[B9766]

I've followed the calculations given with the explanation of Atomic Mass and Atomic Mass Units up to a point.
I follow that the defiinition for Atomic Mass is based upon Carbon-12 and (for C-12): $$Atomic\  Mass = \dfrac{12g \ of\ ^{12}C}{6.022*10^{23} \  of\ ^{12}C\  atoms} = 1.993*10^{-23}g$$
And computing the mass of 1 amu: $$ 1\  amu = \dfrac {1.993*10^{-23}g} {12} = 1.661*10^{-24}g$$
All good. However, when I compute the number of amu's in 1 gram, I get: $$1\ gram = \dfrac{12\ amu} {1.993*10^{-23}} = 6.021*10^{23} amu$$
And that makes sense to me because an amu should be a very, very small thing so there should be a whole lot of them in one gram.
But the book says: $$1\ gram = 6.022*10^{-24}amu$$
What did I get wrong?

[mjc123]

Rounding errors. If you do 12/6.022e23 and store the answer in the memory (assuming you're using a calculator), and then do 12/(number in memory) you will get the right answer. Rounding off 1.99269.. to 1.993 introduces a slight inaccuracy which will propagate in subsequent calculations. The rule is to do your calculations to high precision, and round the answer to the appropriate number of significant figures at the end.

[B9766]

Did you notice the difference in the exponent? The difference between $$10^{23}$$ and $$10^{-24}$$ isn't a rounding error.

[mjc123]

Oops, no I didn't. The book has obviously made an error. Your third equation is just a rearrangement of your first, so if you start with 6.022e23, that's what you should get back.
Textbooks can make mistakes. You'd be surprised (or perhaps not) how often that comes up on this forum.

[B9766]

Thanks mjc123. I really thought this was the case but the last one I posted, where I also thought they made a typo, was actually a mind fart on my part. I just wanted to be sure I didn't miss something major again. I just posted another one. This time I really think it's me but I can't find the error in my calculations. It's "Molecular formula derivation for C and H".