Question

A student creates a sports drink by dissolving 5.0 g of powder in 250 mL of water. The green colour of the drink is due to the presence Green Dye X, which has a lambdamax of 630 nm. A calibration curve (absorbance versus concentration) at lambdamax is created for varying concentrations (mM) of Green Dye X which has the equation y = 26.4x. If 1.0 mL of the sports drink is diluted to 10. mL with water, then the absorbance is measured to be 0.450, what is the mass percent of Green Dye X in the drink powder if the molecular weight is 418.4 g mol-1?

Answer 1

The mass percent of Green Dye X in the drink powder if the molecular weight is 418.4 g/mol is 0.355%

Beer's law explains the relation between the concentration of a solution to its absorbance (A) by using the equation.

• A = ∈×C×L

here:

• ∈ = molar absorptivity
• L = path length

Now from the calibration curve in the question, the plot between the absorbance and the concentration gives a straight line of y = mx.

Here

• y = absorbance
• x = concentration

The equation of the calibration curve is given as:

• y = 26.4x

i.e.

• 0.450 = 26.4x
• $\mathbf{x = \dfrac{0.450}{26.4}}$
• x = 0.017 mM

If the sports drink was 10 times diluted while taking the absorption measurement. Then, the actual concentration of green dye in the sports drink is:

= 0.017 mM × 10

= 0.17 mM

The number of moles of green dye in 250 mL (i.e. 0.25 L) sports drink is estimated as:

= 0.17 mmol/L × 0.25 L

= 0.0425 mmol

However, the mass of the green dye present in the sports drink is

= 0.0425 mmol ×418.4 g/mol

= 17.782 mg

Finally, the mass percent of green dye in the drink powder is:

$\mathbf{ = \dfrac{mass (dye)}{mass (powder)} \times 100\%}$

$\mathbf{ = \dfrac{17.782 \ mg }{5.0 \ g} \times 100\%}$

$\mathbf{ = \dfrac{17.782 \ mg }{5000 \ mg} \times 100\%}$

= 0.355 %

Learn more about mass percent here:

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