What is the empirical formula for titanium chloride with 0.311g of titanium and 0.689g of chloride?

Answer 1

Titanium chloride is TiCl. You must convert those grams given to you into moles first.

So you have .311 g Ti and .689 g Cl.

So....

to convert to moles:

for Ti:

find the molar mass of Ti. This is 47.90g/mol.

0.311g Ti x 1 mol Ti/47.90g/mol Ti= .006mol Ti

For Cl:

The molar mass of Cl is 35.45g/mol.

0.689g x 1mol Cl/35.45g/mol= .019mol Cl

So you've got 0.006mol Ti and 0.019mol Cl. Now you have to divide both by 0.006 to get the empirical formula.

So:

0.006/.006= 1mol Ti

.019/.006= 3mol Cl

Your empirical formula would be TiCl3

Answer 2

This question was in the Chemisty [Additional Science] C2a paper on the Edexcel Exam Board Exam on 16th November 2009

Titanium chloride is TiCl. You must convert those grams given to you into moles first.

So you have .311 g Ti and .689 g Cl.

So....

to convert to moles:

for Ti:

find the molar mass of Ti. This is 47.90g/mol.

0.311g Ti x 1 mol Ti/47.90g/mol Ti= .006mol Ti

For Cl:

The molar mass of Cl is 35.45g/mol.

0.689g x 1mol Cl/35.45g/mol= .019mol Cl

So you've got 0.006mol Ti and 0.019mol Cl. Now you have to divide both by 0.006 to get the empirical formula.

So:

0.006/.006= 1mol Ti

.019/.006= 3mol Cl

Your empirical formula would be TiCl3