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### AuthorTopic: How to calculate the mole of an element in a compound?  (Read 3464 times) !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="https://platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); (function() {var po = document.createElement("script"); po.type = "text/javascript"; po.async = true;po.src = "https://apis.google.com/js/plusone.js";var s = document.getElementsByTagName("script")[0]; s.parentNode.insertBefore(po, s);})();

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#### Kommandante

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##### How to calculate the mole of an element in a compound?
« on: December 09, 2013, 10:44:38 AM »

Hi there,

I have been given the formula for Iron(III) Oxide (Fe2O3) and have been asked to calculate how many moles are present in 24.6g of the compound whilst working out how many moles of Fe are present in this amount.

I have calculated the mole of Fe2O3 at 0.154 mol,
M= (Fe2O3), M= 2(55.8 ) + 3(16.0)
m = 24.6, M = 159.6, n = ?   so, n = m/M, 24.6/159.6 = 0.154mol

I cannot for the life of me grasp how to find out the mol of Fe in 24.6g this compound.
Do I divide the mol of Iron by the weight of the compound? Convert it to a percentage?

Any help would be greatly appreciated
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#### Hunter2

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##### Re: How to calculate the mole of an element in a compound?
« Reply #1 on: December 09, 2013, 07:01:47 PM »

its more easier.  You know the moles of the compound then you know also the moles of each atom. Its given to the formula. Example 1 mol NaCl  contain also 1 mol Na+ and 1 mol Cl-. So how much is it for the iron oxide.
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#### Kommandante

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##### Re: How to calculate the mole of an element in a compound?
« Reply #2 on: December 09, 2013, 07:57:09 PM »

Since the compound is 0.154 mol, does that mean there are 2(0.154)mol Fe?
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#### Hunter2

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##### Re: How to calculate the mole of an element in a compound?
« Reply #3 on: December 09, 2013, 08:19:06 PM »

Exactly. correct.
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#### Kommandante

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##### Re: How to calculate the mole of an element in a compound?
« Reply #4 on: December 09, 2013, 08:53:58 PM »

Thanks for your help,
Greatly appreciated!
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