Stoichiometry and mole concept have been covered in the O levels and is redone during the first few months of Junior College (JC) Chemistry during the first year.
Therefore, many students usually assume that this topic is easy and thus do not spend much time revising it during the A levels Chemistry paper.
Contrary to what many think, there are some Chemistry exam questions that came out during the past year Alevel Chemistry or preliminary examinations, that are not so easy to solve.
We visit one such question here.
H2 Chemistry
A levels 2010 Paper 3
Alcohol J, C_{x}H_{y}OH,is a volatile fungal metabolite whose presence when detected in air can indicate hidden fungal attack on the timbers of a house.
When 0.1 cm^{3} of liquid J was dissolved in an inert solvent and an excess of sodium metaladded, 10.9 cm^{3} of gas (measured at 298 K) was produced.
When 0.1 cm^{3} of liquid J was combusted in an excess of oxygen in an enclosed vessel, the volume of gas (measured at 298 K) was reduced by 54.4 cm^{3}. The addition of an excess of NaOH(aq) caused a further reduction in gas volume of 109 cm^{3}(measured at 298 K). Use these data to calculate values for x and y in the molecular formula C_{x}H_{y}OH for J.
Solutions:
There are 2 ways to solve this question. One of the methods is presented below:
Method 1:
There are two reactions in this question:
 Reaction of the alcohol J with sodium.
 Reaction of the alcohol J with oxygen.
Reaction A:
C_{x}H_{y}OH + Na → C_{x}H_{y}OHa + ½H_{2}
Moles of H_{2} formed at RTP =  10.9 cm^{3}  = 4.542 x 10^{4} moles  

Thus, the moles of C_{x}H_{y}OH reacted = 2(4.542 x 10^{4}) = 9.083 x 10^{4} moles
Since the amount of alcohol J used in reactions A and B are the same,
9.083 x 10^{4} moles of alcohol J was used in Reaction B as well.
Reaction B:
C_{x}H_{y}OH (l) + ( 

) O_{2}(g) → XCO_{2} (g) + ( 

)H_{2}O (l) 
Note that the question mentioned that O_{2} is in excess.
Thus at the end of the reaction, the gases left at RTP are CO_{2} and excess O_{2}.
Since the volume of the gas was reduced by 54.4 cm^{3},
Initial volume of O_{2} (Volume of O_{2} left + Volume of CO_{2} formed) = 54.4 cm^{3}.
(THIS IS THE PART MANY STUDENTS DO NOT KNOW HOW TO INTERPRET!)
⇒ Volume of O_{2} reacted – Volume of CO_{2} formed = 54.4 cm^{3} …(1)
(IMPORTANT DEDUCTION!)
Volume reduction after adding NaOH = 109 cm^{3}.
Since only CO_{2} in the final mixture reacts with
NaOH, volume of CO_{2} formed = 109cm^{3}
Substituting in equation (1):
Volume of O_{2} reacted = 109 + 54.4 = 163.4 cm^{3}
RECAP:
What we have gotten so far:
Moles of C_{x}H_{y}OH reacted = 9.083 x 10^{4}
Moles of O_{2} reacted = 

= 6.808 x 10^{3} 
Moles of CO_{2} formed = 

= 4.541 x 10^{3} 
Moles of C_{x}H_{y}OH  :  moles of CO_{2} 
1  :  X 
9.083 x 10^{4}  :  4.541 x 10^{3} 
X  =  5 
Moles of C_{x}H_{y}OH  :  moles of O_{2}  
1  : 


9.083 x 10^{4}  :  6.808 x 10^{3} 
⇒ 

= 7.495 
2(5) + 

=  2(7.495) + 1  
y  =  11 
Thus J is C_{5}H_{11}OH
Points of discussion:
a. Some students will guess what the alcohol, J is after finding out that X = 5. A natural answer is y = 11, which is the formula for pentanol. However, take note that this is very dangerous, and y should be determined mathematically since y can be less than 11 if
 The alcohol is unsaturated
 The alcohol is alicyclic
 The alcohol is both alicyclic and unsaturated.
Method 2
Question: Can we use volume ratio to solve the question, without once determining any number of reactants and products formed?
Answer: Of course we can!!
Reasons:
 Avogardro’s law states that equal volume of gases (irregardless of the type of gas) at the same temperature and pressure contains the same number of particles (or moles!!).
 But how about alcohol J? Is its volume of 0.1cm^{3} also proportional to the volumes of CO_{2} formed in Reaction A and CO_{2} formed (and O_{2} used) in Reaction B? Is it not a liquid?
For more insights into the 2nd method of solving this question without, even once, needing to calculate moles, find out more at www.FocusChemistry.com NOW!!