How many Cr atoms in 78.82 g potassium dichromate?

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The discussion focuses on calculating the number of chromium (Cr) atoms in 78.82 g of potassium dichromate (K2Cr2O7). The molecular mass of potassium dichromate is determined to be 294.2 g/mol. This leads to the conclusion that 78.82 g corresponds to 0.2680 moles of potassium dichromate, resulting in approximately 3.228 x 1023 chromium atoms, as each mole contains 1.204 x 1024 chromium atoms. Additionally, the discussion touches on challenges with LaTeX formatting and seeks better resources for learning LaTeX.

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How many Cr atoms in 78.82 g potassium dichromate?
Potassium dichromate is \textrm{K}_{2}\textrm{Cr}_{2}\textrm{O}_{7}.

I said:

The molecular mass of potassium dichromate is:
39.1*2+52*2+16*7=294.2

So, in 78.82 g potassium dichromate, there are 78.82/294.2= 0.2680 moles of potassium dichromate. Each mole of potassium dichromate has 6.022~x~10^{23} molecules, each with 2 chromium atoms. So, in one mole, there are 1.204~×~10^{24} chromium atoms. So, in .2680 moles, there are .2680(1.2044~×~10^{24})~=~3.228~×~10^{23}, correct?

As a sidenote, it appears that LaTeX graphical formulas can't be seen in the preview mode -- the only way to check them is to post the message then edit the message if they're wrong? Also, where can I find a better LaTeX tutorial than https://www.physicsforums.com/misc/howtolatex.pdf which just doesn't give enough information? Ah, https://www.physicsforums.com/showthread.php?t=9021 has some pretty good information, but I still don't seem to be doing my LaTeX correctly.
 
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Yes, that looks good. And I don't currently know of any better LaTeX tutorial, I've just been learning it recently too and would also be interested in a more in depth tutorial.
 

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