Stoichiometry Part I: Tackling standard A Level / Prelims examination questions
- February 28, 2017
- Stoichiometry and Mole Concept
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 A-level Chemistry or preliminary examinations, that are not so easy to solve.
We visit one such question here.
A levels 2010 Paper 3
Alcohol J, CxHyOH,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 cm3 of liquid J was dissolved in an inert solvent and an excess of sodium metaladded, 10.9 cm3 of gas (measured at 298 K) was produced.
When 0.1 cm3 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 cm3. The addition of an excess of NaOH(aq) caused a further reduction in gas volume of 109 cm3(measured at 298 K). Use these data to calculate values for x and y in the molecular formula CxHyOH for J.
There are 2 ways to solve this question. One of the methods is presented below:
There are two reactions in this question:
- Reaction of the alcohol J with sodium.
- Reaction of the alcohol J with oxygen.
CxHyOH + Na → CxHyOHa + ½H2
|Moles of H2 formed at RTP =||10.9 cm3||= 4.542 x 10-4 moles|
Thus, the moles of CxHyOH 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.
|CxHyOH (l) + (||
|) O2(g) → XCO2 (g) + (||
Note that the question mentioned that O2 is in excess.
Thus at the end of the reaction, the gases left at RTP are CO2 and excess O2.
Since the volume of the gas was reduced by 54.4 cm3,
Initial volume of O2- (Volume of O2 left + Volume of CO2 formed) = 54.4 cm3.
(THIS IS THE PART MANY STUDENTS DO NOT KNOW HOW TO INTERPRET!)
⇒ Volume of O2 reacted – Volume of CO2 formed = 54.4 cm3 …(1)
Volume reduction after adding NaOH = 109 cm3.
Since only CO2 in the final mixture reacts with
NaOH, volume of CO2 formed = 109cm3
Substituting in equation (1):
Volume of O2 reacted = 109 + 54.4 = 163.4 cm3
What we have gotten so far:
Moles of CxHyOH reacted = 9.083 x 10-4
|Moles of O2 reacted =||
|= 6.808 x 10-3|
|Moles of CO2 formed =||
|= 4.541 x 10-3|
|Moles of CxHyOH||:||moles of CO2|
|9.083 x 10-4||:||4.541 x 10-3|
|Moles of CxHyOH||:||moles of O2|
|9.083 x 10-4||:||6.808 x 10-3|
|=||2(7.495) + 1|
Thus J is C5H11OH
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.
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!!
- 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 0.1cm3 also proportional to the volumes of CO2 formed in Reaction A and CO2 formed (and O2 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!!