TrueTears question thread

[Mao]

1. at a glance, a, b and d are the same (meaning but expressed differently), hence I presume c is correct.

2. the titre is less than what it should be, i.e. the conical flask contains less number of moles of compound than it should have (concentration of aliquot is less than it should be), or the solution in the burette is more concentrated than it should be.

a) increases the number of moles of compound in conical flask --> titre increase
b) decreases concentration in aliquot --> no. moles in conical flask is less --> titre decrease
c) decreases concentration in burette --> titre increase
d) correct procedure
e) correct procedure

[TrueTears]

thanks mao!

[TrueTears]

Could anyone just check if what i did here is correct, these questions are from a prac and these are the results i got from the prac, also the steps to do the prac is listed

Aim: Find the mass of ammonium sulfate(the active ingredient) in a brand of lawn fertiliser by gravimetric analysis.

Mass of fertiliser: 1.015g

1. Add 50mL of deionised water and stir to dissolve as much of the sample as possible. Using a short necked funnel and filter paper, filter the mixture into a 600mL beaker, washing the residue on the filter paper with several portions of deionised water.

2. add 2.5 mL of 2M HCl to the filtered solution (filtrate) and add more distilled water so that the total volume is about 200mL. Use a Bunsen burner to just boil the solution. (Note: the HCl is just used as a catalyst)

3. Add 15mL of the barium chloride solution (an excess) from a measuring cylinder slowly to the hot solution. A white precipitate of barium sulfate will form. Keep the mixture hot and stir continually throughout this process.

(left a few steps out, they just tell you how to use the vacuum filtration, doesn't involve any calculations)

4. Collect the precipitate and weigh.

Ok, the precipitate mass i got was 1.216g

Here's my working to work out the mass of ammonium sulfate








This is correct working out?

[TrueTears]

We then did another prac, but this time we used volumetric analysis(back titration) to do the same thing.

Again can someone check my working out to see if it is right?

Aim: Find the mass of ammonium sulfate in a brand of lawn fertiliser

Mass of lawn fertiliser: 1.301g

Here are the steps to the prac

1. Using a small funnel, carefully transfer the sample of the fertiliser to the volumetric flask (250mL). Add water until it reaches the calibration line (250mL).

2. Using a 20.00mL pipette, dispense aliquots of the fertiliser solution into each of the 3 conical flasks.

3. Thoroughly wash the pipette and rinse it with a small volume of standard sodium hydroxide solution. Place a 20.00mL aliquot of NaOH solution in each of the flasks containing fertiliser solution. The NaOH solution has a concentration of 0.1M

4. Rinse and fill a burette with standard hydrochloric acid, concentration is 0.09933M

5. add 4-6 drops of methyl red indicator to each conical flask containing the fertiliser solution and titrate.

The average titre i got was 16.335mL

Here is my working out:

Original n(NaOH) = = 0.002 mol



= 0.00162454 mol

= 0.00162454 mol

= 0.002 - 0.00162454 = 0.000375458mol

The ionic reaction between ammonium sulfate and NaOH is as follows:















Could someone please check this. Many many thanks!

[Over9000]

Original = 0.1 \times 0.02 = 0.002 mol



= =

= 0.00162454 mol

shouldnt this be amount of NaOH reacted, cos it has 1:1 ratio with HCl, so if there is .002 mol of NaOH, and .00162 mol of HCl, then HCl is limiting, the HCl will fully react, so .00162 mol of HCl and .00162 mol of NaOH will react.

= 0.002 - 0.00162454 = 0.000375458mol

[TrueTears]

yeah sorry by unreacted i mean its NaOH unreacted WITH ammonium sulfate

obviously the NaOH is reacted with HCl

and HCl is in excess not limiting? because you are titrating the unreacted NaOH from the reaction with ammonium sulfate WITH HCl, so you are trying to make all of the NaOH react, therefore you need an excess of HCl.

So i don't see what's wrong??

[Over9000]

K, sorry, it's just your wording was awkward. Dont worry, its all fine then 

[TrueTears]

cool thanks is the first one right too?

[Over9000]

This is correct working out?
[/quote]

Could anyone just check if what i did here is correct, these questions are from a prac and these are the results i got from the prac, also the steps to do the prac is listed

Aim: Find the mass of ammonium sulfate(the active ingredient) in a brand of lawn fertiliser by gravimetric analysis.

Mass of fertiliser: 1.015g

1. Add 50mL of deionised water and stir to dissolve as much of the sample as possible. Using a short necked funnel and filter paper, filter the mixture into a 600mL beaker, washing the residue on the filter paper with several portions of deionised water.

2. add 2.5 mL of 2M HCl to the filtered solution (filtrate) and add more distilled water so that the total volume is about 200mL. Use a Bunsen burner to just boil the solution. (Note: the HCl is just used as a catalyst)

3. Add 15mL of the barium chloride solution (an excess) from a measuring cylinder slowly to the hot solution. A white precipitate of barium sulfate will form. Keep the mixture hot and stir continually throughout this process.

(left a few steps out, they just tell you how to use the vacuum filtration, doesn't involve any calculations)

4. Collect the precipitate and weigh.

Ok, the precipitate mass i got was 1.216g

Here's my working to work out the mass of ammonium sulfate








This is correct working out?

Im pretty sure this is right, I went over the calculations and there arent any mistakes, the method you used also seems valid. As long as the information that you used was accurate, your result should be accurate

[TrueTears]

Thanks man

and also how many s.f would you have to put your answers to?

[/0]

The gravimetric analysis looks right. I eventually had it to 4 s.f., because the original accuracy is 4 s.f. and when you add up to get the molar mass of and , they are also to 4 s.f.

[TrueTears]

but, don't you always choose the number of sig figs according to the info given, in this case it's a prac... so... where do you 'extract' the info from? From the steps of the prac or...?

[/0]

Yeah, all the info given is to 4 s.f.
The recorded mass on the balance is to 4 s.f.
Even though the individual relative atomic masses can be 3 s.f., by adding them together you get a molar mass to 4 s.f.

[TrueTears]

in the first prac ( the grav one) they're not given to 4 sig figs you have like 2 sig figs for some of them, check your prac sheet

[/0]

in the first prac ( the grav one) they're not given to 4 sig figs you have like 2 sig figs for some of them, check your prac sheet

Yeah but those we don't use in calculation. Those are either for excess or just catalysts as you said.