We’ll commonly see milliequivalents (abbreviated mEq) used on labels for electrolytes. Electrolytes are micro-nutrients that dissociate (break down) into ions in water and are typically added to TPN (total parenteral nutrition) solutions. Examples of electrolytes are “normal saline” (a 0.90% W/V solution of sodium chloride or NaCl) and “sodium bicarbonate” (aka baking soda or NaHCO3).

The ions formed by electrolytes carry an electric charge and aid in different bodily functions. In biochemistry, an “equivalent” is generally described as the amount of any chemical required to combine or react with 1 mole (abbreviated mol) of hydrogen ions (approximately 1.00794 grams of hydrogen ions). Since the prefix milli- stands for 1/1,000th then a milliequivalent is 1/1,000th of an equivalent. Furthermore, since 1 equivalent = 1 mole then 1 milliequivalent equals 1 millimole (abbreviated mmol).

If we want to find out how many mg of a chemical are in 1 mEq of it:

- Calculate the combined molecular mass of all the ions.
- Append mg to this total and divide it by the unsigned value (absolute value) of its highest valence ion (charge).

To make this easier I have provided a table (sorted in descending order by charge) of common electrolyte ions, their atomic or molecular mass (rounded to the hundredths place) and their chemical formula. So for example, 1 mEq of normal saline (sodium chloride) contains approximately 58.44 mg of NaCl because it’s combined molecular mass is 22.99 u (Na) + 35.45 u (Cl) and the highest unsigned charge of either ion is 1.

Similarly, we can use this information to calculate percent strength! Let’s use 1 liter of sodium bicarbonate 595 mEq/L as an example. First we find out how many mg of sodium bicarbonate are in 1 mEq of this solution:

$$\frac{\text{22.99 u (Na) + 61.02 (HCO}_\text{3}\text{)}}{\text{|1|}}$$

$$= \frac{\text{84.01 mg}}{\text{|1|}} = \frac{\text{84.01 mg}}{\text{1 mEq}}$$

Then we’ll multiply that ratio by mEq/mL for mg/mL and convert that to g/100 mL for our final % W/V concentration of sodium bicarbonate:

$$\frac{\text{595 mEq}}{\text{1 L}} \times \frac{\text{1 L}}{\text{1,000 mL}} = \frac{\text{0.595 mEq}}{\text{1 mL}} \;\text{;}$$

$$\frac{\text{0.595 mEq}}{\text{1 mL}} \times \frac{\text{84.01 mg}}{\text{1 mEq}} = \frac{\text{50 mg}}{\text{1 mL}} \;\text{;}$$

$$\text{50 mg} \times \frac{\text{1 g}}{\text{1,000 mg}} = \text{0.05 g} \;\text{;}$$

$$\frac{\text{0.05 g}}{\text{1 mL}} \times \frac{\text{100}}{\text{100}} = \frac{\text{5 g}}{\text{100 mL}}$$

$$= \text{5% W/V NaHCO}_\text{3}$$

Charge and Mass | Ion; Formula |
---|---|

+ 3 and 26.98 u | Aluminum; Al |

+ 2 and 40.08 u | Calcium; Ca |

+ 2 and 55.85 u | Iron; Fe |

+ 2 and 24.31 u | Magnesium; Mg |

+ 1 and 1.01 u | Hydrogen; H |

+ 1 and 39.10 u | Potassium; K |

+ 1 and 22.99 u | Sodium; Na |

+ 1 and 18.04 u | Ammonium; ($$\text{NH}_\text{4}$$) |

– 1 and 35.45 u | Chlorine; Cl |

– 1 and 59.04 u | Acetate; ($$\text{C}_\text{2}\text{H}_\text{3}\text{O}_\text{2}$$) |

– 1 and 89.07 u | Lactate; ($$\text{C}_\text{3}\text{H}_\text{5}\text{O}_\text{3}$$) |

– 1 and 195.15 u | Gluconate; ($$\text{C}_\text{6}\text{H}_\text{11}\text{O}_\text{7}$$) |

– 1 and 61.02 u | Bicarbonate; ($$\text{HCO}_\text{3}$$) |

– 1 and 96.99 u | Dihydrogen Phosphate; ($$\text{H}_\text{2}\text{PO}_\text{4}$$) |

– 2 and 60.01 u | Carbonate; ($$\text{CO}_\text{3}$$) |

– 2 and 95.97 u | Hydrogen Phosphate; ($$\text{HPO}_\text{4}$$) |

– 2 and 96.06 u | Sulfate; ($$\text{SO}_\text{4}$$) |

– 3 and 189.10 u | Citrate; ($$\text{C}_\text{6}\text{H}_\text{5}\text{O}_\text{7}$$) |

Lastly, let’s look at how to find mEq/L from percent strength. You’ve probably already guessed that we’d be working backwards to figure this out. Let’s use normal saline as an example. Since it has a percent strength of 0.09% we know that it has 0.9 grams of NaCl per 100 mL of solution or 9 grams per 1,000 L. Since we know that there’s 58.44 mg of NaCl in 1 mEq of normal saline we can convert the 9 grams of NaCl we’re working with to mg first using the unit factor method:

$$\text{9 g} \times \frac{\text{1,000 mg}}{\text{1 g}} = \text{9,000 mg}$$

Then we can multiply this 9,000 mg/L ratio by a unit factor of 1 mEq/58.44 mg to find out how many mEq are in 1 L of normal saline:

$$\frac{\text{9,000 mg}}{\text{1 L}} \times \frac{\text{1 mEq}}{\text{58.44 mg}}$$

$$= \frac{\text{154 mEq NaCl}}{\text{1 L solution}}$$

On a final note, it’s important to know the difference between measurements of mOsmol/L and mmol/L. Pay attention, as always, to what’s written down so you don’t misconstrue one for the other. mOsmol/L deals with osmolarity, where for example the ionic compound NaCl dissociates fully in water to become a Na+ ion and a Cl- ion giving a 1 mol/L NaCl solution an osmolarity of 2 osmol/L (thus NaCl has an osmolarity of 2 x 154 or 308 mOsmol/L).

mmol/L deals with molarity where 1 mmol equals 1 mEq. There is also osmolality which is defined as the number of osmoles of solute per kilogram of solvent so be aware that it too exists. If you’d like to go even deeper you may refer to a general science book.