Question

please help!
A 0.5848 g sample of a pure soluble chloride compound is dissolved in water, and all of the chloride ion is precipitated as AgCl by the addition of an excess of silver nitrate. The mass of the resulting AgCl is found to be 1.5102 g.

What is the mass percentage of chlorine in the original compound?

Answer 1

Cl2 + AgNO3 --> NO3- + AgCl

this is the balanced equation you are dealing with

Answer 2

okay then what would i do

Answer 3

do i need the equation or cant i just go directly into the calculations?

Answer 4

are you there?

Answer 5

wait one sec...thats not what it says

its says a pure soluble chlorine compound...so XCl

XCl + AgNO3 --> XNO3 + AgCl

so you have to work backwards....AgCl=1.5102
Ag=107.9 g
Cl=35.5 g
So 1 mol AgCl=143.4 g
\[{{1.5102~\cancel{g~AgCl}}\over{1}}\times{{1~mol~AgCl}\over{143.4~\cancel{g~AgCl}}}={0.01053~mol~AgCl}\]

since the XCl and AgCl are a 1:1 ratio
0.01053 mol AgCl = 0.01053 mol XCl
let x= molecular weight of X\[{{0.01053~\cancel{mol~XCl}}\over{1}}\times{{(x+35.5)~g~XCl}\over{1~\cancel{mol~XCl}}}=0.5848~g~XCl\]\[0.01053(x+35.5)=0.5848\]\[0.01503x+0.373815=0.5848\]\[0.01503x=0.210985\]\[x=20.04~g\]

So Mass Percent is just the element over the weight of the compound (times 100)
element = Cl = 35.5 g
compound = XCl = 20.04+35.5 = 55.54 g

\[\left({35.5\over55.54}\right)\times100={\text{63.92% Cl}}\]

Answer 6

oh okay wow i get it.. but do you need to have the balanced equation written out or can you just go without it?

Answer 7

no you need the balanced equation to show that you know how the reaction happened

Answer 8

and it also is needed as proof that the ratio is in fact 1:1...its best if you just leave it in there as part of your work

Answer 9

oh okay thank you so much! i hope you can help me change another day