Mole Concept: Class 11 Complete Notes

<header class="bg-gradient-to-r from-blue-600 to-indigo-600 text-white py-12"> <div class="container mx-auto px-4 text-center"> <h1 class="text-4xl font-bold mb-2">Grade 11 Chemistry: Mole Concept</h1> <p class="text-xl">Comprehensive Notes, Formulas, and Solved Problems</p> </div> </header><main class="container mx-auto px-4 py-12"> <article class="bg-white shadow-xl rounded-lg p-8"> <section class="mb-12"> <h2 class="text-3xl font-semibold mb-4 section-header">Mole Concept</h2> <p>A mole is defined as a collection of particles of anything (atoms, ions, electrons, protons, etc) equal in number to the number of atoms present in 1 gm atom of <a href="https://en.wikipedia.org/wiki/Carbon-12" target="_blank" rel="noopener">C-12 isotope</a>.</p> <figure class="text-center my-6"><img src="https://sajhanotes.com/wp-content/uploads/2021/05/mole-concept.png" alt="mole concept illustration"></figure> <p>In terms of atoms,</p> <div class="math-display">\[ \text{number of moles} = \frac{\text{weight in gm}}{\text{atomic weight}} \]</div> <p>In terms of mass,</p> <div class="math-display">\[ \text{number of moles} = \frac{\text{weight in gm}}{\text{molecular weight}} \]</div> <p>In terms of ions,</p> <div class="math-display">\[ \text{number of moles} = \frac{\text{weight in gm}}{\text{ionic weight}} \]</div> <p>In terms of gases,</p> <div class="math-display">\[ \text{number of moles} = \frac{\text{volume at NTP}}{22.4 \, \text{litres}} \]</div> <p>1 mole of H atoms = 6.023 &times; 10²³ atoms = 1.008 gm<br>1 mole of H₂ molecules = 6.023 &times; 10²³ molecules = 2.016 gm<br>1 mole of H₂O = 6.023 &times; 10²³ molecules = 18 gm<br>1 mole of SO₄^&mdash; ions = 6.023 &times; 10²³ ions = 96 gm</p> </section> <section class="mb-12"> <h2 class="text-3xl font-semibold mb-4 section-header">Solved Numericals of Mole Concept</h2> <p>1. Pure water sample having a mass of 0.36 gm is taken, answer the following questions.</p> <p>i. How many moles of water are present?</p> <div class="math-display">\begin{align*} \text{number of moles} &amp;= \frac{\text{weight in gm}}{\text{molecular weight}} \\ &amp;= \frac{0.36}{18} = 0.02 \, \text{mol} \end{align*}</div> <p>ii. How many molecules of water are present?</p> <p>1 mol of water = 6.023 &times; 10²³ molecules<br>0.02 mol of water = 6.023 &times; 10²³ &times; 0.02 = 1.2046 &times; 10²² molecules</p> <p>iii. How many atoms of hydrogen are present?</p> <p>1 mol of water = 2 &times; 6.023 &times; 10²³ atoms<br>0.02 mol of water = 2 &times; 6.023 &times; 10²³ &times; 0.02 = 2.4092 &times; 10²² atoms</p> <p>iv. How many volume of water is present at NTP?</p> <p>1 mol of water at NTP = 22.4 L<br>0.02 mol of water at NTP = 22.4 &times; 0.02 = 0.448 L</p> <p>2. Calculate the weight of 1 molecule of CaCO₃.</p> <div class="math-display">\begin{align*} 1 \, \text{mol of CaCO}_3 &amp;= 100 \, \text{gm} \\ 6.023 \times 10^{23} \, \text{molecules of CaCO}_3 &amp;= 100 \, \text{gm} \\ 1 \, \text{molecule of CaCO}_3 &amp;= \frac{100 \, \text{gm}}{6.023 \times 10^{23}} \\ &amp;= 1.6606 \times 10^{-22} \, \text{gm} \end{align*}</div> <p>3. Calculate the weight of 100 ml of NH3 gas at NTP.</p> <p class="text-center">1 mole of NH₃ = 17 gm<br>22.4 litres of NH₃ at NTP = 17 gm<br>22400 ml of NH₃ at NTP = 17 gm<br>100 ml of NH₃ at NTP = (17 &times; 100) / 22400 = 0.0759 gm</p> <p>4. Which of the following has the largest number of molecules and how? 7 gm of nitrogen or 1 gm of hydrogen.<br>solution:</p> <p>&rarr; In 7 gm of nitrogen<br>Number of moles = 7/28 = 0.25 mol<br>Number of molecules = 0.25 &times; 6.023 &times; 10²³ molecules</p> <p>&rarr; In 1 gm of hydrogen<br>Number of moles = 1/2 = 0.5 mol<br>Number of molecules = 0.5 &times; 6.023 &times; 10²³ molecules</p> <p>So 1 gm of hydrogen has a large number of molecules because its number of moles is large.</p> <p>What volume of CO₂ is produced when 20 gm of 20% CaCO₃ is heated?<br>solution:</p> <div class="math-display">\ce{CaCO3 -&gt;[1 mol (100 g)] CaO + CO2 \ (1 mol, 22.4 L at NTP)} \\ \begin{align*} \text{Given mass of CaCO}_3 &amp;= 20\% \text{ of } 20 \, \text{gm} = 4 \, \text{gm} \\ 4 \, \text{gm of CaCO}_3 &amp;= \frac{22.4}{100} \times 4 = 0.896 \, \text{L at NTP} \\ &amp;= 896 \, \text{cc at NTP} \end{align*}</div> <p>5. What mass of 60% pure sulphuric acid is required to decompose 25 gm of CaCO₃?<br>solution:</p> <div class="math-display">\ce{CaCO3 + H2SO4 -&gt; CaSO4 + H2O + CO2} \\ \text{(1 mol CaCO}_3 \text{ = 100 g, 1 mol H}_2\text{SO}_4 \text{ = 98 g)}</div> <p class="text-center">100 gm of CaCO₃ is decomposed by 98 gm of H₂SO₄<br>25 gm of CaCO₃ is decomposed by 24.5 gm of H₂SO₄<br>Given H₂SO₄ is 60% pure<br>let mass of 60% pure H₂SO₄ = x gm</p> <div class="math-display">x \times \frac{60}{100} = 24.5 \\ \therefore x = 40.833 \, \text{gm}</div> </section> <section class="mb-12"> <h2 class="text-3xl font-semibold mb-4 section-header">Percentage Composition</h2> <p>Percent means certain parts present per hundred parts. Percent of an element in a compound is the number of parts by mass of that element per hundred parts by mass of the compound. The percent composition of a compound gives per cent by the mass of elements present in the compound.</p> <div class="math-display">\% \text{ element} = \frac{\text{mass of element}}{\text{mass of the compound}} \times 100</div> <h3 class="text-2xl font-semibold mt-8 mb-4">Solved Numericals</h3> <p>1. Calculate the percentage composition of Al₂O₃.<br>solution:</p> <div class="math-display">\text{Molecular mass of Al}_2\text{O}_3 = 2 \times 27 + 3 \times 16 = 102 \\ \% \text{ of Al} = \frac{54}{102} \times 100 = 52.94\% \\ \% \text{ of O} = \frac{48}{102} \times 100 = 47.06\%</div> <p>2. Calculate the percentage composition of the water of crystallization in blue vitriol crystal.<br>solution:</p> <p>Molecular weight of CuSO₄.5H₂O = 63.5 + 32 + 4 &times; 16 + 5 &times; 18 = 249.5 gm<br>% of water crystallization = (5 &times; 18 &times; 100)/249.5 = 36.07 %</p> </section> <section class="mb-12"> <h2 class="text-3xl font-semibold mb-4 section-header">Empirical Formula</h2> <p>It is the simplest formula that represents the simplest whole-number ratio of the number of atoms of all elements present in one molecule of the compound. eg. the empirical formula of glucose is CH₂O where its actual formula is C₆H₁₂O₆.</p> </section> <section class="mb-12"> <h2 class="text-3xl font-semibold mb-4 section-header">Molecular Formula</h2> <p>It is the true formula that represents the actual number of atoms of all the elements present in one molecule of the compound.</p> </section> <section class="mb-12"> <h2 class="text-3xl font-semibold mb-4 section-header">Solved Numerical</h2> <p>1. Determine the empirical and molecular formula of a compound having the following percentage composition: 40% carbon, 6.66% hydrogen and 53.44% oxygen. (Mol. mass =60)<br>solution:</p> <table> <thead> <tr> <th>element</th> <th>%</th> <th>At. wt.</th> <th>No. of.<br>moles</th> <th>simplest<br>ratio</th> <th>whole no.<br>ratio</th> </tr> </thead> <tbody> <tr> <td>Carbon</td> <td>40</td> <td>12</td> <td>40/12 = 3.33</td> <td>3.33/3.33=1</td> <td>&nbsp;</td> </tr> <tr> <td>hydrogen</td> <td>6.66</td> <td>1</td> <td>6.66/1 = 6.66</td> <td>6.66/3.33=2</td> <td>1:2:1</td> </tr> <tr> <td>oxygen</td> <td>53.44</td> <td>16</td> <td>53.44/16 = 3.34</td> <td>3.34/3.33=1</td> <td>&nbsp;</td> </tr> </tbody> </table> <p>Empirical formula = CH₂O<br>Empirical formula mass = 12 + 2 + 16 = 30<br>Molecular mass = 60</p> <div class="math-display">\text{Common factor (n)} = \frac{\text{Molecular mass}}{\text{empirical formula mass}} = \frac{60}{30} = 2 \\ \begin{align*} \text{Molecular formula} &amp;= (\text{Empirical formula})_n \\ &amp;= (CH_2O) \\ &amp;= C_2H_4O_2 \end{align*}</div> </section> <section class="mb-12"> <h2 class="text-3xl font-semibold mb-4 section-header">Limiting Reagent</h2> <p>The reagent which is present in the lesser amount gets consumed completely and limits the amount of product formed is limiting reagent.</p> <h3 class="text-2xl font-semibold mt-8 mb-4">Solved Numericals</h3> <p>1. 5 gm of pure CaCO3 is treated with 5 gm of HCl.<br>i. Find the limiting reagent.<br>ii. Calculate the mass of CaCl₂ formed.<br>iii. How many number of water molecules are produced?<br>iv. Calculate the volume of CO2 produced at NTP?<br>Solution:</p> <div class="math-display">\ce{CaCO3 + 2HCl -&gt; CaCl2 + H2O + CO2} \\ \text{(1 mol CaCO}_3 \text{ = 100 g, 2 mol HCl = 73 g, 1 mol CaCl}_2 \text{ = 111 g, 1 mol CO}_2 \text{ = 22.4 L at NTP)}</div> <p>i. 100 gm of CaCO₃ requires 73 gm of HCl<br>5 gm of CaCO₃ requires 3.65 gm of HCl<br>The amount of HCl present is 5 gm and the required is 3.65 gm. So HCl is excess and CaCO₃ is the limiting reagent.</p> <p>ii. 100 gm of CaCO₃ produces 111 gm of CaCl₂<br>5 gm of CaCO₃ produces 5.55 gm of CaCl₂.</p> <p>iii. 100 gm of CaCO₃ produces 6.023 &times; 10²³ water molecules<br>5 gm of CaCO₃ produces 3.011 &times; 10²³ water molecules.</p> <p>iv. 100 gm of CaCO₃ produces 22.4 lit of CO₂ at NTP<br>5 gm of CaCO₃ produces 1.12 lit of CO₂ at NTP</p> <p>2. 10 gm of Fe₂O₃ is reacted with 9 gm of CO.<br>i. Find the limiting reagent.<br>ii. How many moles of unreacted reagent left over?<br>iii. Calculate the mole of CO consumed in the reaction.<br>iv. What mass of NaOH is required to absorb the whole CO₂ produced?<br>Solution:</p> <div class="math-display">\ce{Fe2O3 + 3CO -&gt; 2Fe + 3CO2} \\ \text{(1 mol Fe}_2\text{O}_3 \text{ = 159.7 g, 3 mol CO = 84 g, 3 mol CO}_2\text{)}</div> <p>i. 159.7 gm of Fe₂O₃ reacts with 84 gm of CO<br>10 gm of Fe₂O₃ reacts with 5.25 gm of CO<br>The amount of CO present is 9 gm and the required is 5.25 gm. So CO is excess and Fe₂O₃ is the limiting reagent.</p> <p>ii. Mass of unreacted reagent left over = 9 &ndash; 5.25 = 3.75 gm<br>Moles of unreacted reagent left over = 3.75/28 = 0.133 mol</p> <p>iii. Mole of CO consumed = 5.25/28 = 0.1875 mol</p> <p>iv. 159.7 gm of Fe₂O₃ produces 3 &times; 44 gm of CO₂<br>10 gm of Fe₂O₃ produces 8.26 gm of CO₂</p> <div class="math-display">\ce{2NaOH + CO2 -&gt; Na2CO3 + H2O} \\ \text{(2 mol NaOH = 80 g, 1 mol CO}_2 \text{ = 44 g)}</div> <p>To absorb 44 gm of CO₂, 80 gm of NaOH is required.<br>To absorb 8.26 gm of CO₂, 15 gm of NaOH is required.</p> </section> <section class="mb-12"> <h2 class="text-3xl font-semibold mb-4 section-header">Berzelius Hypothesis</h2> <p>It states that &ldquo;Under the similar condition of temperature and pressure, an equal volume of all gases consists of an equal number of atoms&rdquo;.</p> </section> <section class="mb-12"> <h2 class="text-3xl font-semibold mb-4 section-header">Avogadro&rsquo;s Hypothesis (Law)</h2> <p>It states that &ldquo;Under the similar condition of temperature and pressure, an equal volume of all gases consists of an equal number of molecules&rdquo;.</p> </section> <section class="mb-12"> <h2 class="text-3xl font-semibold mb-4 section-header">Applications of Avogadro&rsquo;s Law</h2> <h3 class="text-2xl font-semibold mt-8 mb-4">1. Determination of Atomicity of Elementary Gases</h3> <p>The number of atoms present in a molecule of elementary gas is called atomicity. One molecule of hydrogen has two atoms. Its atomicity is 2.</p> <p>Q: Prove that hydrogen is diatomic?</p> <figure class="text-center my-6"><img src="https://sajhanotes.com/wp-content/uploads/2021/05/avogardos-law.png" alt=""></figure> <p>We know that,</p> <p>1 molecule of Hydrogen chloride = 1/2 molecule of hydrogen<br>1 molecule of hydrogen chloride = 1 atom of hydrogen<br>1/2 molecule of hydrogen = 1 atom of hydrogen<br>1 molecule of hydrogen = 2 atoms of hydrogen</p> <p>So hydrogen is diatomic molecule.</p> <h3 class="text-2xl font-semibold mt-8 mb-4">2. Determination of the Relationship Between Vapour Density and Molecular Mass</h3> <p>Vapour density is the ratio of the weight of a given volume of gas to the same volume of hydrogen under a similar condition of temperature and pressure.</p> <div class="math-display">\text{Vapour density (VD)} = \frac{\text{Weight of certain volume of gas}}{\text{Weight of same volume of hydrogen gas}}</div> <figure class="text-center my-6"><img src="https://sajhanotes.com/wp-content/uploads/2021/05/vapour-density-and-molecular-weight-reln.png" alt="vapour density and molecular weight reln"></figure> <p>Molecular weight is the ratio of the weight of one molecule of gas to the weight of one atom of hydrogen.</p> <div class="math-display">\begin{align*} \text{Molecular weight} &amp;= \frac{\text{Weight of 1 molecule of gas}}{\text{Weight of 1 atom of hydrogen gas}} \\ \text{VD} &amp;= \frac{\text{Molecular weight}}{2} \\ \text{Hence, Molecular weight} &amp;= 2 \times \text{vapour density} \end{align*}</div> <h3 class="text-2xl font-semibold mt-8 mb-4">3. Derivation of the Relationship Between Gram Molecular Mass and Volume of a Gas</h3> <p>Molecular weight expressed in gram is called gram molecular mass. Experimentally, one gram molecular mass occupies 22.4 litres of a gas at NTP which is explained by Avogadro&rsquo;s law.</p> <p>We have,<br>Mol. wt. = 2 &times; V.D</p> <figure class="text-center my-6"><img src="https://sajhanotes.com/wp-content/uploads/2021/05/gram-molecular-mass-and-vol-of-gas-reln.png" alt="gram molecular mass and vol of gas reln"></figure> <p>Gram molecular weight = 22.4 &times; Weight of 1 litre of gas<br>Gram molecular weight = Weight of 22.4 litres of gas at NTP<br>Gram molecular weight = 22.4 litres at NTP</p> <h3 class="text-2xl font-semibold mt-8 mb-4">4. Derivation of the Relationship Between Gram Molecular Weight and Number of Molecules</h3> <p>we know that,</p> <p>6.023 &times; 10²³ molecules = gram molecular weight<br>1 molecule = gram molecular weight / 6.023 &times; 10²³</p> </section> <section class="mb-12"> <h2 class="text-3xl font-semibold mb-4 section-header">Some Important Questions</h2> <h3 class="text-2xl font-semibold mt-8 mb-4">Mole Concept</h3> <ol> <li>A pure water sample having a mass of 0.18 gm is taken.<br>a. How many moles of water are present? (0.1)<br>b. How many molecules of water are present? (6.023 &times; 10²²)<br>c. How many atoms of hydrogen are present? (1.2044 &times; 10²³)<br>d. How many gram molecules of water are present? (0.1)<br>e. Calculate the volume occupied at NTP? (2.24 L)</li> <li>Which of the following has a larger number of molecules and why? 7 gm of nitrogen or 1 gm of hydrogen. (H₂)</li> <li>How many molecules are contained in 0.35 moles of N₂? (2.1 &times; 10²³)</li> <li>What volume of CO₂ is produced at NTP when 20 gm of 20% pure CaCO₃ is completely heated? (896 cc)</li> <li>How many moles of hydrogen are left when 3&times;10²¹ molecules of hydrogen are removed from a vessel containing 40 mg of hydrogen? (0.015 mol)</li> <li>What mass of 60% pure sulphuric acid is required to decompose 25 gm of CaCO₃? (40.833 gm)</li> <li>Calculate the mass of 120 cc of nitrogen at NTP. How many molecules are present in it? (0.15 gm, 3.22 &times; 10²¹)</li> <li>34.2 gm of sucrose (C₁₂H₂₂O₁₁) is dissolved in 180 gm of water. Calculate the number of oxygen atoms in the solution. (6.684 &times; 10²⁴)</li> <li>Calculate the number of hydrogen atoms present in 25.6 gm of urea. (1.03 &times; 10²⁴)</li> <li>Sulphuric acid is manufactured through the following reactions.<br>S + O₂ &rarr; SO₂<br>2SO₂ + O₂ &rarr; 2SO₃<br>SO₃ + H₂O &rarr; H₂SO₄<br>What mass of oxygen is needed to produce 4.9 gm of H₂SO₄? (2.4)</li> </ol> <h3 class="text-2xl font-semibold mt-8 mb-4">Limiting Reagent</h3> <ol> <li>5 gm of pure CaCO₃ if treated with 5 gm of HCl to produce CaCl₂, H₂O and CO₂.<br>a. Find the limiting reagent. (CaCO₃)<br>b. Calculate the mass of CaCl₂ formed. (5.55 gm)<br>c. How many water molecules are produced? (3.011&times;10²²)<br>d. Calculate the volume of CO₂ produced at NTP. (1.12 L)</li> <li>10 gm of pure zinc reacts with excess dilute sulphuric acid to yield zinc sulphate and hydrogen.<br>a. Calculate the number of moles of H₂SO₄ consumed. (0.153)<br>b. Calculate the mass of ZnSO₄ formed. (24.70 gm)<br>c. What volume of hydrogen is evolved at NTP? (3430.32 cc)</li> <li>10.6 gm of pure Na₂CO₃ if treated with 7.9 gm of HCl to produce NaCl, H₂O and CO₂.<br>a. Find the limiting reagent. (Na₂CO₃)<br>b. Calculate the mole of unreacted reagent left over. (0.0164)<br>c. What volume of CO₂ is produced at NTP? (2.24 L)<br>d. Calculate the mass of NaCl formed. (11.7 gm)</li> <li>200 gm of 90% pure CaCO₃ is completely reacted with excess HCl to produce CaCl₂, H₂O and CO₂.<br>a. Calculate the mass of CaCl₂ formed. (199.8 gm)<br>b. How many moles of water are formed? (1.8)<br>c. What volume of CO₂ is produced if the reaction is carried out at 27oC and 760 mm of Hg? (44.30 L)</li> <li>17 gm of ammonia is completely reacted with 45 gm of oxygen to produce NO and H₂O.<br>a. Which is limiting reagent? (NH₃)<br>b. Calculate the number of moles of unreacted reagent left over. (0.156)<br>c. What volume of NO is produced at NTP? (22.4 L)<br>d. Calculate the mass of water produced. (27 gm)</li> <li>A chemical reaction was carried out by mixing 25 gm of pure CaCO₃ and 0.75 moles of pure HCl.<br>a. Which one is limiting reagent? (CaCO₃)<br>b. Calculate the mass of CaCl₂ produced. (27.75 gm)<br>c. How many water molecules are formed? (1.505&times;10²³)<br>d. What mass of NaOH is required to absorb the whole CO₂ produced in the reaction? (20 gm)</li> </ol> <h3 class="text-2xl font-semibold mt-8 mb-4">Percentage Composition</h3> <ol> <li>A compound contains 75% carbon and 25% hydrogen. Determine its empirical formula. (CH₄)</li> <li>Butyric acid contains C, H and O. 4.24 gm sample of butyric acid is completely burnt. It gives 8.45 gm of CO₂ and 3.46 gm of H₂O. If molecular mass of butyric acid is 88, determine its molecular formula. (C₄H₈O₂)</li> <li>A sample of pure oxide of sulphur contains 50.1 % sulphur and 49.9% oxygen by mass. What is the simplest formula of compound. (SO₂)</li> <li>7gm of nitrogen is combined with 12 gm of oxygen. What is the empirical formula. (N₂O₃)</li> <li>In hydrocarbon, the ratio of hydrogen to carbon is 2:1 and the molecular weight is 42. Find its molecular formula. (C₃H₆)</li> <li>During construction of Palpa-Butwal road, a new compound was suspected. It contains 4.6 gm sodium, 6.4 gm sulphur and 4.8 gm oxygen.<br>a. Calculate the percentage composition.<br>b. What is the empirical formula of compound. (Na₂S₂O₃)</li> </ol> </section> </article> </main><footer class="bg-gray-800 text-white py-6"> <div class="container mx-auto px-4 text-center"> <p>&copy; 2025 Sajha Notes. All rights reserved.</p> </div> </footer>