After creating her standard curve for absorption versus the concentration (M) of

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 2736x + 0.007 Her Kool-Aid sample had an absorbance of 0.386. If 0.532 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? Do not include the percentage sign in your answer. Pay attention to units. You are trying to find the mass percentage of red dye, so you need to know each variable in the equation in the manual

Explanation / Answer

Ans. Step 1: Determine [FD&C] in 8-fl oz cup: It’s assumed that the abs of 8fl oz sample is 0.386.

The trendline equation for the absorbance vs concertation graph is in the form of y = m x + b.

#b. In the graph, Y-axis indicates absorbance and X-axis depicts concentration. That is, according to the trendline (linear regression) equation y = 2736x + 0.007 obtained from the graph, 1 absorbance unit (1 Y = Y) is equal to 2736 units on X-axis (concentration) plus 0.0079.

# Given: Abs of sample = 0.386

Now, putting Y = 0.386 in trendline equation-

            0.386 = 2736x + 0.007

            Or, X = (0.386 – 0.007) / 2736 = 1.385 x 10-4

Therefore, [FD&C] in 8-fl oz cup sample = 1.385 x 10-4 M

# Step 2: Total moles of dye in 8-fl oz sample = Molarity x Vol in liters

                                                = 1.385 x 10-4 M x (8 oz x 29.5735 mL oz-1)

                                                = 1.385 x 10-4 M x (236.588 mL)

                                                = 1.385 x 10-4 M x 0.236588 L

                                                = 3.2773 x 10-5 mol

# Mass of dye in sample 8-fl oz sample = Moles x MW

                                                = 3.2773 x 10-5 mol x (496.42 g/ mol)

                                                = 0.0163 g

# Step 3: Since 8fl oz sampe is prepared from 0.532 g Kool-Aid powder, the dye content in both samples must be the same.

So,

            Dye content of 0.532 g Kool-Aid powder = 0.0163 g

Now,

            % dye = (mass of dye / Mass of sample) x 100

                        = (0.0163 g / 0.532 g) x 100

                        = 3.06 %

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