Endocrinology Math Problem #2

[FamilyGuy0395]

Problem: Testosterone at a physiologic concentration is 30 nmol/L. Molar mass (or molecular weight) of testosterone is 288.42 g/mol. Rob Gronkowski is 6'6" and 265 lbs. Assuming he is normally hydrated (8% of body weight is water), how much testosterone does he have in his system? (Provide your answer to the nearest whole number with appropriate units. For example, don't provide 137,000 g; instead, use the nearest reasonable metric unit of 137 kg. Similarly, don't answer 0.000002 g; instead, use 2 μg.)

What I've Tried: Given both the physiologic concentration and molar mass measurements use the unit mol, I converted 30 nmol/L to 3E-8 mol/L. Furthermore, I converted 265 lbs to kg, which roughly = 120.46 kg. I also calculated the water weight to be roughly = 9.64 kg.

 

[Dr.Peterson]

That's good (to two significant figures, since you used 2.2 lb/kg).

Now how many liters of water is that? How many mols of testosterone will be in it? How many grams is that?

 

[FamilyGuy0395]

@Dr.Peterson After redoing some of my calculations, here's what I'm getting for the answer:

m_Rob = 265 lbs. ≈ 120.2 kg

m_H2O = 8% of 120.2 kg = 0.08(120.2 kg) ≈ 9.62 kg

ρ_H20 = m/V
1 kg/L = 9.62 kg/V
V = 9.62 L

mol_test. = (3E-8 mol/L)(9.62 L) ≈ 2.88E-7 mol

m_test. = (288.42 g/mol)(2.88E-7 mol) ≈ 8.3205E-5 g ≈ 83 μg

 

[Dr.Peterson]

Yes, that's what I got.