1 / 23

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.

eloise
Download Presentation

Chemical calculations used in medicine part 1

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related