# Calculate the number of Carbon atoms in .0400g C7H5NO3S

I understand how to get from grams to moles and from moles to molecules with Avagadro's number. That gives us 1.31x10^20 molecules of C7H5NO3S. Now we take that and multiply by 7 to get the individual Carbon atoms. This is where I get confused because my calculator says 9.17x10^20 but my book says 9.20x10^21. Please help explain!

• Based on the mass and molecular formula you've given, I get $9.21\times 10^{20}~\mathrm{C~atoms}$, essentially in agreement with your number. – hBy2Py Feb 25 '16 at 22:08
• Yes, the book answer is wrong. – orthocresol Feb 25 '16 at 22:08
• Are you sure that $\ce{C7H5NO3S}$ is the molecular formula? Could it instead be the empirical formula? – hBy2Py Feb 25 '16 at 22:09
• My book doesn't actually specify, I was just assuming that this was the molecular formula. – Ashley Feb 25 '16 at 22:16
• It was originally 40.0mg. – Ashley Feb 25 '16 at 22:23

## 1 Answer

.0400g is a really non-chemical way to write that number.

I expect the text meant 0.400g, which is 10 times higher than the number actually printed, and this would explain why the book's answer is 10 times higher than the actual answer: they are working from the number they didn't print.