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Topic: Mole Ratio  (Read 5669 times)

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Offline Joules23

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Mole Ratio
« on: February 09, 2007, 02:21:20 PM »
This molecule contains 33% water, find X
(MnI2)(XH2O)

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heres what i did

100g Compound(Xg water/X+309g) = 33g water

100(X/X+309) = 33

.33=(x/x+308.74)
x=152
152/18 = 8.4

Is this right??
« Last Edit: February 11, 2007, 01:38:06 AM by Joules23 »
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Offline Borek

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Re: Mole Ratio
« Reply #1 on: February 09, 2007, 05:18:41 PM »
Are you sure your X is the same X you are asked for?
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Offline Joules23

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Re: Mole Ratio
« Reply #2 on: February 10, 2007, 02:09:23 AM »
fixed it..
Did i solve this correctly?
« Last Edit: February 11, 2007, 01:38:40 AM by Joules23 »
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Offline Joules23

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Re: Mole Ratio
« Reply #3 on: February 11, 2007, 02:01:20 AM »
any1?
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Offline Borek

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Re: Mole Ratio
« Reply #4 on: February 11, 2007, 07:25:50 AM »
Looks OK at first sight.
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Offline vhpk

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Re: Mole Ratio
« Reply #5 on: February 11, 2007, 11:11:14 PM »
I want to ask if the percent is 33% or 33,...%, give us the correct number
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Offline Joules23

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Re: Mole Ratio
« Reply #6 on: February 15, 2007, 01:19:49 AM »
sorry for the delay, i had to wait to get the test back... it is 33.13%
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Offline AWK

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Re: Mole Ratio
« Reply #7 on: February 15, 2007, 03:16:02 AM »
Quote from: Joules23 on February 09, 2007, 02:21:20 PM

152/18 = 8.4

Is this right??
Result is OK. Use brackets in denominator to remove ambiguity ie .33=x/(x+308) where x is mass of water in 1 mole of hydrate.

Note,  capital X is a number of water molecules in hydrate.

Quote
100(X/X+309) = 33
.33=(x/x+308.74)

Do not mix X and x in formulas above
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