1. Danielle DelVillano, Pharm.D.
Pharmaceutical Calculations: Electrolyte Solutions – Milliequivalents, Millimoles, Milliosmols
Danielle DelVillano, Pharm.D.

2. Objectives
Calculate the milliequivalent weight from an atomic or formula weight
Convert between milligrams and milliequivalents
Calculate problems involving milliequivalents, millimoles, and milliosmols

3. 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)

4. Equivalent Values

5. Molecular Weights To KnowNa Cl K Ca
NaCl
KCl
CaCl2
Dextrose
1, 16, 23, 35.5, 39, 40
58.5, 74.5, , 180

6. Equations
mEq = mg substance * Valence Atomic Weight mg = mEq substance * Atomic weight Valence # mg/mL = mEq/mL substance * Atomic weight

7. 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

8. 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 x = 535 mg = g per 100 mL
10 mEq x mg or 0.535%

9. 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

10. Problem 4
How 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 = mEq
100 mL 15 mL
x = 1.5 g = 1500 mg

11. 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

12. Keep In Mind Eq wt of a substance = MW / valance
Moles of a substance = MW

13. 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 mg
x = mmol

14. 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
Assume complete dissolution

15. Equation and Definitions
mOsmol/L = g/L substance * # species * 1000 MW
Osmolarity – mOsmol/L solution
Osmolality – mOsmol/kg of solvent

16. 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

17. 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 * 1000
40 g
Answer = 2.5 mOsmol

18. 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
mL/kg for infants

19. Equation Plasma Osmolality (mOsmol/kg) =
2(Na + K) plasma + BUN + Glucose
Na and K measured in mEq/L
BUN and glucose measured in mg/100mL (or mg/dL)

20. 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

21. 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( )
Answer 298 mOsmol/kg

22. Questions

23. Reference
Ansel, H. C. (2009) Phamaceutical Calculations (13th Ed.). Philadelphia:Lippincott Williams & Wilkins, and Wolters Kluwer Publishers

24. 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 50 g
NaCl g
KCl g
Water for injectoin ad mL

25. Chapter 12 Page 197 Na 77 mEq K 20 mEq Cl 97 mEq Dextrose 278 mmol
472 mOsmol

26. Chapter 12 Page 191
How many mEq of Na+ would be contained in a 30 mL dose of the following solution?
Disodium hydrogen phosphate 18 g
Sodium biphosphate g
Purified water ad mL
(Disodium hydrogen phosphate MW 268)
(Sodium biphosphate MW 138)

27. Chapter 12 Page 191

28. 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?
4 mEq/mL

29. 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?
1092 mL

30. Chapter 12 Problem 56
What is the osmolarity of an 8.4% w/v solution of sodium bicarbonate?
2000 mOsmol/L

31. 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?
100 mmol
200 mEq
300 mOsmol

32. 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