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= 2,896x + -0.004. Her Kool-Aid sample had an absorbance of 0.842. If 0.484 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

percent by mass of FD&C Red 40 = 7.09 %

Explanation

The equation for best fit line is : y = 2896x - 0.004

absorbance of sample = 0.842 = y

from the equation of best fit line, x = (y + 0.004) / 2896

x = (0.842 + 0.004) / 2896

x = 2.92 x 10^-4

concentration of FD&C Red 40 = 2.92 x 10^-4 M

volume of solution = 8 fl oz

volume of solution = 8 fl oz * (1 L / 33.814 fl oz)

volume of solution = 0.2366 L

moles FD&C Red 40 = (concentration of FD&C Red 40) * (volume of solution in Liter)

moles FD&C Red 40 = (2.92 x 10^-4 M) * (0.2366 L)

moles FD&C Red 40 = 6.908 x 10^-5 mol

mass FD&C Red 40 = (moles FD&C Red 40) * (molar mass FD&C Red 40)

mass FD&C Red 40 = (6.908 x 10^-5 mol) * (496.42 g/mol)

mass FD&C Red 40 = 0.0343 g

mass percent FD&C Red 40 = (mass FD&C Red 40 / mass sample) * 100

mass percent FD&C Red 40 = (0.0343 g / 0.484 g) * 100

mass percent FD&C Red 40 = (0.0709) * 100

mass percent FD&C Red 40 = 7.09 %