Question

Red food dye (FD&C Red 40, also known as allura red AC) has the formula C18H14N2Na2O8S2.
1. What is the molar mass of FD&C Red 40?

To make a solution you take 0.0097 g of FD&C Red 40.
2. How many moles of FD&C Red 40 are in your sample?

You dissolve the sample of FD&C Red 40 in enough water to make 250.0 mL of solution. We call this the stock solution.
3. What is the concentration of the stock solution in terms of molarity?


You make two dilutions. You take 90.40 mL of the stock solution and dilute to 500.0 mL to make Solution #1. You then add 64.68 mL of Solution #1 to 30.32 mL of water to make Solution #2.
4. What is the concentration of the Solution #1 in terms of molarity?
5. What is the concentration of the Solution #2 in terms of molarity?

You measure the absorbance of each solution (at ?max = 524 nm) and obtain the following results:

Solution Absorbance
Stock solution 2.87
Solution #1 0.527
Solution #2 0.352
6. Determine the average epsilon for FD&C Red 40.

You are given an unknown solution of FD&C Red 40. You measure the absorbance and find it to be 0.416 at 524 nm.
7. Determine the concentration of the unknown solution in terms of molarity.

You have a solution of a yellow dye. You measure its absorbance at 524 nm and find it to be 0.018.
8. Determine the absorbance of a solution made by mixing 10.0 mL of Solution #1 of the red dye and 10.0 mL of the yellow dye.

You have 1.000 L of a unknown solution made by mixing Solution #1 and the yellow dye. The absorbance of this unknown solution is 0.426.
9. Determine the volume of Solution #1 used to make 1.000 L of the unknown solution.
10. Determine the volume of the yellow dye used to make 1.000 L of the unknown solution.

Answer 1

Concepts and reason

The problem is based on the concepts of molar mass, moles and molarity.

Molar mass is the sum of atomic mass of the elements present in the compound. Molarity defines the number of moles of solute in a liter of solution.

Fundamentals

Molar mass $\left( {\rm{M}} \right)$ is a physical quantity. It is determined by adding the atomic weights of the element present in the compound.

Moles $\left( n \right)$ are calculated by dividing the weight $\left( w \right)$ of the compound to its molar mass $\left( {\rm{M}} \right)$ . Number of moles is calculated by using the formula:

$n = \frac{w}{{\rm{M}}}$ …… (1)

Molarity $\left( m \right)$ expressed the concentration of the solution. It is calculated by using the formula as:

$m = \frac{n}{V} \times 1000$ …… (2)

${\rm{Red }}40$ or Allura Red AC is represented by the chemical formula ${{\rm{C}}_{{\rm{18}}}}{{\rm{H}}_{{\rm{14}}}}{{\rm{N}}_{\rm{2}}}{\rm{N}}{{\rm{a}}_{\rm{2}}}{{\rm{O}}_{\rm{8}}}{{\rm{S}}_{\rm{2}}}$ .

Chemical formula of ${\rm{Red }}40$ or Allura Red AC is ${{\rm{C}}_{{\rm{18}}}}{{\rm{H}}_{{\rm{14}}}}{{\rm{N}}_{\rm{2}}}{\rm{N}}{{\rm{a}}_{\rm{2}}}{{\rm{O}}_{\rm{8}}}{{\rm{S}}_{\rm{2}}}$

Molar mass, ${\rm{M = }}\left( {{\rm{18}} \times {\rm{C}}} \right){\rm{ + }}\left( {{\rm{14}} \times {\rm{H}}} \right){\rm{ + }}\left( {{\rm{2}} \times {\rm{N}}} \right){\rm{ + }}\left( {2 \times {\rm{Na}}} \right) + \left( {8 \times {\rm{O}}} \right) + \left( {2 \times {\rm{S}}} \right)$

Substitute $12{\rm{ g mo}}{{\rm{l}}^{ - 1}}$ for C, $1{\rm{ g mo}}{{\rm{l}}^{ - 1}}$ g for H, $14{\rm{ g mo}}{{\rm{l}}^{ - 1}}$ for N, $23{\rm{ g mo}}{{\rm{l}}^{ - 1}}$ for Na, $16{\rm{ g mo}}{{\rm{l}}^{ - 1}}$ for O and $32{\rm{ g mo}}{{\rm{l}}^{ - 1}}$ for S in the above expression.

$\begin{array}{c}\\{\rm{M = }}\left( {18 \times 12{\rm{ g mo}}{{\rm{l}}^{ - 1}}} \right) + \left( {14 \times 1{\rm{ g mo}}{{\rm{l}}^{ - 1}}} \right) + \left( {2 \times 14{\rm{ g mo}}{{\rm{l}}^{ - 1}}} \right) + \\\\\left( {2 \times 23{\rm{ g mo}}{{\rm{l}}^{ - 1}}} \right) + \left( {8 \times 16\;{\rm{g mo}}{{\rm{l}}^{ - 1}}} \right) + \left( {2 \times 32\;{\rm{g mo}}{{\rm{l}}^1}} \right)\\\\{\rm{M = 496 g mo}}{{\rm{l}}^{ - 1}}\\\end{array}$

The number of moles is calculated as from the equation (1).

$n = \frac{w}{{\rm{M}}}$

Substitute 0.0097 g for $w$ and $496{\rm{ g mo}}{{\rm{l}}^{ - 1}}$ for ${\rm{M}}$ /

$\begin{array}{l}\\n = \frac{{0.0097{\rm{ g}}}}{{496\;{\rm{g mo}}{{\rm{l}}^{ - 1}}}}\\\\n = 1.956 \times {10^{ - 5}}{\rm{ mol}}\\\end{array}$

Molarity of solution is calculated from the equation (2).

$m = \frac{n}{V} \times 1000$

Substitute, $1.956 \times {10^{ - 5}}{\rm{ mol}}$ for n, $250.0{\rm{ mL}}$ for V in the equation:

$\begin{array}{c}\\m = \frac{{1.956 \times {{10}^{ - 5}}{\rm{ mol}}}}{{250.0{\rm{ mL}}}} \times 1000\\\\ = 7.824 \times {10^{ - 5}}{\rm{ mol }}{{\rm{L}}^{ - 1}}\\\end{array}$

Ans:

The molar mass of ${\rm{Red }}40$ or Allura Red AC is $496{\rm{ g mo}}{{\rm{l}}^{ - 1}}$ .

The number of moles of ${\rm{Red }}40$ or Allura Red AC are $1.956 \times {10^{ - 5}}{\rm{ mol}}$ .

Molarity of the solution is $7.824 \times {10^{ - 5}}{\rm{ mol }}{{\rm{L}}^{ - 1}}$