Question

Lead poisoning is treated by adding chelating agents which bind to metal ions to the blood, which allows them to be removed by the kidneys into the urine. One reagent used this way is succimer (C_4H_6O_4S_2); each mole of succimer binds to 1 mole of lead. You are designing a treatment for a patient whose blood lead levels are 45 mu g/dL (1 dL = 100 mL), and you estimate the total blood volume to be 5.0 L. How many milligrams of succimer should you add to bind all of the lead in the patient's bloodstream?

Answer 1

given

concentration of lead = 45 ug / dL

we know that

1 ug = 10-6 g

1 dL = 100 ml

so

concentration of lead = 45 x 10-6 / 100 g/ml

concentration of lead = 4.5 x 10-7 g/ml

now

volume of blood = 5 L

we know that

1 L = 1000 ml

so

volume of blood = 5000 ml

now

mass of lead = concentration in g/ ml x volume of blood (ml)

mass of lead = 4.5 x 10-7 x 1000 g

mass of lead = 4.5 x 10-4 g

now

moles = mass / molar mass

molar mass of lead = 207.2 g/mol

so

moles of lead = 4.5 x 10-4 / 207.2

moles of lead = 2.172 x 10-6

given

moles of succimer required = moles of lead

so

moles of succimer required = 2.172 x 10-6

now

mass = moles x molar mass

molar mass of succimer = 182.22 g/mol

so

mass of succimer required = 2.172 x 10-6 x 182.22 g

mass of succimer required = 3.9575 x 10-4 g

now

1 g = 1000 mg

so

mass of succimer required = 3.9575 x 10-4 x 1000 mg

mass of succimer required = 0.3957 mg

so

0.3957 mg of succimer is required