Chemical calculations used in medicine (concentration, dilution) Pavla Balínová
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 Prefixes for units
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 Substance amount (n) in moles (mol) n = m/mw (m = mass (g)) Also used mmol, µmol, nmol, pmol,
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
Molar concentration - examples 1) 17.4 g NaCl in 300 ml of solution, Mr (NaCl) = 58 c =? 2) 4.5 g glucose in 2.5 L of solution, Mr (glucose) = 180 c =? 3) Solution of glycine, c = 3 mm, V = 100 ml, Mr (glycine) = 75 m =? mg of glycine in the solution
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 Oncotic pressure = is a form of osmotic pressure exerted by proteins in blood plasma Osmosis = the movement of a 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!
Osmolarity Osmolarity (osmol/l) is a number of moles of a substance that contribute to osmotic pressure of solution The osmolar concentration of body fluids is typically reported in mosmol/l. Osmolarity of blood plasma is 290 300 mosmol/l π = i c R T The figure is found at http://en.wikipedia.org/wiki/osmotic_pressure
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 to 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 to 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
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 (in mol/l) of 900 mosmol/l solution of MgCl 2?
Percent concentration is expressed as part of solute per 100 parts of total solution (%) % = mass of solute x 100 mass of solution % concentration 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
Percent concentrations - examples 1) 600 g 5% NaCl,? mass of NaCl,? mass of H 2 O 2) 16 g of glucose is administrated to the patient intravenously as first aid during hypoglycemia. How many ml of 40% infusion solution of glucose will be administrated to patient? 3) How many ml of 95% ethanol (Spiritus absolutus) and how many ml of H 2 O do you need for preparation 250 ml 60% ethanol solution (Spiritus dilutus)? 4) Saline is 150 mm solution of NaCl. Calculate the percent concentration by mass of this solution. Mr(NaCl) = 58.5
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/cm3 or g/ml - density of water = 1 g / cm 3 - density of 95% ethanol = 0.79 g/cm 3 - density of lead (Pb) = 11.34 g/cm 3
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 m = 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
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
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,? %
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) useful equation derived from the definition of dilution n 1 = n 2 V 1 x c 1 = V 2 x c 2
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) 200 ml of 0.2 M acid solution was diluted to the final volume of 0.8 litre. Molarity of final solution?
The dose of drug - problem 1) Nasal drops used to treat sinusitis include antibiotic drugs neomycine (1.32 g/l) and bacitracine (100 IU/mL). In the pharmacy these drops are prepared by dilution of solution Pamycon which contains 0.33% (w/v) of neomycine and 250 IU/mL of bacitracine. Calculate how many ml of Pamycon solution is necessary to prepare 10 ml nasal drops mentioned above. 2) AMOCLEN 250 syrup contains 250 mg of amoxicillin in 5 ml of syrup. The usual daily dose for children is 50 mg of amoxicillin per 1 kg of body weight. The total daily dose is divided into 3 individual doses to be administered every 8 hours. Calculate the volume of a single dose for a child weighing 12 kg.