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Topic: Chemistry mole question.  (Read 2489 times)

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Offline Starbright

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Chemistry mole question.
« on: February 09, 2015, 08:48:44 PM »
How many grams of hydrogen have the same number of atoms as 99.0 g of oxygen?

99.0 g O2 x1 mole/31.98 g O2 x (6.02 x 10^23)/1 mole = 1.86 molecules of O2.    Not sure if this is how to begin but now I am lost.  Please help.

Offline Hunter2

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Re: Chemistry mole question.
« Reply #1 on: February 10, 2015, 01:54:51 AM »
Convert the Mas of Oxygen int to mol and convert back to the mass of hydrogen. You dont nee the Avogadro number, because its defined over the mole already.

Offline themonk

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Re: Chemistry mole question.
« Reply #2 on: February 10, 2015, 05:18:41 AM »
So a mole is basically the amount of substance you have and is an SI unit. In one mole you will have Avogadro's number (Avogadro constant) of atoms. You convert mass to moles as you did in the first step but 1 mole of oxygen has the same number of atoms as one mole of hydrogen. So you can convert simply to moles of oxygen and then moles of hydrogen to get the grams of hydrogen.

[itex]99 g\,  O_{2}\cdot \left (\frac{1\, mol\, }{2 \cdot Y \, g} \ O_{2}  \right ) \cdot \left [ \left (\frac{6.022*10^{24} atoms}{mol} \ O_2   \right ) \cdot \left (\frac{mol}{6.022*10^{24} atoms} \ H_2   \right )\right ]\cdot \frac{2 \cdot X \, g}{1\, mol\, } \ H_{2}
[/itex]

Non-latex

The part in the square brackets [] you do not need.
Edit: removed final answer and some specifics.

Offline Starbright

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Re: Chemistry mole question.
« Reply #3 on: February 10, 2015, 02:14:47 PM »
Thank you so much but I am not sure what the ( 2Y g) or the (2 X g) represents.    Thank you.  I am very confused and look forward to your answer.

Offline themonk

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Re: Chemistry mole question.
« Reply #4 on: February 11, 2015, 04:03:56 AM »
X and Y are variables that represent molecular weights. You actually do not need the 2 in front of them now that I am looking at it again (my mistake) since I included the diatomic molecules.

Offline Starbright

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Re: Chemistry mole question.
« Reply #5 on: February 14, 2015, 07:28:43 AM »
Thank you so much.  I understand it now  :)

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