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 cm ^{3 }of a gaseous hydrocarbon was exploded with 90 cm^{3} of O_{2} (excess). There was a decrease in volume of 30 cm^{3} (measured at 298K). On adding aqueous NaOH the volume drops to 40 cm^{3}. What is the molecular formula of the hydrocarbon?**

**Worked Solution:**

The volume of H_{2}O can be ignored since it exists as a liquid at room temp.

The initial volume was 100 cm^{3} and the final volume is 70 cm^{3} since there was a decrease in volume of 30 cm^{3}

When NaOH is added, the volume drops again to 40 cm³. Since NaOH absorbs the CO_{2}, the decrease of 30 cm³ corresponds to the acidic CO_{2} gas. Hence there is 40 cm^{3} of O_{2} remaining after combustion. There is no hydrocarbon remaining as it is limiting and O_{2} 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 C_{x}H_{y} produces x moles of CO_{2}

and 10 cm^{3} of C_{x}H_{y} reacts to produce 30 cm^{3} of CO_{2}, we can solve for x.

Since 1 mole of C_{x}H_{y} reacts with (x+y/4) moles of O_{2}

and 10 cm_{3} of C_{x}H_{y} reacts with 50 cm^{3} of O_{2}, we can solve for y.

The molecular formula of the hydrocarbon is C_{3}H_{8}.

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This is another difficult question that stumps many students. It is a question from the 2010 A Level Paper 3__Question 2:__

**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 0.10 cm ^{3} of liquid J was dissolved in an inert solvent and an excess of sodium metal added, 10.9 cm^{3} of gas (measured at 298K) was produced.**

**When 0.10 cm ^{3} 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 cm^{3}.The addition of an excess of NaOH(aq) caused a further reduction in gas volume of 109 cm^{3} (measured at 298K). Use the data to calculate value of x and y in the molecular formula C_{x}H_{y}OH for J.**

**Worked Solution:**

C_{x}H_{y}OH (l) + Na (s) → C_{x}H_{y}O^{−}Na^{+} (l) + 1/2H_{2} (g)

C_{x}H_{y}OH (l) + zO_{2} (g) → xCO_{2} (g) + (y+1)/2 H_{2}O (l)

10.9 cm^{3} refers to the amount of H_{2} produced.

109 cm^{3} refers to the amount of CO_{2} 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 volum**e of the reaction. Since there was a reduction by 54.4 cm^{3} 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 C_{5}H_{11}OH.

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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, CaCO_{3} 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 CaCO_{3} in this case.

1. A known amount of excess HCl (x mol) is added to react with all the CaCO_{3}.

2. (Back) titrate with NaOH (y mol) to determine the amount of remaining HCl.

3. Calculate the amount of HCl that has reacted with CaCO_{3} (x – y mol).

4. CaCO_{3} + 2HCl → CaCl_{2} + CO_{2} + H_{2}O

Since CaCO_{3} and HCl are in a ratio of 1 : 2, the amount of CaCO_{3} in the sample is ½(x – y) mol.

__Question 3:__

**A 0.5 g sample of impure CaCO _{3} was dissolved in 50.0 cm^{3} 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 cm^{3} of NaOH was required for the reaction. Find the % purity of CaCO_{3}.**

**Worked Solution:**

CaCO_{3} + 2HCl → CaCl_{2} + H_{2}O + CO_{2}

HCl + NaOH → NaCl + H_{2}O

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 CaCO_{3}

= Amount of NaOH

= 0.125 x 48/1000^{ }

= 0.006 mol **(this is the y)**

Amount of HCl reacted with the CaCO_{3}

= 0.0100 – 0.00600

= 0.00400 mol **(this is the x-y)**

CaCO_{3} + 2HCl → CaCl_{2} + CO_{2} + H_{2}O

2 mol HCl reacts with 1 mol CaCO_{3}

0.00400 mol HCl reacts with 0.00200 mol CaCO_{3} **[this is the 0.5(x − y)]**

Mass of CaCO_{3} present in sample = 0.0020 x 100 g = 0.200 g

Hence, % purity of CaCO_{3} = (0.200/0.5) x 100 % = 40.0 %