Mathematical Manipulations for drug dosage - how to calculate molar concentration from mass/kg bodyweight (Sep/10/2007 )
Hello, I am very much confused about some calculations and would appreciate if anyone out there can bail me out.
How can one calculate the Molar concentration of a drug if one is asked to inject an animal with 5mg/Kg body weight ( assuming the drug is Aluminum sulphate).
If the animal weighs 60g and one is to inject it with 5mg/Kg bodyweight of Aluminum sulphate, what is the molar concentration of the amount of Aluminum sulphate to be injected.
I would very much appreciate your response. Thanks in advance.
Okay, this is how I would do it & then I would get someone else to double check my calculations!
First workout how many mg you would need to inject for 5mg/kg and a 60g animal
5mg/1000g = A mg/60g
(5/1000) x 60 = A mg
A = 0.3 mg
MW of Ammonium sulphate = 132.14
1M = 132.14g per 1L (132.14/0.3 = 440.47)
1M/ 440.47 = 0.3g/L = 2.27mM, then divide by 1000 to get mg/L
0.3mg of Ammonium sulphate = 2.27uM/L or 2.27nM/mL
I hope this makes sense, and is correct! 
Quantity of drug = 5*60\1000= 0.3 mg per animal. What is about route of drug administration and injection volume. In general Molar concenration =(0.3\M(Al2(SO4)3)* 1000)/V(ml) mM.
Thanks for your response. the route of administration is intra-peritoneal. I hope this would not have any effect on the calculation of the dose and it molar concentration
First workout how many mg you would need to inject for 5mg/kg and a 60g animal
5mg/1000g = A mg/60g
(5/1000) x 60 = A mg
A = 0.3 mg
MW of Ammonium sulphate = 132.14
1M = 132.14g per 1L (132.14/0.3 = 440.47)
1M/ 440.47 = 0.3g/L = 2.27mM, then divide by 1000 to get mg/L
0.3mg of Ammonium sulphate = 2.27uM/L or 2.27nM/mL
I hope this makes sense, and is correct!
Thanks for your response. I really appreciate it. it makes some sense. i also hope it is correct but would never the less try it for now except you have another method in mind.