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April 27, 2015

Posted by **Jake** on Wednesday, January 30, 2008 at 6:05pm.

here's what i have below, your help is much appreciated.

so i have X= g AgNO3

0.953-x = Ba(NO3)2

therefore .953-X-NO3=?

so far I caluclated the % of Ag and Ba in their nitrate salts:

% Ag in Ag2NO3= 107.87/161.91

= .666X

% Ba in Ba(NO3)2 = 137.33/261.32

= .526(.953-X)

i got this far, but now i'm stuck.

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**Jake**, Wednesday, January 30, 2008 at 6:07pmsorry, it would probably hep you to know that the objective is to calculate the percentage of silver and barium in the nitrate salt mixture. i know how to find the percentage once i get the grams, but i have no idea how to get the grams from this point.

thanks :]

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**DrBob222**, Wednesday, January 30, 2008 at 6:33pmA couple of starting points for you but not finished.

First, you are ok with X = AgNO3 (note you wrote Ag2NO3 a few lines below that but apparently you didn't use that formula throughout.

Second, you have avoided the need for a second variable by using 0.953-X and that's ok, too, but you still need a second equation.

Third, 0.953 - X - NO3 I don't get. If you want to get rid of the nitrate ion, which apparently is what you are driving at, it's easier to let X = Ag and make the conversions as I did for Catie.

Fourth, your factor AgNO3 (written with a typo of Ag2NO3) is not correct. Ag is ok at 107.87 but AgNO3 is closer to 170 although I don't remember the exact number. I THINK it is 169.87 so your factor of 0.666 is not correct.

Fifth, you need a second equation utilizing the BaCrO4.

I'll leave it at that to let your work through some of this but I'll be glad to help later.

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**Jake**, Wednesday, January 30, 2008 at 6:58pmYes, I didn't want to use a second variable because my teacher wants us to use different approaches.

i checked the molar mass of AgNo3 and it is 169.87, so thank you. I didn't notice that.

the new factor would be:

107.87/169.87 = .635

i'm not quite sure how to remove the nitrate, that was just a note i threw down and wasn't sure how to go about it.

i'm still really confused as to how the calculation works for this :/

i thank you so much, i promise i'm not looking for you to do answer the ? for me,i'm just really really confused, i've been tearing my hair out over this calculation!!!

i wrote the balanced equations out.

i thought it might help me figure it out.

i know there's 2*more AgCro4 in the product.

2Ag + CrO4 (-2) ---> Ag2CrO4

Ba (2+) + CrO4 (-2) ---> BrCrO4

but how would that help me make up a second equation? would it be the % of Ba in BaCrO4? Would this work if i followed a similar method?

% Ba in BaCrO4 = 137.33/261.33

= .526

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**DrBob222**, Wednesday, January 30, 2008 at 8:29pmTwo or three points here.

If you let X = mass AgNO3

then 0.953 - X = mass Ba(NO3)2.

First, if we set this up with just this information, the following is what you get.

X + 0.953 - X = 0.953

and we end up with 0 = 0, which of course, is true but it doesn't help us much if the idea was to solve for X. :-)).That's why you need a second equation. That's where the mass BaCrO4 comes in. You must understand that the whole idea behind doing the BaCrO4 thing is to ppt ALL of the Ag and ALL of the Ba as BaCrO4.

Second, back to the X and 0.953 - X thing, if you multiply X by the percent Ag in AgNO3 and 0.953 - X by the percent Ba in Ba(NO3)2, you will have converted AgNO3 to Ag and Ba(NO3)2 to Ba and that, eventually, is what you want in order to calculate percent Ag. If you will look at the equations I made for Catie, you will see those factors are what I did, also, except I started with Ag and Ba so I converted to AgNO3 and Ba(NO3)2. You are using a different set up so that won't help you. So on to the BaCrO4.

You want to convert g AgNO3 (that's what X is) to Ag2CrO4 and you want to convert 0.953-X [that's the Ba(NO3)2] to BaCrO4, add those together and the sum should be 0.944 grams BaCrO4.

No matter how you slice it though, if my algebra stuck together last night, I tried solving the equations and I came up with a negative number for one of the salts and a positive number for the other one. A negative number can't be; therefore, I concluded that either I made an error OR there was an error in the procedure. I think my algebra is ok. I think the mass BaCrO4 or the mass of the sample (or both) are in error due to experimental error.

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**Jake**, Thursday, January 31, 2008 at 3:01amso the second equation would be:

0.635X + 0.526(0.953-X)=0.944 ?????

0.635X + 0.5013-.526X=0.944

0.109X/.109=0.4427/.109

x=4.061????

to three sig figs=4.06 is that the grams of BaCrO4???

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**Jake**, Thursday, January 31, 2008 at 3:03amthat couldn't be right though because i only started out with .953 g and conservation of mass applies

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