Question

\[Al^{3+}+3C_{9}H_{6}(OH)N:\rightarrow Al[C_{9}H_{6}(O^{-})N:]_{3}\downarrow + 3H^{+}\]

Answer 1

Calculate the formula weight of the Al [C9H6ON]3 complex that will be formed in this experiment.

A student performed a gravimetric analysis of an unknown solid sample by precipitating the aluminum using 8-HQ, forming the Al(HQ)3 complex as you will do in this lab. If the precipitate, after subtracting the crucible and microfilter, weighed 1.2176 g, what is the mass (in g) of aluminum in the precipitate?

Now suppose this student told you that she had sissolved the unknown in 250.00 mL, but only used 10.00 mL of that 250.00 mL solution to do the precipitation with 8-HQ. How many grams of Al were in the ORIGINAL 250.00 mL solution (i.e., in the original unknown sample)?

Supposed another student told you he calculated the number of grams of Al in the original 250.00 mL solution to be 1.788 g, and that he prepared that solution by dissolving 4.7400 g of a solid unknown in water. What is the wt % Al3+ in the original solid sample? which is which?

Answer 2

1)=26.928+3(9(12.011)+6(1.0079)+15.999+14.007)
=459.385g/mols

Answer 3

2) \[\frac{1.2176g of Al[C_{9}H_{6}(O^{-})N:]_{3}}{1}*\frac{1mole of Al[C_{9}H_{6}(O^{-})N:]_{3}}{459.3852g of Al[C_{9}H_{6}(O^{-})N:]_{3}}\]
\[*\frac{1mole Al}{3moles of Al[C_{9}H_{6}(O^{-})N:]_{3}}\]
= \[8.83*10^{-4}moles\]
\[8.83*10^{-4}moles*\frac{26.982gof Al}{1mole}=.0238gof Al\]

Answer 4

@kittykitty04 any good at this stuff?

Answer 5

Ahh sorry Kkutie7, I can't quite give you help with this one. It was something I did last year but I would need to review from my notes. Good luck!

Answer 6

@kittykitty04 ah too bad thank you any how

Answer 7

3) \[\frac{?molesAl}{.250L}*\frac{.01L}{1}\]
This is how I've been thinking about this. I would get the moles of Al in the 10mL solution right?
\[\frac{?molesAl}{.250L}*\frac{.01L}{1}=?moles of Al *\frac{26.982gof Al}{1mol of Al}=g of Al\]
Does this look right? I'm trying to work backwards here.

Answer 8

kkutie7 I showed you how to do that previous question the way you did it is wrong

Answer 9

Or this question rather

Answer 10

i don't think so I saw your link and it doesn't look like what we are doing in class.

Answer 11

Well graviemtric factor is what I learned in school, I can probably find my old course notes that show you it explicitly.

All the gravimetric factor is, is taking a ratio of an element to compound and using that to determine the mass that element takes up in grams of compound you have

Answer 12

I never learned that.

Answer 13

Show me what they taught you or the formula provided.

Answer 14

what I have on here was what was provided. no formula. its more of a stoichiometry approach.the only equation we have really been given in class is M1V1=M2V2

Answer 15

That isn't applicable to this kind of problem, that is having to do with solutions

Answer 16

Just looking at your method I can tell you will get strange units

Answer 17

I know

Answer 18

I'm a laboratory technologist and I have a degree in biochemistry

Answer 19

read my explanation of the formula

Answer 20

I know It was just something I was trying out... I stopped because it didn't work. good for you. I'm just saying that I wasn't taught or given a certain formula and if I go into class and show work with what my professor didn't teach he is gonna think I copied it from somewhere.

Answer 21

You can say you used logic to solve this,

\[\frac{Molecular\; Mass\; Al\;}{Molecular\; Mass\; Of Al(C_9H6(O)N)_3}* 100 = Percent\; Al\; IN\; Al(C_9H6(O)N)_3 \]

Answer 22

Most profs expect you to read ahead, or to figure things out for yourself (if you are in uni).

Answer 23

I'm just trying to be helpful I'm not trying to gloat I'm just trying to establish the fact that I know how to solve this problem

Answer 24

I'm not doing percent mass yet I'm still trying to wrap my head around the third problem.

Answer 25

I wasn't talking about mass percent I was talking about gravimetric factor and trying to show you how it works.

Answer 26

oh I got confused.

Answer 27

you did this right?
459.4317g/mol for the compound Al has a molecular mass of 26.98g/mol so using gravimetric formula: (26.98g/mol/459.4317g/mol)*1.2176g = 0.0715g

Answer 28

yes

Answer 29

Its like saying,

A metal has a mass of 1g and it is composed of 0.2g of sodium. how much sodium is in 3grams of this metal

thus,

(0.2g/1g)*3g = your answer

Answer 30

\[\frac{26.98g}{mol}\frac{mol}{459.4317g}*1.2176g=.0715g\]
I understand the units. I didn't understand why you used 1.2176.

Answer 31

that is the mass of the precipitate

Answer 32

that ratio is essentially telling you that you have 0.0715g of aluminum in 1.2176g of precipitate assuming the precipitate is 100% Al [C9H6ON]3

Answer 33

I get that part now. you did this too..
0.0238g*(250.0mL/10.0mL) = Grams of Al in original solution though

Answer 34

Ok good

Answer 35

this would be .0715g*250/10 right?

Answer 36

Yes that is correct if you had the ratio upside down you would get a smaller number which you should identify immediately as wrong.

Answer 37

so the answer to number 2 is .0715g right and the answer to number 3 is 1.7875g?

Answer 38

So for the second part of the problem,

By conservation of mass law the amount mass of Al will stay the same (electrons do have a mass but it is extremely small so we can just assume the weight of Al3+ is the same as Al present in Al [C9H6ON]3)



\[wt\% = (\frac{Grams}{Volume\; of\; Solution})*100\]

Answer 39

At first I thought this:
\[\frac{1.788}{4.7400}*100\]
but now I dont know about the solution do I just use 250mL?

Answer 40

yes you use ml

Answer 41

Remember water pretty much a density of 1g/mL

so 1g of water = 1mL of water

Answer 42

what about the 4.7400g?

Answer 43

that is just the amount of unknown compound all you care about is Al3+ in it

Answer 44

ok so
\[\frac{1.788}{250}*100=.7152\]
should I be using mL ro convert to L

Answer 45

No you should always use mL in %wt

Answer 46

ok so that is it?

Answer 47

%wt = ((Grams(g)/Volume(mL))*100

Answer 48

Isn't that what I did?

Answer 49

Yeah that looks good but include a % at the end of it

Answer 50

0.7152%wt

Answer 51

well actually just write 0.7152%

Answer 52

OK will do. Thank you.