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Chemical calculations used in medicine part 1

Pavla Balínová. Chemical calculations used in medicine part 1. Prefixes for u nits. 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.

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Chemical calculations used in medicine part 1

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  1. Pavla Balínová Chemical calculations used in medicine part 1

  2. Prefixes for units giga- G 109 mega- M 106 kilo- k 103 deci- d 10-1 centi- c 10-2 milli- m 10-3 micro- μ10-6 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 1023 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 concentrationormolarity (c) – number of moles of a substance per liter of solution • unit: mol/L= mol/dm3 = 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 Na2HPO4. How many mmol Na+ ions are found in 20 mL of this solution? Mr (Na2HPO4) = 142 • Substance amount of Na2HPO4 in 2 L of solution: n = 142/142 = 1 mol • 1 mol of Na2HPO4 in 2 L • 0.5 mol of Na2HPO4 in 1 L→0.5 mol Na2HPO4 gives 1 molof Na+and 0.5 mol of HPO42- • 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 CaCl2 solution is 0.1 M. Calculate the volume of solution containing 4 mmol of Cl-. • 0.1 M CaCl2 = 0.1 mol in 1 L • 0.1 mol of CaCl2 gives 0.1 mol of Ca2+ 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 pressureπis 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 CaCl2 solution contains 3 osmol of solute per liter of solution. CaCl2→ 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. • C6H12O6→ C6H12O6 • 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) MgCl2 c) Na2HPO4 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 CaCl2 c) 300 mM KCl d) 0.15 M NaH2PO4 3. What is molarity of 900 mosmol/l solution of MgCl2 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 it has 3 forms: 1. weight per weight (w/w) 10% of KCl = 10g of KCl + 90 g of H2O = 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 H2O • 2) 250 g 8% Na2CO3, ? mass of Na2CO3 (purity 96%) • 3) 250 mL 39% ethanol solution; ? mL of ethanol, ? mL of H2O • 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/cm3or g/mL • density of water = 1 g / cm3 • density of lead (Pb) = 11.34 g/cm3

  14. Conversions of concentrations (% and c) with density • What is a percent concentration of 2 M HNO3 solution? Density (HNO3) = 1,076 g/ml, Mr (HNO3) = 63,01 ? Conversion of molar concentration to % concentration? 2 M HNO3 solution means that 2 mol of HNO3 are dissolved in 1 L of solution Mass of HNO3 = n x Mr = 2 x 63.01 = 126.02 g of HNO3 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 xρ(g/cm3) • ● Conversion of percent concentration to molarity • c = % x 10 xρ(g/cm3) • Mr

  16. Conversions of concentrations (% and c) with density - examples • 1) ? % (w/w) of HNO3; ρ = 1.36 g/cm3, if 1dm3 of solution contains 0.8 kg of HNO3 • 2) c (HNO3) = 5.62 M; ρ = 1.18 g/cm3, MW = 63 g/mol, ? % • 3) 10% HCl; ρ = 1.047 g/cm3, MW = 36.5 , ? c (HCl)

  17. Dilution • =concentration of a substance lowers, number of moles of the substance remains the same!1) mix equation:m1x p1 + m2x p2 = p x( m1 + m2 ) • m = mass of mixed solution, p = % concentration2) expression of dilutionIn 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: c1 = 0.25M (original concentration) x 1/5 = 0.05 M (final concentration c2) • 3) useful equation n1 = n2V1 x c1 = V2x c2

  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) H2O • 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 CO2 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 (Io) 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 < I0)→the transmittance (T)ofa solution is defined as the proportion: T = I / I0 • ● transmittance can be expressed in percentage: T(%) = (I/I0) 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 theabsorbance(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 absorptioncoefficient (tabelatedvalue) Relationship between A and T:A = log (1/T)= -log T

  23. Calculations in spectrophotometry - examples • Ast = 0.40, cst = 4 mg/L • Asam = 0.25, csam = ? 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

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