Question

A 67 kg patient is in need of intravenous iron supplementation. She was ordered to receive a single intravenous 7 mg/kg dose of iron (III) sulfate (ferric sulfate - Fe2(SO4)3 - MW=399.88 g/mol). Iron (III) sulfate was prepared as a stabilized aqueous solution for this patient by dissolving 875 mg of stabilized iron sulfate salt in water to produce 10 mL of solution. How many milliequivalents of iron (Fe 3+) will the patient receive as a single dose? Assume the salt completely disassociates in solution.

Answer 1

Order.

Iron sulfate.

Weight of the patient= 67kg.

Needed dose. = 7 mg/ kg in single dose.

Total dose that is needed is.

= 69*7mg/kg body weight.

=469 mg iron sulfate is needed.

Available total solution is 875 mg in 10 ml.

= Desired dose/ dose on hand *quantity of dilution.

= 469 mg/ 875 mg *10 ml

=. 5.36 ml ~ 5.40 ml aqueous solution is needed.

Molecular weight = 399.88 gm/ mol.

Weight of drug needed is= 469 mg.

Valency of iron (iii) sulphate is 3 ( three)

Formula mEq=

mEq= Weight in grams   *valence / molecular weight.

mEq= 469 mg * 3 / 399.88 gm/ mol.

As,molecular weight is given in grams / mol ,convert it to mg/ mmol

To convert 399.88gm/ mol *1000mg/ 1g * 1mol /1000mmol

399.88 mg/ mmol

mEq= weight in mg * valence / molecular weight.

mEq= 469 mg *3/ 399.880 mg/ mmol

mEq= 1407 / 399.880

= 3.51 Meq ( milliequivalents) will be received by the patient in a single dose.