Transcript ppt

Chemical calculations used in medicine
part 1
Pavla Balínová
Prefixes for units
gigamegakilodecicentimillimicronanopicofemtoatto-
G
M
k
d
c
m
μ
n
p
f
a
109
106
103
10-1
10-2
10-3
10-6
10-9
10-12
10-15
10-18
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
Molar concentration or molarity (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
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
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 mol of Na+ and
0.5 mol of HPO421 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 Cl0.2 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 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
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
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+ + Cl1 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 Cl1 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 - 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
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 = 10 g 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
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 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 / cm3
- density of lead (Pb) = 11.34 g/cm3
Conversions of concentrations (% and c)
with density
1)
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 density
●
Conversion of molarity to percent concentration
% = c (mol/L) x Mr
10 x ρ (g/cm3)
●
Conversion of percent concentration to molarity
c = % x 10 x ρ (g/cm3)
Mr
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
= concentration of a substance lowers, number of moles
of the substance remains the same!
1) mix equation: m1 x p1 + m2 x p2 = p
x
( m1 + m2 )
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: c1 = 0.25 M (original concentration) x 1/5 =
0.05 M (final concentration c2)
3) useful equation
n1 =
n2
V1 x c1 = V2 x c2
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
• 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
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
●
●
●
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) of a solution is defined as
the proportion: T = I / I0
●
transmittance can be expressed in percentage: T(%) = (I/I0)
●
values of the measured transmittance are found from 0 - 1 = 0 - 100%
x
100
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
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
ε
Calculations in spectrophotometry examples
1) 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