Question

After creating her standard curve for absorption versus the concentration (M) of FD&C Red 40, a student found that her best fit linear line for FD&C Red 40 was y = 2800x + -0.003. Her Kool-Aid sample had an absorbance of 0.74. If 0.558 grams of Kool-Aid powder was used to prepare an 8-fl oz cup of her assigned flavor, what is the percent by mass of FD&C Red 40 in her 8-fl oz cup?

Answer 1

Given that,

linear line equation , y = 2800x + -0.003

Let us consider Beer's Law,  

Absorbance = e * l *C where l = path length of light , e=absorptivity of species

A callibration curve between Abs and concentration is linear of the form y = mx+c,

e=2387 M^-1 for l=1cm ,c=intercept = 0.003

If Abs=0.74, Actual Abs= 0.74 - 0.003

= 0.737

Concentration of FD&C Red 40 = Abs / e

= 0.737 / 2800 M^-1

= 0.000263

= 2.63 * 10^-4M

mol of FD&C Red 40 = (8 fl oz)*(29.573ml/1fl oz) *(1L/1000ml)*(2.63*10^-4mol/L)

= 6.22 *10^-5 mol

mass of FD&C Red 40 = mol*molar mass of of FD&C Red 40

= (6.22 *10^-5 mol)*(496.42g/mol)

= 0.03087 g

% FD&C Red 40 = (mass of FD&C Red 40 / mass of Kool-Aid powder)*100

= (0.03087 / 0.558)*100

= 5.53 %