Calculation Questions! (1 Viewer)

[faisalabdul16]

Post some of the hard calculation questions you guys have come across in papers!

Here's one to start off with -

A student carried out an investigation to analyse lawn fertiliser. The student weighed out 1.00 g of fertiliser containing 24.0 % sulfur (S) and dissolved it in water. 100 mL of 0.20 mol L1 barium chloride solution was then added and a precipitate formed.

(a) Calculate the theoretical percentage by mass of sulfate in the fertiliser.

After you've answered a question please post up another one.

 

[zhertec]

Post some of the hard calculation questions you guys have come across in papers!

Here's one to start off with -

A student carried out an investigation to analyse lawn fertiliser. The student weighed out 1.00 g of fertiliser containing 24.0 % sulfur (S) and dissolved it in water. 100 mL of 0.20 mol L1 barium chloride solution was then added and a precipitate formed.

(a) Calculate the theoretical percentage by mass of sulfate in the fertiliser.

Is it actually 24% of sulfur or sulfate?

 

[SuchSmallHands]

72% ?

 

[faisalabdul16]

Is it actually 24% of sulfur or sulfate?

Sulfur

72% ?

Correct! Post questions pls

 

[SuchSmallHands]

Only calcs? Because I'm doing environment stuff at the minute

 

[enigma_1]

Post some of the hard calculation questions you guys have come across in papers!

Here's one to start off with -

A student carried out an investigation to analyse lawn fertiliser. The student weighed out 1.00 g of fertiliser containing 24.0 % sulfur (S) and dissolved it in water. 100 mL of 0.20 mol L1 barium chloride solution was then added and a precipitate formed.

(a) Calculate the theoretical percentage by mass of sulfate in the fertiliser.

pls help, what equation do we use for this?

 

[SuchSmallHands]

pls help, what equation do we use for this?

You just work out how many moles of sulfur you have and you have to have the same number of moles of SO42-. Then it's just a mass to mole and a mass to percentage

 

[zhertec]

You just work out how many moles of sulfur you have and you have to have the same number of moles of SO42-. Then it's just a mass to mole and a mass to percentage

Lol I realised why I couldnt get the answer, I did moles of sulfur x (32.07 x (16x4)). Eugh I hope that doesnt occur in the exam lol

 

[enigma_1]

You just work out how many moles of sulfur you have and you have to have the same number of moles of SO42-. Then it's just a mass to mole and a mass to percentage

Ohh ok thanks!! ))

 

[90atarpls]

Post some of the hard calculation questions you guys have come across in papers!

Here's one to start off with -

A student carried out an investigation to analyse lawn fertiliser. The student weighed out 1.00 g of fertiliser containing 24.0 % sulfur (S) and dissolved it in water. 100 mL of 0.20 mol L1 barium chloride solution was then added and a precipitate formed.

(a) Calculate the theoretical percentage by mass of sulfate in the fertiliser.

After you've answered a question please post up another one.

can som1 post the working for this? im lost

 

[fatima96]

Question:
Citric acid, the predominant acid in lemon juice, is a weak triprotic acid.
A student titrated 25.0 mL samples of lemon juice with 0.55 M NaOH. The mean titration value was 29.5 mL. What was the concentration of C6H8O7 in the lemon juice?

 

[SuchSmallHands]

can som1 post the working for this? im lost

It's just 24% = 0.24g
0.24g (S) = 0.00748 mol
0.00748 mol (SO4) = 0.7189g
= 72%

 

[fatima96]

can som1 post the working for this? im lost

Mass of sulfur = 24% of 1g
thus m(S) = 0.24 gram.
n(S) = 0.24/32.07 = 0.007483.......mol

n(S) = n(SO42-)

m(SO42-) = n x M
= 0.0074836 x (32.07 + 16 + 16 + 16 + 16)
= 0.7189....g

Thus, precentage = 0.718 g sulfate / 1 g fertiliser = 72%

 

[90atarpls]

It's just 24% = 0.24g
0.24g (S) = 0.00748 mol
0.00748 mol (SO4) = 0.7189g
= 72%

wtf does barium have to do with it then

 

[faisalabdul16]

can som1 post the working for this? im lost

<img src="http://latex.codecogs.com/gif.latex?\text{You&space;have&space;0.24g&space;of&space;sulfur.&space;Moles&space;of&space;sulfur}=&space;\\&space;\frac{0.24}{32.07}&space;\\&space;\\&space;\text{Moles&space;of&space;sulfur&space;=&space;moles&space;of&space;sulfate&space;ions&space;(1&space;to&space;1&space;ratio)}\\&space;\\&space;\frac{0.24x96x100}{32.07}=&space;~72%" title="\text{You have 0.24g of sulfur. Moles of sulfur}= \\ \frac{0.24}{32.07} \\ \\ \text{Moles of sulfur = moles of sulfate ions (1 to 1 ratio)}\\ \\ \frac{0.24x96x100}{32.07}= ~72%" />

 

[SuchSmallHands]

wtf does barium have to do with it then

That bit was irrelevant, it relates to other parts of the question I'm assuming

 

[enigma_1]

can som1 post the working for this? im lost

n of sulphur = 1/32.07 = 0.031181789834737 =n of SO4 2-

m of S04 2- = n x M = something

Then mass of (sulphate/1g of fertiliser) x100 = answer

 

[90atarpls]

ok well im fucked for tomorrow

 

[faisalabdul16]

Question:
Citric acid, the predominant acid in lemon juice, is a weak triprotic acid.
A student titrated 25.0 mL samples of lemon juice with 0.55 M NaOH. The mean titration value was 29.5 mL. What was the concentration of C6H8O7 in the lemon juice?

0.22

 

[fatima96]

0.22

That's correct.