In an experiment, 20 cm of a gaseous hydrocarbon Z requires 80 cm' of oxygen for complete combustion to produce 60 cm of carbon dioxide at room conditions.
Which of the following formula represents Z?

A C2H2
B C2H4
C C3H4
D C3H8

Today, we will discuss on the Determination of Molecular Formula of Hydrocarbons using Combustion Data. This is a new concept for those that making their transition from GCE O-Levels to GCE A-Levels and thus will be one of the key questions to be asked in GCE A-Level H1 and H2 Chemistry Examinations.

Let’s look at the method involved.

Hydrocarbons burns in excess oxygen based on the following equation:

CxHy (g) + (x + y/4) O2 (g) –> xCO2 (g) + y/2 H2O (l)

At room temperature conditions or s.t.pm the water product is actually a liquid. Thus, the volume is negligible compared to volumes of CxHy (g), O2 (g) and CO2 (g).

Using Volume Ratio;

Hence, if 1cm3 of is completely burned in oxygen,

Volume of oyxgen used = (x + y/4) cm3

Volume of carbon dioxide produced = x cm3

Volume of water produced (as liquid) = 0 cm3

Let’s check out an exam-based question to see how we can make use of the above to solve question.

Example:

10cm3 of a gaseous hydrocarbon required 20 cm3 of oxygen for complete combustion. 10 cm3 of carbon dioxide was produced. Calculate the molecular formula of the hydrocarbon.

and

Suggested Solution:

Using the general equation and applying volume ratio, we have

CxHy (g) + (x + y/4) O2 (g) –> xCO2 (g) + y/2 H2O (l)

10cm3…… 20cm3 ……………….10cm3

1mol 2 mol 1 mol

Based on the above comparision, we have

x = 1 and (x + y/4) = 2

Solving it gives y = 4

Hence, the molecular formula of the hydrocarbon is CH4.

Hope the above explanation is useful to you. Do note that this is a very important concept when you are in JC1 and will still be tested when you are in JC2. I realised that many of my students in my GCE A-Level H2 Chemistry Tuition Classes are not well trained in their Junior Colleges when it comes to this basic concept.

Mark me as brainliest

stupid i wont let you get a chance

explanation:

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.