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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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prefixes for u nits
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

basic terms
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, …
concentration amount of a substance in specified final volume
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
molar concentration examples
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
number of ions in a certain volume
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
osmotic pressure
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
osmolarity
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

osmolarity examples
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
osmolarity examples10
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?

percent concentration
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

percent concentrations examples
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
density
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
conversions of concentrations and c with density
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

conversions of concentrations and c with density15
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
conversions of concentrations and c with density examples
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)
dilution
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
dilution examples
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
conversions of units
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
conversions of units examples
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.
calculations in spectrophotometry
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%
calculations in spectrophotometry22
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

calculations in spectrophotometry examples
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