Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: britbro19 on September 20, 2008, 05:42:15 PM
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I need some help with 3 homework questions. We're really behind due to missing so much school because of Hurrican Gustav and Hurricane Ike; therefore, I'm having a hard time catching up. Any help would be greatly appreciated.
1. Calculate the percent yield of an analysis if 1.9365g of AgCl (MM 143.32g/mol) was precipitated in a gravimetric experiment if 1.2000g of a soluble sample that contained 40.25% Cl- was used as the starting material.
--Ok, I know for gravimetric analysis you need to know the starting material and ending material which would be 1.2g of sample and 1.9365 g AgCl product. After this step, I'm lost.
2.A mixture containing only Al2O3 (MM 101.96 g/mol) and Fe2O3 (MM 159.69 g/mol) weighs 2.0190g. When heated under a stream of H2, Al2O3 is unchanged, but Fe2O3 decomposes to Fe(s) and water vapor. If the residue weighs 1.7740g, what is the %Fe2O3 in the original sample?
Fe2O3--->2Fe(s) + 3H20(g)
Hint: Think about what's left in the residue and where the change in mass comes from!
--This one has me 100% lost. I don't even know where to start.
3. When a 100.0mL portion of a solution containing 0.5000g of AgNO3 (MM 169.87 g/mol) is mixed with 100.0 mL of a solution containing 0.3000g of K2CrO4 (MM 194.19 g/mol), a bright red precipitate of Ag2CrO4 (MM 331.73 g/mol) forms. Calculate the molar concentration of the excess reactant that remains in solution.
--I have no idea what to do if there's 2 starting materials.
Thanks for all of your time :)
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1. How many moles of Cl- in the starting sample? How much AgCl should be precipitated with that amount of chlorides?
2. Hint you were given is a very good one. You started with Fe2O3/Al2O3 mixture. What is left in the mixture after the reaction?
3. This is a limiting reagent question. Start with the reaction equation.
http://www.chembuddy.com/?left=balancing-stoichiometry&right=limiting-reagents
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Ok, for #1 I know there are 0.013512 mol Cl-...I'm still lost after that.
I'm still lost for the rest, I know I'm really slow in chemistry, sorry.
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Ok, for #1 I know there are 0.013512 mol Cl-
Write balanced reaction equation.
Do you know how to read these equations? Check here:
http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations
Now, assuming you started with 0.013512 moles of Cl-, how many moles of AgCl should precipitate? What mass should it have?
Don't forget about significant digits when writing final answer.
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OK:
Ag + Cl ---> AgCl
.4025 g 1.9365g
Is this right? I convert the percent they gave me to grams? Or do I put them both in moles?
Ag + Cl -----> AgCl
.013512mol .013512mol
Or is this right?
Then what do I do?
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Ag + Cl -----> AgCl
.013512mol .013512mol
This is how the reaction proceeds - you should end with 0.013512 mole of AgCl. Now, calculate its mass - that'll be theoretical yield. You know real yield (that was given) - use both numbers to calculate percent yield.
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Wait...if I change 0.013512 moles of AgCl to grams of AgCl that gives me 1.9365 g of AgCl, which is the same number the problem gives...so that'd be 100% yield.
I think I'm misunderstanding what you said. Can you explain it in a different way. What two numbers am I using for percent yield? Do I even need to use the 40.25% Cl- in the problem? ???
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Pretty close, but on the closer scrutiny numbers you list are a little bit strange.
Where did you get 0.013512 from?
What two numbers am I using for percent yield?
percent yield = practical yield/theoretical yield * 100%
Do I even need to use the 40.25% Cl- in the problem?
Yes, to calculate theoretical yield.
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I had did...
1.9365 g AgCl x 1 mol of AgCl x 1 mol Cl
143.321 g AgCl 1mol AgCl = 0.013512 mol Cl
What's the formula I need to use to find the percent yield? And where does that 40.25% come into play?
I did figure out how to do #3 though, YAY!
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I'm sorry, I meant what's the formula to find theoretical yield?
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40.25% is for calculation of the MASS of chlorides.
Then you are trying to convert this MASS of chlorides to MOLES of chlorides. OK. But you don't use molar mass of AgCl for that.
Theoretical yield is the amount you get from stoichiometric calculations. You know mass/moles of chlorides, calculate how much AgCl should be produced if chlorides react completely. That will be your theoretical yield.
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How do I use 40.25% to find the mass of chlorides?
.4025 mol Cl x 35.452 g Cl
1 mol Cl = 14.27 g Cl
Is this correct?
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You are told that 1.2 g of the sample contains 40.25% percent of chlorides. What does it mean?
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I have no clue. I'm so confused ???
.4025 g Cl
1.2 g sample X 100 = 33.54 % Cl in sample?
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Stop guessing blindly.
What is percentage definition?
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For percentage we always did g of what we wanted divided by g of the sample times 100%
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You want mass of chlorides - solve for that.
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1.2 g of sample is 440.25% chlorine.
1.2 x .4025 = .483 g of chlorine in the sample
.483 g Cl x 1 mol Cl
35.452 g Cl = .013624 mol Cl
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OK, now moles af AgCl, mass of AgCl and you are ready.
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but I thought the problem already gave the mass of AgCl?
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Practical, you need a theoretical one as well to calculate percent yield.
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So,
100-40.25 Cl=59.75% Ag
1.2 x .5975=.717 g AgCl
.717 g AgCl x 1 mol AgCl
143.32 g AgCl =.005003 mol AgCl
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No. Take a look at the reaction equation. How many moles of AgCl produced for one mole of Cl? How many moles of Cl (hint: don't calculate it again, you already did it just a moment ago).
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Ag+ + Cl- ----> AgCl
Therefore it's a 1:1 ration, so there's the same number of moles of Ag and Cl? Ok. Now what?
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Now you reread the whole thread and do what I told you several times earlier. Sorry, but it is up to you to solve the question, not up to me to spoonfeed you for hours.
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Yes, I know that. Thank-you for being so rude. But as I've told you before, we've never done this before because we've missed so much school due to hurricanes and chemistry is not the sort of subject that just clicks.
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Yes, I know that. Thank-you for being so rude. But as I've told you before, we've never done this before because we've missed so much school due to hurricanes and chemistry is not the sort of subject that just clicks.
Funny that the local time in your profile puts your location as somewhere in the Mid-Pacific (at GMT-11.0). Unless I'm mistaken Hurricanes Gustav and Ike never got within 4 time zones of you.
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Funny that the local time in your profile puts your location as somewhere in the Mid-Pacific (at GMT-11.0). Unless I'm mistaken Hurricanes Gustav and Ike never got within 4 time zones of you.
That's default time zone of everyone that has not fully configured its account. IP puts britbro19 in New Orleans.
Interestingly, same IP was used over a year ago by user BRIT, asking for a... theoretical yield.
http://www.chemicalforums.com/index.php?topic=17049
What a coincidence.