Qualification > Sciences
ALL CIE CHEMISTRY DOUBTS HERE !!
moon:
plz help in Nov.2009 ppr22 Q2(v) &
also Nov.2009 ppr42 Q8(c) (ii), d(ii),(iv)
thanx in advanc.
moon:
can u plz another question?
complete combustion of a hydrocarbon gave 0.352gm of CO2 & 0.072gm of water. What is the formula of hydrocarbon? ??? ???
elemis:
--- Quote from: moon on October 09, 2010, 01:01:15 am ---can u plz another question?
complete combustion of a hydrocarbon gave 0.352gm of CO2 & 0.072gm of water. What is the formula of hydrocarbon? ??? ???
--- End quote ---
Your question is incomplete I need the mass of the original hydrocarbon.
But the methodology is as follows :
If there are 2 grams of Hydrogen in 18 grams of water THEN
there are X grams of Hydrogen in 0.072 grams of water
Hence, X = 8*10-3
If there are 12 grams of Carbon in 44 grams of CO2 THEN
there are Y grams of Carbon in 0.352 grams of CO2
Hence, Y = 0.096
At this point you would add X and Y and subtract that value from the mass of the Hydrocarbon to find the mass of oxygen.
Then you would have the mass of C H and O
You could then calculate the Empirical formula.
Deadly_king:
--- Quote from: moon on October 09, 2010, 01:01:15 am ---can u plz another question?
complete combustion of a hydrocarbon gave 0.352gm of CO2 & 0.072gm of water. What is the formula of hydrocarbon? ??? ???
--- End quote ---
Nan Ari......the data provided is sufficient to find the answer.
CxHy + (x + y/4)O2 ----> xCO2 + (y/2) H2O
Number of moles CO2 = Mass / Mr = 0.352/44 = 0.008
Number of moles of H2O = 0.072/18 = 0.004
1 mole of the hydrocarbon gives 0.008 moles of CO2 and 0.004 moles of H2O
Therefore from equation above we can note that x = 0.008 while y/2 = 0.004
Empirical formula will hence be C8H8
If you don't understand, let me know :)
elemis:
Ah, damn. I didnt think of it in that way. :-X
+rep.
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