Tetrahydrocannabinol (thc), the active agent in marijuana, is composed of 80.16% carbon, 9.63% hydrogen and 10.17% oxygen by mass. What is the empirical formula of thc?

Question

The answer I got was C7H10O, however I think I am missing a step since my answer was incorrect.

This is what I did

80.16g C (1mol/12.01g)C =7 mol C

9.63g H (1mol/1.008g)H =10mol H

10.17g O (1mol/16.00g)O =0.6= 1 mol O

therefore said the empirical formula was => C7H10O

can someone please tell me what I'm doing wrong?

Answer 1

Hi Yolanda J.,

You're rounding your numbers too soon!

Calculate exactly, then find the best ratio: hint: usually the oxygens are the least, so try to make them an integer.

So: 80.16/12.01 = 6.67443796836

9.63/1.008 = 9.55357142857

10.17/16.00 = 0.635625

So try oxygens = 1 as the first ratioing (or the equivalent):

6.6744/0.6356 = 10.50

9.5536/0.6356 = 15.03

0.6356/0.6356 = 1

Well, you can see that you just need to double these values, to be nicely integral.

So: C₂₁H₃₀O₂ .

And by the way, marijuana contains oodles of chemicals, many of them active in one way or another. There's quite an industry built around CBD (cannabidiol), for example.

-- Cheers, -- Mr. d.