Percent morphine in a sample of opium?

Question

Morphine has the formula C17H19NO3. It is a base and accepts one proton per molecule. It is isolated from opium. A 0.678-g sample of opium is found to require 8.93 mL of a 1.19×10−2 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.

Any help would be appreciated. Thanks.

Answer 1

First, you need a balanced equation: in this case, it is a little ambigous as they don't say if both protons of the H2SO4 are being reacted, I am assuming both do (so one H2SO4 reacts with 2 morphines.

So the formula would be:

2 Morphine + H2SO4 --> 2 morphine(H) + SO4-2

Now all you need to do is calculate how many moles of H2SO4 was used:

1.19 x 10^-2 x 0.00893L = 0.000106 moles H2SO4

From the equation, we know that one H2SO4 reacts with 2 morphines so:

0.000106mole x (2morphine/1H2SO4) = 0.000213 moles morphine

Now we need to calculate the molecular weight for morphine:

(17 x 12.01) + (19 x 1.01) + 14.01 + (3 x 16.00) = 285.37g/mole

Now we can calculate how many grams of morphine was in the sample:

285.37g/mole x 0.000213mole = 0.0608g

Now we can get the percentage of morphine inthe sample:

100% x (0.0608/0.678) = 8.97% morphine

This will be good practice if you ever want to be drug dealer.