Burberry's Chem Question Thread

[burbs]

Not really sure on the deal with question threads, but I'm sure I'll have plenty so I thought I'd put it here. If I'm out of line let me know.

1. When taking sigfigs, is it only the sigfigs in the question or including from the data book?
Edit: and if I have a whole number, how do sigfigs exactly work, do I put it in scientific notation? like would 14304 be 14*10^3 to 2 sigfigs?

2. If a question asks for an answer in number of particles, does that just mean number of mol*Avogadro's constant?

3. An impure sample of iron(II) sulfate, weighing 1.545 g, was treated to produce a precipitate of Fe₂O₃. If the mass of the dried precipitate was 0.315 g, calculate the percentage of iron in the sample.
 - Unsure of what the reaction would be for this, and how exactly I should know what is being used.

Thanks, undoubtedly more to come.

[vea]

1. From the question.
2. Yes
3. I'm also unsure of the reaction here so I'll also be waiting to see the answer. :\

[Greatness]

1) Usually questoins would specify the no. of sig figures, otherwise use whichever is the lowest.
2) Yes that is correct
3) ill work on that now

[burbs]

Thanks guys!

[Milkshake]

3. Is the answer 14% iron?

[Greatness]

Yeah im unsure on the reaction of question 3 as well..

[burbs]

3. Is the answer 14% iron?

Says 14.3% so yeah. How did you do it?
 

[Milkshake]

Don't think you need the actual reaction. I just used percentages.
M(Fe2)/M(Fe2O3) x0.315g = mass of iron = 0.2203g
(0.2203g/1.545) x 100 = 14.26%?

[burbs]

Oh right thanks! So unless its obvious, questions should always give the reaction?

[Pixon]

Pretty much. There are certain reactions which you need to recognise from compound names such as acid+base, acid + metal etc.
But for a question like this, some people find it easier to imagine the FeSO4 is being treated with X. But you quickly realise this is pointless because with mole ratios (and enough information) it doesn't help. You just need to make sure no Iron is lost from the transition of sample to precipitate.

[burbs]

Cheers.

Another question:

 A 2.203 g sample of an organic compound was extracted from a plant. When it was burned in oxygen, the hydrogen in the compound was converted to 1.32 g of water and the carbon was oxidised to 3.23 g of carbon dioxide.
 a)   Find the empirical formula of the compound.
 b)   Another sample was analysed in a mass spectrometer.
The mass spectrum produced showed that the molar mass of the compound was 60.0 g mol⁻¹. What is its molecular formula?

[Pixon]

a)m(C)= 12/44 * 3.23         m(H)= 2/18 * 1.32         <--- Refer to Milkshake's post for this method
      = 0.881g                        = 0.147g

therefore:
m(O)=2.203-0.881-0.147
       = 1.175g
              
n(C)= 0.881 / 12        n(H)=0.147/1      n(O)=1.175/16
      = 0.0734mol              =0.147mol          = 0.0734mol

Ratio is 1:2:1

therefore EF is CH20

b) M(CH20)=12+1*2+16
               =30g/mol

Molar mass found by mass spectrum is 60.0

therefore MF=60/30 * EF
                 =C2H4O2

(sorry I don't know that Latex or whatever you call it thing)

[burbs]

Possibly a slightly trivial question but:

During the preparation of the standard solution show in 3.4 (Basically just has the steps of 1) place weighed sample in flask, 2) half fill with water and shake, 3) add water to calibration line) why is water added to the level of the calibration line after the solid has dissolved rather than below?


Also, why can't you express concentration of carbohydrates as molarity?

[Mao]

First question: I'm assuming this is a volumetric flask. I generally half-fill the flask, as it is easier to swirl/shake to dissolve the solids. It's more a matter of practicality due to the design of the flasks.

As for carbohydrates, they are a class of biomolecules, there is no set molecular mass because they are polymers with varying numbers of monomers. Thus it will be pointless to express the concentration as molarity.

[burbs]

Yeah it was a volumetric flask. Thanks!