# Question 8d963

Apr 14, 2017

${\text{AlCl}}_{3}$

#### Explanation:

The idea here is that all the chloride anions that make up the chloride will combine with the silver(I) cations to form silver chloride, an insoluble solid that precipitates out of solution.

${\text{Ag"_ ((aq))^(+) + "Cl"_ ((aq))^(-) -> "AgCl}}_{\left(s\right)} \downarrow$

You know that the reaction produced $\text{4.05 g}$ of silver chloride, so your first goal here will be to figure out exactly how much chlorine, $\text{Cl}$, in the form of chloride anions, it contains.

The molar mass of silver chloride is equal to

M_ ("M AgCl") = "143.32 g mol"^(-1)

The molar mass of chlorine is

M_ ("M Cl") = "35.453 g mol"^(-1)

You know that every mole of aluminium chloride contains $1$ mole of silver cations and $1$ mole of chloride anions. This is equivalent to saying that every $\text{143.32 g}$ of silver chloride contains $\text{35.453 g}$ of chlorine.

You can thus say that $\text{4.305 g}$ of aluminium chloride will contain

4.305 color(red)(cancel(color(black)("g AgCl"))) * "35.453 g Cl"/(143.32 color(red)(cancel(color(black)("g AgCl")))) = "1.065 g Cl"

This means that initial sample contained $\text{1.065 g}$ of chlorine and

$\text{1.335 g " - " 1.065 g" = "0.270 g Al}$

To find the empirical formula of the chloride, you must determine the smallest whole number ratio that exists between the number of moles of each element in the compound.

Use the molar masses of the two elements to calculate the number of moles of each present in the sample

1.065 color(red)(cancel(color(black)("g Cl"))) * "1 mole Cl"/(35.453color(red)(cancel(color(black)("g Cl")))) = "0.03004 moles Cl"

0.270 color(red)(cancel(color(black)("g Al"))) * "1 mole Al"/(26.982color(red)(cancel(color(black)("g Al")))) = "0.01001 moles Al"

Divide both values by the smallest one to get

"For Al: " (0.01001 color(red)(cancel(color(black)("moles"))))/(0.01001color(red)(cancel(color(black)("moles")))) = 1

"For Cl: " (0.3004color(red)(cancel(color(black)("moles"))))/(0.01001color(red)(cancel(color(black)("moles")))) = 3.001 ~~ 3#

Therefore, the empirical formula for this compound is

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{Al"_ 1"Cl"_ 3 implies "AlCl}}_{3}}}}$