Pharmaceutical Calculations:
Electrolyte Solutions –
Milliequivalents, Millimoles,
Milliosmols
Danielle DelVillano, Pharm.D.
Objectives
• Calculate the milliequivalent weight from an
atomic or formula weight
• Convert between milligrams and
milliequivalents
• Calculate problems involving milliequivalents,
millimoles, and milliosmols
Milliequivalents
• A 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 #
Equivalent Values
Molecular Weights To Know
•
•
•
•
•
•
H
O
Na
Cl
K
Ca
•
•
•
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NaCl
KCl
CaCl2
Dextrose
Equations
mEq =
mg substance * Valence
Atomic Weight
mg =
mEq substance * Atomic weight
Valence #
mg/mL = mEq/mL substance * Atomic weight
Valence #
Problem 1
• What 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
mg/mL = 149 mg/mL
Problem 2
• What 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
10 mEq
x mg
x = 535 mg = 0.535 g per 100 mL
or 0.535%
Problem 3
• A 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/L
Problem 4
• How many mEq of KCl are represented in a 15
mL dose of 10% w/v KCl elixir?
10% = 10g/100 mL
10 g = x g
100 mL
15 mL
x = 1.5 g = 1500 mg
mEq = 1500 mg * 1
74.5 mg
Answer = 20.13 mEq
Millimoles and Micromoles
• Mole - MW of a substance in grams
• Millimole - MW of a substance in mg
• Micromile - MW of a substance in mcg
• Measure representing the combining power of
a species
– Monovalent species: mEq = mmol
Keep In Mind
• Eq wt of a substance = MW / valance
• Moles of a substance = MW
Problem 5
• How 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 mmol
Osmolarity
• Osmotic pressure is proportional to the total
number of particles in a solution
• Measured 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 particles
Equation and Definitions
• mOsmol/L = g/L substance * # species * 1000
MW
• Osmolarity – mOsmol/L solution
• Osmolality – mOsmol/kg of solvent
Problem 6
• What 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/L
Problem 7
• A 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
40 g
Answer = 2.5 mOsmol
* 1000
Clinical Considerations of Water and
Electrolyte Balance
• Total body water for an adult male is 55-65%
body weight
– Females 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 infants
Equation
• Plasma 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)
Problem 8
• Calculate 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 mL
Problem 9
• Estimate 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)
Answer 298 mOsmol/kg
+ 14 + 90
2.8
18
Questions
Reference
• Ansel, H. C. (2009) Phamaceutical Calculations
(13th Ed.). Philadelphia:Lippincott Williams &
Wilkins, and Wolters Kluwer Publishers
Chapter 12 Page 197
• Calculate he mEq of sodium, potassium, and
chloride, the millimoles of anhydrous
dextrose, and the osmolarity of the following
paerenteral fluid.
– Dextrose, anhydrous
– NaCl
– KCl
– Water for injectoin
ad
50 g
4.5 g
1.49 g
1000 mL
Chapter 12 Page 197
Chapter 12 Page 191
• How many mEq of Na+ would be contained in
a 30 mL dose of the following solution?
– Disodium hydrogen phosphate
– Sodium biphosphate
– Purified water
ad
18 g
48 g
100 mL
(Disodium hydrogen phosphate MW 268)
(Sodium biphosphate MW 138)
Chapter 12 Page 191
Chapter 12 Problem 3
• A 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?
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?
Chapter 12 Problem 56
• What is the osmolarity of an 8.4% w/v
solution of sodium bicarbonate?
Chapter 12 Problem 58
• How 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?
Equations; Relating mOsmol, mmol
and mEq
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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