Earlier, we have discussed on the strategy of Determining the Molecular Formula of Compounds using Composition by Mass.

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:

C_xH_y (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 C_xH_y (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

C_xH_y (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.

PS: Let me know how you find it. Drop me a comment. I would love to hear from you.

Many substances consist of Molecules. A molecule consists of a small number of atoms joined together by covalent bonds.

The Molecular Formula shows all the atoms of each element contained in one molecule.

Empirical Formula shows the simplest ratio of the different types of elements present in a molecule.

For example, Butane has Molecular Formula of C₄H₁₀ and Empirical Formula of C₂H₅.

As such, Molecular Formula is a simple multiple of the Empirical Formula:

Molecular Formula   =   n   X   Empirical Formula

where n = 1, 2, 3, etc

The molecular formula can be calculated from:

1. Empirical Formula (we have discussed how to obtain it)
2. Relative Molecular Mass, Mr of Compound

Let’s take a look at an example.

Example:

Glucose has empirical formula CH₂O and relative molecular mass of 180. Find the molecular formula of glucose.

(given: H = 1.0; C = 12.0; O = 16.0)

Suggested Answer:

Relative mass of Empirical Formula, CH₂O = 12 + 2(1) + 16 = 30

Relative molecular mass   =   n   X   relative mass of the empirical formula

thus, n = Relative molecular mass / relative mass of the empirical formula

= 180 / 30 = 6

Therefore, Molecular Formula of Glucose =   6   X   CH₂O   =   C₆H₁₂O₆

Let me know how you feel about this blogposts. Drop me a comment.

As mentioned in earlier post on Atoms, Molecules & Stoichiometry – there aren’t many differences in the Mole Calculations questions between GCE A-Level (Advanced Level) Chemistry & GCE O-Level (Ordinary Level) Chemistry, in terms of formulae and approach.

However, in GCE A-Level Chemistry (H1/H2), the examiners expect you to have a good grasp of Basic Chemistry knowledge i.e. topics in GCE O-Level Chemistry should be at your finger tips.

Take the following question that i recently gave it to my A-Level H2 Chemistry Weekly Class.

Question:

What is the mass of Zn obtained when 50g of Zinc Oxide is reduced by 50g of charcoal?

In this Mole Calculations question, students are expected to be able to write the balanced chemical equation based on the information given. Half of my students can’t write the balanced chemical equations! Because they have forgotten about the topic of Metals & Reactivity Series in GCE O-Levels!

Warning: Do not throw away your GCE O-Level Chemistry notes just yet. You might need it at times for quick revision.

Let’s take a look at the approach to solve this question.

Suggested Answer:

Recalling what we have learned in GCE O-Level Chemistry, when ZnO is reduced by charcoal (which is essentially Carbon) – we will obtain the Zn metal and Carbon Monoxide as the products.

ZnO (s)   +   C (s)   –>   Zn (s)   +   CO (g)

50g                50g

No. of moles of ZnO = Mass of ZnO / Mr of ZnO = 50 / (65.4 + 16) = 0.614 mol

No. of moles of C = Mass of C / Ar of C = 50 / 12 = 4.17 mol

Therefore,  ZnO is the Limiting Reagent & C is the Excess Reagent

No. of moles of Zn formed = No. of moles of ZnO = 0.614 mol

[Limiting Reagent determines the amount of product(s) formed]

Therefore, Mass of Zn formed = Mole of Zn X Ar of Zn = 0.614 X 65.4 = 40.2g (3 sig. fig)

Hope the above example gives you an idea on some of the expectations you need when making transition from GCE O-Level Chemistry to GCE A-Level Chemistry.

Let’s try out one question below.

Quick Check 1:

Calculate the volume of CO₂ produced (at r.t.p) by decomposing 15g of calcium carbonate.

PS: Try out the question and leave your answers in the comment section below.