# If a 2.60g sample of titanium metal chemically combines with chlorine gas to form 10.31g of a titanium chloride, what is the empirical formula product?

Oct 20, 2015

${\text{TiCl}}_{4}$

#### Explanation:

The idea here is that the difference between the mass of the titanium chloride and the mass of titanium metal will represent the mass of chlorine.

This means that you can find how much chlorine was needed to form that much titanium chloride by

${m}_{\text{titanium chloride" = m_"chlorine" + m_"titanium}}$

${n}_{\text{chlorine" = "10.31 g" - "2.60 g" = "7.71 g Cl}}$

So, your titanium chloride contains $\text{2.60 g}$ of titanium and $\text{7.71 g}$ of chlorine. To get the compound's empirical formula you need to first establish how many moles of each element you get in that sample.

To do that, use the two elements' respective molar masses

$\text{For Ti: " (2.60color(red)(cancel(color(black)("g"))))/(47.867color(red)(cancel(color(black)("g")))/"mol") = "0.054317 moles Ti}$

and

$\text{For Cl: " (7.71color(red)(cancel(color(black)("g"))))/(35.453color(red)(cancel(color(black)("g")))/"mol") = "0.21747 moles Cl}$

To find the mole ratio that exists between the two elements in the sample, divide both numbers of moles by the smallest of the two

"For Ti: " (0.054317color(red)(cancel(color(black)("moles"))))/(0.054317color(red)(cancel(color(black)("moles")))) = 1

"For Cl: " (0.21747color(red)(cancel(color(black)("moles"))))/(0.054317color(red)(cancel(color(black)("moles")))) = 4.0037 ~~ 4

This means that the empirical formula will be

${\text{Ti"_1"Cl}}_{4}$, or $\textcolor{g r e e n}{{\text{TiCl}}_{4}}$