Chemistry 3/4 2013 Thread

[Bad Student]

How much water did you dilute the HCl with?

[Ancora_Imparo]

As with most of these questions that require multiple steps, it's best to write out what you know and what you want to find. For this gravimetric question, you know the mass of AgCl precipitate, and you want to find the amount of Cl- ions in the brick cleaner.

So, you would proceed like this:
m(AgCl) = 2.945g
Find the amount of AgCl in moles using n(AgCl) = m/M
The number of moles of Cl- ions in the 20mL sample would equal the number of moles of AgCl just found, that is, n(Cl- in 20mL diluted sample) = n(AgCl)
You didn't actually state the volume that you diluted the 5mL sample of brick cleaner to. Let's say it was x mL. You would then do n(Cl- in 5mL undiluted sample) = x/20 * n(Cl- in 20mL sample).

For mass of HCl in sample, you know that n(HCl) = n(Cl-). So, m(HCl) = n(HCl) * M(HCl).
You will then be asked to use this mass to determine the actual %HCl in the brick cleaner and compare the value you obtained with the 32% HCl that is on the label.

Hope this clears things up!

[academicbulimia]

As with most of these questions that require multiple steps, it's best to write out what you know and what you want to find. For this gravimetric question, you know the mass of AgCl precipitate, and you want to find the amount of Cl- ions in the brick cleaner.

So, you would proceed like this:
m(AgCl) = 2.945g
Find the amount of AgCl in moles using n(AgCl) = m/M
The number of moles of Cl- ions in the 20mL sample would equal the number of moles of AgCl just found, that is, n(Cl- in 20mL diluted sample) = n(AgCl)
You didn't actually state the volume that you diluted the 5mL sample of brick cleaner to. Let's say it was x mL. You would then do n(Cl- in 5mL undiluted sample) = x/20 * n(Cl- in 20mL sample).

we diluted in a 100ml volumetric flask so 95 ml? But yeah makes a lot more sense!!

For mass of HCl in sample, you know that n(HCl) = n(Cl-). So, m(HCl) = n(HCl) * M(HCl).
You will then be asked to use this mass to determine the actual %HCl in the brick cleaner and compare the value you obtained with the 32% HCl that is on the label.

your exactly right, thank you so much!!!!!!!!!!!!!!!!! Just practice some of these and I'll be right for the SAC i hope

[Ancora_Imparo]

we diluted in a 100ml volumetric flask so 95 ml?

Though you only added 95mL of water to the 5mL sample, you consider the volume of the total solution when performing calculations, as the Cl- ions will be evenly distributed throughout all 100mL of solution, not just the 95mL added. So x=100mL.

your exactly right, thank you so much!!!!!!!!!!!!!!!!! Just practice some of these and I'll be right for the SAC i hope

Couldn't have said it any better. There are only a few different types of questions out there, so if you cover them, you'll be set to score very well. Keep it up!

[academicbulimia]

You didn't actually state the volume that you diluted the 5mL sample of brick cleaner to. Let's say it was x mL. You would then do n(Cl- in 5mL undiluted sample) = x/20 * n(Cl- in 20mL sample).

Hang on, one question is this to find the mol of chloride ions after dilution ?

and thanks i will!!

[Ancora_Imparo]

Quote from: Ancora_Imparo on 34 minutes  ago
You didn't actually state the volume that you diluted the 5mL sample of brick cleaner to. Let's say it was x mL. You would then do n(Cl- in 5mL undiluted sample) = x/20 * n(Cl- in 20mL sample).

Hang on, one question is this to find the mol of chloride ions after dilution ?

Remember that the number of moles of chloride ions before and after dilution is still the same. Adding water does not change the amount of Cl-. So, in the above step, you are finding the amount of mol of Cl- in the 100mL diluted solution, which is in fact that same as the amount of mol of Cl- in the 5mL undiluted solution.

[academicbulimia]

Remember that the number of moles of chloride ions before and after dilution is still the same. Adding water does not change the amount of Cl-. So, in the above step, you are finding the amount of mol of Cl- in the 100mL diluted solution, which is in fact that same as the amount of mol of Cl- in the 5mL undiluted solution.

oh yeah your right, thanks

the actual percentage was 12.47% wow

[Stick]

I just need confirmation with my answer to part d, since I don't have a copy of the extended solutions on me and the answer isn't at the back of the book. Thanks.

[Bad Student]

If water was used as a solvent, the appearance of the chromatogram would not likely have a similar appearance because the solubility of the pigments in the solvent determines how much they are separated by. Since water is a polar substance and the original solvent was non-polar, the pigments in the plant extract would probably dissolve in the water to a different degree.

[Ancora_Imparo]

If water was used as a solvent, the appearance of the chromatogram would not likely have a similar appearance because the solubility of the pigments in the solvent determines how much they are separated by. Since water is a polar substance and the original solvent was non-polar, the pigments in the plant extract would probably dissolve in the water to a different degree.

Just to add to the excellent answer, try including terms like 'mobile phase', 'stationary phase', 'adsorption', 'desorption', 'R_f value' etc as much as you can in these types of questions. Examiners like them. The only term that would really apply here is 'mobile phase', as the solvent is the mobile phase in TLC.

[Stick]

I also said that the order of the pigments would most likely be inverted. Yay or nay?

[thushan]

I also said that the order of the pigments would most likely be inverted. Yay or nay?

Nah, not necessarily. Water and silica are both polar; you can't make that conclusion.

[Stick]

Ah, fair enough. I thought the question wanted us to be really specific - otherwise, my answer would have been a watered down response of Bad Student's response. Thanks for the help everyone.

[clıppy]

Can somebody help me out with this gravimetric question:

A 1.00g sample of steel wool was dissolved in sulfuric acid and then treated with ammonia in order to precipitate the iron as iron(III) hydroxide. This reaction can be represented by the chemical equation:
Fe³⁺_(aq) + 3H₂O_(l) + 3NH_3(aq) -> Fe(OH)_3(s) + 3NH₄⁺
The precipitate formed was collected by filtration, washed and then heated. During this process the iron(III) hydroxide was converted to iron(III) oxide as described by the chemical equation:
2Fe(OH)_3(s) -> Fe₂O_3(s) + 3H₂O_(g)
The mass of iron(III) oxide formed after heating to constant mass was 1.358g. Calculate the percentage of iron by mass in the steel wool

The main problem i'm having is figuring out the moles of Fe (Fe³⁺) because it goes through so many stages and has different amounts in both equations. Any help?

[Ancora_Imparo]

Calculate the moles of Fe₂O₃ using .
From the second equation, n(Fe(OH)₃) = 2 * n(Fe₂O₃).
From the first equation, n(Fe³⁺) = n(Fe(OH)₃).
Calculate the mass of Fe using .
Calculate the %mass of Fe using %mass = .