Question

A sample of adenosine triphosphate (ATP) (MW \(507, \varepsilon=14,700 M^{-1} \mathrm{~cm}^{-1}\) at \(257 \mathrm{~nm}\) ) is dissolved in \(5.0 \mathrm{~mL}\) of buffer. A \(250-\mu \mathrm{L}\) aliquot is removed and placed in a \(1 \mathrm{~cm}\) cuvette with sufficient buffer to give a total volume of \(2.0 \mathrm{~mL}\). The absorbance of the sample at \(257 \mathrm{~nm}\) is \(1.15\). Calculate the weight of ATP in the original \(5.0 \mathrm{~mL}\) sample.

Short answer

Answer: The weight of ATP in the original 5.0 mL sample is approximately \(1.59 × 10^{-4}\) grams.

Step-by-step solution

Calculate the concentration of ATP in the cuvette

To calculate the concentration of ATP, use the Beer-Lambert Law equation: A = ε × l × c Where: A = Absorbance (1.15) ε = Molar absorptivity (\(14,700 M^{-1} cm^{-1}\)) l = Pathlength of the cuvette (1 cm) c = Concentration of ATP (unknown) Rearrange the equation to solve for the concentration of ATP: c = A / (ε × l) Substitute the given values: c = 1.15 / (14,700 × 1) c ≈ \(7.82 × 10^{-5} M\)

Calculate the moles of ATP in the cuvette

Now we have the concentration of ATP in the cuvette, we can calculate the moles of ATP using the following formula: moles = concentration × volume Remember that the volume of the aliquot in the cuvette was 250 μL, which was diluted to a total volume of 2.0 mL. Therefore, we will use the volume of 2.0 mL in the equation: moles = \(7.82 × 10^{-5} M × 0.002 L\) moles ≈ \(1.564 × 10^{-7} mol\)

Calculate the moles of ATP in the original sample

We need to determine the moles of ATP in the original 5.0 mL sample, we can set up a proportion equation: (\(1.564 × 10^{-7} mol\) in 2.0 mL cuvette) / (\(x\) mol in 5.0 mL sample) = (0.25 mL aliquot) / (5.0 mL sample) \(x\) = (\(1.564 × 10^{-7} mol\) in 2.0 mL cuvette) × (5.0 mL sample / 0.25 mL aliquot) \(x\) ≈ \(3.13 × 10^{-7} mol\)

Calculate the weight of ATP in the original sample

Now that we have the moles of ATP in the original sample, we can calculate the weight of ATP using its molecular weight (MW = 507): weight = moles × molecular weight weight = \(3.13 × 10^{-7} mol\) × 507 weight ≈ \(1.59 × 10^{-4} g\) Therefore, the weight of ATP in the original 5.0 mL sample is approximately \(1.59 × 10^{-4}\) grams.