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Topic: Moles and Molecules  (Read 8321 times)

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sundrops

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Moles and Molecules
« on: January 21, 2005, 02:14:49 AM »
The molecular formula of a substance is C16H17O8.
The molar mass is 337.3g/mol

How many moles of C16H17O8 molecules are in 111mg of this compund?
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sundrops

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Re:Moles and Molecules
« Reply #1 on: January 21, 2005, 02:16:05 AM »
I would like to know HOW I would find this information - I need help. It's onl like question #2 in my homework and I can't solve it!?!
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Offline Mitch

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Re:Moles and Molecules
« Reply #2 on: January 21, 2005, 02:32:57 AM »
Start by determining how many moles of the molecule you have.
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sundrops

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Re:Moles and Molecules
« Reply #3 on: January 21, 2005, 02:43:00 AM »
what do u mean? well there's 337.3 g/mol.
so that would be 337300 mg/mol

I don`t really understand what to do.  ??? I mean how would I figure it out?
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Offline Mitch

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Re:Moles and Molecules
« Reply #4 on: January 21, 2005, 02:46:52 AM »
How can you multiply or divide those 2 numbers(the 111mg and the 337.3 mg/mol) to end up with a number with units of moles?
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sundrops

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Re:Moles and Molecules
« Reply #5 on: January 21, 2005, 02:51:20 AM »
so would it be 111g of C16H17O8 * 1mol of C16H17O8 / 337.3g of C16H17O8 = 3.29*10^-4
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sundrops

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Re:Moles and Molecules
« Reply #6 on: January 21, 2005, 02:54:20 AM »
yay! that was right! thanks!!


except next part is that I need to find the mass of 0.100 mol of this compund.

so would I just go: 0.100 mol C16H17O8 * 337.296g C16H17O8 / 1 mol C16H17O8 = 33.7296g

did I do that right?
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Offline Mitch

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Re:Moles and Molecules
« Reply #7 on: January 21, 2005, 02:59:02 AM »
looks good, I didn't actually take out a calculator though.
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sundrops

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Re:Moles and Molecules
« Reply #8 on: January 21, 2005, 03:03:37 AM »
yay! that was good too!

I'm always scared to submit my answers (it's all done online) and you only get 5 tries. So you have to make sure you really understand it and that you have the proper units and proper sig figs - very very annoying! lol

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sundrops

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Re:Moles and Molecules
« Reply #9 on: January 21, 2005, 03:12:26 AM »
thanks for all your help mitch - i really appreciate it.

but there's one more question.

How many hydrogen atoms are in 1.78 pg of this compound?

here's my reasoning...

(1.78pg) * (1g / 10E^-12pg) * (1mol / 337.296g) * (6.02E23 / 1 mol) = 3.1769E33

how does that look?
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sundrops

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Re:Moles and Molecules
« Reply #10 on: January 21, 2005, 03:40:45 AM »
I then multiplyed the 3.17797E13 molecules by 17 hydrogen atoms / molecule = 5.40E14 atoms of hydrogen

this turned out to be wrong - can u point out my mistake?
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