Stoichiometry comes from the Greek word “Stoichen” meaning element and the word “Metron” which means measure. The study of relative amounts of the substances involved in the chemical reaction is known as stoichiometry.

Stoichiometry is an integral part of the chemical industry and has a wide range of practical applications.

For example, the chemical composition of everyday products such as soap, detergent, toothpaste, shampoos, cosmetics, medicines, and fertilizers is determined by stoichiometry.

Methanol burns according to the following equation.

2CH₃OH + 3O₂(g) ⟶ 2CO₂(g) + 4H₂O(g)

If 3.50 moles of methanol are burnt in oxygen, calculate:

1. How many moles of oxygen are used?
2. How many moles of water are produced?

Solution:

(a) Moles of methanol = 3.50 moles

Moles of oxygen = ?

According to the balanced chemical equation:

2 moles of CH₃OH = 3 moles of O₂

1 mole of CH₃OH = 3/2 moles of O₂

3.5 moles of CH₃OH = (3.50 × 3)/2 moles of O₂

= 5.25 moles of O₂

So the number of moles of O₂ consumed = 5.25 moles.

(b) No. of moles of CH₃OH = 3.5 moles.

No. of moles of H₂O = ?

According to the balanced chemical equation:

2 moles of CH₃OH = 4 moles of H₂O

1 mole of CH₃OH = 4/2 moles of H₂O

3.5 moles of CH₃OH = (3.50 × 4)/2 moles of O₂

= 7.00 moles of H₂O

So the number of moles of H₂O produced = 7.00 moles.

It is obtained by the combination of N₂ and H₂ as shown by the following balanced equation:

N₂(g) + 3H₂(g) ⟶ 2NH₃(g)

How many moles of the following are required to manufacture 8.0 moles of NH₃?

1. Nitrogen
2. Hydrogen

Solution:

The balanced chemical equation is:

N₂(g) + 3H₂(g) ⟶ 2NH₃(g)

(a) Nitrogen (N₂):

From the balanced equation, 1 mole of N₂ produces 2 moles of NH₃. To find out how many moles of N₂ are required to produce 8.0 moles of NH₃, we can set up a ratio:

moles of N₂ / moles of NH₃ = 1/2

So, moles of N₂ = 1/2 × moles of NH₃

moles of N₂ = 1/2 × 8.0 moles = 4.0 moles

Therefore, 4.0 moles of nitrogen are required.

(b) Hydrogen (H₂):

From the balanced equation, 3 moles of H₂ produce 2 moles of NH₃.

To find out how many moles of H₂ are required to produce 8.0 moles of NH₃, we can set up a ratio:

moles of H₂ / moles of NH₃ = 3/2

So, moles of H₂ = 3/2 × moles of NH₃

moles of H₂ = 3/2 × 8.0 moles = 12.0 moles

Therefore, 12.0 moles of hydrogen are required.

Answer:

(a) 4.0 moles of Nitrogen (N₂)

(b) 12.0 moles of Hydrogen (H₂)

Iron can be produced from iron ore Fe₂O₃ by reacting the ore with carbon monoxide (CO). What mass of iron can be formed from 425 g of iron ore?

Solution:

The balanced equation can be written as:

Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂

Mass of iron ore = 425 g (given mass)

No. of moles of iron ore = mass / molecular mass = 425 g / (159.6 g mol⁻¹)

= 2.66 moles of Fe₂O₃

According to the balanced chemical equation:

1 mole of Fe₂O₃ = 2 moles of Fe

2.66 moles of Fe₂O₃ = 2.66 × 2 moles of Fe = 5.32 moles of Fe

Mass of Fe produced = no. of moles of Fe × molar mass of iron

= 5.32 × 55.9 g

Mass of iron produced = 297.388 g

2H₂(g) + O₂(g) ⟶ 2H₂O(g)

How many grams of oxygen are needed to burn 100 g of hydrogen completely?

Solution:

The balanced equation is:

2H₂(g) + O₂(g) ⟶ 2H₂O(g)

This balanced equation tells us that 2 moles of hydrogen (H₂) react with 1 mole of oxygen (O₂).

First, we need to convert the mass of hydrogen to moles:

The molar mass of H₂ is approximately 2.016 g/mol.

moles of H₂ = mass of H₂ / molar mass of H₂

moles of H₂ = 100 g / 2.016 g/mol ≈ 49.6 mol

Now, we use the stoichiometry of the balanced equation to find the moles of O₂ required:

From the balanced equation, 2 moles of H₂ react with 1 mole of O₂.

So, moles of O₂ = 1/2 × moles of H₂

moles of O₂ = 1/2 × 49.6 mol = 24.8 mol

Finally, convert the moles of O₂ to grams:

The molar mass of O₂ is approximately 32.00 g/mol.

mass of O₂ = moles of O₂ × molar mass of O₂

mass of O₂ = 24.8 mol × 32.00 g/mol = 793.6 g

Therefore, approximately 793.6 grams of oxygen are needed to burn 100 g of hydrogen completely.

2KClO₃(s) ⟶ 2KCl(s) + 3O₂(g)

Solution: The balanced equation is:

2KClO₃(s) ⟶ 2KCl(s) + 3O₂(g)

This tells us that 2 moles of potassium chlorate (KClO₃) decompose to produce 3 moles of oxygen (O₂).

First, we need to convert the mass of KClO₃ to moles:

The molar mass of KClO₃ = 39.1 + 35.5 + 16 × 3

= 39.1 + 35.5 + 48 = 122.55 g/mol.

moles of KClO₃ = mass of KClO₃ / molar mass of KClO₃

moles of KClO₃ = 49 g / 122.55 g/mol ≈ 0.40 mol

Now, we use the stoichiometry of the balanced equation to find the moles of O₂ produced:

From the balanced equation, 2 moles of KClO₃ produce 3 moles of O₂.

So, moles of O₂ = 3/2 × moles of KClO₃

moles of O₂ = 3/2 × 0.40 mol = 0.60 mol

Finally, convert the moles of O₂ to grams:

The molar mass of O₂ is approximately 32.00 g/mol.

mass of O₂ = moles of O₂ × molar mass of O₂

mass of O₂ = 0.60 mol × 32.00 = 19.2 g

Therefore, approximately 19.2 grams of oxygen are produced by the thermal decomposition of 49 grams of KClO₃.

Determine the volume that 2.5 moles of chlorine molecules occupy at RTP.

Solution:

We know that:

24 dm³ of Cl₂ (Chlorine) at RTP = 1 mole

Or 1 mole of Cl₂ occupies a volume of 24 dm³ at RTP.

2.5 mole of Cl₂ occupy a volume of 24 dm³ × 2.5 = 60 dm³

Concept Assessment Exercise 3.3

Solution:

Given:

• Volume of oxygen gas (V) = 50.0 dm³
• Molar volume at RTP (V_m) = 24 dm³/mol

To find the number of moles (n), we can use the formula:

n = V / V_m

Plugging in the values:

n = 50.0 dm³ / (24 dm³/mol)

n ≈ 2.08 mol

Therefore, there are approximately 2.08 moles of oxygen gas in 50.0 dm³ at RTP.

(b) What volume does 0.80 mole of N₂ gas occupy at RTP?

Solution:

Given:

• Number of moles of N₂ (n) = 0.80 mol
• Molar volume at RTP (V_m) = 24 dm³/mol

To find the volume (V), we can use the formula: V = n × V_m.

Plugging in the values:

V = 0.80 mol × 24 dm³/mol

V = 19.2 dm³

Therefore, 0.80 moles of N₂ gas occupies a volume of 19.2 dm³ at RTP.

Percentage Composition:

The relative amounts of each element in a compound are expressed as the percentage composition.

For example, the percentage composition by mass in MgO is as follows. To determine the percentage composition of a known compound:

1. Calculate the molar mass of the compound.
2. Calculate the percentage of each element in one mole of the compound. This is done by dividing the mass of each element in one mole of the compound by the molar mass multiplied by 100.

% of an element = (mass of element in 1 mole of compound / molar mass of compound) × 100

Molar mass of MgO = 24 + 16 = 40 g/mol

% of Mg = (24/40) × 100 = 60.3

% of O = (16/40) × 100 = 39.7

The sum of percentages of all the elements in a compound must be equal to 100.

Concept Assessment Exercise 3.4

1. Water H₂O
2. Sulphuric acid H₂SO₄
3. Glucose C₆H₁₂O₆
4. Ammonium nitrate NH₄NO₃

Solution:

1. Water (H₂O)

Molar mass of H₂O = (2 × 1.008) + 16.00 = 18.016 g/mol

% of H = ((2 × 1.008)/18.016) × 100 = 11.19%

% of O = (16.00/18.016) × 100 = 88.81%

2. Sulfuric Acid (H₂SO₄)

Molar mass of H₂SO₄ = (2 × 1.008) + 32.07 + (4 × 16.00) = 98.086 g/mol

% of H = ((2 × 1.008)/98.086) × 100 = 2.06%

% of S = (32.07/98.086) × 100 = 32.70%

% of O = ((4 × 16.00)/98.086) × 100 = 65.25%

3. Glucose (C₆H₁₂O₆)

Molar mass of C₆H₁₂O₆ = (6 × 12.01) + (12 × 1.008) + (6 × 16.00) = 180.156 g/mol

% of C = ((6 × 12.01)/180.156) × 100 = 40.00%

% of H = ((12 × 1.008)/180.156) × 100 = 6.71%

% of O = ((6 × 16.00)/180.156) × 100 = 53.29%

4. Ammonium Nitrate (NH₄NO₃)

Molar mass of NH₄NO₃ = (2 × 14.01) + (4 × 1.008) + (3 × 16.00) = 80.042 g/mol

% of N = ((2 × 14.01)/80.042) × 100 = 35.01%

% of H = ((4 × 1.008)/80.042) × 100 = 5.04%

% of O = ((3 × 16.00)/80.042) × 100 = 59.97%

Difference between limiting and non limiting reactants:

Limiting and Non-Limiting Reactants:

A limiting reactant is the reactant that is completely consumed in a reaction. A limiting reactant can also be defined as the reactant that produces the lowest moles of products in a reaction. The reactant in excess is the reactant left unutilised or unreacted after the reaction is complete. It is also referred to as a Non-Limiting reactant.

Identification of a Limiting Reactant in a Reaction:

A limiting reactant can be recognised by calculating the number of moles of products formed from data of the given amounts of the reactants, using a balanced chemical equation.

The reactant, that produces the least number of moles of products, is the limiting reactant. For example, 10 moles of H₂ and 10 moles of O₂ were reacted to produce H₂O. Which one of the reactants is the limiting reactant? We can calculate as follows:

The reaction is 2H₂(g) + O₂(g) → 2H₂O(l)

According to the balanced chemical equation:

2H₂ = 2H₂O

2 moles of H₂ give = 2 moles of H₂O

10 moles of H₂ will give = (2/2) × 10 = 10 moles of H₂O

O₂ = 2H₂O

1 mole of O₂ gives = 2 moles of H₂O

7 moles of O₂ = 2 × 7 = 14 moles of H₂O

Since H₂ gives the least number of moles of H₂O, i.e. 10 moles, so H₂ is the limiting reactant.

200 g of Zn were reacted with 200g dil. H₂SO₄. Determine:

1. Limiting Reactant
2. Mass of hydrogen produced

Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)

Solution:

(a) Mass of Zn = 200g

Molar mass of Zn = 63.5 g mol⁻¹

Number of moles of Zn = 200/63.5 = 3.15 mol

Mass of H₂SO₄ = 200 g/mol

Molar mass of H₂SO₄ = 2 + 32 + 16 × 4 = 98 g/mol

No of moles of H₂SO₄ = 200/98 = 2.04 mol

According to the balanced chemical equation:

1 mole of Zn produces = 1 mol of H₂

3.15 mole of Zn = 1 × 3.15 mol of H₂ = 3.15 mol of H₂

1 mole of H₂SO₄ = 1 mol of H₂

2.04 mole of H₂SO₄ = 1 × 2.04 = 2.04 mol of H₂

As H₂SO₄ is producing lesser moles of product so, H₂SO₄ is the limiting reactant.

Mass of hydrogen produced = Number of moles of hydrogen × molar mass of hydrogen

= 2.04 × 2 = 4.08 g

Mg(s) + S(s) → MgS(s)

Solution:

Calculate the moles of each reactant:

Moles of Mg = mass / molar mass = 1.50g/24.31g/mol = 0.0617mol

Moles of S = mass / molar mass = 1.50g/32.07g/mol = 0.0468 mol

Determine the limiting reactant:

The balanced equation (Mg + S → MgS) shows a 1:1 stoichiometric ratio between Mg and S. This means that one mole of Mg reacts with one mole of S.

Since we have fewer moles of S (0.0468 mol) than Mg (0.0617 mol), sulfur is the limiting reactant. This means that all of the sulfur will be used up, and the amount of magnesium sulfide produced will be determined by the initial amount of sulfur.

Calculate the moles of product (MgS) formed:

Since the reaction is 1:1, the moles of MgS produced will be equal to the moles of the limiting reactant (S).

Moles of MgS = 0.0468 mol

Calculate the mass of MgS produced:

Mass of MgS = moles × molar mass

= 0.0468 mol × (24.31 + 32.07) g/mol

= 0.0468 mol × 56.38 g/mol = 2.64 g

Answer: 2.64 grams of magnesium sulfide (MgS) can be made.

(a) The limiting reactant.

(b) The mass of Zinc Sulphide produced.

(a) Determine the limiting reactant:

Calculate the moles of each reactant:

Moles of Zn = mass / molar mass = 10.0 g/65.38 g/mol = 0.153 mol

Moles of S = mass / molar mass = 10.0 g/32.07 g/mol = 0.312 mol

Determine the limiting reactant based on the stoichiometry of the reaction:

The balanced equation (Zn + S ⟶ ZnS) shows a 1:1 stoichiometric ratio between Zn and S. This means that one mole of Zn reacts with one mole of S.

Since we have fewer moles of Zn (0.153 mol) than S (0.312 mol), zinc is the limiting reactant.

(b) Calculate the mass of Zinc Sulphide produced:

Calculate the moles of product (ZnS) formed:

Since the reaction is 1:1, the moles of ZnS produced will be equal to the moles of the limiting reactant (Zn).

Moles of ZnS = 0.153 mol

Calculate the mass of ZnS produced:

Mass of ZnS = moles × molar mass

= 0.153 mol × (65.38 + 32.07) g/mol

= 0.153 mol × 97.45 g/mol = 14.91 g

Answer:

• (a) The limiting reactant is Zinc (Zn).
• (b) The mass of Zinc Sulphide produced is 14.91 g.

2Al(s) + 3Br₂(l) ⟶ 2AlBr₃(s)

(a) The limiting reactant

(b) The mass of AlBr₃ produced

Calculate the moles of each reactant:

Aluminum (Al):

Molar mass of Al = 26.98 g/mol

Moles of Al = mass / molar mass = 54g/26.98 g/mol = 2.00 moles

Bromine (Br₂):

Molar mass of Br₂ = 2 × 79.90 g/mol = 159.8 g/mol

Moles of Br₂ = mass / molar mass = 200g/159.8 g/mol = 1.25 moles

Determine the limiting reactant:

The balanced equation is:

2Al(s) + 3Br₂(l) ⟶ 2AlBr₃(s)

This means 2 moles of Al react with 3 moles of Br₂.

To react completely with 2.00 moles of Al, you would need 3/2 × 2 moles of Br₂ = 3.00 moles of Br₂. Since you only have 1.25 moles of Br₂, bromine is the limiting reactant.

Alternatively, to react completely with 1.25 moles of Br₂, you would need 3/2 × 1.25 moles of Al = 0.833 moles of Al. Since you have 2.00 moles of Al, bromine is the limiting reactant.

Therefore, Bromine (Br₂) is the limiting reactant.

Calculate the mass of AlBr₃ produced:

Since Br₂ is the limiting reactant, we use the moles of Br₂ to calculate the mass of AlBr₃ produced.

From the balanced equation, 3 moles of Br₂ produce 2 moles of AlBr₃.

Moles of AlBr₃ produced = 2/3 moles of Br₂

2/3 × 1.25 moles = 0.833 moles

Molar mass of AlBr₃ = 26.98g/mol + 3 × 79.90 g/mol

= 266.68g/mol

Mass of AlBr₃ produced = moles × molar mass

= 0.833 moles × 266.68g/mol = 222.16g

Therefore, the mass of AlBr₃ produced is approximately 222.16 g.

Final Answers:

• (a) The limiting reactant: Bromine (Br₂)
• (b) The mass of AlBr₃ produced: approximately 222.16 g

Difference between Theoretical Yield, Actual yield and Percent yield:

The amount of product as calculated from a balanced chemical equation is known as the theoretical yield of a chemical equation. The actual yield is the amount of product actually produced in the reaction. The percent yield of a chemical reaction can be calculated as follows:

Formula: Percentage Yield = $\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$

For many reactions, the theoretical yield is close to the actual yield. These reactions are quantitative and can be used for chemical analysis.

This is because of the following reasons:

1. Side reactions may result in by-products.
2. Some reactions may be reversible.
3. Mechanical loss can occur as a result of filtration or distillation.
4. Separation by a funnel, washing or crystallization can also cause mechanical loss.

In an industry copper metal was prepared by the following reaction:

Fe_(s) + CuSO_4(aq) ⟶ FeSO_4(aq) + Cu_(s)

1.274 g of CuSO₄ when reacted with excess of Fe metal a yield of 0.392 g of Cu metal is obtained. Calculate the percentage yield.

Solution:

Mass of CuSO₄ = 1.274 g

Molar mass of CuSO₄ = 63.5 + 32 + 64 = 159.5 g/mol

159.5 g of CuSO₄ = 1 mole

1.274 g of CuSO₄ = $\frac{1}{159.5} \times 1.274$

= $7.982 \times 10^{-3}$ mol of Cu

1 mol of CuSO₄ = 1 mole of Cu

$7.98 \times 10^{-3}$ moles of CuSO₄ = $7.982 \times 10^{-3}$ mol of Cu

As, 1 mole of Cu = 63.5 g

so, $7.982 \times 10^{-3}$ moles of Cu = 63.5 × $7.982 \times 10^{-3}$

= 0.5072 g of Cu.

Hence, Theoretical yield = 0.5072 g

Actual yield = 0.392 g

So, % yield = $\frac{0.392}{0.5072} \times 100$ = 77.3%

= $\frac{0.392}{0.5072} \times 100$ = 77.3%

Concept Assessment Exercise 3.6

In a reaction, 2.00 moles of CH₄ were reacted with an excess of Cl₂. As a result, 177.0 g of CCl₄ is obtained. What is the,

1. theoretical yield
2. actual yield
3. % yield of this reaction?

CH_4(g) + 4Cl_2(g) ⟶ CCl_4(l) + 4HCl_(g)

Solution:

a) Theoretical Yield

The balanced equation is: CH_4(g) + 4Cl_2(g) ⟶ CCl_4(l) + 4HCl_(g)

From the balanced equation, 1 mole of CH₄ produces 1 mole of CCl₄.

Given 2.00 moles of CH₄, the theoretical yield of CCl₄ in moles is also 2.00 moles. To convert this to grams, we need the molar mass of CCl₄.

Molar mass of CCl₄ = 12.01 + 4(35.45)

= 12.01 + 141.8 = 153.81 g/mol

Theoretical yield in grams = 2.00 mol × 153.81 g/mol = 307.62 g

So, the theoretical yield is 307.62 g.

b) Actual Yield

The actual yield is given in the problem: 177.0 g of CCl₄ is obtained.

c) Percent Yield

Percent Yield = <span>Actual Yield / Theoretical Yield</span> × 100

Percent Yield = (177.0 g / 307.62 g) × 100

Percent Yield ≈ 57.53%

Final Answers:

• (a) Theoretical yield: 307.62 g
• (b) Actual yield: 177.0 g
• (c) Percent yield: 57.53%

2P_(s) + 3Br_2(l) ⟶ 2PBr_3(s)

Solution:

Calculate the moles of each reactant:

Moles of Br₂:

Molar mass of Br₂ = 2 × 79.90 g/mol = 159.80 g/mol

Moles of Br₂ = 35 g / 159.80 g/mol ≈ 0.219 mol

Moles of P:

Molar mass of P = 30.97 g/mol

Moles of P = 27.9 g / 30.97 g/mol ≈ 0.901 mol

Determine the limiting reactant:

From the balanced equation:

2 mol P : 3 mol Br₂

Calculate the required moles of Br₂ to react with 0.901 mol of P:

Moles of Br₂ = 3/2 × 0.901 mol ≈ 1.352 mol

Since we only have 0.219 mol of Br₂, therefore Br₂ is the limiting reactant.

Calculate the theoretical yield of PBr₃:

From the balanced equation:

3 mol Br₂ : 2 mol PBr₃

Calculate the moles of PBr₃ produced from 0.219 mol of Br₂:

Moles of PBr₃ = 2/3 × 0.219 mol ≈ 0.146 mol

Calculate the mass of PBr₃ (theoretical yield):

Molar mass of PBr₃ = 30.97 + 3(79.90) = 30.97 + 239.70 = 270.67 g/mol

Theoretical yield = 0.146 mol × 270.67 g/mol ≈ 39.52 g

4. (b) Actual yield:

The actual yield is given as 27.9 g.

Calculate the percent yield:

Percent yield = (Actual yield / Theoretical yield) × 100

Percent yield = (27.9 g / 39.52 g) × 100 ≈ 70.60%

Final Answers:

• (a) Theoretical yield: 39.52 g
• (b) Actual yield: 27.9 g
• (c) Percent yield: 70.60%

Determining Empirical and Molecular Formulas:

The empirical formula of a compound tells you the simplest ratio of elements in the compound. On the other hand, the molecular formula tells you the actual ratio of elements in the compound.

Determining the empirical formula of a compound involves the following steps:

Step 1.

• Find the number of moles of each element from its mass or percentage composition.

Step 2.

Divide each element’s molar amount by the smallest or least molar amount to get the simplest whole-number ratio of the compound.

Step 3.

Write the empirical formula using the simplest ratio as subscripts.

Determining the molecular formula of a compound involves the following steps:

Step 1.

Find the molar mass of the empirical formula.

Step 2.

Divide the compound's molar mass by the empirical formula's molar mass. This will give you a whole number (n).

Step 3.

Multiply all the subscripts in the empirical formula by n to get the molecular formula.

A compound has the following composition:

Carbon = 40%, hydrogen = 6.7% and oxygen = 53.3%

The molar mass of the compound is 180 g/mol. Determine its empirical and molecular formulas.

Solution:

Moles of elements

C = 40/12 = 3.33

H = 6.7/1 = 6.7

O = 53.3/16 = 3.33

Mole ratio:

Empirical formula = C₁H₂O₁ = CH₂O

Molar mass of empirical formula = 1 × 12 + 2 × 1 + 1 × 16

= 12 + 2 + 16 = 30 g/mol

n = Molar mass / Empirical formula molar mass = 180/30 = 6

Therefore molecular formula = (CH₂O)₆ = C₆H₁₂O₆

26.4% Carbon, 3.3% Hydrogen, 70.3% Oxygen; molar mass: 91.0 g/mol.

Solution:

Determine the moles of each element assuming 100 g of compound:

Carbon (C):
Mass: 26.4 g
Molar mass: 12.01 g/mol
Moles: 26.4 g / 12.01 g/mol ≈ 2.20 mol

Hydrogen (H):
Mass: 3.3 g
Molar mass: 1.008 g/mol
Moles: 3.3 g / 1.008 g/mol ≈ 3.27 mol

Oxygen (O):
Mass: 70.3 g
Molar mass: 16.00 g/mol
Moles: 70.3 g / 16.00 g/mol ≈ 4.39 mol

Find the simplest whole number ratio of moles:

Divide each mole value by the smallest mole value (2.20 mol):

Carbon: 2.20/2.20 = 1

Hydrogen: 3.27/2.20 ≈ 1.49 (approximately 1.5)

Oxygen: 4.39/2.20 ≈ 2

To get whole numbers, multiply all ratios by 2:

Carbon: 1 × 2 = 2
Hydrogen: 1.5 × 2 = 3
Oxygen: 2 × 2 = 4

So, the empirical formula is C₂H₃O₄.

Determine the empirical formula mass:

Empirical formula mass = 2(12.01) + 3(1.008) + 4(16.00)

= 24.02 + 3.024 + 64 = 91.044 g/mol

Determine the molecular formula:

Divide the molar mass of the compound by the empirical formula mass:

91.0 g/mol / 91.044 g/mol ≈ 1

Since the ratio is approximately 1, the molecular formula is the same as the empirical formula.

Final Answer:

• Empirical formula: C₂H₃O₄
• Molecular formula: C₂H₃O₄

Solution:

Calculate the number of moles of each element:

Carbon (C):
Mass: 18.24 g

Molar mass of carbon: 12.01 g/mol

Moles: $\frac{18.24\text{ g}}{12.01\text{ g/mol}} = 1.52\text{ mol}$

Hydrogen (H):
Mass: 0.51 g
Molar mass: 1.008 g/mol

Moles: $\frac{0.51\text{ g}}{1.008\text{ g/mol}} \approx 0.51\text{ mol}$

Fluorine (F):
Mass: 16.91 g
Molar mass: 19.00 g/mol

Moles: $\frac{16.91\text{ g}}{19.00\text{ g/mol}} = 0.89\text{ mol}$

Find the simplest whole number ratio of moles:

Divide each mole value by the smallest mole value (0.51 mol):

Carbon: $\frac{1.52}{0.51} = 2.98 \approx 3$

Hydrogen: $\frac{0.51}{0.51} = 1$

Fluorine: $\frac{0.89}{0.51} = 1.75 = \frac{7}{4}$

To get a whole number ratio, we can multiply by 4:

C: 3 × 4 = 12
H: 1 × 4 = 4
F: 1.75 × 4 = 7

Thus, the empirical formula is C₁₂H₄F₇.

Calculate the empirical formula mass:

Empirical formula mass = 12(12.01) + 4(1.008) + 7(19.00)
= 144.12 + 4.032 + 133 = 281.152 g/mol

Determine the molecular formula:

Divide the molar mass of the compound by the empirical formula mass:

$\frac{562.0\text{ g/mol}}{281.152\text{ g/mol}} \approx 2$

Multiply the subscripts in the empirical formula by 2 to get the molecular formula:

C_12×2H_4×2F_7×2 = C₂₄H₈F₁₄

Final Answer:

• Empirical formula: C₁₂H₄F₇
• Molecular formula: C₂₄H₈F₁₄

Percentage purity:

The amount of the pure substance in a sample represented as a percentage of the sample's total mass is known as percentage purity.

You can determine the purity of a compound by comparing the amount of pure compound compared to the total mass of a sample. The formula for percentage purity is:

Percentage purity = $\frac{\text{Actual mass of compound}}{\text{Total mass of sample}} \times 100$

A student synthesized a compound and obtained a sample weighing 6.3 g. Then he performed chemical analysis and found that the sample contains 4.2 g of pure compound. Calculate the percentage purity of the compound.

Solution:

Actual mass of compound = 4.2 g
The total mass of the sample = 6.3 g

Percentage purity = $\frac{\text{Actual mass of compound}}{\text{Total mass of sample}} \times 100$

$= \frac{4.2 \times 100}{6.3} = 66.67\%$

Concept Assessment Exercise 3.8

Solution:

Understand the concept of purity:

Purity refers to the percentage of the desired substance (in this case, calcium carbonate) in a sample.

Calculate the percentage purity:

Percentage purity is calculated using the following formula:

Percentage Purity = $\frac{\text{Mass of pure substance}}{\text{Mass of impure sample}} \times 100\%$

Apply the formula to the given data:

Mass of pure calcium carbonate recovered: 56 g
Mass of impure sample: 70 g

Percentage Purity = $\frac{56\text{ g}}{70\text{ g}} \times 100\%$

Percentage Purity = 0.8 × 100% = 80%

Final Answer:

The percentage purity of the calcium carbonate is 80%.

The quantity of a solute present in a given amount of solvent or solution is called the concentration of a solution.

Molarity is the concentration unit in which the amount of solute is expressed in moles and the quantity of solution in dm³.

Molarity is defined as the number of moles of solute dissolved per dm³ of solution.

Mathematically,

$M = \frac{\text{moles of solute}}{\text{dm}^3\text{ of solution}}$

Urea is a white solid used as fertilizer and starting material for synthetic plastic. A solution contains 40 g urea dissolved in 500 cm³ of solution. Calculate the molarity of this solution.

Solution:

Mass of urea = 40 g

Molar mass of urea = 14 + 1 × 2 + 12 + 16 + 14 + 1 × 2
= 60 g/mol

Moles of urea = $\frac{40}{60}$ = 0.667 mol

Volume of solution = 500 cm³ = $\frac{500}{1000}$ = 0.5 dm³

Now,

Molarity = $\frac{\text{moles of solute}}{\text{dm}^3\text{ of solution}}$

Molarity = $\frac{0.667}{0.5}$ = 1.334 M

Calculating molarity from moles of solute and volume of solution.

Potassium permanganate (KMnO₄) is a dark blue-black compound. When it is dissolved in water, it forms a bright purple solution. It is used as a disinfectant in water tanks. It is also known as pinky. A solution contains 0.05 moles of (KMnO₄) in 600 cm³ of solution. Calculate the molarity of this solution.

Solution:

Moles of (KMnO₄) = 0.05

Volume of solution = 600 cm³ = $\frac{600}{1000}$ = 0.6 dm³

Now,

Molarity = $\frac{\text{moles of solute}}{\text{dm}^3\text{ of solution}}$

Molarity = $\frac{0.05}{0.6}$ = 0.083 M

Solution:

(a) 1.5 moles of KClO₃ dissolved in 250 cm³ of solution

Molarity (M) is defined as moles of solute per dm³ of solution.

First, convert the volume from cm³ to dm³:

250 cm³ = $\frac{250}{1000}$ = 0.250 dm³

Now, calculate the molarity:

Molarity = $\frac{\text{moles of KClO}_3}{\text{volume of solution in dm}^3}$ = $\frac{1.5\text{ moles}}{0.250\text{ dm}^3}$ = 6.0 M

(b) 75 g of KClO₃ dissolved to produce 1.25 dm³ of solution

Next, calculate the number of moles of KClO₃.

The molar mass of KClO₃ is:

= 39.10 (K) + 35.45 (Cl) + 3 × 16.00 (O)
= 39.10 + 35.45 + 48.00 = 122.55 g/mol

Now, calculate the moles of KClO₃:

Moles of KClO₃ = $\frac{\text{mass}}{\text{molar mass}}$ = $\frac{75\text{ g}}{122.55\text{ g/mol}}$ ≈ 0.612 moles

Finally, calculate the molarity:

Molarity = $\frac{\text{moles of KClO}_3}{\text{volume of solution in dm}^3}$ = $\frac{0.612\text{ moles}}{1.25\text{ dm}^3}$ ≈ 0.490 M

(c) Molarity of a 50 cm³ sample that yields 0.25 g residue after evaporation

First, convert the volume from cm³ to dm³:

50 cm³ = 0.050 dm³

Next, calculate the number of moles of KClO₃ in the residue:

Moles of KClO₃ = mass / molar mass = 0.25 g / 122.55 g/mol ≈ 0.00204 moles

Finally, calculate the molarity:

Molarity = moles of KClO₃ / volume of solution in dm³ = 0.00204 moles / 0.050 dm³ ≈ 0.0408 M

Final Answers:

• (a) Molarity = 6.0 M
• (b) Molarity ≈ 0.490 M
• (c) Molarity ≈ 0.0408 M

Strength of a Solution:

The strength of a solution refers to the concentration of solute in a given volume of solution. It is expressed in terms of grams of solute per unit volume of solution. It is typically represented as:

• g/dm³ (grams per cubic decimeter), or
• g/cm³ (grams per cubic centimeter).

These units express how much solute is dissolved in a specified volume of solution.

Mathematically:

Strength of solution = Mass of solute (g) / Volume of solution (dm³ or cm³)

Mass of solute is measured in grams (g). Volume of solution is measured in cubic decimeters (dm³) or cubic centimeters (cm³). Unit conversion:

1 dm³ = 1000 cm³

So, to convert between g/dm³ and g/cm³:

1 g/dm³ = 0.001 g/cm³

A solution contains 20 g of salt dissolved in 2 dm³ of solution. Calculate the strength of the solution.

Solution:

Mass of solute (g) = 20 g

Volume of solution = 2 dm³

Strength of solution = Mass of solute (g) / Volume of solution (dm³)

Strength of solution = 20 g / 2 dm³ = 10 g/dm³

A solution contains 25 g of sugar dissolved in 500 cm³ of solution. Calculate the strength of the solution.

Solution:

Mass of solute = 25 g

Volume of solution = 500 cm³

Strength of solution = Mass of solute (g) / Volume of solution (cm³)

Strength of solution = 25 g / 500 cm³ = 0.05 g/cm³

A solution has a strength of 50 g/dm³. Express this strength in g/cm³.

Solution:

Strength of solution = 50 g/dm³.

Strength of solution in g/cm³ = ?

1 g/dm³ = 1/1000 = 0.001 g/cm³

50 g/dm³ = 50 × 0.001 = 0.05 g/cm³

Converting the molarity of a solution into its concentration in g/dm³ and g/cm³

A flask contains 0.25 M NaOH solution. What mass of NaOH is present per dm³ of solution?

Solution:

Molarity = 0.25 M

Moles of NaOH = 0.25

Volume of solution = 1 dm³

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mole

Mass of NaOH = ?

Moles of solute = mass of solute / molar mass of solute (g/mole)

0.25 moles = mass of NaOH / 40

Mass of NaOH = 0.25 moles × 40 g/mole = 10 g

Thus, the solution contains 10 g NaOH per dm³.

Now, as 1 dm³ = 1000 cm³:

The solution contains 10 g/dm³ of solution, or the solution contains 10 g of NaOH per 1000 cm³.

So, 1000 cm³ of solution contains = 10 g NaOH

1 cm³ of solution contains = 10 g/1000 = 0.01 g NaOH

Therefore, concentration of solution is = 0.01 g/cm³.

Converting concentration in g/dm³ into molarity.

Potassium hydroxide (KOH) is used in the manufacturing of shaving creams, paints and varnish. An analyst makes up a solution by dissolving 5.8 g of KOH in one dm³ of solution. Calculate the molarity of this solution.

Solution:

Mass of KOH dissolved in one dm³ of solution = 5.6 g

Molar mass of KOH = 39 + 16 + 1 = 56 g/mol

Moles of KOH = mass of KOH / molar mass of KOH = 5.6/56 = 0.1 mol

Thus, the solution contains 0.1 moles of KOH in one dm³ of the solution, so the molarity of the solution is 0.1 M.

Concept Assessment Exercise 3.10

Solution:

Molarity = ?

Mass of NaOH = 25 g

Volume of solution = 1 dm³

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mole

Moles of solute = mass of solute / molar mass of solute

= 25 g / 40 g/mole = 0.625 moles

Molarity = moles of solute / dm³ of solution = 0.625 / 1 dm³ = 0.625 M

Volume of solution = 1 dm³ = 1000 cm³

Molarity (M) = 1.2 M

Molar mass of solute NaOH = 23 + 16 + 1 = 40 g

Mass of solute = ?

Molarity (M) = (mass of solute / molar mass of solute) × (1000 / volume of solution in cm³)

1.2 = (mass of solute / 40) × (1000 / 1000)

1.2 = mass of solute / 40

Mass of solute in g/dm³ = 1.2 × 40 = 48 g/dm³

As 1 dm³ = 1000 cm³, the solution contains 48 g/dm³ of solution, or the solution contains 48 g of NaOH per 1000 cm³.

So, 1000 cm³ of solution contains = 48 g NaOH

1 cm³ of solution contains = 48 g/1000 = 0.048 g NaOH

Therefore, concentration of solution is = 0.048 g/cm³.

Mass of solute in g/cm³ = 0.048 g/cm³

Mass of solute haemoglobin = 10 g

Molar mass of solute haemoglobin = 6.51 × 10⁴ g/mole

Volume of solution = 1 dm³ = 1000 cm³

Molarity (M) = ?

Molarity (M) = (mass of solute / molar mass of solute) × (1000 / volume of solution in cm³)

Molarity (M) = (10 / (6.51 × 10⁴)) × (1000 / 1000)

Molarity (M) = 10 / (6.51 × 10⁴) = 1.54 / 10⁴ M = 1.54 × 10⁻⁴ M

Dilution of Solutions

Dilution is a process by which you make a solution less concentrated by adding more of the liquid (usually water). By using the dilution formula, you can determine how much of a concentrated solution is to be mixed with water to get the desired concentration and volume.

To understand how to dilute a solution, we use a simple formula:

$$M_1 \times V_1 = M_2 \times V_2$$

Where:

• $M_1$ is the concentration of the original (concentrated) solution.
• $V_1$ is the volume of the original solution you use.
• $M_2$ is the concentration you want after dilution.
• $V_2$ is the total volume of the final solution.

Plug the known values into the formula to find out how much of the concentrated solution you need. Mix the calculated amount of the concentrated solution with enough water to reach the final volume you want.

If you have 150 cm³ of a 4 M potassium nitrate (KNO₃) solution, what volume of this solution will be needed to prepare 500 cm³ of a 1 M potassium nitrate solution?

Solution:

The formula for dilution is:

$$M_1 \times V_1 = M_2 \times V_2$$

Where:

$M_1$ is the concentration of the initial (concentrated) solution of KNO₃ = 4 M. $V_1$ is the volume of the initial solution of KNO₃ that you need to find. $M_2$ is the concentration of the final (diluted) solution = 1 M. $V_2$ is the volume of the final solution = 500 cm³.

Substitute the known values into the formula:

$$4\text{ M} \times V_1 = 1\text{ M} \times 500$$

$$V_1 = \frac{1\text{ M} \times 500\text{ cm}^3}{4\text{ M}}$$

$$V_1 = 125\text{ cm}^3$$

To prepare 500 cm³ of a 1 M potassium nitrate solution, you need to use 125 cm³ of the 4 M potassium nitrate solution. Mix 125 cm³ of the 4 M solution with enough water to make a total of 500 cm³ to achieve the desired concentration.

$$\mathrm{HCl_{(aq)} + NaOH_{(aq)} \longrightarrow NaCl_{(aq)} + H_2O_{(l)}}$$

Calculate (i) the molarity of NaOH solution (ii) the amount of NaOH present per cm³ of solution.

Solution:

(i) Calculate the molarity of the NaOH solution:

Use the balanced equation:

The balanced equation is: $$\mathrm{HCl_{(aq)} + NaOH_{(aq)} \rightarrow NaCl_{(aq)} + H_2O_{(l)}}$$

This shows that 1 mole of HCl reacts with 1 mole of NaOH.

Calculate the moles of HCl used: (since 1 dm³ = 1000 cm³)

Volume of HCl solution = 9.8 cm³ = 0.0098 dm³

Molarity of HCl solution = 0.1 M

Moles of HCl = Molarity × Volume

= 0.1 mol/dm³ × 0.0098 dm³ = 0.00098 mol

Determine the moles of NaOH reacted:

Since the reaction is 1:1, the moles of NaOH reacted are equal to the moles of HCl used:

Moles of NaOH = 0.00098 mol

Calculate the molarity of the NaOH solution:

Volume of NaOH solution = 10 cm³ = 0.010 dm³

Molarity of NaOH = moles of NaOH / volume of NaOH solution (in dm³)

= 0.00098 mol / 0.010 dm³ = 0.098 M

(ii) Calculate the amount of NaOH present per cm³ of solution:

Convert molarity to grams per liter:

Molarity of NaOH = 0.098 M

Molar mass of NaOH ≈ 40.00 g/mol

Grams of NaOH per liter = Molarity × Molar mass

= 0.098 mol/dm³ × 40.00 g/mol = 3.92 g/dm³

Convert grams per liter to grams per cm³:

1 dm³ = 1000 cm³

Grams of NaOH per cm³ = grams of NaOH per liter / 1000

= 3.92 / 1000 cm³ = 0.00392 g/cm³

Final Answers:

1. Molarity of NaOH solution = 0.098 M
2. Amount of NaOH present per cm³ of solution = 0.00392 g/cm³

Solution:

Use the dilution formula:

The dilution formula is: $$M_1V_1=M_2V_2$$

Where:

• M₁ = initial molarity
• V₁ = initial volume
• M₂ = final molarity
• V₂ = final volume

Identify the knowns and unknowns:

• $M_1 = 6\text{ M}$
• $V_1 =$ unknown (what we want to find)
• $M_2 = 2\text{ M}$
• $V_2 = 200\text{ cm}^3$

Solve for $V_1$:

$$6\text{ M} \times V_1 = 2\text{ M} \times 200\text{ cm}^3$$

$$V_1 = \frac{2\text{ M} \times 200\text{ cm}^3}{6\text{ M}} = \frac{400}{6}\text{ cm}^3 \approx 66.67\text{ cm}^3$$

Final Answer:

The chemist should use approximately $66.67\text{ cm}^3$ of the original $6\text{ M}$ hydrochloric acid solution and mix it with water to achieve the desired concentration of $2\text{ M}$ in a total volume of $200\text{ cm}^3$.