Mole Concept and Stoichiometry

There are 2 main types of questions in The Mole Concept and Stoichiometry (H2 Chemistry) that students struggle with.

1. Determine the molecular formula of a hydrocarbon that using the combustion of hydrocarbon general equation.
2. Back Titration

Let’s take a look!

Question 1:

10 cm3 of a gaseous hydrocarbon was exploded with 90 cm3 of O2 (excess). There was a decrease in volume of 30 cm3 (measured at 298K). On adding aqueous NaOH the volume drops to 40 cm3. What is the molecular formula of the hydrocarbon?

Worked Solution:

The volume of H2O can be ignored since it exists as a liquid at room temp.

The initial volume was 100 cm3 and the final volume is 70 cm3 since there was a decrease in volume of 30 cm3

When NaOH is added, the volume drops again to 40 cm³. Since NaOH absorbs the CO2, the decrease of 30 cm³ corresponds to the acidic CO2 gas. Hence there is 40 cm3 of O2 remaining after combustion. There is no hydrocarbon remaining as it is limiting and O2 is in excess.

We will use the reacting volume (change in volume) to find the coefficients in the balanced chemical equation. We will fill up the change in volume values using the initial and final volume values.

​Mole ratio is equivalent to volume ratio for gases.

Since 1 mole of CxHy produces x moles of CO2
and 10 cm3 of CxHy reacts to produce 30 cm3 of CO2, we can solve for x.

Since 1 mole of CxHy reacts with (x+y/4) moles of O2
and 10 cm3 of CxHy reacts with 50 cm3 of O2, we can solve for y.

The molecular formula of the hydrocarbon is C3H8.


This is another difficult question that stumps many students. It is a question from the 2010 A Level Paper 3

Question 2:

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 0.10 cm3 of liquid J was dissolved in an inert solvent and an excess of sodium metal added, 10.9 cm3 of gas (measured at 298K) was produced.

When 0.10 cm3 of liquid J was combusted in an excess of oxygen in an enclosed vessel, the volume of gas (measured at 298K) 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 298K). Use the data to calculate value of x and y in the molecular formula CxHyOH for J.

Worked Solution:

CxHyOH (l) + Na (s) → CxHyONa+ (l) + 1/2H2 (g)
CxHyOH (l) + zO2 (g) → xCO2 (g) + (y+1)/2 H2O (l)

10.9 cm3 refers to the amount of H2 produced.
109 cm3 refers to the amount of CO2 produced which was absorbed by NaOH.

The following ratios can be obtained from the above equations and the AMOUNT OF LIQUID J USED WAS THE SAME IN BOTH REACTIONS, we can integrate the ratios into a single line.

Using mole ratio being equivalent to volume ratio for gases, we can solve for x.
x/(1/2) = 109/10.9
⇒ x = 5

Now let’s sub the value of x =5 into equation and calculate the value of y.

You can add the change in volume of reactants and products and that will equate to the net change in volume of the reaction. Since there was a reduction by 54.4 cm3 after the combustion reaction, we can create an equation.

−b + 109 = -54.4
b = 163.4

Mole ratio is equivalent to volume ratio for gases.

z/5 = 163.4/109
z/5 = 1.5
z = 7.5

Balance oxygen on both sides of equations

The formula for J is C5H11OH.


The second type of question is involving Back Titration.

Back titrations are usually employed when the determination of the amount of a substance poses some difficulty in the direct titration method, e.g. solid substances.

Say for example, CaCO3 in toothpaste is insoluble in water hence it is not possible to do a direct titration. The following steps will enable one to determine the amount of CaCO3 in this case.

1. A known amount of excess HCl (x mol) is added to react with all the CaCO3.
2. (Back) titrate with NaOH (y mol) to determine the amount of remaining HCl.
3. Calculate the amount of HCl that has reacted with CaCO3 (x – y mol).
4. CaCO3 + 2HCl → CaCl2 + CO2 + H2O
Since CaCO3 and HCl are in a ratio of 1 : 2, the amount of CaCO3 in the sample is ½(x – y) mol.

Question 3:

A 0.5 g sample of impure CaCO3 was dissolved in 50.0 cm3 of 0.200 mol dm-3 HCl. The acid remaining was then titrated with NaOH containing 0.125 mol dm-3 NaOH. In the titration, 48.0 cm3 of NaOH was required for the reaction. Find the % purity of CaCO3.

Worked Solution:

CaCO3 + 2HCl → CaCl2 + H2O + CO2
HCl + NaOH → NaCl + H2O

Amount of HCl added = 50/1000 x 0.200 = 0.0100 mol (this is the x)

Since 1 mole of NaOH reacts with 1 mole of HCl
Amount of HCl that remaining after adding CaCO3
= Amount of NaOH
= 0.125 x 48/1000
= 0.006 mol (this is the y)

Amount of HCl reacted with the CaCO3
= 0.0100 – 0.00600
= 0.00400 mol (this is the x-y)

CaCO3   + 2HCl → CaCl2  +  CO2  +  H2O

2 mol HCl reacts with 1 mol CaCO3
0.00400 mol HCl reacts with 0.00200 mol CaCO3 [this is the 0.5(x − y)]

Mass of CaCO3 present in sample = 0.0020 x 100 g = 0.200 g

Hence, % purity of CaCO3 =  (0.200/0.5)  x 100 % =  40.0 %