Sample Questions and Answers on Amount of Substance and the Mole

Understanding the concept of the amount of substance and the mole is fundamental in mastering chemistry. These topics form the backbone of stoichiometry, allowing students to relate macroscopic quantities to the molecular scale. To help solidify your understanding, this post provides sample questions and answers on the amount of substance and the mole. Whether you're preparing for exams or simply seeking to enhance your grasp of these concepts, these examples will guide you through practical applications and common problem-solving techniques.

Questions and Answers on Amount of Substance and the Mole

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Question 1:

The atomic mass unit (amu) of carbon-12 is $1.6605 \times 10^{-24} \, \text{g}$. If the mass of an atom is $5.313\times 10^{-23}\text{g, determine the relative mass of the atom.}$

Solution:

The relative mass of an atom is calculated as the ratio of the mass of the atom to the atomic mass unit:

$$\text{Relative Mass}=\frac{\text{Mass of the atom}}{\text{Atomic Mass Unit (amu)}}$$

Substituting the given values:

• Mass of the atom: $5.313 \times 10^{-23} \, \text{g}$
  
• Atomic mass unit: $1.6605 \times 10^{-24} \, \text{g}$
  

$$\text{Relative Mass}=\frac{5.313\times 10^{-23}}{1.6605\times 10^{-\mathrm{24}}}$$

Perform the division:

Divide the coefficients:

$$\frac{5.313}{1.6605}\approx 3.2$$

Subtract the exponents of 10:

$$10^{-23}-10^{-24}=10^{-23+24}=10^1$$

Combine the results:      

                                3.2 × 10¹ = 32

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Question 2:

How many atoms are there in 0.3 mol of sodium? Given $L = 6.0 \times 10^{23} \, \text{mol}^{-1}$ (Avogadro's constant).

Solution:

The number of atoms in a substance can be calculated using the formula:

$$\text{Number of atoms}=\text{moles}\times L$$

Where:

• $\text{moles}=0.3\text{mol}$
• $L = 6.0 \times 10^{23} \, \text{mol}^{-1}$
  

Substituting the values:

$$\text{Number of atoms}=0.3\times 6.0\times 10^{23}$$

Perform the calculation:

1. Multiply the coefficients:

$$0.3\times 6.0=1.8$$

2. Include the power of 10:

$$1.8\times 10^{23}$$

The number of atoms in $0.3 \, \text{mol}$ of sodium is $1.8\times 10^{23}\text{atoms.}$

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Question 3:

Calculate the amount of oxygen in moles in $1.505 \times 10^{23}$ molecules of oxygen gas. Given $L = 6.02 \times 10^{23} \, \text{mol}^{-1}$ (Avogadro's constant).

Solution:

The number of moles can be calculated using the formula:

$$\text{Moles} = \frac{\text{Number of molecules}}{\text{Avogadro's constant (L)}}$$

Where:

• $\text{Number of molecules} = 1.505 \times 10^{23}$
  
• $L = 6.02 \times 10^{23} \, \text{mol}^{-1}$
  

Substituting the values:

$$\text{Moles} = \frac{1.505 \times 10^{23}}{6.02 \times 10^{23}}$$

Perform the calculation:

1. Divide the coefficients:

$$\frac{1.505}{6.02} \approx 0.25$$

2. Subtract the exponents of $10^{23}$:

$$10^{23} - 10^{23} = 10^0 = 1$$

Combine the results:

$$0.25 \times 1 = 0.25$$

The amount of oxygen is 0.25 moles.

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Question 4:

Calculate the number of oxygen atoms in $0.20 \, \text{mol}$ of oxygen gas. Given $L = 6.02 \times 10^{23} \, \text{mol}^{-1}$.

Solution:

Step 1: Relationship between moles of $O_2$ molecules and oxygen atoms

Oxygen gas exists as diatomic molecules ($O_2$), so, each molecule contains 2 oxygen atoms. The total number of oxygen atoms can be found using:

$$\text{Number of atoms}=\text{moles of }{O}_2\times L\times 2$$

Where:

• $\text{moles of } O_2 = 0.20 \, \text{mol}$
  
• $L = 6.02 \times 10^{23} \, \text{mol}^{-1}$
  
• $2 = \text{atoms per molecule of } O_2$

Step 2: Substitute the values

$$\text{Number of atoms}=0.20\times 6.02\times 10^{23}\times 2$$

Step 3: Perform the calculation

1. Multiply the coefficients:

   $$0.20\times 6.02=1.204$$

2. Multiply by 2:

   $$1.204 \times 2 = 2.408$$
   

3. Include the power of 10:

   $$2.408 \times 10^{23}$$

The number of oxygen atoms in 0.20 mol of oxygen is 2.408 × 10²³ atoms.

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Question 5:

Calculate:

(i)  the number of molecules contained in $0.6 \, \text{mol}$ of $\text{H}_2\text{S}$.
(ii) the number of atoms of H and S present in $0.6 \, \text{mol}$ of $\text{H}_2\text{S}$.

Solution:

Part (i): 

The number of molecules in a substance is calculated using the formula:

$$\text{Number of molecules}=\text{moles}\times L$$

Where:

• $\text{moles} = 0.6 \, \text{mol}$
  
• $L = 6.02 \times 10^{23} \, \text{mol}^{-1}$ (Avogadro's constant)

Substituting the values:

$$\text{Number of molecules}=0.6\times 6.02\times 10^{23}$$

Perform the calculation:

$$0.6\times 6.02=3.612$$

Include the power of $10^{23}$:

$$\text{Number of molecules}=3.612\times 10^{23}$$

Answer for (i): The number of $\text{H}_2\text{S}$ molecules is $3.612 \times 10^{23}$.

Part (ii): 

1. Atoms per molecule of $\text{H}_2\text{S}$:

   • Each $\text{H}_2\text{S}$ molecule contains:
     • 2 hydrogen ($H$) atoms
     • 1 sulfur ($S$) atom

2. Total number of atoms of each type:

   • Hydrogen atoms:

     $$\text{Number of H atoms}=\text{Number of molecules}\times 2$$

     Substituting:

     $$\text{Number of H atoms}=3.612\times 10^{23}\times 2=7.224\times 10^{23}$$

   • Sulfur atoms:

     $$\text{Number of S atoms} = \text{Number of molecules} = 3.612 \times 10^{23}$$
     

Final Answer:

(i) The number of $\text{H}_2\text{S}$ molecules is $3.612 \times 10^{23}$.
(ii) The number of hydrogen ($H$) atoms is $7.224 \times 10^{23}$, and the number of sulfur ($S$) atoms is $3.612 \times 10^{23}$.

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Question 6:

Calculate the number of nitrogen ($N$) atoms in $0.25 \, \text{mol}$ of nitrogen (I) oxide ($\text{N}_2\text{O}$). Given $L = 6.02 \times 10^{23} \, \text{mol}^{-1}$.

Solution:

Step 1: Relationship between moles, molecules, and atoms

Nitrogen (I) oxide ($\text{N}_2\text{O}$) contains 2 nitrogen atoms per molecule.

The total number of $N$ atoms can be calculated using:

$$\text{Number of atoms of }N=\text{moles of }{\text{N}}_2\text{O}\times L\times \text{number of }N\text{ atoms per molecule}\mathrm{.}$$

Step 2: Substituting the values

• Moles of ${\text{N}}_2\text{O}=0.25\text{mol}$
• $L = 6.02 \times 10^{23} \, \text{mol}^{-1}$
  
• Number of $N$ atoms per molecule = 2

$$\text{Number of atoms of } N = 0.25 \times 6.02 \times 10^{23} \times 2$$

Step 3: Perform the calculation

• Multiply the coefficients:

  $$0.25 \times 6.02 = 1.505$$
  

$$1.505 \times 2 = 3.01$$

• Include the power of $10^{23}$:

  $$3.01 \times 10^{23}$$

Final Answer:

The number of nitrogen ($N$) atoms in $0.25\text{mol of }$$\text{N}_2\text{O}$ is $3.01 \times 10^{23} \, \text{atoms}$.

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Question 7:

Calculate the mass of $0.2 \, \text{mol}$ of magnesium atoms. Given the molar mass of magnesium ($\text{Mg}$) = $24\text{g/mol.}$

Solution:

The mass of a substance can be calculated using the formula:

$$\text{Mass}=\text{Moles}\times \text{Molar Mass}$$

Where:

• $\text{Moles} = 0.2 \, \text{mol}$
  
• $\text{Molar Mass} = 24 \, \text{g/mol}$
  

Substituting the values:

$$\text{Mass}=0.2\times 24$$

Perform the calculation: \( \text{Mass} = 4.8 \, \text{g} \)

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Question 8:

Calculate the molar mass of carbon (IV) oxide ($\text{CO}_2$) if $0.25 \, \text{mol}$ has a mass of $11 \, \text{g}$.

Solution:

The molar mass ($M$) of a substance can be calculated using the formula:

$$M=\frac{\text{Mass}}{\text{Moles}}$$

Where:

• $\text{Mass} = 11 \, \text{g}$
  
• $\text{Moles} = 0.25 \, \text{mol}$
  

Substituting the values:

$$M = \frac{11}{0.25}$$

Perform the calculation:

$$M = 44 \, \text{g/mol}$$

Final Answer:

The molar mass of carbon (IV) oxide ($\text{CO}_2$) is 44 g/mol.

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Question 9:

Calculate the amount of oxygen in moles in $8 \, \text{g}$ of oxygen gas ($O_2$).

Solution:

The number of moles is calculated using the formula:

$$\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$$

Step 1: Determine the molar mass of oxygen gas ($O_2$) 

The atomic mass of oxygen ($O$) = $16 \, \text{g/mol}$. Since oxygen gas is diatomic ($O_2$):

$$\text{Molar Mass of } O_2 = 16 \times 2 = 32 \, \text{g/mol}.$$

Step 2: Substitute the values into the formula:

• $\text{Mass} = 8 \, \text{g}$
  
• $\text{Molar Mass of } O_2 = 32 \, \text{g/mol}$
  

$$\text{Moles} = \frac{8}{32}$$

Step 3: Perform the calculation:

$$\text{Moles} = 0.25 \, \text{mol}.$$

Final Answer:

The amount of oxygen gas in $8 \, \text{g}$ is 0.25 moles.

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Question 10:

What mass of sodium is contained in 53 g of sodium carbonate (\( \text{Na}_2\text{CO}_3 \))?
Given that: \[ \text{Na} = 23, \quad \text{O} = 16, \quad \text{C} = 12. \]

Solution:

Step 1: Calculate the molar mass of sodium carbonate 

The molar mass of $\text{Na}_2\text{CO}_3$ can be found by adding the atomic masses of its components:

$$\text{Molar Mass of } \text{Na}_2\text{CO}_3 = (2 \times 23) + 12 + (3 \times 16)$$
$$\text{Molar Mass of } \text{Na}_2\text{CO}_3 = 46 + 12 + 48 = 106 \, \text{g/mol}$$

Step 2: Determine the mass of sodium in

There are 2 sodium atoms in each molecule of sodium carbonate, and the molar mass of sodium is $23 \, \text{g/mol}$. So, the mass of sodium in one mole of $\text{Na}_2\text{CO}_3$ is:

$$\text{Mass of Sodium}=2\times 23=46\text{g}\mathrm{.}$$

Step 3: Use the proportion to find the mass of sodium in 53g

We know that $106 \, \text{g}$ of $\text{Na}_2\text{CO}_3$ contains $46 \, \text{g}$ of sodium. We can set up the following proportion:

$$\frac{\text{Mass of Sodium}}{\text{Mass of } \text{Na}_2\text{CO}_3} = \frac{46}{106}$$

Now, to find the mass of sodium in $53 \, \text{g}$ of $\text{Na}_2\text{CO}_3$:

$$\text{Mass of Sodium}=\left(\frac{46}{106}\right)\times 53$$

Step 4: Perform the calculation:

The mass of sodium can be calculated as follows:

\[ \text{Mass of Sodium} = \frac{46 \times 53}{106} = \frac{2438}{106} \approx 23 \, \text{g}. \]

Final Answer:

The mass of sodium in $53 \, \text{g}$ of sodium carbonate is 23 g.

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Question 11

Find the value of \( x \) in \( \text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O} \) if \( 0.375 \, \text{mol} \) of the compound has a mass of \( 107.25 \, \text{g} \).

Solution;

Step 1: Write the formula for the molar mass of \( \text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O} \)

The molar mass of \( \text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O} \) is given by:

\[ M = M_{\text{Na}_2\text{CO}_3} + x \cdot M_{\text{H}_2\text{O}} \] Where:

• \( M_{\text{Na}_2\text{CO}_3} = 106 \, \text{g/mol} \) (calculated as \( 2 \times 23 + 12 + 3 \times 16 \))
• \( M_{\text{H}_2\text{O}} = 18 \, \text{g/mol} \)

Step 2: Use the molar mass formula

From the question, the total mass of \( 0.375 \, \text{mol} \) is \( 107.25 \, \text{g} \). The molar mass of the compound can be calculated as:

\[ M = \frac{\text{Mass}}{\text{Moles}} \] Substitute the values:

\[ M = \frac{107.25}{0.375} = 286 \, \text{g/mol}. \]

Thus:

\[ 286 = 106 + x \cdot 18 \]

Step 3: Solve for \( x \)

Rearrange the equation to isolate \( x \):

\[ 286 - 106 = x \cdot 18 \]

\[ 180 = x \cdot 18 \]

Divide through by 18:

\[ x = \frac{180}{18} = 10 \]

Final Answer:

The value of \( x \) is 10.

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Question 12

Calculate the number of molecules in \( 1.7 \, \text{g} \) of ammonia (\( \text{NH}_3 \)). Given:

• \( \text{N} = 14 \, \text{g/mol} \), \( \text{H} = 1 \, \text{g/mol} \)
• Avogadro's constant \( L = 6.02 \times 10^{23} \, \text{mol}^{-1} \)

Solution

Step 1: Calculate the molar mass of ammonia (\( \text{NH}_3 \))

The molar mass of ammonia is:

\[ M_{\text{NH}_3} = 14 + (3 \times 1) = 17 \, \text{g/mol}. \]

Step 2: Determine the number of moles of ammonia

The number of moles (\( n \)) is given by:

\[ n = \frac{\text{Mass}}{\text{Molar Mass}}. \]

Substitute the values:

\[ n = \frac{1.7}{17} = 0.1 \, \text{mol}. \]

Step 3: Calculate the number of molecules

The number of molecules is calculated using Avogadro's constant:

\[ \text{Number of molecules} = n \times L. \]

Substitute the values:

\[ \text{Number of molecules} = 0.1 \times 6.02 \times 10^{23} = 6.02 \times 10^{22}. \]

Final Answer:

The number of molecules in \( 1.7 \, \text{g} \) of ammonia is \( 6.02 \times 10^{22} \, \text{molecules} \).

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Question 13

Calculate the amount of substance (in moles) in:

1. 7 g of nitrogen gas (\( \text{N}_2 \))
2. 6 g of ethanoic acid (\( \text{CH}_3\text{COOH} \))

Solution

Part (i): 7 g of nitrogen gas (\( \text{N}_2 \))

Step 1: Calculate the molar mass of \( \text{N}_2 \)

The atomic mass of nitrogen (\( \text{N} \)) is \( 14 \, \text{g/mol} \). Since \( \text{N}_2 \) is diatomic:

\[ \text{Molar Mass of } \text{N}_2 = 2 \times 14 = 28 \, \text{g/mol}. \]

Step 2: Calculate the number of moles

The number of moles is given by:

\[ n = \frac{\text{Mass}}{\text{Molar Mass}}. \]

Substituting the values:

\[ n = \frac{7}{28} = 0.25 \, \text{mol}. \]

Part (ii): 6 g of ethanoic acid (\( \text{CH}_3\text{COOH} \))

Step 1: Calculate the molar mass of \( \text{CH}_3\text{COOH} \)

The molar mass of ethanoic acid is calculated as:

\[ M_{\text{CH}_3\text{COOH}} = (2 \times 12) + (4 \times 1) + (2 \times 16) = 24 + 4 + 32 = 60 \, \text{g/mol}. \]

Step 2: Calculate the number of moles

Substituting into the formula:

\[ n = \frac{\text{Mass}}{\text{Molar Mass}}. \]

Substituting the values:

\[ n = \frac{6}{60} = 0.1 \, \text{mol}. \]

Final Answer:

• The amount of substance in 7 g of nitrogen gas (\( \text{N}_2 \)) is \( 0.25 \, \text{mol} \).
• The amount of substance in 6 g of ethanoic acid (\( \text{CH}_3\text{COOH} \)) is \( 0.1 \, \text{mol} \).

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Question 14:

Calculate the number of atoms in 18.5 g of magnesium metal. Given:

• Molar mass of magnesium (\( \text{Mg} \)) = 24.3 g/mol
• Avogadro's number (\( L \)) = \( 6.0 \times 10^{23} \, \text{mol}^{-1} \)

Solution:

Step 1: Calculate the number of moles of magnesium.

The number of moles (\( n \)) is given by:

\( n = \frac{\text{Mass}}{\text{Molar Mass}} \)

Substitute the values:

\( n = \frac{18.5}{24.3} \approx 0.761 \, \text{mol} \)

Step 2: Calculate the number of atoms.

The number of atoms is given by:

Number of atoms = \( n \times L \)

Substitute the values:

Number of atoms = \( 0.761 \times 6.0 \times 10^{23} \)

Perform the calculation:

Number of atoms = \( 4.566 \times 10^{23} \)

Final Answer:

The number of atoms in 18.5 g of magnesium is \( 4.57 \times 10^{23} \, \text{atoms} \).

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Question 15:

Calculation of Chloride Ions in 65.0g of Calcium Chloride

Given:

• Calcium (Ca) = 40 g/mol
• Chlorine (Cl) = 35.5 g/mol
• Avogadro's number (\( L \)) = \( 6.02 \times 10^{23} \, \text{particles/mol} \)

Solution:

Step 1: Calculate the molar mass of calcium chloride (\( \text{CaCl}_2 \))

The molar mass of \( \text{CaCl}_2 \) is calculated as:

\( M_{\text{CaCl}_2} = M_{\text{Ca}} + 2 \times M_{\text{Cl}} = 40 + (2 \times 35.5) = 40 + 71 = 111 \, \text{g/mol} \)

Step 2: Calculate the number of moles of \( \text{CaCl}_2 \)

The number of moles (\( n \)) is given by:

\( n = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{65.0}{111} \approx 0.586 \, \text{mol} \)

Step 3: Determine the number of chloride ions

Each formula unit of \( \text{CaCl}_2 \) contains two chloride ions. Therefore, the total number of chloride ions is:

Number of chloride ions = \( n \times L \times 2 = 0.586 \times 6.02 \times 10^{23} \times 2 \)

Perform the calculation:

Number of chloride ions = \( 7.06 \times 10^{23} \, \text{ions} \)

Final Answer:

The number of chloride ions in 65.0 g of calcium chloride is \( 7.06 \times 10^{23} \, \text{ions} \).

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Question 16:

Calculation of Chlorine Atoms in 0.2g of DDT

Given:

• Carbon (C) = 12 g/mol
• Hydrogen (H) = 1 g/mol
• Chlorine (Cl) = 35.5 g/mol
• Avogadro's number (\( L \)) = \( 6.02 \times 10^{23} \, \text{particles/mol} \)

Solution:

Step 1: Calculate the molar mass of DDT (\( \text{C}_{14}\text{H}_9\text{Cl}_5 \))

The molar mass of DDT is calculated as:

\( M_{\text{C}_{14}\text{H}_9\text{Cl}_5} = (14 \times 12) + (9 \times 1) + (5 \times 35.5) = 168 + 9 + 177.5 = 354.5 \, \text{g/mol} \)

Step 2: Calculate the number of moles of DDT

The number of moles (\( n \)) is given by:

\( n = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.2}{354.5} \approx 5.64 \times 10^{-4} \, \text{mol} \)

Step 3: Determine the number of chlorine atoms

Each molecule of DDT contains 5 chlorine atoms. Therefore, the total number of chlorine atoms is:

Number of chlorine atoms = \( n \times L \times 5 = 5.64 \times 10^{-4} \times 6.02 \times 10^{23} \times 5 \)

Perform the calculation:

\( 5.64 \times 10^{-4} \times 6.02 \times 10^{23} = 3.397 \times 10^{20} \)

\( 3.397 \times 10^{20} \times 5 = 1.698 \times 10^{21} \)

Final Answer:

The number of chlorine atoms in 0.2 g of DDT is \( 1.70 \times 10^{21} \, \text{atoms} \).

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Question 17:

Calculation of Hydrogen Atoms in 3.7g of Diethyl Ether

Given:

• Carbon (C) = 12 g/mol
• Hydrogen (H) = 1 g/mol
• Oxygen (O) = 16 g/mol
• Avogadro's number (\( L \)) = \( 6.02 \times 10^{23} \, \text{particles/mol} \)

Solution:

Step 1: Determine the molecular formula and molar mass of diethyl ether (\( \text{C}_4\text{H}_{10}\text{O} \))

The molar mass of diethyl ether is calculated as:

\( M_{\text{C}_4\text{H}_{10}\text{O}} = (4 \times 12) + (10 \times 1) + (1 \times 16) = 48 + 10 + 16 = 74 \, \text{g/mol} \)

Step 2: Calculate the number of moles of diethyl ether

The number of moles (\( n \)) is given by:

\( n = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{3.7}{74} \approx 0.05 \, \text{mol} \)

Step 3: Determine the number of hydrogen atoms in one molecule of diethyl ether

Each molecule of diethyl ether contains 10 hydrogen atoms.

Step 4: Calculate the number of hydrogen atoms

The number of hydrogen atoms is given by:

Number of hydrogen atoms = \( n \times L \times \text{Number of H atoms per molecule} = 0.05 \times 6.02 \times 10^{23} \times 10 \)

Perform the calculation:

\( 0.05 \times 6.02 \times 10^{23} = 3.01 \times 10^{22} \)

\( 3.01 \times 10^{22} \times 10 = 3.01 \times 10^{23} \)

Final Answer:

The number of hydrogen atoms in 3.7 g of diethyl ether vapor is \( 3.01 \times 10^{23} \, \text{atoms} \).

SHS Chemistry Topics

• Preparation of Standard Solution: A Complete Guide
• Concept of solution in chemistry
• Hydrogen Bonding
• Intermolecular Bonding: A Complete Guide to Understanding Forces Between Molecules
• Van der Waals Forces: A Comprehensive Overview
• Understanding Amount of Substance and the Mole: A Comprehensive Guide
• Chemical Bonding and Hybridization Notes
• Electronegativity
• Ionization Energy
• Atomic Radius
• Questions and answers on periodic table
• Periodic Chemistry notes
• The Structure of the Atom
• Quantum Mechanical Model of the Atom
• Rutherford’s Nuclear Model
• Bohr’s Model of the Atom
• The Discovery of the Electron by J.J. Thomson
• Dalton's Atomic Theory
• Atoms, Ions, and Molecules Notes: The Building Blocks of Matter
• Solubility in Chemistry: A Comprehensive Guide
• Senior High School (SHS) Chemistry Topics: Comprehensive Guide
• Acid-Base Titrations: Understanding the Process
• Understanding Buffer Solutions: A Comprehensive Guide for High School Students
• pH of Strong Acids and Strong Bases
• Understanding Base Dissociation Constant ( Kb ) and pKb
• Understanding Acid Dissociation Constant (Ka) and pKa