Question

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)?

Answer 1

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

Answer 2

Not sure what it means by 8-HQ is that what the actual question says?

what you need to realize is that concentration is consistent, for example if you take a portion of a solution that portion will have the same concentration as its parent solution.

Answer 3

8-HQ is the same as \[3C_{9}H_{6}(OH)N:\]

Answer 4

I also have this information:
Aluminum precipitated= .0238grams of Al
and
Al(HQ)3 complex precipitated= 1.2176 g

Answer 5

You didnt provide enough information to solve this problem, you need a concentration or a quantity of 8-HQ that you used to cause complete precipitation of the unknown compound in the 10mL solution of it to figure out the unknowns concentraiton

Answer 6

I don't have that

Answer 7

Then this question cannot be answered

Answer 8

The experiment has not been completed thus there is no data to allow you to determine your answer

Answer 9

It has to be. its on my prelab. this is all I have:
Calculate the formula weight of the Al [C9H6ON]3 complex that will be formed in this experiment.=459.385.
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?= .0238grams of Al

Answer 10

@Australopithecus

Answer 11

Ok you do have enough information to answer this

Answer 12

phew! good. show me how to go about this please

Answer 13

Is 0.0238 your guess as to how much aluminium is in your precipitate?

Answer 14

thats is what I calculated yes

Answer 15

So you know how much product you produced:
1.2176 g

Using,

Gravimetric formula
\[Grams\; of\; compound*\frac{Molecular\; Mass\; of\; Element}{Molecular\; Mass\; Of\; Compound} = Grams\; of\; Element\]


If this is what you did you are fine

Answer 16

no I did this:
\[\frac{1.2176g of Al[C_{9}H_{6}(O^{-})N:]_{3}}{1}*\frac{1mole of Al[C_{9}H_{6}(O^{-})N:]_{3}}{459.3852g of Al[C_{9}H_{6}(O^{-})N:]_{3}}*\frac{1mole Al}{3moles of Al[C_{9}H_{6}(O^{-})N:]_{3}} \]

Answer 17

0.0238g*(250.0mL/10.0mL) = Grams of Al in original solution

Make up a ratio in this case 250mL/10mL = the amount you need to scale up

Notice that the number gets larger which is what you expected.

Answer 18

\[\frac{1.2176g of Al[C_{9}H_{6}(O^{-})N:]_{3}}{1}*\frac{1mole of Al[C_{9}H_{6}(O^{-})N:]_{3}}{459.3852g of Al[C_{9}H_{6}(O^{-})N:]_{3}}\]
\[*\frac{1mole Al}{3moles of Al[C_{9}H_{6}(O^{-})N:]_{3}}\]

Answer 19

Whenever a solution is diluted or a portion of it taken from the stock solution you just need to create a ratio to figure out the amount in the original solution

Answer 20

ok I get that 250is divided by 10

Answer 21

yeah, makes sense right?

Answer 22

Let me check your calculation

Answer 23

ok

Answer 24

so you are saying that the answer would be .595 grams of Al?

Answer 25

http://www.webqc.org/molecular-weight-of-Al(C9H6ON)3.html
so, 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

http://www.wolframalpha.com/input/?i=%2826.98g%2Fmol%2F459.4317g%2Fmol%29*1.2176g+%3D+

Answer 26

gravimetric formula?

Answer 27

why do you multiply by 1.2176?
whats wrong with the way I did it?

Answer 28

also need help with this I know its weight of solute/ weight of solution*100
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 29

I have no idea what you did, mine just uses the ratio of Al to compound, that percentage can be used to figure out how many grams of compound are made up by that specific element.

here is to show that other people use this formula
https://ca.answers.yahoo.com/question/index?qid=20100202072330AAJTPqC

I wish I could help you but I have to eat and spend time with friends etc. I will be able to help you later tonight maybe if I can. I recommend opening a new question though.

I hope I was helpful