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# Chapter 09: Mole Concept & Chemical Calculations

## Chapter 9: Mole Concept & Chemical Calculations  Question 1
How many C atoms does 0.5 mol of CH₃COOH contain?

number of C atoms
= no. of mol x Avagaro's constant x no. of C atoms in formula
= 0.5 mol x 6 10²³ x 2
= 6.00 x 10²³ (3sf)

Now that we know what mol represents, let’s look at the important formulas to determine number of mol.

• 1. Molar mass (g/mol)

Molar mass is defined as the mass of 1 mol of substance. This means, if a student wants to weigh one mol of Carbon, they need to weigh out 12 g of Carbon. If a student wants to weigh one mol of sodium hydroxide, they need to weigh out 40 g of NaOH. Where do we get these values?

The molar mass of a substance will have the same value as the relative atomic mass of an element, or relative molecular mass of a compound. The only difference between relative molecular mass and molar mass is that molar mass has unit (which is g/mol) while relative molecular mass does not.

Substance Relative molecular mass Molar mass
Magnesium metal 24 24 g/mol
Sulfur dioxide 64 64 g/mol
Substance
Magnesium metal
Relative molecular mass
24
Molar mass
24 g/mol
Substance
Sulfur dioxide
Relative molecular mass
64
Molar mass
64 g/mol

What is the formula to find number of mol related to mass?  • 2. Molar volume (24 dm³) of gases

Molar volume is the volume occupied by 1 mole of gas, and at room temperature and pressure (r.t.p.), this volume is 24 dm³.

When the volume of gas is given, always check that the units is the same as the onein molar volume, which is in dm³. Under the topic of molar volume, we need to look at Avogadro’s law, which states that equal volumes of gas under the same conditions of temperature and pressure, have equal number of particles.

To illustrate the example, we have two balloons containing 100cm³ of methane and carbon dioxide respectively. Although the balloons contain different gases, both balloons contain the SAME number of mole and molecules of gases as the VOLUME is the same. However, mass of the balloons will be different as their MOLAR MASSES are different.

 Substance  No. of mol 100cm³ 24 000cm³ =4.17 x 10³ mol 100cm³ 24 000cm³ =4.17 x 10³ mol No. of molecules 0.00417 × 6 × 10²³ = 2.50 × 10²¹ CO₂ molecules 0.00417 × 6 × 10²³ = 2.50 × 10²¹ CH₄ molecules Mass of gas 0.00417𝑚𝑜𝑙 × 44𝑔/𝑚𝑜𝑙 = 0.183g 0.00417𝑚𝑜𝑙 × 16𝑔 /𝑚𝑜l = 0.0667g
 Substance No. of mol 100cm³ 24 000cm³ =4.17 x 10³ mol No. of molecules 0.00417 × 6 × 10²³ = 2.50 × 10²¹ CH₄ molecules Mass of gas 0.00417𝑚𝑜𝑙 × 44𝑔/𝑚𝑜𝑙 = 0.183g

Question 2

40 cm³ of ethane, C₂H₆, are reacted with 150 cm³ of oxygen to form carbon dioxide and water.

The equation for the reaction is shown.

2C₂H₆(g) + 5O₂(g) → 4CO₂(g) + 6H₂O(l)

All volumes are measured at r.t.p.

What is the total volume of gas remaining at the end of the reaction?

• A. 80 cm³
• B. 120 cm³
• C. 130 cm³
• D. 200 cm³

Substance 2C₂H₆ + 5O₂ 4CO₂ + 6H₂O
Initial
volume
40 150 0 NA
Change in
volume
-40 -100 +80 NA
Final
volume
0 50 80 NA

The total volume of gas remaining
= 50cm³ + 80cm³
= 130cm³

 Substance No. of mol 100cm³ 24 000cm³ =4.17 x 10³ mol No. of molecules 0.00417 × 6 × 10²³ = 2.50 × 10²¹ CH₄ molecules Mass of gas 0.00417𝑚𝑜𝑙 × 16𝑔 /𝑚𝑜l = 0.0667g  • 3. Molar concentration of solutions (mol/dm³)

An aqueous solution consists of solute(s) dissolved in water. Concentration of solution is defined as amount of solute dissolved per unit volume. We can qualitatively describe a solution as diluted or concentrated based on the amount of solute dissolved per unit volume.

To make a solution more diluted, add more water. To make a solution more concentrated, add more solute or remove water.

Calculations using Chemical Equations

Stoichiometry is the method which allows scientists to study the relative quantities of reactants and products in chemical reactions.

These are the important steps in stoichiometric calculations:

1. Construct a balanced chemical equation
2. Convert the mass or volume of a gas or concentration and volume of a solution to the number of moles.
3. Compare mol ratio between what was found in step 2 to the substance we need to find to determine the no. of mol of unknown substance.
4. Convert number of moles of unknown substance to mass or volume or concentration, demanded by the question.

Question 3

25.0 cm³ of 2.00 mol/dm³ aqueous sodium hydroxide solution reacted completely
with ammonium chloride. What is the volume of gas produced at room
temperature and pressure?

• 1. Construct a balanced chemical equation

NaOH (aq) + NH₄Cl (aq) NaCl (aq) + H₂O (l) + NH₃ (g)

• 2. Convert the concentration and volume of NaOH to number of moles.
• No. of mol of NaOH
• = concentration x volume
• = 2.00 mol/dm³ x 25/1000 dm³
• = 0.05 mol
• 3. Compare mol ratio between NaOH to NH₃ to find the no. of mol of NH₃.
Refer to balanced chemical equation and compare the coefficient (or the
number) before the compound. No number written means coefficient = 1.
• Mol ratio between NaOH : NH₃ = 1 : 1
Since 1 mol of NaOH produces 1 mol of NH₃ gas,
0.05 mol of NaOH will produce 0.05 mol of NH₃ gas.
• 4. Convert number of moles of unknown substance to mass or volume or
concentration, demanded by the question.

• No. of mol of NH₃ = 0.05 mol
Volume of gas
= no. of mole x molar volume
= 0.05mol x 24dm³/mol
= 1.20 dm³

Question 4
Compound X has the molecular formula Fe₂O₃.
Which statement is incorrect?

A. The empirical formula of compound X is Fe₂O₃.
B. One mole of compound X contains 112g of iron and 48g of oxygen
C. Iron is present in compound X as the Fe²⁺ ion.
D. Three moles of oxygen molecules are needed to make two moles of X.

C. Iron is present in compound X as the Fe²⁺ ion.

Limiting reactants

To prepare a chicken burger, we will need two buns and a chicken patty. However, what happens if I have 8 buns and 5 patties? How many chicken burgers of the same recipe can I create?

We can only create 4 burgers, with 1 chicken patty in excess. The amount of product, the chicken burger in this case, depends on the number of buns. The buns are what we call the limiting reagent. In chemistry, the limiting reactant is the one that is completely consumed in the reaction, and it determines the amount of product formed.

2 buns + 1 patty → 1 burger
Based on the equation above;
Buns : Patty
2 : 1
If I have 8 buns and 5 patties, I can only use 4 patties to make 4 burgers.
8 buns + 4 patty → 4 burgers
Buns : Patty
2:1
As all buns are used up, the buns are the limiting reactant.
8
:
4
This means that 1 patty will be in excess. We call this the excess reageant.

The limiting reactant will determine the amount of product.
Based on the 1ˢᵗ equation:
Buns : burger
2 : 1
8 : 4
2 buns + 1 patty → 1 burger
Based on the equation above;
Buns : Patty
2 : 1
If I have 8 buns and 5 patties, I can only use 4 patties to make 4 burgers.
8 buns + 4 patty → 4 burgers
Buns : Patty
2:1
As all buns are used up, the buns are the limiting reactant.
8
:
4
This means that 1 patty will be in excess. We call this the excess reageant.

The limiting reactant will determine the amount of product.
Based on the 1ˢᵗ equation:
Buns : burger
2 : 1
8 : 4

When do we need to find the limiting reagent?
• When there are 2 information of at least 2 reactants provided in question.

The steps to solving questions involving limiting reagents are similar to that for stoichiometric calculations.

1. Construct a balanced chemical equation
2. Find the number of mol of both reactants.
3. Compare mol ratio between the reactants to determine limiting reagent.
4. Once limiting reagent is identified, compare mol ratio of limiting reagent to product of interest.
5. Convert number of moles of product of interest to mass or volume or concentration, demanded by the question.

Question 5

A student mixed 25.0 cm³ of 1.00 mol/dm³ hydrochloric acid with 25.0 g of calcium carbonate. What is the volume of carbon dioxide gas that could be collected at room temperature and pressure?

• 1. Construct a balanced chemical equation

CaCO₃(s) + 2HCl (aq) CaCl ₂ (aq) + H₂O(l) + CO₂(g)

• 2. Find no. of mole of both reactants (CaCO3 and HCl)
• No. of mol of HCl
= concentration x volume
= 1.00 mol/dm³ x 25/1000 dm³
= 0.0250 mol
• No. of mol of CaCO₃
= mass / molar mass
= 25.0 g / (40 + 12 + 3(16)) g/mol
= 0.250 mol
• 3. Compare mol ratio between CaCO3 and HCl to find the limiting reactant.
Refer to balanced chemical equation and compare the coefficient (or the
number) before the compound. No number written means coefficient = 1.

• Mol ratio between CaCO₃ : HCl = 1 : 2
Since 1 mol of CaCO₃ reacts with 2 mol of HCl,
0.250 mol of CaCO₃ will react with 0.500 mol of HCl.
• As only 0.0250 mol of HCl is given, which is lesser than 0.500 mol required, HCl is the limiting reactant.
• 4. Compare mol ratio between limiting reactant (HCl ) and product of interest
(CO₂)

• Mol ratio between HCl : CO₂ = 2 : 1
• Since 2 mol of HCl reacts with 1 mol of CaCO3 to produce 1 mol of CO₂,
• 0.0250 mol of HCl will react with 0.0125 mol of CaCO₃ to produce
• 0.0125 mol of CO₂.
• 5. Convert number of moles of product of interest to mass or volume or concentration, demanded by the question.
• Volume of CO₂
= no. of mol x molar volume
= 0.0125 mol x 24 dm³/mol
= 0.300 dm³

Question 6
In an experiment, 4.0 cm³ of 1.0 mol/dm³ aqueous copper(II) sulfate and 4.0 cm³ of 0.5 mol/dm³ aqueous sodium carbonate are mixed.

What does the reaction vessel contain once the reaction is complete?

• A. A colourless solution only
• B. A green precipitate and a blue solution
• C. A green precipitate and a colourless solution
• D. A white precipitate and a colourless solution