Milliequivalents Millimoles Milliosmols Danielle DelVillano PharmD Objectives Calculate the milliequivalent weight from an atomic or formula weight Convert between milligrams and ID: 525568
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Slide1
Pharmaceutical Calculations: Electrolyte Solutions – Milliequivalents, Millimoles, Milliosmols
Danielle DelVillano, Pharm.D.Slide2
ObjectivesCalculate the milliequivalent weight from an atomic or formula weight
Convert between milligrams and
milliequivalents
Calculate problems involving
milliequivalents
,
millimoles
, and
milliosmolsSlide3
MilliequivalentsA chemical unit: mEq
Related to the total number of ionic charges in solution, and takes note of valence of ions
1
mEq
=
atomic weight (mg)
valence #
1
Eq
=
atomic weight (
g
)
valence #Slide4
Equivalent ValuesSlide5
Molecular Weights To KnowHONa
Cl
K
Ca
NaCl
KCl
CaCl2
Dextrose Slide6
EquationsmEq = mg substance * Valence
Atomic
Weight
m
g =
mEq
substance * Atomic weight
Valence #
mg/
mL
=
mEq/mL
substance * Atomic weight
Valence #Slide7
Problem 1What is the concentration, in mg/mL, of a solution containing 2
mEq
of potassium chloride (
KCl
) per
mL
?
mg/
mL
=
2
mEq/mL
* 74.5
1
m
g/
mL
= 149 mg/
mLSlide8
Problem 2What is the percent (w/v) concentration of a solution containing 100
mEq
of ammonium chloride per liter? (MW = 53.5)
1
mEq
=
53.5 mg
= 53.5 mg
1
100
mEq
=
x
mEq
x
= 10
mEq
1000
mL
100
mL
1
mEq
=
53.5 mg
x
= 535 mg = 0.535
g
per 100
mL
10
mEq
x
mg or 0.535%Slide9
Problem 3A solution contains 10 mg/100 mL Ca++ ions. Express this concentration in terms of
mEq
/L
10 mg/ 100
mL
* 10/10 = 100 mg/1000
mL
mEq
/L =
100 mg/L * 2
40 mg
Answer = 5
mEq
/LSlide10
Problem 4How many mEq of
KCl
are represented in a 15
mL
dose of 10%
w/v
KCl
elixir?
10% = 10g/100
mL
mEq
=
1500 mg * 1
74.5 mg
10
g
=
x
g
Answer = 20.13
mEq
100
mL
15
mL
x
= 1.5
g
= 1500 mgSlide11
Millimoles and MicromolesMole - MW of a substance in gramsMillimole
- MW of a substance in mg
Micromile
- MW of a substance in mcg
Measure representing the combining power of a species
Monovalent
species:
mEq
=
mmolSlide12
Keep In MindEq wt of a substance = MW / valanceMoles of a substance = MWSlide13
Problem 5How many millimoles of monobasic sodium phosphate (MW 138) are present in 100
g
of a substance?
1
mmol
=
138 mg
x
mmol
100000 mg
x
= 724.6
mmolSlide14
OsmolarityOsmotic pressure is proportional to the total number of particles in a solutionMeasured in mOsmol
Example
1
mmol
dextrose = 1
mOsmol
total particles
1
mmol
NaCl
= 2
mOsmol
total particles
1
mmol
CaCl2 = 3
mOsmol
total particles
1
mmol
Na3C6H5O7 = 4
mOsmol
total particlesSlide15
Equation and DefinitionsmOsmol/L = g
/L substance
* # species * 1000
MW
Osmolarity
–
mOsmol
/L solution
Osmolality
–
mOsmol
/kg of solventSlide16
Problem 6What is the ideal osmolarity of a 0.9% sodium chloride injection?
0.9g/100
mL
* 10/10 = 9g/1000mL
mOsmol
/L =
9g/1L
* 2 species * 1000
58.5
g
Answer = 308
mOsmol
/LSlide17
Problem 7A solution contains 10 mg% of Ca++ ions. How many milliosmols are represented in 1 liter of the solution?
10 mg/100
mL
*10/10 = 100 mg/1000mL
mOsmol
/L =
0.1
g
/L
* 1 species * 1000
40
g
Answer = 2.5
mOsmolSlide18
Clinical Considerations of Water and Electrolyte BalanceTotal body water for an adult male is 55-65% body weightFemales about 10% lower
Newborn infants 75%
Daily requirement equations
1500
mL
per square meter of BSA
32
mL
/kg for adults
100-150
mL
/kg for infantsSlide19
EquationPlasma Osmolality (mOsmol
/kg) =
2(Na + K) plasma +
BUN
+
Glucose
2.8 18
Na and K measured in
mEq
/L
BUN and glucose measured in mg/100mL (or mg/dL)Slide20
Problem 8Calculate the estimated daily water requirement for a healthy adult with a body surface area of 1.8 m2.
1 m2
=
1500
mL
1.8 m2
x
mL
x
= 2700
mLSlide21
Problem 9Estimate the plasma osmolality from the following data: sodium 135
mEq
/L, potassium 4.5
mEq
/L, BUN 14 mg/dL, glucose 90 mg/dL
mOslmol
/kg = 2(135+4.5) +
14
+
90
2.8 18
Answer 298
mOsmol
/kgSlide22
QuestionsSlide23
ReferenceAnsel, H. C. (2009) Phamaceutical Calculations
(13th Ed.).
Philadelphia:Lippincott
Williams & Wilkins, and
Wolters
Kluwer
PublishersSlide24
Chapter 12 Page 197Calculate he mEq of sodium, potassium, and chloride, the
millimoles
of anhydrous dextrose, and the
osmolarity
of the following
paerenteral
fluid.
Dextrose, anhydrous 50
g
NaCl
4.5
g
KCl
1.49
g
Water for
injectoin
ad 1000
mLSlide25
Chapter 12 Page 197Slide26
Chapter 12 Page 191How many mEq of Na+ would be contained in a 30
mL
dose of the following solution?
Disodium hydrogen phosphate 18
g
Sodium
biphosphate
48
g
Purified water ad 100
mL
(Disodium hydrogen phosphate MW 268)
(Sodium
biphosphate
MW 138)Slide27
Chapter 12 Page 191Slide28
Chapter 12 Problem 3A 10-mL ampule of potassium chloride contains 2.98
g
of potassium chloride (
KCl
). What is the concentration of the solution in terms of
milliequivalents
per milliliter? Slide29
Chapter 12 Problem 40 A patient has a sodium deficit of 168 mEq. How many milliliters of isotonic sodium
chloride solution (0.9%
w/v
) should be administered to replace the
deficit
? Slide30
Chapter 12 Problem 56What is the osmolarity of an 8.4% w/v solution of sodium bicarbonate? Slide31
Chapter 12 Problem 58How many (a) millimoles, (b)
milliequivalents
, and (
c
)
milliosmoles
of calcium chloride (CaCl2⋅2H2O—
m.w
. 147) are represented in 147
mL
of a 10%
w/v
calcium
chloride solution? Slide32
Equations; Relating mOsmol, mmol and mEq
mEq
= (mg substance * valance) / MW
mEq
= MW (mg) / valance
mmol
= MW (mg)
mOsmol
= (
Weight(g
) /
MW(g
)) * species * 1000
mOsmol
/kg = 2(Na + K) + (BUN/2.8) + (glucose/18)
mOsmol
=
mmol
* # species
mOsmol
=
(
mEq
* # species)
valence