Some progressive hair coloring products marketed to men, such as ...

Question

Some progressive hair coloring products marketed to men, such as Grecian Formula 16, contain lead acetate Pb(CH3CO2)2. As the coloring solution is rubbed on the hhair, the Pb2+ ions react with the sulfur atoms in hair proteins to give lead(II) sulfide (PbS), which is black. A typical coloring solution contains 0.3 mass% Pb(CH3CO2)2, and about 2 mL of the solution is used per application. 
(a) Assuming that 30% of the Pb(CH3CO2)2 is converted to PbS, how many milligrams of PbS are formed per application of the coloring solution?

Accepted Answer

Welcome back everyone. We're told that the bleaching of hair usually involves the reaction of hydrogen peroxide with melanin given by this formula here, C 18 H 10 N 204 as the bleaching solution is rubbed into the hair melon and is oxidized by hydrogen peroxide. In the following reaction, where we have melon reacting with 34 moles of peroxide to produce two moles of ammonia, 36 moles of water and 18 moles of carbon dioxide. We're told that a bleaching solution contains 9% of peroxide by mass. If 20 g of the bleaching solution is used to calculate the mass and milligrams of melanin that becomes oxidized. So what we first want to make note of is our mass of peroxide. Using that info of our percent by mass, which we are interpreting the 9% as a decimal being 0.9. We're going to take that 0.9 and we're going to multiply it by the mass of the bleaching solution being 20 g because we know our bleaching solution contains hydrogen peroxide in the formula. This is gonna give us our mass of peroxide equal to a value of 1.8 g. Now that we have our massive peroxide, we also want to make note of its molar mass from the periodic table Which we see has a molar mass of 34.01 g per mole. Now going to our molar mass of our melanin, which is c 18 H 10 n 204. We see from the periodic table that Melanin has a molar mass equal to 318.28 g per mole. And so now we want to go ahead and figure out the mass of our melanin that reacts When the bleaching is done to the hair. So beginning with that mass of peroxide that we calculated, we have 1.8 g of peroxide To get rid of g of peroxide. We're going to plug in that molar mass of peroxide where we determined we have 34.01 g of peroxide equivalent to one mole of peroxide. We can get rid of the units of grams of peroxide. And now we can focus on getting into melanin in our study geometry here by looking at our balanced equation where we see we have according to our coefficients, one mole of our peroxide reacting with 34 moles. Or sorry, one mole of melanin reacting with 34 moles of peroxide. So we have in our denominator are one mole of melanin, which is and rather not melanin. But we're going to plug in the 34 moles of peroxide in the denominator so that we can cancel out moles of peroxide. We will have our one mole of melanin in the numerator C 18 H 10 N 204. Now we want to plug in our molar mass of melanin as our next conversion factor. Where in our denominator, we understand that we have a mass. Or sorry, for one mole of melanin C 18 H 10 n 204. That is equal to our mass of 318.28 g of melanin. So now we can cancel out moles of melanin. And again our final answer according to the prompt should be in milligrams. So we're going to go from grams to milligrams of melanin. And so we would recognize that our prefix milli tells us that we have to the negative third power of our base unit grams. This allows us to get rid of grams and now we're left with milligrams of melanin as our final unit which tells us our massive melanin that reacts or that is oxidized according to our prompt. And this is going to result in the value Of 495.45 mg. And this would be our final answer to complete this example as our mass of melanin in milligrams. That is oxidized. I hope everything I explained was clear. If you have any questions, leave them down below and I will see everyone in the next practice video

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