Morphine has the formula C17H19NO3. It is a base and accepts
one ...

Question

Morphine has the formula C17H19NO3. It is a base and accepts
one proton per molecule. It is isolated from opium. A 0.682-g
sample of opium is found to require 8.92 mL of a 0.0116 M solution
of sulfuric acid for neutralization. Assuming that morphine
is the only acid or base present in opium, calculate the
percent morphine in the sample of opium.

Accepted Answer

Hey everyone today, we're being asked to find the mass percent of amphetamine, A weak base that is present within a 30 mg drug sample that has been titrate. Id with sulfuric acid. So there are a few key takeaways from the question stem right off the bat. The drug sample is indeed 30 mg, 30 mg. The sulfuric acid is in a concentration of 0.01 molar And it took 6.95 ml of sulfuric acid to completely neutralize all the amphetamine in the drug sample. So before moving on we can go ahead and write out the balanced equation For this reaction. So we have amphetamine C9 h. 13 and reacts with sulfuric acid H two s. 0. four to form some products. Now a key thing to note is that sulfuric acid is die product. It has two acidic protons. Which means that when ty traded acidic protons will titrate in succession. And in the first line of the question, we can see that amphetamine is a weak base that can accept only one hydrogen Ion per molecule can only accept a one H plus. Which means if we want to completely react with the sulfuric acid, we're going to need two equivalents amphetamine. So this will give us C9 h. 13 and H plus we'll have two of those plus the leftover sulfate two minutes. So from here we can see that we have a molar ratio that for every two amphetamines for every two amphetamines We have one sulfuric acid. So keep this in mind for later. Now to find Masterson. To find Master sent, we need to take the mass of amphetamine divided by the total mass of the drug sample of mass of the drug sample. Let's write these out in a little different colors. The massive amphetamine massive amp divided by The mass of the truck sample which we know to be 30 mg. So we need to convert something in order to get the massive amphetamine easiest way to do this would be actually by taking the volume of sulfuric acid. Remember It took 6.95 mm. Took 6.95. Let's write that in the page Took 6. ml. No sulfuric acid to completely neutralize and react with the amphetamine. President of the drug sample. Which means we can use this volume to convert to amphetamine. The massive amphetamine. We want to convert this first into leaders And recall that for every one Middle leader there are 10 to the -3 leaders. So milliliters will cancel out. And now because we are given the polarity of sulfuric acid which is nothing but molds divided by leaders. We can use that as another conversion. We have 0.01 moles Each two s. 0. four Divided by one leader and he will cancel out. So now that we have moles of H. Two S. 04. We can use the molar ratio found in the balanced equation to arrive at molds of amphetamine. So for every two moles for every two moles of amphetamine, which I'll shorten to AMp We have one mole of H two S. 04. So these values will cancel each other out. And to convert moles to g. We simply multiply by the molar mass of amphetamine. Amphetamine has nine hydrogen. Or sorry, nine carbons, each of which are 12 grams from more 13 Hydrogen, which are each one g per mole, And a single nitrogen, which is 14 g from all. So overall the molar mass of amphetamine will be 1:35 g per mole. And this will give us a final total value of zero point zero 188 grams of amphetamine. However, we're not done here. We still need to get this in terms of milligrams to compare it to the drug sample. And this is really simple. zero point will write this in red, 0.188, 0188 grams times. We have 10 to the third milligrams for every one g. So this will cancel out. And we will be left with 18 0.8 mg of amphetamine. Look at that in blue. Sorry that is our massive amphetamine 18.8 milligrams of amp. So from here we can go ahead and put that back into the equation. We wrote up here, we will have that the mass percent mass percent percent is equal to 18.8 milligrams of amp per 30 mg a drug the total sample multiply this value by as this is a percentage we're dealing with And we get 62.7%. Therefore, the mass percent of amphetamine in the drug sample after filtration Is 62.7%. I hope this helps. And I look forward to seeing you in the next one.

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