My Chemistry 3/4 thread

[Jason12]

i'm stuck on a few review questions. Anyone able to help?

Q1: A 2.203 g sample of an organic compound that contains oxygen, Cx Hy Oz , 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?

Q2: A precipitate of Fe2O3, of mass 1.43 g, was obtained
by treating a 1.5 L sample of bore water. What was the
concentration of iron, in mol L−1, in the water?

Q3: A 0.500 g sample of sodium sulfate (Na2SO4) and a 0.500 g
sample of aluminium sulfate (Al2(SO4)3) were dissolved in
a volume of water and excess barium chloride was added to
precipitate barium sulfate. What was the total mass of barium
sulfate produced?

[mad_maxine]

 Only have time to do the first one, hope this helps
 1 a)

n(H) = n(H2O) X 2 = 1.32/18  = 0.15 mol
m(H) = 0.15 g

n(C) = n(CO2) = 3.23/44  = 0.0734 mol
m(C) = 0.0734 × 12 = 0.88g

Therefore m(O) = 2.203 – 0.88 – 0.15 = 1.173g

             C   :       H        :   O
         0.88   :   0.15       :   1.17.               
   0.88/12    :   0.15/1    :   1.17/16.         
       0.073   :   0.15.       :   0.073
             1   :        2        :   1

ef is therefore CH2O

Hope this made sense and helped a bit!! Just did a very similar qn an hour ago lol 

[2NE1]

Q2: A precipitate of Fe2O3, of mass 1.43 g, was obtained
by treating a 1.5 L sample of bore water. What was the
concentration of iron, in mol L−1, in the water?

n(Fe2O3)=1.43/159.69=0.0089549
n(Fe)=2x0.0089549=0.0179097 (as there are two molecules of Fe)
hence 0.0179097 moles in 1.5L

and  hence 0.011939 mol L-1 or 0.0119 mol L-1 of iron in water if you wanted it to the correct significant figures.

[2NE1]

Q3: A 0.500 g sample of sodium sulfate (Na2SO4) and a 0.500 g
sample of aluminium sulfate (Al2(SO4)3) were dissolved in
a volume of water and excess barium chloride was added to
precipitate barium sulfate. What was the total mass of barium
sulfate produced?

Basically you want to know all the sulfate there was originally as it all precipitated to barium sulfate.
so you start with Na2SO4
n(Na2SO4)= 0.500/142.0=0.00352 moles
n(SO4)=n(Na2SO4)=0.00352 moles

now for Al2(SO4)3
n(Al2(SO4)3)= 0.500/342.2=0.00146 moles
n(SO4)=0.00146 x 3 as there are 3 molecules of SO4 in the compound=0.00438 moles

now we add the total moles of SO4, so 0.00352+0.00438 =0.00790 moles
now we want the n(BaSO4)=0.00790
m(BaSO4)=0.00790 x 233.4= 1.84 grams produced!

[Snorlax]

From how I see it, these are all Hienamann questions...
You can get the worked solutions which are helpful

[Jason12]

Easiest way to convert to ppm? Examples would be helpful

[Yacoubb]

Easiest way to convert to ppm? Examples would be helpful

ppm is parts per million:
* mg/L
* mg/kg
* mL/L
* mL/kg

Molarity to ppm:
- Molarity to g/L
- g/L x 1000 = mg/L

The most common form for ppm is mg/L. So, convert g/L to mg/L (i.e. ppm) by multiplying the concentration in g/L by 1000.

[Jason12]

i have a prac to design and perform a redox titration analysis to determine the iron content (as soluble iron (II) ions) of a lawn fertiliser.

- for the reaction of iron (II) with permanganate ions, write
a) a half equation for the oxidation reaction
b) half equation for the reduction reaction
c) overall equation

for part c) I got this: 2Mn(S) + 7Fe2+(aq) + 8H2O(aq) -> 2MnO4-(aq) + 7Fe(s) + 16H+(aq)
can anyone confirm this is correct?

Also Q2: 20ml of 1M sulfuric acid should be added to the fertiliser solution before it is titrated with the potassium permanganate solution. Explain why the acid is added.

I know that sulfuric acid is a strong acid so it could be to do with getting a sharp equivalence point but I'm not too confident with acid-base curves.

[RKTR]

erm permanganate ions are MnO4 -
i got 5Fe2+(aq)+MnO4-(aq)+8H+(aq)-->5Fe3+（aq)+Mn2+(aq)+4H2O(l)

acid donates hydrogen ions which are needed for the reduction of permanganate ions.

[Jason12]

erm permanganate ions are MnO4 -
i got 5Fe2+(aq)+MnO4-(aq)+8H+(aq)-->5Fe3+（aq)+Mn2+(aq)+4H2O(l)

acid donates hydrogen ions which are needed for the reduction of permanganate ions.

what were your 2 half equations?

[RKTR]

Fe2+(aq)-->Fe3+(aq) +e-
MnO4- (aq)+8H+(aq)+5e---->Mn2+(aq)+4H2O(l)