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Offline bbcnp

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Chemistry Book Problem
« on: August 28, 2007, 05:52:46 PM »
I have been trying to do this book problem for hours and can't seem to figure it out. I know its simple but I just can't do it? Any help would be greatly appreciated.

Indium Oxide contains 4.784g of indium for every 1.000g of oxygen. In 1869, when Mendeleev first presented his version of the periodic table, he proposed the formula In2O3 for Indium Oxide. Before that time, it was thought that the formula was InO. What values for the atomic mass of indium are obtained using these two formulas? Assume that oxygen has an atomic mass of 16.00.

The answers to this question are....
InO, atomic mass of In = 76.54
In2O3, atomic mass of In= 114.8

Can someone please show me how to do this problem. Thanks in advance.

Offline DevaDevil

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Re: Chemistry Book Problem
« Reply #1 on: August 28, 2007, 07:06:17 PM »
well, first of all transfer the mass of oxygen into moles.

then calculate how many moles of In there would have to be by using the ratio given in the hypothetical formulas.

you have then the mass (given) and amount of moles of In, so you have the molar mass.

Offline bbcnp

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Re: Chemistry Book Problem
« Reply #2 on: August 28, 2007, 08:10:44 PM »
First of all, thanks for replying.

I don't really understand step two, can you type it out so I can see it visually? Also I am trying to find atomic mass and not molar mass so your method doesn't really work unless there is something I am not seeing.

Thanks anyway.

I still need help on this problem, any help will be greatly appreciated.

Offline Yggdrasil

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Re: Chemistry Book Problem
« Reply #3 on: August 28, 2007, 08:20:00 PM »
Atomic mass is numerically the same as molar mass.

Lets say you have 30 mol of oxygen.  Based on the formula InO, you would expect 1 mole of indium per mole of oxygen, so you would expect 30 mol of indium.  Based on the formula In2O3, you would expect 2 mol of indium for every 3 mol of oxygen, so you would expect 20 mol of indium.

Offline bbcnp

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Re: Chemistry Book Problem
« Reply #4 on: August 28, 2007, 08:34:50 PM »
Thank you so much, I finally figured it out.

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