In medicine, concentrations are often expressed as milliequivalents per liter (mEq/L) or millimoles per liter (mmol/L). But what’s the difference?
A mole is the International System of Units (SI) base unit for the amount of a substance (just like a second for time, meter for length, kilogram for mass, etc.) There are ~6.022 x 1023 (Avogadro’s constant) quantifiable “things” (molecules, atoms, ions, etc.) in one mole of a sample. The same sample’s molar mass is determined by adding its constituents from the periodic table. For example, potassium chloride (KCl) has a molar mass of ~74.6 grams/mole (~39.1 g/mol for K, ~35.5 g/mol for Cl). A mmol is one-thousandth of a mole.
Similarly, a mEq is one-thousandth of an equivalent. Without delving into the chemistry definitions, mEq/L is often used for plasma electrolyte concentrations. mEq is related to the number (mmol) and electrical charge of ions in solution. Univalent ions like potassium (K+), sodium (Na+), and chloride (Cl–) have the same number of moles as equivalents in solution. For example, a normal plasma sodium range could be 135 – 145 mEq/L or mmol/L. In contrast, divalent ions like calcium (Ca2+) or magnesium (Mg2+) produce twice the equivalents (e.g., 1 mmol/L of calcium = 2 mEq/L).
In summary, “mEq/L” is most likely encountered when dealing with electrolyte concentrations. In the case of univalent ions, “mEq/L” can be replaced with “mmol/L.” A number without a unit is meaningless!
Drop me a comment with your questions!
I have a question to further my understanding on mEq in relation to mmols. You stated:
“Univalent ions like potassium (K+), sodium (Na+), and chloride (Cl–) have the same number of moles as equivalents in solution. For example, a normal plasma sodium range could be 135 – 145 mEq/L or mmol/L. In contrast, divalent ions like calcium (Ca2+) or magnesium (Mg2+) produce twice the equivalents (e.g., 1 mmol/L of calcium = 2 mEq/L).”
Does that mean that ions with higher valency would also alter the equivalency of mEq to mmols? For example, and I apologize if this seems uneducated, but you have the best explanation I have heard regarding this and it is actually connecting in my head now,
If you have a trivalent ion would the relation be the same in regards to 1 mEq= about .33mmol or does the concept completely change after uni- and di-valent rules?
Hello I read this. Is this correct?
The milliequivalent concentration is the number of equivalent weights of the substance in the volumetric water basis times 1000 meq/eq.
If I’m reading that correctly, it’s basically defining what the metric prefix “milli” means – one thousandth of the base unit. So for example, 1000 milliequivalents = 1 equivalent.
Thank you Rishi,as a chemistry student I have learnt something I was not aware of.
Right on! Thanks for the comment! 🙂
Hello Rishi,
This is similar to the previous question and I think they have made a mistake. But I would like to check my logic before sending a post to the forum.
Calculate the mass (mg) of ferrous sulfate [molar mass = 278 g/mol] required to prepare 12 mL of a solution containing 2.5 mEq of ferrous ion.
mEq/L = (mg/L * valence)/molecular weight.
Let x mg be the weight of FeSO4. Valence of Fe and S04 are both divalent 2.
2.5 mEq/L = (x mg/L * 2 )/278 mg/mol
Solving for x gives.
x mg = (2.5 mEq/L * 278 mg)/2 L
x mg = 695 mg/ 2 = 347.5 mg. (They claim this answer is correct)
Which is the answer that they have given. But the question is asking mg in 12 mL.
but this is FeSO4 in one liter of solution. However, if we want 12 mL then
y mg = 12 mL
347.5 mg = 1000 mL
∴ y mg = 4170 mg/1000 mL = 4.17 mg
Hey Peter, this is similar to the previous question except that given the divalent nature of the iron moeity and sulfate anion, one mole of ferrous sulfate will produce two equivalents of Fe(II) and two equivalents of sulfate. Again, the question seems to only be asking about the mass rather than the osmolarity, so there’s no need to consider the volume.
Because of the divalency, 2.5 mEq of ferrous ion comes from only 1.25 mmol of ferrous sulfate.
1.25 mmol FeSO4 = 0.00125 mol FeSO4
(278 grams/mol FeSO4) x (0.00125 mol FeSO4) = 0.3475 grams of FeSO4
0.3475 grams FeSO4 = 347.5 mg FeSO4
Hope this helps!
Hello Rishi,
I am a second year pharamacy student and I pretty sure I have a mistake in two of the quiz questions. Before I write to the forum post could you please verify that my logic for these two questions are correct. I have given the answer which the quiz claims to be correct. I also give my soution which I think correctly shows the there is a mistake.
The first question is:
You have been asked to prepare 60 mL of a solution containing 78 mEq of sodium ion. Calculate the mass (g) of sodium chloride [molar mass = 58.4 g/mol] required.
The correct answer is: 4.56 g (They claim this to be the correct answer and I am disputing this answer)
Using the definition of mEq
mEq = (xmg/L * 1000 * valence)/ MW.
We are given that the solution is 78 mEq of NaCl. Let x mg be the NaCl. Then
78 mmol/L = (x mg/L * 1) *1000 / 58.4 g, 78 mmol/L = x mg/ 58.4 mg/mmol.
x mg = 78 mmol * 58.4 mg/ mmol
x mg = 4,555.2 mg = 4.56 g
But this is in 1 L of solution. The question is asking for 60 mL of solution.
Let y g be the amount of NaCl in 60 mL then we have
y g = 60 mL
4.56 g = 1000 mL
Therefore y g = 4.56 g * 60 mL/1000 mL
y g = 0.273 g for a 60 mL solution. This answer is wrong.
The correct answer they say is 4.56 g but that is for 1L?
I think there is a mistake in the answer.
Many thanks,
I will post the second question separately as there is a limited number of characters per post.
The question never said anything about making a molar solution (mol/L) or anything about the volume. It’s simply asking for a mass. From my understanding, the question could have been rewritten: “Calculate the mass (g) of NaCl to give 78 mEq of sodium ions in solution.”
NaCl will dissociate into 1 mEq of sodium and 1 mEq of chloride ions. In other words, 1 mol NaCl will dissociate into 1 mol sodium and 1 mol chloride ions.
78 mEq = 78 mmol = 0.078 mol
(58.4 grams/mol) x (0.078 mol) = 4.56 grams of NaCl
your answer is right ie 0.273g/60ml will give 78 mEq of Na+ ions in 60ml of solution