Hercules Chem questions :D

[Mao]

A sample of contaminated hydrated copper sulfate CuSO4.5H20 was tested for purity. A 15.0 gram sample was dissolved in water and filtered to remove insoluble impurities. The sulfate ions were precipitated by the addition of excess barium chloride and the resulting precipitated was collected, dried and weighed. If the final mass is 4.95 grams find the percentage of purity of the sample.

n(BaSO4) = 4.95/(137.3+32+16*4) = 0.0212 mol
n(CuSO4.5H2O) = n(BaSO4) = 0.0212 mol
Mr(CuSO4.5H2O) = 63.5 + 32 + 16*4 + 5*18 = 249.5 g/mol
m(CuSO4.5H2O) = 0.0212 mol * 249.5 g/mol = 5.29 g = 35.3%

[BlueSky_3]

Oh I thought we were supposed to ignore hydrated part and just work out the percentage of copper sulfate. So would the hydrated part just mean crystals then? And is that why we would have to consider the mass entire compound? Thanks for your solution Mao! 

[HERculina]

but what did the (x+1) before the H2O mean? And where did the Na from NaOH go?

You determined x=5, and thus on the RHS there are 6 waters.

The Na+ is a spectator ion, for all intents and purposes we can ignore them. (They're present as Na+(aq) on both sides of the equation. A chemical equation is just like a math equation, you can cancel things out which are on both sides)

ah i see. So is the equation that you wrote an ionic one? And i still don't get how you managed to figure out the (x+1) bit when you didn't know what x was at the start. OH and with gravimetric analysis involving precipitation, will it always be a 1:1 ratio of the precipiate to the original sample? :/

[Mao]

And i still don't get how you managed to figure out the (x+1) bit when you didn't know what x was at the start.

If we ignore the hydration, the equation will simply be:



With x degrees of hydration:

OH and with gravimetric analysis involving precipitation, will it always be a 1:1 ratio of the precipiate to the original sample? :/

Not necessarily 1:1, it depends on the charge of the salts.  E.g. if you want to determine the silver content in a solution, you could choose to add sulfate, which will precipitate Ag+ in a 2:1 ratio to form . You should write a chemical equation for every question you do.

[HERculina]

oh wow thanks Mao
ok so how would you approach a question like this:

The amount of calcium carbonate (CaCO3; molar mass = 100.1 g mol-1) in the ore dolomite can be determined by gravimetric analysis. The dolomite sample is dissolved in acid and the calcium ions (Ca2+) present are precipitated as calcium oxalate (CaC2O4; molar mass = 128.1 g mol-1). The calcium oxalate is filtered, dried and strongly heated to form calcium oxide (CaO; molar mass = 56.1 g mol-1).

Question 8
In one analysis the mass of dolomite used was 3.72 g. The mass of calcium oxide formed was found to be
1.24 g.
The percentage of calcium carbonate in the dolomite sample is closest to
A. 26.9
B. 33.3
C. 56.0
D. 59.5

Why is it that n(CaCO3)= n(CaO)  

[b^3]

When the CaC₂O₄ is heated to form CaO, 1 molecule of CaC₂O₄ will produce 1 molecule of CaO. In 1 mol of CaC₂O₄ there is 1 mol of Calcium and so since in 1 mol of CaO there is 1 mol of calcium,(ignore this but, now that I look at the the reasoning wasn't entierly correct although it gave the right answer) then the ratio is 1:1. Hint: try writing out an equation

So in doing the maths.
m(CaO)=1.24g


n(CaC₂O₄)=n(CaO)
n(CaCO₃)=n(CaO)
n(CaCO₃)=0.0221 mol
m(CaCO₃)=0.0221*100.1
=2.2125g
% by mass of CaCO₃=
=59.5 %
So the answer would be D.

I hope there aren't any mistakes in there

EDIT: Crossed out e.t.c

[HERculina]

When the CaC₂O₄ is heated to form CaO, 1 molecule of CaC₂O₄ will produce 1 molecule of CaO. In 1 mol of CaC₂O₄ there is 1 mol of Calcium and so since in 1 mol of CaO there is 1 mol of calcium, then the ratio is 1:1. (Hint: try writing out an equation)

hm you know how you say since theres 1:1 ration of Ca in CaO and CaC₂O₄, why did you use Ca. Like why wasn't it 1:4 ratio of O in CaO compared to O in  CaC₂O₄? (i hope i make sense xD)

and YUP, D is correct

[b^3]

When the CaC₂O₄ is heated to form CaO, 1 molecule of CaC₂O₄ will produce 1 molecule of CaO. In 1 mol of CaC₂O₄ there is 1 mol of Calcium and so since in 1 mol of CaO there is 1 mol of calcium, then the ratio is 1:1. (Hint: try writing out an equation)

hm you know how you say since theres 1:1 ration of Ca in CaO and CaC₂O₄, why did you use Ca. Like why wasn't it 1:4 ratio of O in CaO compared to O in  CaC₂O₄? (i hope i make sense xD)

and YUP, D is correct

You've got to write out the equation.
CaC₂O₄ (s) --> CaO (s) + CO (g) + CO₂
From there you can see that CaC₂O₄ and CaO are in a 1:1 ratio.

EDIT: That would be the decomposition reaction.

[HERculina]

How did you manage to figure out that equation? is it from the general rule; heating metal carbonate --> metal oxide + CO2 (but is CaC2O4 a metal carbonate?)
AND
would you have to also do an equation for  the precipitation of Calcium carbonate too - an ionic one?

[b^3]

How did you manage to figure out that equation? is it from the general rule; heating metal carbonate --> metal oxide + CO2 (but is CaC2O4 a metal carbonate?)
AND
would you have to also do an equation for  the precipitation of Calcium carbonate too - an ionic one?

To be fair I had a go at it writring it and then googled to check it (since its not a "normal" equation that you would come across).

On the precipitation equation, you could write it out also, but if you think about it, firstly you are taking the Ca²⁺ ions out ion a 1:1 ratio then precipitating them forming  CaC2O4  in a 1:1 ratio, writing out the equation here will help show you.

[HERculina]

OHHH ok. i don't htink the part you crossed out was wrong. essentials wrote something similar:

Worked example 2.4b
The content of saccharine (C7H7NO3S) in diet sweetener tablets can be determined by oxidising
the sulfur to sulfate and precipitating it as barium sulfate (BaSO4). A 0.607 g sample yields
0.3196 g barium sulfate. What was the percentage of saccharine in the sample?
Solution
n(BaSO4) =
m(BaSO4)
M(BaSO4)
=
0.3196 g
233.4 g mol−1 = 0.001 369 mol
1 mole of C7H7NO3S yields 1 mole of BaSO4 (as the number of sulfur atoms in each compound
must be the same).
So, the ratio
n(C7H7NO3S)
n(BaSO4)
=
1/1
n(C7H7NO3S) = n(BaSO4) = 0.001 369 mol
m(C7H7NO3S) = n(C7H7NO3S) × M(C7H7NO3S)
= 0.001 369 mol × 185.2 g mol−1 = 0.2535 g
So, the 0.607g sample contains 0.2535 g of C7H7NO3S. Therefore,
% C7H7NO3S =
0.2535 g × 100
0.607 g
= 41.76%
The percentage of saccharine in the tablets is 41.8% (to three signifi cant fi gures).

and then it says:

Notice that no equation has been written for this
reaction. Apart from the fact that it would not be
easy to write an equation using the information
supplied, all we really need to know is the mole
ratio between C7H7NO3S and BaSO4.

i don't understand the bolded part though

[b^3]

OHHH ok. i don't htink the part you crossed out was wrong. essentials wrote something similar:

Worked example 2.4b
The content of saccharine (C7H7NO3S) in diet sweetener tablets can be determined by oxidising
the sulfur to sulfate and precipitating it as barium sulfate (BaSO4). A 0.607 g sample yields
0.3196 g barium sulfate. What was the percentage of saccharine in the sample?
Solution
n(BaSO4) =
m(BaSO4)
M(BaSO4)
=
0.3196 g
233.4 g mol−1 = 0.001 369 mol
1 mole of C7H7NO3S yields 1 mole of BaSO4 (as the number of sulfur atoms in each compound
must be the same).
So, the ratio
n(C7H7NO3S)
n(BaSO4)
=
1/1
n(C7H7NO3S) = n(BaSO4) = 0.001 369 mol
m(C7H7NO3S) = n(C7H7NO3S) × M(C7H7NO3S)
= 0.001 369 mol × 185.2 g mol−1 = 0.2535 g
So, the 0.607g sample contains 0.2535 g of C7H7NO3S. Therefore,
% C7H7NO3S =
0.2535 g × 100
0.607 g
= 41.76%
The percentage of saccharine in the tablets is 41.8% (to three signifi cant fi gures).

and then it says:

Notice that no equation has been written for this
reaction. Apart from the fact that it would not be
easy to write an equation using the information
supplied, all we really need to know is the mole
ratio between C7H7NO3S and BaSO4.

i don't understand the bolded part though

Yeh what I said wasn't wrong, it just won't work for every situation. What the bolded part is trying to say is that we know that the number of sulfur atoms is staying the same, so the amount of sulfur atoms in the first compound must equal the amount of sulfur atoms in the second compound. So 1 mol of the first compound contains 1 mol of sulfur atoms (i.e. the subscript is 1) and 1 mol of the second compound contains 1 mol of sulfur (i.e. the subscript is 1), then 1 mol of the first compound will become 1 mol of the second compound. i.e. a 1:1 ratio.

Lets take another example where it isn't a 1:1 ratio.
Say we have HCl and Zn. Which will form ZnCl₂. In 1 mol of HCl there is 1 mol of Chlorine (the subscript is 1), in 1 mol of ZnCl₂ there is 2 mols of Chlorine (the subscript is 2 (for Cl that is)). i.e. when The number of chlorine atoms must stay the same (so the number of mol of chlorine atoms is the same), so 2 mols of HCl will be required to provide 2 mols of Chlorine and so 2 mols of HCl will be needed to make 1 mol of ZnCl₂. i.e. that ratio is 2:1. *you can check this by writing out the equation and balancing it).

Now to answer you're question regarding the O₄ and why it isn't 1:4 ratio. If we write out the full equation (this may not always be possible at vce level) then we can see that like the example above there is other products bering formed like CO and CO₂, that are taking the other Oxygen atoms.

So basically when you do this, if you can't write the equation then try to thinking intuitvely and find the ratio. Sorry about all this above, I kinda haven't explained it well. Good Luck

[HERculina]

Ohhhh Ok then.  THANKS for exaplaining it I think i get it now