The drug dobutamine (fw=301.38g/mol, c18h23no3c18h23n03) has a molar absorptivity of 703 at 262nm. one tablet is dissolved in water and diluted to a volume of 2l. if the solution exhibits an absorbance of 0.687 in the uv region at 262nm in a 1- cm cell, how many grams to dobutamine are contained in the tablet? a. 0.5277g b. 0.58889 c. 9.77x 10g d. 97.778ge. 9.7778g

0.589 g

Explanation:

The absorbance (A) of an analyte is related to its concentration (c) through the Beer-Lambert's law.

A = ε . l . c

where,

ε: molar absorptivity

l: optical path length

A = ε . l . c

0.687 = 703 M⁻¹.cm⁻¹ . 1 cm . c

c = 9.77 × 10⁻⁴ M

The molar mass is 301.38 g/mol and the volume of solution is 2 L. The mass of Dobutamine is:

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.