1. Chemical calculations I
Vladimíra Kvasnicová

2. Expression of concentration
molar concentration
percent concentration
conversion of units
Osmotic pressure, osmolarity
Dilution of solutions
Calculation of pH
strong and weak acids and bases
buffers
Calculation in a spectrophotometry
Calculation in a volumetric analysis

3. Important terms
solute = a substance dissolved in a solvent in forming a solution
solvent = a liquid that dissolves another substance or substances to form a solution
solution = a homogeneous mixture of a liquid (the solvent) with a gas or solid (the solute)
concentration = the quantity of dissolved substance per unit quantity of solution or solvent

4. Important terms
density () = the mass of a substance per unit of volume (kg.m-3 or g.cm-3)  = m/V
mass m = n x MW (in grams)
amount of substance (n) = a measure of the number of entities present in a substance (in moles)
Avogadro constant (NA) = the number of entities in one mole of a substance (NA = 6.022x1023)
molar weight (MW) = mass of one mole of a substance in grams

5. Important terms
relative molecular mass (Mr) = the ratio of the average mass per molecule of the naturally occurring form of an element or compound to 1/12 of the mass of 12C atom
Mr = sum of relative atomic masses (Ar) of all atoms that comprise a molecule
MW (in grams) = Mr
dilution = process of preparing less concentrated solutions from a solution of greater concentration

6. Expression of concentration
Molarity (c) (mol x l-1 = mol x dm-3 = M ) = number of moles per liter of a solution
c = n / V 
number of moles / 1000 mL of solution
DIRRECT PROPORTIONALITY

7. ! DIRRECT PROPORTIONALITY !
1M NaOH MW = 40g /mole => 1M solution of NaOH = 40g of NaOH / 1L of solution 0,1M solution of NaOH = 4g of NaOH / 1L of solution
Preparation of 500 mL of 0,1M NaOH:
0,1M solution of NaOH = 4g of NaOH / 1 L of solution
2g of NaOH / 0.5 L of solution
! DIRRECT PROPORTIONALITY !

8. Exercises 1) 17,4g NaCl / 300mL, MW = 58g/mol, C = ? [1M]
2) Solution of glycine, C = 3mM, V = 100ml.
? mg of glycine are found in the solution?
[22,5mg]
3) Solution of CaCl2, C = 0,1M.
Calculate volume of the sol. containing 4 mmol of Cl-.
[20ml]

9. Normality (N)
= concentration in terms of equivalent weights of substance (reflect the number of combining or replaceable units).
It is not in common use!
1M HCl = 1N HCl
1M H2SO4 = 2N H2SO4
1M H3PO4 = 3N H3PO4
1M CaCl2 = 2N CaCl2
1M CaSO4 = 2N CaSO4

10. = concentration in moles of substance per 1 kg of solvent
Molality (mol.kg –1)
= concentration in moles of substance per
1 kg of solvent
Osmolality ( mol.kg –1 or osmol.kg -1)
= concentration of osmotic effective particles
(i.e. particles which share in osmotic pressure of solution)
it is the same (for nonelectrolytes) or higher (for electrolytes: they dissociate to ions) as molality of the same solution
Osmolarity (osmoles / L)
= osmolality expressed in moles or osmoles per liter

11. the passage of a solvent through a semipermeable membrane is called osmosis
the semipermeable membrane separates two solutions of different concentrations

12. osmotic pressure
osmolarity = molarity of all particles dissolved in a solution (= osmotic active particles)

13. Exercise Describe dissociation of the salts: KNO3 → K2CO3 →
Na3PO4 →
Na2HPO4 →
NaH2PO4 →
NH4HCO3 →
What is the osmolarity of the 1M solutions?
K+ + NO3- Σ 2 ions
2 K+ + CO32- Σ 3 ions
3 Na+ + PO43- Σ 4 ions
2 Na+ + HPO42- Σ 3 ions
Na+ + H2PO4- Σ 2 ions
NH4+ + HCO3- Σ 2 ions

16. Osmotic pressure (Pa)
π = i x c x R x T
i = 1 (for nonelectrolytes)
i = number of osmotic effective particles
(for strong electrolytes)
isotonic solutions
= solutions with the same value of the osmotic pressure (c.g. blood plasma x saline )
Oncotic pressure
= osmotic pressure of coloidal solutions, e.g. proteins

17. Exercises 4) ? osmolarity of 0,15mol/L solution of : a) NaCl b) MgCl2
c) Na2HPO4
d) glucose
5) Saline is 150 mM solution of NaCl.
Which solutions are isotonic with saline? [= 150 mM = 300 mosmol/l ]
a) 300 mM glucose
b) 50 mM CaCl c) 300 mM KCl d) 0,15 M NaH2PO4
[0,30 M]
[0,45 M]
[0,15 M]
[300]
[150]
[600]

18. Percent concentrations
generally expressed as parts of solute per 100 parts of total solution (percent or „per one hundred“)
three basic forms:
a) weight per unit weight (W/W) g/g of solution
10% NaOH
→ 10g of NaOH + 90g of H2O = 100g of sol.
10% KCl
→ 10g of KCl/100g of solution

19. b) volume per unit volume (V/V) ml/100ml of sol.
5% HCl = 5ml of HCl / 100ml of sol.
c) weight per unit volume (W/V) g/100 ml
(g/dl; mg/dl; μg/dl; g % )
the most frequently used expression in medicine
20% KOH = 20g of KOH / 100 ml of sol.

20. Exercises 6) 600g 5% NaCl, ? mass of NaCl, mass of H2O
[30g NaCl + 570g H2O]
7) 250g 8% Na2CO3, ? mass of Na2CO3 (purity 96%)
[20,83g {96%}]
8) Normal saline solution is 150 mM. What is its percent concentration?
[ 0,9%]

21. Exercises 9) 14g KOH / 100ml MW = 56,1g; C = ? [ 2,5M ]
10) C(HNO3) = 5,62M; ρ = 1,18g/cm3 (density),
MW = 63g/mol, ? %
[ 30% ]
11) 10% HCl; ρ = 1,047g/cm3 , MW = 36,5
? C(HCl)
[ 2,87M ]

22. Conversion of units
pmol/L ‹ nmol/L ‹ mol/L ‹ mmol/L ‹ mol/L mol/L
g ‹ mg ‹ g g
L ‹ mL ‹ dL ‹ L L
1L = 1dm3 1mL = 1 cm3
Exercise
12) cholesterol (MW = 386,7g/mol) 200 mg/dl = ? mmol/L
[5,2 mM]

23. Conversion of units
pressure = the force acting normally on unit area of a surface (in pascals, Pa) 1 kPa = 103 Pa
Dalton´s law = the total pressure of a mixture of gasses or vapours is equal to the sum of the partial pressures of its components
partial pressure = pressure of one gas present in a mixture of gases

24. Conversion of units Air composition: Air pressure: 1 mmHg = 0,1333 kPa
78% N2 21% O2 1% water, inert gases, CO2 (0,04%)
Air pressure:
1 atm = Pa (~ 101 kPa) = 760 Torr (= mmHg)
1 mmHg = 0,1333 kPa
1 kPa = 7,5 mmHg

25. Exercise
13) Partial pressures of blood gases were measured in a laboratory:
pO2 = 71 mmHg
pCO2 = 35 mmHg
Convert the values to kPa.
pO2 = 9,5 kPa
pCO2 = 4,7 kPa

26. Conversion of units energy content of food: 1 kcal = 4,2 kJ
1 kJ = 0,24 kcal
14) A snack - müesli bar (30g) was labelled: 100g = 389 kcal. Calculate an energy intake (in kJ) per the snack.
490kJ / 30g

27. Dilution of solutions
= concentration of a substance lowers, substance amount remains the same
1) useful equation
n1 = n2
V1 x C1 = V2 x C2
2) mix rule
% of sol.(1) parts of sol.(1)
% of final sol.
% of sol.(2) parts of sol.(2)

28. 3) expression of dilution 1 : 5 or 1 / 5
1 part (= sample) + 4 parts (= solvent)
= 5 parts = total volume
c1 = 0,25 M (= concentration before dilution)
dilution 1 : 5 ( five times diluted sample )
→ c2 = 0,25 x 1/5 = 0,05 M (= final concentration )
4) mix equation (m1 x p1) + (m2 x p2) = p x (m1 + m2)
m = mass of mixed solution, p = % concentration

29. Exercises
15) final solution: 190g 10% sol. ? mass (g) of 38% HCl + ? mass (g) H2O
[50g HCl]
16) dilute 300g of 40% to 20% sol.
[1+1 = 300g of H2O]
17) 20g 10% solution of NaOH → 20% sol. ? m (g) of NaOH
[2,5g of NaOH]

30. Exercises 18) ? prep. 250ml of 0,1M HCl from stock 1M HCl
19) 10M NaOH was diluted 1: 20, ? final concentration
[0,5M]
20) 1000mg/l glucose was diluted 1: 10 and then 1 : ? final concentration
[50mg/l]
21) what is the dilution of serum in a test tube containing μl of serum μl of saline μl of reagent
[1 : 5]