1. Chemical calculations used in medicine part 1 Pavla Balínová

2. Prefixes for units giga-G 10 9 mega- M 10 6 kilo-k 10 3 deci-d 10 -1 centi-c 10 -2 milli-m 10 -3 micro- μ 10 -6 nano-n 10 -9 pico-p 10 -12 femto-f 10 -15 atto-a 10 -18

3. Basic terms MW = molecular weight (g/mol) = mass of 1 mole of substance in grams or relative molecular weight Mr Avogadro´s number N = 6.022 x 10 23 particles in 1 mol n = substance amount in moles (mol) n = m/MW (m = mass (g)) Also used mmol, µmol, nmol, pmol, …

4. Concentration – amount of a substance in specified final volume Molar concentration or molarity (c) – number of moles of a substance per liter of solution unit: mol/L = mol/dm 3 = M c = n (mol) / V (L) Molality (mol/kg) – concentration of moles of substance per 1 kg of solvent

5. Molar concentration - examples 1) 17.4 g NaCl in 300 mL of solution, MW (NaCl) = 58 c = ? 2) 4.5 g glucose in 2.5 L of solution, MW (glucose) = 180 c = ? 3) Solution of glycine, c = 3 mM, V = 100 mL, MW (glycine ) = 75 m = ? mg of glycine in the solution

6. Number of ions in a certain volume Problem 1: 2 litres of solution contain 142 g of Na 2 HPO 4. How many mmol Na + ions are found in 20 mL of this solution? Mr (Na 2 HPO 4 ) = 142 Substance amount of Na 2 HPO 4 in 2 L of solution: n = 142/142 = 1 mol 1 mol of Na 2 HPO 4 in 2 L 0.5 mol of Na 2 HPO 4 in 1 L → 0.5 mol Na 2 HPO 4 gives 1 mol of Na + and 0.5 mol of HPO 4 2- 1 mol of Na + in 1 L X mol of Na + in 0,02 L X = 0.02/1 x 1 = 0.02 mol = 20 mmol Problem 2: Molarity of CaCl 2 solution is 0.1 M. Calculate the volume of solution containing 4 mmol of Cl -. 0.1 M CaCl 2 = 0.1 mol in 1 L 0.1 mol of CaCl 2 gives 0.1 mol of Ca 2+ and 0.2 mol of Cl - 0.2 mol of Cl - in 1 L 0.004 mol of Cl - in X L X = 0.004/0.2 x 1 = 0.02 L = 20 mL

7. Osmotic pressure Osmotic pressure π is a hydrostatic pressure produced by solution in a space divided by a semipermeable membrane due to a differential in the concentrations of solute. unit: pascal Pa Π = i x c x R x T Osmosis = the movement of solvent from an area of low concentration of solute to an area of high concentration ! Free diffusion = the movement of solute from the site of higher concentration to the site of lower concentration ! Oncotic pressure = is a form of osmotic pressure exerted by proteins in blood plasma

8. Osmolarity Osmolarity is a number of moles of a substance that contribute to osmotic pressure of solution (osmol/L) The concentration of body fluids is typically reported in mosmol/L. Osmolarity of blood is 290 – 300 mosmol/L The figure is found at http://en.wikipedia.org/wiki/Osmotic_pressure

9. Osmolarity - examples Example 1: A 1 M NaCl solution contains 2 osmol of solute per liter of solution. NaCl → Na + + Cl - 1 M does dissociate 1 osmol/L 1 osmol/L 2 osmol/L in total Example 2: A 1 M CaCl 2 solution contains 3 osmol of solute per liter of solution. CaCl 2 → Ca 2+ + 2 Cl - 1 M does dissociate 1 osmol/L 2 osmol/L 3 osmol/L in total Example 3: The concentration of a 1 M glucose solution is 1 osmol/L. C 6 H 12 O 6 → C 6 H 12 O 6 1 M does not dissociate → 1 osmol/L

10. Osmolarity - examples 1. What is an osmolarity of 0.15 mol/L solution of: a) NaCl b) MgCl 2 c) Na 2 HPO 4 d) glucose 2. 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 2 c) 300 mM KCl d) 0.15 M NaH 2 PO 4 3. What is molarity of 900 mosmol/l solution of MgCl 2 in mol/L?

11. Percent concentration expressed as part of solute per 100 parts of total solution (%) % = mass of solute x 100 mass of solution mass of solution it has 3 forms: 1. weight per weight (w/w) 10% of KCl = 10 g of KCl + 90 g of H 2 O = 100 g of solution 2. volume per volume (v/v) 5% HCl = 5 mL HCl in 100 mL of solution 3. weight per volume (w/v) the most common expression 0.9% NaCl = 0.9 g of NaCl in 100 mL of solution

12. Percent concentrations - examples 1) 600 g 5% NaCl, ? mass of NaCl, ? mass of H 2 O 2) 250 g 8% Na 2 CO 3, ? mass of Na 2 CO 3 (purity 96%) 3) 250 mL 39% ethanol solution; ? mL of ethanol, ? mL of H 2 O 4) Saline is 150 mM solution of NaCl. Calculate the percent concentration by mass of this solution. Mr(NaCl) = 58.5

13. Density ρ - is defined as the amount of mass per unit of volume ρ = m/V → m = ρ x V and V = m / ρ - these equations are useful for calculations units: g/cm 3 or g/mL - density of water = 1 g / cm 3 - density of lead (Pb) = 11.34 g/cm 3

14. Conversions of concentrations (% and c) with density 1)What is a percent concentration of 2 M HNO 3 solution? Density (HNO 3 ) = 1,076 g/ml, Mr (HNO 3 ) = 63,01 ? Conversion of molar concentration to % concentration? 2 M HNO 3 solution means that 2 mol of HNO 3 are dissolved in 1 L of solution Mass of HNO 3 = n x Mr = 2 x 63.01 = 126.02 g of HNO 3 Mass of solution = ρ x V = 1.076 x 1000 = 1076 g W = 126,02 x 100 = 11,71% 1076 2) What is the molarity of 38% HCl solution? Density (38% HCl) = 1.1885 g/ml and Mr(HCl) = 36.45 ? Conversion of percent by mass concentration to molar concentration? 38% HCl solution means that 38 g of HCl in 100 g of solution. One liter of solution has a mass m = V x ρ = 1000 x 1.1885 = 1188.5 g. 38 g HCl -------> 100 g of solution x g HCl -------> 1188,5 g of solution x = 451.63 g HCl in 1 L of solution → n = m / M = 451.63 / 36.45 = 12.4 mol of HCl c(HCl) = n / V = 12.4 / 1 = 12.4 mol/L

15. Conversions of concentrations (% and c) with density ● Conversion of molarity to percent concentration % = c (mol/L) x Mr 10 x ρ (g/cm 3 ) ● Conversion of percent concentration to molarity c = % x 10 x ρ (g/cm 3 ) Mr

16. Conversions of concentrations (% and c) with density - examples 1) ? % (w/w) of HNO 3 ; ρ = 1.36 g/cm 3, if 1dm 3 of solution contains 0.8 kg of HNO 3 2) c (HNO 3 ) = 5.62 M; ρ = 1.18 g/cm 3, MW = 63 g/mol, ? % 3) 10% HCl; ρ = 1.047 g/cm 3, MW = 36.5, ? c (HCl)

17. Dilution = concentration of a substance lowers, number of moles of the substance remains the same! 1) mix equation: m 1 x p 1 + m 2 x p 2 = p x ( m 1 + m 2 ) m = mass of mixed solution, p = % concentration 2) expression of dilution In case of a liquid solute, the ratio is presented as a dilution factor. For example, 1 : 5 is presented as 1/5 (1 mL of solute in 5 mL of solution) Example: c 1 = 0.25 M (original concentration) x 1/5 = 0.05 M (final concentration c 2 ) 3) useful equation n 1 = n 2 V 1 x c 1 = V 2 x c 2

18. Dilution - examples 1) Mix 50 g 3% solution with 10 g 5% solution, final concentration = ? (%) 2) Final solution: 190 g 10% sol. ? m (g) of 38% HCl + ? m (g) H 2 O 3) Dilute 300 g of 40% solution to 20 % solution. ? g of solvent do you need? 4) ? preparation of 250 mL of 0.1 M HCl from stock 1 M HCl 5) 10 M NaOH is diluted 1 : 20, ? final concentration 6) 1000 mg/L glucose is diluted 1 : 10 and then 1 : 2. ? final concentration

19. Conversions of units concentration: i.e. g/dL → mg/L i.e. mol/L ↔ osmol/L units of pressure used in medicine 1 mmHg (millimeter of mercury) = 1 Torr 1 mmHg = 133.22 Pa units of energy 1 cal = 4.1868 joule (J) 1 J = 0.238 cal calorie (cal) is used in nutrition

20. Conversions of units – examples 1) Concentration of cholesterol in patient blood sample is 180 mg/dL. Convert this value into mmol/L if MW of cholesterol is 387 g/mol. 2) Partial pressure of CO 2 is 5.33 kPa. Convert this value into mmHg. 3) Red Bull energy drink (1 can) contains 160 cal. Calculate the amount of energy in kJ.

21. Calculations in spectrophotometry ● spectrophotometry is an analytical method based on the interaction between an electromagnetic radiation of a known intenzity (I o ) and the analyzed sample ● a part of the radiation is absorbed by the analyzed substance found in the of sample ● the intensity of the radiation which passed (I) through the solution is detected (I < I 0 ) → the transmittance (T) of a solution is defined as the proportion: T = I / I 0 ● transmittance can be expressed in percentage: T(%) = (I/I 0 ) x 100 ● values of the measured transmittance are found from 0 - 1 = 0 - 100%

22. Calculations in spectrophotometry How to calculate a concentration of substance in analyzed sample?? ● in the laboratory it is more convenient to calculate concentration from values of the absorbance (A) which is directly proportional to the concentration than from the transmittance Calculation from the Beer´s law: A = c x l x ε c = molar concentration (mol/L) l = inner width of the cuvette in centimeters ε = molar absorption coefficient (tabelated value) Relationship between A and T: A = log (1/T)= -log T

23. Calculations in spectrophotometry - examples 1)A st = 0.40, c st = 4 mg/L A sam = 0.25, c sam = ? mg/L 2) Standard solution of glucose (conc. = 1000 mg/L) reads a transmittance 49 %. Unknown sample of glucose reads T = 55 %. Calculate the concentration of glucose in the sample in mg/L and mmol/L. Mr (glucose) = 180