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Topic: Empirical Formula  (Read 4545 times)

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Offline H Mac

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Empirical Formula
« on: March 12, 2008, 09:05:28 PM »
The question is ...
A compound containing titanium (Ti) and chlorine is analyzed by converting all the titanium into 1.20g of TiO2 and all the chlorine into 6.45g of AgCl.  What is the simplest formula for the original compound?

This is what I did, but any hints of where I may have gone wrong, would be greatly appreciated! (Even better is if I don't go wrong ;) )

moles of TiO2 = 1.20 g / 79.9 g/mol = 0.015 mol
moles of AgCl = 6.45 g /  144 g/mol = 0.045 mol

Therefore the moles of Ti atoms is 0.015 mol and th moles of Cl atoms is 0.045 mol.

mass of Ti = 0.015 mol * 47.9 g/mol = 0.719 g
mass of O = 1.20 g - 0.719 g = 0.481 g

mass of Cl = 0.045 mol * 35.5 g/mol = 1.60 g
mass of Ag = 6.45 g - 1.60 g =  4.85 g

moles of O : 0.481 g / 16.0 g/mol = 0.030 mol
moles of Ag : 4.85 g / 107 g/mol = 0.042 mol

Therefore Ti : O : Ag : Cl  as
0.015 : 0.030 : 0.042 : 0.045
0.015   0.015   0.015    0.015
=1 : 2 : 2 : 3

Therefore the empirical formula for the original compound is TiCl3

Thank you again for your *delete me*

Offline Borek

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Re: Empirical Formula
« Reply #1 on: March 13, 2008, 04:17:37 AM »
Therefore the moles of Ti atoms is 0.015 mol and th moles of Cl atoms is 0.045 mol.

No idea what you did later, but you have the answer ready - just look at these numbers :)

Oh, BTW - we don't say "moles of atoms of Ti" - just "moles of Ti".
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Offline AWK

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Re: Empirical Formula
« Reply #2 on: March 13, 2008, 08:27:03 AM »
Quote
moles of TiO2 = 1.20 g / 79.9 g/mol = 0.015 mol
moles of AgCl = 6.45 g /  144 g/mol = 0.045 mol
moles TiO2 = moles Ti
moles AgCl = moles Cl

The answer for titanium chloride is:
AWK

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