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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.

amount of substance and the mole

Questions and Answers on Amount of Substance and the Mole


Question 1:

The atomic mass unit (amu) of carbon-12 is 1.6605×10−24 g1.6605 \times 10^{-24} \, \text{g}. If the mass of an atom is 5.313×10−23 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:

Relative Mass=Mass of the atomAtomic Mass Unit (amu)​

Substituting the given values:

  • Mass of the atom: 5.313×10−23 g5.313 \times 10^{-23} \, \text{g}
  • Atomic mass unit: 1.6605×10−24 g1.6605 \times 10^{-24} \, \text{g}
Relative Mass=5.313×10−231.6605×10−24​

Perform the division:

Divide the coefficients:

5.3131.6605≈3.2

Subtract the exponents of 10:

10−23−10−24=10−23+24=101

Combine the results:      
                                3.2 × 101 = 32


Question 2:

How many atoms are there in 0.3 mol of sodium? Given L=6.0×1023 mol−1L = 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:

Number of atoms=moles×L

Where:

  • moles=0.3 mol
  • L=6.0×1023 mol−1L = 6.0 \times 10^{23} \, \text{mol}^{-1}


Substituting the values:

Number of atoms=0.3×6.0×1023


Perform the calculation:

1. Multiply the coefficients:

0.3×6.0=1.8

2. Include the power of 10:

1.8×1023

The number of atoms in 0.3 mol0.3 \, \text{mol} of sodium is 1.8×1023 atoms.



Question 3:

Calculate the amount of oxygen in moles in 1.505×10231.505 \times 10^{23} molecules of oxygen gas. Given L=6.02×1023 mol−1L = 6.02 \times 10^{23} \, \text{mol}^{-1} (Avogadro's constant).


Solution:

The number of moles can be calculated using the formula:

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

Where:

  • Number of molecules=1.505×1023\text{Number of molecules} = 1.505 \times 10^{23}
  • L=6.02×1023 mol−1L = 6.02 \times 10^{23} \, \text{mol}^{-1}


Substituting the values:

Moles=1.505×10236.02×1023\text{Moles} = \frac{1.505 \times 10^{23}}{6.02 \times 10^{23}}


Perform the calculation:

1. Divide the coefficients:

1.5056.02≈0.25\frac{1.505}{6.02} \approx 0.25


2. Subtract the exponents of 102310^{23}:

1023−1023=100=110^{23} - 10^{23} = 10^0 = 1

Combine the results:
0.25×1=0.250.25 \times 1 = 0.25
The amount of oxygen is 0.25 moles.



Question 4:

Calculate the number of oxygen atoms in 0.20 mol0.20 \, \text{mol} of oxygen gas. Given L=6.02×1023 mol−1L = 6.02 \times 10^{23} \, \text{mol}^{-1}.


Solution:

Step 1: Relationship between moles of O2 molecules and oxygen atoms

Oxygen gas exists as diatomic molecules (O2O_2), so, each molecule contains 2 oxygen atoms. The total number of oxygen atoms can be found using:

Number of atoms=moles of O2×L×2

Where:

  • moles of O2=0.20 mol\text{moles of } O_2 = 0.20 \, \text{mol}
  • L=6.02×1023 mol−1L = 6.02 \times 10^{23} \, \text{mol}^{-1}
  • 2=atoms per molecule of O22 = \text{atoms per molecule of } O_2


Step 2: Substitute the values

Number of atoms=0.20×6.02×1023×2


Step 3: Perform the calculation

  1. Multiply the coefficients:

    0.20×6.02=1.204
  2. Multiply by 2:

    1.204×2=2.4081.204 \times 2 = 2.408
  3. Include the power of 10:

    2.408×10232.408 \times 10^{23}

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


Question 5:

Calculate:

(i)  the number of molecules contained in 0.6 mol0.6 \, \text{mol} of H2S\text{H}_2\text{S}.
(ii) the number of atoms of H and S present in 0.6 mol0.6 \, \text{mol} of H2S\text{H}_2\text{S}.


Solution:

Part (i): 

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

Number of molecules=moles×L

Where:

  • moles=0.6 mol\text{moles} = 0.6 \, \text{mol}
  • L=6.02×1023 mol−1L = 6.02 \times 10^{23} \, \text{mol}^{-1} (Avogadro's constant)

Substituting the values:

Number of molecules=0.6×6.02×1023

Perform the calculation:

0.6×6.02=3.612

Include the power of 102310^{23}:

Number of molecules=3.612×1023

Answer for (i): The number of H2S\text{H}_2\text{S} molecules is 3.612×10233.612 \times 10^{23}.


Part (ii): 

  1. Atoms per molecule of H2S\text{H}_2\text{S}:

    • Each H2S\text{H}_2\text{S} molecule contains:
      • 2 hydrogen (HH) atoms
      • 1 sulfur (SS) atom
  2. Total number of atoms of each type:

    • Hydrogen atoms:

      Number of H atoms=Number of molecules×2

      Substituting:

      Number of H atoms=3.612×1023×2=7.224×1023
    • Sulfur atoms:

      Number of S atoms=Number of molecules=3.612×1023\text{Number of S atoms} = \text{Number of molecules} = 3.612 \times 10^{23}

Final Answer:

(i) The number of H2S\text{H}_2\text{S} molecules is 3.612×10233.612 \times 10^{23}.
(ii) The number of hydrogen (HH) atoms is 7.224×10237.224 \times 10^{23}, and the number of sulfur (SS) atoms is 3.612×10233.612 \times 10^{23}.



Question 6:

Calculate the number of nitrogen (NN) atoms in 0.25 mol0.25 \, \text{mol} of nitrogen (I) oxide (N2O\text{N}_2\text{O}). Given L=6.02×1023 mol−1L = 6.02 \times 10^{23} \, \text{mol}^{-1}.


Solution:

Step 1: Relationship between moles, molecules, and atoms

Nitrogen (I) oxide (N2O\text{N}_2\text{O}) contains 2 nitrogen atoms per molecule.

The total number of NN atoms can be calculated using:

Number of atoms of N=moles of N2O×L×number of N atoms per molecule.

Step 2: Substituting the values

  • Moles of N2O=0.25 mol
  • L=6.02×1023 mol−1L = 6.02 \times 10^{23} \, \text{mol}^{-1}
  • Number of NN atoms per molecule = 2

Number of atoms of N=0.25×6.02×1023×2\text{Number of atoms of } N = 0.25 \times 6.02 \times 10^{23} \times 2

Step 3: Perform the calculation

  • Multiply the coefficients:

    0.25×6.02=1.5050.25 \times 6.02 = 1.505

1.505×2=3.011.505 \times 2 = 3.01
  • Include the power of 102310^{23}:

    3.01×10233.01 \times 10^{23}

Final Answer:

The number of nitrogen (NN) atoms in 0.25 mol of N2O\text{N}_2\text{O} is 3.01×1023 atoms3.01 \times 10^{23} \, \text{atoms}.



Question 7:

Calculate the mass of 0.2 mol0.2 \, \text{mol} of magnesium atoms. Given the molar mass of magnesium (Mg\text{Mg}) = 24 g/mol.


Solution:

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

Mass=Moles×Molar Mass

Where:

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


Substituting the values:

Mass=0.2×24

Perform the calculation: Mass=4.8g


Question 8:

Calculate the molar mass of carbon (IV) oxide (CO2\text{CO}_2) if 0.25 mol0.25 \, \text{mol} has a mass of 11 g11 \, \text{g}.


Solution:

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

M=MassMoles​

Where:

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


Substituting the values:

M=110.25M = \frac{11}{0.25}

Perform the calculation:

M=44 g/molM = 44 \, \text{g/mol}

Final Answer:

The molar mass of carbon (IV) oxide (CO2\text{CO}_2) is 44 g/mol.



Question 9:

Calculate the amount of oxygen in moles in 8g8 \, \text{g} of oxygen gas (O2O_2).


Solution:

The number of moles is calculated using the formula:

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

Step 1: Determine the molar mass of oxygen gas (O2) 

The atomic mass of oxygen (OO) = 16 g/mol16 \, \text{g/mol}. Since oxygen gas is diatomic (O2O_2):

Molar Mass of O2=16×2=32 g/mol.\text{Molar Mass of } O_2 = 16 \times 2 = 32 \, \text{g/mol}.

Step 2: Substitute the values into the formula:

  • Mass=8 g\text{Mass} = 8 \, \text{g}
  • Molar Mass of O2=32 g/mol\text{Molar Mass of } O_2 = 32 \, \text{g/mol}
Moles=832\text{Moles} = \frac{8}{32}

Step 3: Perform the calculation:

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

Final Answer:

The amount of oxygen gas in 8 g8 \, \text{g} is 0.25 moles.



Question 10:

What mass of sodium is contained in 53 g of sodium carbonate (Na2CO3)?
Given that: Na=23,O=16,C=12.


Solution:

Step 1: Calculate the molar mass of sodium carbonate 

The molar mass of Na2CO3\text{Na}_2\text{CO}_3 can be found by adding the atomic masses of its components:

Molar Mass of Na2CO3=(2×23)+12+(3×16)\text{Molar Mass of } \text{Na}_2\text{CO}_3 = (2 \times 23) + 12 + (3 \times 16)
Molar Mass of Na2CO3=46+12+48=106 g/mol\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 g/mol23 \, \text{g/mol}. So, the mass of sodium in one mole of Na2CO3\text{Na}_2\text{CO}_3 is:

Mass of Sodium=2×23=46 g.


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

We know that 106 g106 \, \text{g} of Na2CO3\text{Na}_2\text{CO}_3 contains 46 g46 \, \text{g} of sodium. We can set up the following proportion:

Mass of SodiumMass of Na2CO3=46106\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 g53 \, \text{g} of Na2CO3\text{Na}_2\text{CO}_3:

Mass of Sodium=(46106)×53


Step 4: Perform the calculation:

The mass of sodium can be calculated as follows:

Mass of Sodium=46×53106=2438106≈23g.


Final Answer:

The mass of sodium in 53 g53 \, \text{g} of sodium carbonate is 23 g.



Question 11

Find the value of x in Na2CO3â‹…xH2O if 0.375mol of the compound has a mass of 107.25g.


Solution;

Step 1: Write the formula for the molar mass of Na2CO3â‹…xH2O

The molar mass of Na2CO3â‹…xH2O is given by:

M=MNa2CO3+xâ‹…MH2O Where:

  • MNa2CO3=106g/mol (calculated as 2×23+12+3×16)
  • MH2O=18g/mol


Step 2: Use the molar mass formula

From the question, the total mass of 0.375mol is 107.25g. The molar mass of the compound can be calculated as:

M=MassMoles Substitute the values:

M=107.250.375=286g/mol.

Thus:

286=106+xâ‹…18


Step 3: Solve for x

Rearrange the equation to isolate x:

286−106=x⋅18

180=xâ‹…18

Divide through by 18:

x=18018=10


Final Answer:

The value of x is 10.



Question 12

Calculate the number of molecules in 1.7g of ammonia (NH3). Given:

  • N=14g/mol, H=1g/mol
  • Avogadro's constant L=6.02×1023mol−1


Solution

Step 1: Calculate the molar mass of ammonia (NH3)

The molar mass of ammonia is:

MNH3=14+(3×1)=17g/mol.


Step 2: Determine the number of moles of ammonia

The number of moles (n) is given by:

n=MassMolar Mass.

Substitute the values:

n=1.717=0.1mol.


Step 3: Calculate the number of molecules

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

Number of molecules=n×L.

Substitute the values:

Number of molecules=0.1×6.02×1023=6.02×1022.


Final Answer:

The number of molecules in 1.7g of ammonia is 6.02×1022molecules.



Question 13

Calculate the amount of substance (in moles) in:

  1. 7 g of nitrogen gas (N2)
  2. 6 g of ethanoic acid (CH3COOH)


Solution

Part (i): 7 g of nitrogen gas (N2)

Step 1: Calculate the molar mass of N2

The atomic mass of nitrogen (N) is 14g/mol. Since N2 is diatomic:

Molar Mass of N2=2×14=28g/mol.


Step 2: Calculate the number of moles

The number of moles is given by:

n=MassMolar Mass.

Substituting the values:

n=728=0.25mol.


Part (ii): 6 g of ethanoic acid (CH3COOH)

Step 1: Calculate the molar mass of CH3COOH

The molar mass of ethanoic acid is calculated as:

MCH3COOH=(2×12)+(4×1)+(2×16)=24+4+32=60g/mol.

Step 2: Calculate the number of moles

Substituting into the formula:

n=MassMolar Mass.

Substituting the values:

n=660=0.1mol.


Final Answer:

  • The amount of substance in 7 g of nitrogen gas (N2) is 0.25mol.
  • The amount of substance in 6 g of ethanoic acid (CH3COOH) is 0.1mol.


Question 14:

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

  • Molar mass of magnesium (Mg) = 24.3 g/mol
  • Avogadro's number (L) = 6.0×1023mol−1


Solution:

Step 1: Calculate the number of moles of magnesium.

The number of moles (n) is given by:

n=MassMolar Mass

Substitute the values:

n=18.524.3≈0.761mol


Step 2: Calculate the number of atoms.

The number of atoms is given by:

Number of atoms = n×L

Substitute the values:

Number of atoms = 0.761×6.0×1023

Perform the calculation:

Number of atoms = 4.566×1023


Final Answer:

The number of atoms in 18.5 g of magnesium is 4.57×1023atoms.



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) = 6.02×1023particles/mol


Solution:

Step 1: Calculate the molar mass of calcium chloride (CaCl2)

The molar mass of CaCl2 is calculated as:

MCaCl2=MCa+2×MCl=40+(2×35.5)=40+71=111g/mol


Step 2: Calculate the number of moles of CaCl2

The number of moles (n) is given by:

n=MassMolar Mass=65.0111≈0.586mol


Step 3: Determine the number of chloride ions

Each formula unit of CaCl2 contains two chloride ions. Therefore, the total number of chloride ions is:

Number of chloride ions = n×L×2=0.586×6.02×1023×2

Perform the calculation:

Number of chloride ions = 7.06×1023ions


Final Answer:

The number of chloride ions in 65.0 g of calcium chloride is 7.06×1023ions.



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) = 6.02×1023particles/mol


Solution:

Step 1: Calculate the molar mass of DDT (C14H9Cl5)

The molar mass of DDT is calculated as:

MC14H9Cl5=(14×12)+(9×1)+(5×35.5)=168+9+177.5=354.5g/mol


Step 2: Calculate the number of moles of DDT

The number of moles (n) is given by:

n=MassMolar Mass=0.2354.5≈5.64×10−4mol


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×L×5=5.64×10−4×6.02×1023×5

Perform the calculation:

5.64×10−4×6.02×1023=3.397×1020

3.397×1020×5=1.698×1021


Final Answer:

The number of chlorine atoms in 0.2 g of DDT is 1.70×1021atoms.



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) = 6.02×1023particles/mol


Solution:

Step 1: Determine the molecular formula and molar mass of diethyl ether (C4H10O)

The molar mass of diethyl ether is calculated as:

MC4H10O=(4×12)+(10×1)+(1×16)=48+10+16=74g/mol


Step 2: Calculate the number of moles of diethyl ether

The number of moles (n) is given by:

n=MassMolar Mass=3.774≈0.05mol


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×L×Number of H atoms per molecule=0.05×6.02×1023×10

Perform the calculation:

0.05×6.02×1023=3.01×1022

3.01×1022×10=3.01×1023


Final Answer:

The number of hydrogen atoms in 3.7 g of diethyl ether vapor is 3.01×1023atoms.


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