVOGADRO’s number NA
NA = x 1023 /mol
number of entities = n x NA
analogy: „counting units“
1 pair = dozen = gross = 144
1 mol = x 1023

21.
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 x 1023 = x 1024

22. Calculate the number of protons released
Calculate the number of protons released
during complete dissociation of 2 mmol H3PO4.
H3PO > 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 x = 3.61 x 1018

23. Calculate the number of C atoms in 0.350 mol of glucose.
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 x = 1.26 x 1024

24.
Calculate the amount of substance: 2.71 x 1024 molecules of NaCl n = number of NaCl / NA n = 2.71 x 1024 / ( x 1023 ) = 4.5 mol

25. Molar mass ( M ) - the mass of 1 mol of a substance - unit: g / 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
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 = x g
Ar = matom / u Mr = mmolecule / u
Ar ( 126C ) = Ar ( H ) = 1.008
relative molecular mass - no „true“ unit in biochemistry: Dalton ( Da )
e.g. protein 55 kDa

26. What is a molar mass of glucose ?
What is a molar mass of glucose ?
C6H12O6
Ar (C) = 12,0
Ar (H) = 1,0
Ar (O) = 16,0
M = 6 x x x 16 = 180 g/mol

27.
Often we know the mass and need to calculate the amount of substance and vice versa! n = m / M m = n x M

28.
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

29.
Calculate the mass of mol of calcium nitrate. M = g/mol m = n x M m = x = 71.1 g

30. How many molecules of glucose are in 5.23 g C6H12O6 ? M = 180 g/mol
How many molecules of glucose are in
5.23 g C6H12O6 ? M = 180 g/mol
n = m / M
n = 5.23 / 180 = mol
number of molecules = n x NA
number of molecules = x x 1023
= 1.75 x 1022

31. Solution composition - „Concentration“
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.

32. Solution composition - „Concentration“
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

33. Molar concentration ( c ) (substance concentration, MOLARITY)
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

34.
Calculate molar concentration of NaCl solution, if 250 mL contain 0.1 mol NaCl. c = n / V c = 0.1 / = 0.4 mol/L units ! 250 mL = L

35.
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 ) = mol/L

36.
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

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

38.
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

39. Interconverting substance concentration ( c )
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

40.
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

41. 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

42.
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 = % solution

43. Volume fraction ( j ) analogy of mass fraction j = Vsolute / Vsolution
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 % vol.

44. „Percent concentration“ - summary
„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

45.
Calculate %(w/w) concentration of a solution prepared from 15 g NaOH and 80 mL of water. w = mNaOH / msolution w = 15 / ( ) = ( i.e % ) note: density of water 1.0 g/cm3

46. Interconverting substance concentration (c) and mass fraction (w)
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 =

47.
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 =
= mol/L

48. ( 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

49. Calculate the concentration of a solution %(w/w)
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 x 0.2 = ( ) w
310 = 800 w
w = 310 / 800 = ( that is % )

50. = 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

51. a) H2SO4 + 2 NaOH --> Na2SO4 + 2 H2O n(H2SO4) / n(NaOH) = 1 / 2
a) H2SO NaOH --> Na2SO H2O
n(H2SO4) / n(NaOH) = 1 / 2
b) 2 AgNO3 + K2CrO4 --> 2 KNO3 + Ag2CrO n(AgNO3) / n(K2CrO4) = 2 / 1
c) 2 KMnO (COOH) H2SO4 --> K2SO4 + 2 MnSO CO2 + 8 H2O
n(KMnO4) / n( (COOH)2 ) = 2 / 5
MANGANOMETRY !

52. How many moles of NaOH are necessary for
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

53. 2 HCl + Na2CO3 --> 2 NaCl + CO2 + H2O
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

54. How many grams of gold are in a 160 g piece marked 18 carat ?
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 = 120 g
note: „purity“ of gold: CARAT ( pure gold: 24 carat )
THOUSANDS ( pure gold: 1000 / 1000 )
18-carat gold = 18 / 24 = 750 / 1000 = i.e. 75% gold