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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Noor on December 05, 2007, 12:44:58 AM

Title: URGENT PLZ! Weight percent
Post by: Noor on December 05, 2007, 12:44:58 AM

a sample of MgCl2 is known to be contaminated with NaCl. A 3.47 g portion of the sample is dissolved in water and treated with an excess of AgNO3 solution. The precipitate is filtered off, dried and weighed. The dried precipitate weights 10.3 g. Calculate the amount of MgCl2 in the original sample (in weight percent)

Ok the answer is 92.3% MgCl2 (by mass) , but the way I do it doesn't give me the right answer!!


I used Weight percent= weight of solute/weight of solution X 100%

10.3 g/ 10.3 + 3.47    = 74 %


What am I doing wrong??? I don't get this q. plz help me asap!!!!!!!!!!

Title: Re: URGENT PLZ! Weight percent
Post by: AWK on December 05, 2007, 01:19:21 AM

Write down reactions.

Title: Re: URGENT PLZ! Weight percent
Post by: Noor on December 05, 2007, 09:02:08 AM

I don't know how!!!!!!!! that's another thing that's bugging me..


BUMPPPP...please people! anyone??

Title: Re: URGENT PLZ! Weight percent
Post by: Borek on December 05, 2007, 09:38:34 AM

AgCl precipitates.

Title: Re: URGENT PLZ! Weight percent
Post by: Noor on December 05, 2007, 10:11:33 AM

Then???


Just help me with it if you know how to do it! Why post point form?

Title: Re: URGENT PLZ! Weight percent
Post by: Borek on December 05, 2007, 10:30:59 AM

Nothing more happens, precipitation is a key to both balanced reaction equations and calculations. Think what you have in solution, what is the solid substance that you weight after filtration and how its mass depends on the number of moles of MgCl₂ and NaCl in the original sample.

Title: Re: URGENT PLZ! Weight percent
Post by: Noor on December 05, 2007, 10:57:54 AM

The thing is, I don't get how to write the equation for this specific mixture!!

MgCl2 + NaCl + AgNO3 ---> AgCl + NaNO3 + MgNO3

??
ok wait, so if I do,

3.47 g/153.65 g/mol [MgCl2+ NaCl]
= 0.0225 moles

and if I do the ratio thing,

10.3 g x  0.0225 moles/3.47g
= 0.06678 moles of the dried precipitate

mass of MgCl2 =  0.06678 moles x 95.21 g/mol
                   = 6.358

100 - 6.358 = 93.642 g ???? it's not 92.3 AHHHHHHH!!

Title: Re: URGENT PLZ! Weight percent
Post by: Borek on December 05, 2007, 11:57:57 AM

There are two separate reaction equations, one for NaCl, one for MgCl₂.

You have two unknowns - amount of NaCl and amount of MgCl₂.

Precipitation gives you one equation combining these unknowns.

Mass of the original sample gives you second equation.

Write these (simultaneous) equations and solve.

Title: Re: URGENT PLZ! Weight percent
Post by: Noor on December 05, 2007, 04:58:17 PM

Ok Borek, here's what I got so far


moles of AgCl

10.3 g/ 142.35 g/mol = 0.07235 moles

0.07235 = (3.47- x / 58.44) +  2(x/95.21)

4.288 = 3.47 - x + 1.2273 x

x = 3.33

mass of MgCl2 by percentage = 3.33/3.47 X 100
                                         = 95.96 % !!!!!!!!!!!!!!!!


 :-\

Title: Re: URGENT PLZ! Weight percent
Post by: Borek on December 05, 2007, 05:43:02 PM

You are very close, just few details:

Molar mass of AgCl is not 142.35.

Quote from: Noor on December 05, 2007, 04:58:17 PM

0.07235 = (3.47- x / 58.44) +  2(x/95.21)

Missed parenthesis:

0.07236 = ((3.47- x)/ 58.44) +  2(x/95.21)

I've got 92.42%

Title: Re: URGENT PLZ! Weight percent
Post by: Noor on December 05, 2007, 06:13:57 PM

yay!!!!!!! AT LAST!!!

THANK YOU SO MUCH!!

Title: Re: URGENT PLZ! Weight percent
Post by: Borek on December 05, 2007, 06:15:37 PM

You see - after "I don't know" at the very beginning you did it finally by yourself :)