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EDIT 9/2/2012:

Acutally the question is mistaken. There is actually 595mg of potassium gluconate per tablet of which 99mg of that is potassium (the other 496mg is gluconate). Therefore, there is actually 2.54mEq of potassium (K+) per tablet. See below for calculation.

99mg K x (1 mmol K/39.1mg K) x (1mEq K/1mmol K) = 2.54mEq K

{By plugging in 595mg in the below calculation instead of 99mg you will also come up with 2.54mEq K}

Sorry, with all due respect, the previous answer (see below) is mistaken. I've Never bothered write an answer before, but since this is medical and potentially important to someone's health, let me try a brief explanation.

A "mEq" or milli-equivalent, refers to how much absolute potassium, measured as moles (avagadro's number of total molecules (atoms in this case), something like 6.022x10^23 potassium ions (or atoms)), is delivered in a given amount or dose. Since potassium will always be delivered as a salt, the amount (mass) of salt required to deliver a mEq of potassium will depend on the other ion in the salt. Potassium chloride is one popular form of potassium, which has a different (much less) weight (mass) than another popular form of potassium, potassium gluconate.

The molecular weight of potassium gluconate is reported to be 234.25, meaning a mole (or 6.022x10^23 molecules of potassium gluconate) will weight 234.25g. Each molecular of the salt delivers one molecule of potassium so 234.25g will deliver one molar equivalent of potassium. 234.25 milligrams of potassium gluconate will deliever one milli-equivalent of potassium.

SO... 99mg potassium gluconate divided by 234.25mg/mol potassium gluconate = X mEq of potassium gluconate, which equals about 0.423 mEq.

Hope this helps, for all my faults of explanation.

Previous Answer:

[That depends on the concentration strength of the potassium gluconate. Just knowing the mass of the drug is not enough information to determine the mEq strength. We must also know the volume and contents of diluent as well.]

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