Chapter 06 Notes: Mole Concept and Related Calculations

The Mole and Avogadro's Number

• The mole is the unit for measuring the amount of a chemical substance.

• A mole contains exactly $N_A = 6.02\times 10^{23}$ entities (atoms, molecules, ions, etc.).

• 1 mole of a substance contains $N_A$ particles and has a mass equal to its molar mass in grams.

  • Examples:

  • 1 mole of H$<em>2$O contains $N</em>A$ molecules and has a mass of its molar mass (18 g for H$_2$O).

  • 1 mole of CO$<em>2$ has a molar mass of 44 g, i.e. $1\ ext{mol CO}</em>2 = 44\text{ g}$ and $6.02\times 10^{23}$ molecules.

  • 1 mole of Na = 23 g; $1\text{ mole Na} = 6.02\times 10^{23}$ atoms.

  • 1 mole of C = 12 g; 1\text{ mole C} = 6.02\times 10^{23}$ atoms.

  • 1 mole of NaCl = 58.5 g; contains 6.02\times 10^{23}$formula units.</p></li></ul></li><li><p>Avogadro's number (NA) is the number of constituent particles in one mole of a substance:$N_A = 6.02\times 10^{23}$.</p></li></ul><div data-type="horizontalRule"><hr></div><h3 id="422f29d4-681c-4e27-8358-112ac52f95db" data-toc-id="422f29d4-681c-4e27-8358-112ac52f95db" collapsed="false" seolevelmigrated="true">Molar Volume and STP</h3><ul><li><p>The volume occupied by 1 mole of a gaseous substance is called the molar volume.</p></li><li><p>At STP (Standard Temperature and Pressure), the molar volume is$V_m = 22.4\ \text{L mol}^{-1}$.</p></li><li><p>STP conditions:</p><ul><li><p>Temperature: 0°C (273.15 K)</p></li><li><p>Pressure: 1 atm (101.3 kPa)</p></li></ul></li></ul><div data-type="horizontalRule"><hr></div><h3 id="ef5b6ae6-7fe5-4784-8870-55623ba5e008" data-toc-id="ef5b6ae6-7fe5-4784-8870-55623ba5e008" collapsed="false" seolevelmigrated="true">Molar Mass and Unit Conversions</h3><ul><li><p>Molar mass, often denoted$M_m$, is the mass per mole of a substance (g mol$^{-1}$).</p></li><li><p>Key relation:</p><ul><li><p>$n = \frac{m}{M_m}$where$n$is amount in moles and$m$is mass in grams.</p></li><li><p>For gases at STP,$V = n \times V_m = n \times 22.4\ \text{L}$.</p></li></ul></li><li><p>Examples:</p><ul><li><p>1 mole CO$2$has mass 44 g (since$Mm(\mathrm{CO_2}) = 12 + 2\times 16 = 44\,\text{g mol}^{-1}$)).</p></li><li><p>1 mole NaCl has mass 58.5 g (Na = 23 g, Cl = 35.5 g).</p></li><li><p>1 mole Na = 23 g; 1 mole C = 12 g; 1 mole H$2$SO$4$= 98 g (H = 1, S = 32, O$_4$= 64).</p></li></ul></li></ul><div data-type="horizontalRule"><hr></div><h3 id="dcc62245-1aa4-406c-96a4-ad4fa96c3d45" data-toc-id="dcc62245-1aa4-406c-96a4-ad4fa96c3d45" collapsed="false" seolevelmigrated="true">Sample Calculations: Atoms, Molecules, and Volumes</h3><ul><li><p>Number of atoms in 5 g of H$2$SO$4$:</p><ul><li><p>M(H$2$SO$4$) = 98 g mol$^{-1}$.</p></li><li><p>Moles in 5 g:$n = \frac{5}{98} \approx 0.0510\,\text{mol}.$</p></li><li><p>Each molecule contains 7 atoms, so total atoms:</p></li><li><p>$N = n \times N_A \times 7 = 0.0510 \times 6.02\times 10^{23} \times 7 \approx 2.15\times 10^{23}\text{ atoms}.$</p></li></ul></li><li><p>Volume of water vapour corresponding to 100 g H$_2$O at STP:</p><ul><li><p>M(H$_2$O) = 18 g mol$^{-1}$.</p></li><li><p>Moles in 100 g:$n = \frac{100}{18} \approx 5.556\,\text{mol}.$</p></li><li><p>Volume:$V = n \times V_m = 5.556 \times 22.4 \approx 124.4\ \text{L}.$</p></li></ul></li><li><p>Number of molecules in 500 L of O$_2$:</p><ul><li><p>1 mole O$_2$at STP occupies 22.4 L.</p></li><li><p>Moles in 500 L:$n = \frac{500}{22.4} \approx 22.32\,\text{mol}.$</p></li><li><p>Molecules:$N = n \times N_A \approx 22.32 \times 6.02\times 10^{23} \approx 1.34\times 10^{25}\text{ molecules}.$</p></li></ul></li></ul><div data-type="horizontalRule"><hr></div><h3 id="31b3f147-3e27-4729-99ad-9f8fd3cc68f0" data-toc-id="31b3f147-3e27-4729-99ad-9f8fd3cc68f0" collapsed="false" seolevelmigrated="true">Molarity and Solutions</h3><ul><li><p>Molarity is the amount of solute per liter of solution:$M = \frac{n}{V}$with$n = \frac{m}{M_m}$.</p></li><li><p>Therefore,$M = \frac{m}{Mm \times V}$where$m$is mass of solute (g),$Mm$is molar mass (g mol$^{-1}$), and$V$is volume of solution (L).</p></li><li><p>Example:</p><ul><li><p>6 g NaOH dissolved in 200 mL (0.200 L) of solution.</p></li><li><p>Molar mass:$M_m(\mathrm{NaOH}) = 23 + 16 + 1 = 40\,\text{g mol}^{-1}$.</p></li><li><p>$M = \frac{6}{40 \times 0.200} = \frac{6}{8} = 0.75\,\text{M}.$</p></li></ul></li></ul><div data-type="horizontalRule"><hr></div><h3 id="5d7e4708-2887-48ff-85c7-e9166ab9a6a7" data-toc-id="5d7e4708-2887-48ff-85c7-e9166ab9a6a7" collapsed="false" seolevelmigrated="true">Limiting Reactant and Excess Reagents</h3><ul><li><p>Limiting reactant: the reactant that is consumed completely first; it limits the amount of product formed.</p></li><li><p>Steps to identify:<br>1) Convert all reactants to moles.<br>2) Use the balanced equation to find the mole ratio.<br>3) Determine which reactant produces the smallest amount of product (i.e., is limiting).</p></li><li><p>Example: HNO$3$+ NaOH → NaNO$3$+ H$_2$O</p><ul><li><p>Suppose 25 g HNO$_3$(M = 63 g/mol) and 25 g NaOH (M = 40 g/mol) are used.</p></li><li><p>Moles:$n{HNO3} = \frac{25}{63} \approx 0.398\,\text{mol}$;$n_{NaOH} = \frac{25}{40} = 0.625\,\text{mol}$.</p></li><li><p>Stoichiometry 1:1 → Limiting reactant is HNO$_3$.</p></li><li><p>Theoretical yield of NaNO$3$(molar mass 85 g/mol):$m{theo} = n{HNO3} \times 85 \approx 0.398 \times 85 \approx 33.8\,\text{g}.$</p></li><li><p>Excess NaOH remaining:$n_{excess} = 0.625 - 0.398 = 0.227\,\text{mol}$; mass leftover ≈$0.227 \times 40 \approx 9.1\,\text{g}.$</p></li></ul></li></ul><div data-type="horizontalRule"><hr></div><h3 id="6a122b5f-6c7f-40a0-bcbe-c2a09ba5d38a" data-toc-id="6a122b5f-6c7f-40a0-bcbe-c2a09ba5d38a" collapsed="false" seolevelmigrated="true">Percentage Yield</h3><ul><li><p>Definition: percentage yield = (actual yield / theoretical yield) × 100$3$→ CaO + CO$2$.</p></li><li><p>If 100 g of CaCO$3$decomposes completely, theoretical yields are: CaO (mass) = 56 g and CO$2$= 44 g (since molar masses: CaCO$3$≈ 100 g/mol; CaO ≈ 56 g/mol; CO$2$≈ 44 g/mol).</p></li><li><p>Suppose the actual amount of CaO obtained is 29 g.</p></li><li><p>$\%\text{yield} = \frac{29}{56} \times 100\% \approx 51.8\%.$</p></li></ul></li><li><p>Note: The theoretical yield depends on the limiting reagent; real processes often have yields < 100$3$-CH$2$-OH (ethanol) has molecular mass$M(\mathrm{C2H6O}) = 2\times 12 + 6\times 1 + 16 = 46\,\text{g mol}^{-1}$.</p><ul><li><p>Percentages for C, H, and O can be calculated similarly using atomic masses.</p></li></ul></li></ul><div data-type="horizontalRule"><hr></div><h3 id="58ce45f3-3324-41c5-9c00-bfe12f7e6fb2" data-toc-id="58ce45f3-3324-41c5-9c00-bfe12f7e6fb2" collapsed="false" seolevelmigrated="true">Empirical Formula vs Molecular Formula</h3><ul><li><p>Empirical formula: simplest whole-number ratio of atoms in a compound.</p></li><li><p>Molecular formula: actual number of each type of atom in a molecule; may be the same as the empirical formula or a multiple of it.</p></li><li><p>When to use:</p><ul><li><p>Empirical formula is obtained from percent composition or elemental analysis.</p></li><li><p>Molecular formula requires the molecular (molar) mass to determine the multiplier from the empirical formula.</p></li></ul></li><li><p>Key distinctions:</p><ul><li><p>An empirical formula may correspond to many different molecular formulas (isomers) with the same empirical formula but different molecular formulas.</p></li><li><p>Except for isobaric species, different compounds can share the same empirical formula but have different molecular formulas.</p></li></ul></li></ul><div data-type="horizontalRule"><hr></div><h3 id="81bf519f-e38b-4ad0-ab79-2be4c4a6e56b" data-toc-id="81bf519f-e38b-4ad0-ab79-2be4c4a6e56b" collapsed="false" seolevelmigrated="true">Isomerism (Isomers)</h3><ul><li><p>Isomers: Compounds with the same molecular formula but different structural arrangements.</p></li><li><p>Examples:</p><ul><li><p>Glucose: C$6$H${12}$O$6$(empirical formula CH$2$O, since the empirical ratio is 1:2:1 but actual molecule is C$6$H${12}$O$_6$).</p></li><li><p>Acetylene: C$2$H$2$(empirical formula CH).</p></li></ul></li><li><p>Takeaway:</p><ul><li><p>It is possible for different compounds to have the same empirical formula but different molecular formulas and arrangements; this is a key reason to determine molecular formula via molar mass in addition to empirical formula.</p></li></ul></li></ul><div data-type="horizontalRule"><hr></div><h3 id="c9b9dfa5-dc2c-4b2c-a523-7603a4c138fe" data-toc-id="c9b9dfa5-dc2c-4b2c-a523-7603a4c138fe" collapsed="false" seolevelmigrated="true">From Percent Composition to Empirical Formula and Molecular Formula</h3><ul><li><p>Given a compound with percent composition:</p><ul><li><p>Example 1: 36.8$_n$with$n = \frac{\text{Molecular mass}}{\text{Empirical mass}}.$</p></li></ul></li><li><p>Example outcome:</p><ul><li><p>If empirical formula is N$2$O$3$with empirical mass$M{emp} = (2\times 14) + (3\times 16) = 76\,\text{g mol}^{-1}$and the molecular mass is also 76 g/mol, then the molecular formula is N$2$O$_3$(x = 1).</p></li></ul></li></ul><div data-type="horizontalRule"><hr></div><h3 id="9f0d6ba4-689c-4f5d-8287-d2a411a14a9c" data-toc-id="9f0d6ba4-689c-4f5d-8287-d2a411a14a9c" collapsed="false" seolevelmigrated="true">A Worked Empirical Formula from Percent Composition</h3><ul><li><p>Problem: A sample contains 20 g C, 0.33 g H, and the rest O. Determine empirical formula.</p></li><li><p>Steps: 1) Assume a 100 g sample, so masses become: C = 20 g, H = 0.33 g, O = 100 - (20 + 0.33) = 79.67 g. 2) Convert to moles:</p><ul><li><p>n(C) = 20 / 12 ≈ 1.667 mol</p></li><li><p>n(H) = 0.33 / 1 ≈ 0.330 mol</p></li><li><p>n(O) = 79.67 / 16 ≈ 4.979 mol<br>3) Divide by the smallest number of moles (0.330):</p></li><li><p>C ≈ 1.667 / 0.330 ≈ 5.06 → ~5</p></li><li><p>H ≈ 0.330 / 0.330 = 1</p></li><li><p>O ≈ 4.979 / 0.330 ≈ 15.1 → ~15<br>4) Empirical formula ≈ C$5$H$1$O${15}$→ simplified to C$5$H$2$O${15}$is not ideal; in practice we seek simplest whole-number ratio; if needed, adjust rounding to nearest whole numbers that yield a simple ratio.</p></li></ul></li><li><p>Note: The rounding step may yield a simpler ratio such as C$5$H${10}$O$_{15}$if a consistent reduction is found; always verify by reducing the ratio to the smallest integers possible.</p></li></ul><div data-type="horizontalRule"><hr></div><h3 id="0d303364-d2c9-4937-ad9f-5eee875083c6" data-toc-id="0d303364-d2c9-4937-ad9f-5eee875083c6" collapsed="false" seolevelmigrated="true">Quick Reference: Key Formulas</h3><ul><li><p>Amount (moles) from mass:$n = \dfrac{m}{M_m}$.</p></li><li><p>Mass from amount:$m = n \times M_m$.</p></li><li><p>Molarity:$M = \dfrac{n}{V} = \dfrac{m}{M_m \ times V}$where$V$is volume in liters.</p></li><li><p>Molar Volume at STP:$V_m = 22.4\ \text{L mol}^{-1}$.</p></li><li><p>Avogadro's Number:$N_A = 6.02\times 10^{23}$.</p></li><li><p>Percentage yield:$\%\text{yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100\%.$</p></li><li><p>Percentage composition:</p><ul><li><p>$\%\text{element} = \dfrac{\text{mass of element in compound}}{\text{molar mass of compound}} \times 100\%.$</p></li></ul></li><li><p>Empirical vs Molecular formulas:</p><ul><li><p>Empirical formula gives the simplest whole-number ratio of atoms.</p></li><li><p>Molecular formula gives the actual number of each type of atom in a molecule; related to empirical by a multiplier$x$, where$M{ ext{mol}} = x \times M{ ext{emp}}.$</p></li></ul></li></ul><div data-type="horizontalRule"><hr></div><h3 id="64b766ad-b0c7-4ec3-802f-d10d21c57248" data-toc-id="64b766ad-b0c7-4ec3-802f-d10d21c57248" collapsed="false" seolevelmigrated="true">Summary of Key Concepts from Chapter 6</h3><ul><li><p>The mole and Avogadro's constant are the foundation for converting between microscopic particles and macroscopic masses.</p></li><li><p>Molar volume links gas volume to amount of substance at STP.</p></li><li><p>Molarity provides a way to express concentration in terms of moles per liter.</p></li><li><p>Stoichiometry (limiting reactant and yield) governs how much product is formed in reactions.</p></li><li><p>Percentage composition and empirical vs molecular formulas show how to deduce a compound’s composition and structure from data.</p></li><li><p>Isomers are compounds with the same molecular formula but different structures, leading to different properties.</p></li><li><p>Practical problems often require combining these concepts: mass, moles, volume, concentration, and formula calculations all in one.</p></li></ul><div data-type="horizontalRule"><hr></div><h3 id="e63adc58-94e8-4986-a7fd-af6541d1d7ad" data-toc-id="e63adc58-94e8-4986-a7fd-af6541d1d7ad" collapsed="false" seolevelmigrated="true">Quick Practice Problem Set (Selected Examples)</h3><ul><li><p>Calculate the number of atoms in 5 g of H$2$SO$4$.</p><ul><li><p>$Mm$(H$2$SO$4$) = 98 g/mol;$n = 5/98 \,\text{mol}$; atoms per molecule = 7; total atoms ≈$n \times NA \times 7 \approx 2.15\times 10^{23}$.</p></li></ul></li><li><p>Calculate the volume of 100 g of water vapour at STP.</p><ul><li><p>$n = 100/18 \approx 5.556$mol;$V = n \times 22.4 \approx 124.4$L.</p></li></ul></li><li><p>How many molecules are in 500 L of O$_2$at STP?</p><ul><li><p>$n = 500/22.4 \approx 22.32$mol; molecules ≈$n \times N_A \approx 1.34\times 10^{25}$.</p></li></ul></li><li><p>If 6 g NaOH is dissolved in 200 mL of solution, what is the concentration?</p><ul><li><p>$M_m$of NaOH = 40 g/mol.</p></li><li><p>$M = 6/(40 \times 0.200) = 0.75$M.</p></li></ul></li><li><p>In a reaction with limiting reactant, determine the theoretical yield and the mass of product formed when given masses of reactants.</p></li><li><p>For a decomposition of CaCO$3$with 100 g starting material, calculate the theoretical mass of CaO and CO$2$$ produced and the % yield if the actual CaO recovered is 29 g.

  The notes above consolidate the major and supporting ideas from the transcript, with key formulas rendered in LaTeX for clarity and exam readiness. You can use these as a consolidated study guide for Mole concepts and related quantitative techniques.