Page-by-Page Chemistry Notes: Atomic Theory to Percent Yield

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• Stoichiometry Topics Discussed: Atomic Theory, Average atomic mass, Mass Spectrometry, Avogadro's number, % composition, Empirical and molecular formulas, Stoichiometric calculations, limiting and excess reactants and percentage yield.

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Basic Atomic Theory

Basic Atomic Theory is a fundamental concept in chemistry and physics that explains the nature of matter in terms of atoms, the smallest units of an element that retain the properties of that element.

The three fundamental particles that make up an atom are the proton, neutron and electron.

a.m.u. = atomic mass units.

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Key Concepts and Terms in Atomic Theory

• Mass Number (A): The total number of protons and neutrons in the nucleus.

• Atomic Number (Z): The number of protons in the nucleus of an atom, which defines the element.

• Isotopes: Variants of elements with the same number of protons but different numbers of neutrons. Example: Carbon-12 and Carbon-14. Also noted: Copper-65 isotope as another example.

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Isotopes explained in detail

• Isotopes of carbon are similar in chemical properties because they have the same number of protons, but differ in neutron count which changes mass.

• Neutron count = Mass number - Number of protons.

Isotope data example:

• Protons: 6, 6

• Electrons: 6, 6

• Mass number: 12, 14

• Neutrons: 6, 8

This illustrates how two isotopes of the same element can have identical chemical behavior but different masses and neutron counts.

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What information can be determined from the periodic table?

• The number of protons and electrons for Mercury is 80 (equal to the atomic number).

• The Element Symbol and name.

• The average atomic mass is found by a calculation (to be learned later).

• Some periodic tables list the oxidation state, but this is not needed at this time.

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Atomic Structure example and try it problems

• Name, Nuclear Symbol, Atomic Number, Atomic Mass, # of Protons, # of Electrons, # of Neutrons, Isotopic Symbol.

• Examples given: Ti-48, Ca-42, etc.

• For illustration in the slide: Titanium-48 has Atomic Number 22 and Atomic Mass 48; Calcium-42 has Atomic Number 20 and Atomic Mass 42; Chlorine-35 has Atomic Number 17 and Atomic Mass 35; Gold-197 has Atomic Number 79 and Atomic Mass 197.

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Atomic Structure example and try it problems (Answers)

• Ti-48: Atomic Number 22, Atomic Mass 48, Protons 22, Electrons 22, Neutrons 26, Isotopic Symbol Ti-48.

• Ca-42: Atomic Number 20, Atomic Mass 42, Protons 20, Electrons 20, Neutrons 22, Isotopic Symbol Ca-42.

• Cl-35: Atomic Number 17, Atomic Mass 35, Protons 17, Electrons 17, Neutrons 18, Isotopic Symbol Cl-35.

• Au-197: Atomic Number 79, Atomic Mass 197, Protons 79, Electrons 79, Neutrons 118, Isotopic Symbol Au-197.

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Average Atomic Mass

• Average atomic mass is the weighted average of the atomic masses of the naturally occurring isotopes of an element. It considers both the mass and the relative abundance of each isotope.

• Example: The average atomic mass of Zinc is 65.38 amu, accounting for isotopes $^{64}$Zn, $^{66}$Zn, $^{67}$Zn, $^{68}$Zn, and $^{70}$Zn.

Formally:
$\bar{M} \,=\, \,\sum<em>i f</em>i M<em>i$ where $fi$ is the fractional (relative) abundance of isotope $i$ and $M_i$ is its mass.

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Relative abundance and weighted averages

• Relative abundance refers to the percentage of a naturally occurring isotope in nature.

• The percentages must add to 100%.

• If an element has 3 or more isotopes, the average mass is the sum of each $(\text{abundance})\times(\text{mass of isotope})$ terms.

• Example fragment from the table: Generally, the mass spectrum shows isotopes at different masses due to varying neutron numbers; peaks correspond to isotope masses with heights corresponding to relative abundances.

Note: The page includes a set of isotopes and their masses for various elements; the key idea is that the average atomic mass is a weighted sum based on isotopic abundances.

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Average atomic mass problem: Einsteinium

Es has two main isotopes:

• Es-252: mass = $252.08$ amu, relative abundance = $0.473$

• Es-255: mass = $255.10$ amu, relative abundance = $0.527$
  Calculate the average atomic mass:
  $\text{Average} = (0.473)(252.08) + (0.527)(255.10) = 253.67\ \,\text{amu}.$

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Average atomic mass problem (Answer)

• Reiterates the two-isotope calculation for Einsteinium:
  $\bar{M}_{Es} = (0.473)(252.08) + (0.527)(255.10) = 253.67\ \text{amu}.$

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Average atomic mass try it problem (Unknown X)

• Isotopes: X-107 with mass 106.905 amu, abundance 51.8%; X-109 with mass 108.905 amu, abundance 48.2%

• Calculate the average atomic mass:
  $\bar{M}_X = (0.518)(106.91) + (0.482)(108.91) = 107.87\ \text{amu}.$

• Identify the element on the periodic table: 107.87 amu corresponds to Silver (Ag).

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Average atomic mass try it problem (Answer)

• Final result: ${\bar{M}} = 107.87\ \text{amu}$ and the element is Silver (Ag).

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Mass Spectrometry What does a mass spectrometer do?

• It measures mass more accurately than other techniques, especially for isotopes of elements.

• It can provide information about chemical structures.

• Schematic of a Mass Spectrometer (described): a modern mass spectrometer separates ions by mass-to-charge ratio and detects isotopes.

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What information can a mass spectrum tell us?

• A mass spectrum reveals the presence and relative abundance of isotopes in a sample.

• Peaks appear at different masses due to different numbers of neutrons in isotopes.

• Example: In a given spectrum, 60% of the sample has mass 69 and 40% has mass 71 (illustrative for Ga).

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Mass spectra overview and reference information

• The mass spectra of elements cover the entire periodic table; the slide includes a dense figure showing isotopic peaks for many elements and their detection limits.

• The key takeaway is that mass spectra provide isotope masses and relative abundances, enabling us to read off isotopic composition from peaks.

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AP exam questions on mass spectra

• The three common types of questions: (1) choose a mass spectrum for a given element, (2) identify the element from a mass spectrum, (3) identify an isotope from a mass spectrum.

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Practice AP MCQ #1 (Mass spectrum interpretation)

• Question: The mass spectrum of Sr is most likely represented by which figure (Option A–D).

• The concept tested: matching isotope masses and relative abundances to an element.

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Practice AP MCQ #1 Answer

• Answer: The correct option is the one that corresponds to Sr’s natural isotopic pattern (interpretation of spectra).

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Practice AP MCQ #2

• Question: Which of the following elements is represented by the spectrum: Y, Zr, Nb, Mo.

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Practice AP MCQ #2 Answer

• The correct element is determined by the isotope pattern shown in the spectrum.

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Practice AP MCQ #3

• Question: The mass spectrum of a pure element shows a peak at 63 amu. This represents an isotope of which element? Options: Eu (63 neutrons), Gd (64 neutrons), Cu (34 neutrons), Zn (33 neutrons).

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Practice AP MCQ #3 Answer

• Answer corresponds to the element whose isotope mass aligns with 63 amu and neutron count matches. (See slide for exact mapping.)

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FRQ Try it Problem 2025 Exam Q#1 part (a)

1. Magnesium mass spectrum problem (incomplete spectrum is shown):

• Given: ${}^{24}$Mg abundance = 79%; other two isotopes Mg-25 and Mg-26 having approximately equal abundances.
  i) Complete the spectrum by drawing lines for Mg-25 and Mg-26.
  ii) Describe the difference in atomic structure that accounts for the mass difference between Mg-25 and Mg-26.

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FRQ Try it Problem 2025 Exam Q#1 part (a) ANSWERS

• Part (i): The lines for Mg-25 and Mg-26 should be at masses around 25 and 26 amu with relative abundances roughly between 10 and 11% each.

• Part (ii): Mg-26 has one more neutron than Mg-25, accounting for the mass difference.

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Avogadro’s number and the mol

• Avogadro’s number: $N_A = 6.02 \times 10^{23}\;
  mol^{-1}$.

• Defined as the number of particles in 1 mole of a substance.

• Named after Amedeo Avogadro (1776-1856).

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Learning to use Avogadro’s number (moles to particles)

Example: Calculate the number of atoms in 3.04 moles of solid Lithium (Li).

• Process: Moles to particles using $N_A$.

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Avogadro’s number (moles to particles) – Answer

• Calculation steps and final result for the example in Page 27.

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Learning to use Avogadro’s number (moles to particles) – Try it

• Problem: Calculate the number of molecules in 0.732 moles of CO₂ gas.

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Avogadro's number – Try it Answer

• Solution: $0.732\ ext{mol}\ \mathrm{CO_2} \times 6.02\times 10^{23}\ ext{molecules/mol} = 4.41\times 10^{23}$ molecules of CO₂.

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Learning to use Avogadro’s number (particles to moles)

• Try It Problem: Calculate the number of moles in $7.95\times 10^{23}$ atoms of Zn.

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Avogadro’s number (particles to moles) – Answer

• Calculation: $7.95\times 10^{23} \text{ atoms Zn} \times \dfrac{1\text{ mol Zn}}{6.02\times 10^{23}\text{ atoms}} = 1.32\ \text{moles Zn}$

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2-Step problems involving Avogadro’s number and molar mass flow charts

• Mass to particles: mass → mol × molar mass → particles via $N_A$.

• Particles to mass: particles → mol × molar mass.

• Includes: Molar mass can be a formula mass; 1 mol = $6.02\times10^{23}$ particles.

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2-Step problems involving Avogadro’s number and molar mass – Try It

Problem: Calculate the number of grams in a sample of $1.41\times 10^{24}$ atoms of Na.

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2-Step problems – Try It Answer

• Calculation steps yield grams of Na corresponding to the given number of atoms, using:

  • $1\text{ mol Na} = 6.02\times 10^{23}\text{ atoms}$ and molar mass of Na = 22.99 g.

• Result: 53.8 g Na.

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AP MCQ Practice involving Avogadro's number

• Question: Which numerical expression gives the number of particles in 15.0 g of P?

• Options include combinations of Avogadro's number, molar mass, and mass.

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AP MCQ Practice involving Avogadro's number – Answer

• Correct expression is the one that multiplies the mass-to-moles ratio by Avogadro’s number, i.e., $\dfrac{15.0\text{ g}}{30.97\text{ g/mol}} \times 6.02\times10^{23}$.

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AP FRQ Practice – Neon isotopes (information learned thus far)

• The average atomic mass of naturally occurring neon is 20.18 amu.

• Table shows two common isotopes: Ne-20 (mass 19.99 amu) and Ne-22 (mass 21.99 amu).

• (a) Determine the number of protons and neutrons in Ne-22.

• (b) Use information to calculate the percent abundances of each isotope.

• (c) Calculate moles of Ne-22 in a 12.55 g sample.

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AP FRQ Practice – Neon Answers (part a)

• Ne-22 has 10 protons and 12 neutrons (Z = 10; N = 12).

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AP FRQ Practice – Neon Answers (part b)

• Percent abundances to be calculated using the weighted average: let $a$ be the abundance of Ne-20 and $b$ be that of Ne-22, with $a+b=1$ and $19.99a + 21.99b = 20.18$.

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AP FRQ Practice – Neon Answers (part c)

• Moles of Ne-22 in 12.55 g sample: $n = \dfrac{12.55\text{ g}}{20.18\text{ g/mol}}$ is the total moles; You then multiply by the Ne-22 fraction to get moles of Ne-22.

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Percent Composition by Mass

• Definition: Percent Composition by mass shows how much of an element is in a compound in percentages; also used to determine the mass of an element in a sample.

• General formula:
  $\%\text{ Composition} = \frac{\text{mass of part}}{\text{mass of whole}} \times 100$

• Note: This formula is not on the AP equation sheet.

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Percent Composition by Mass – Problem solving steps

1) Determine the molar mass of the compound (the mass of whole).
2) Apply the general formula for each element using the total mass of the element in the compound.
3) Determine the mass of an element in a sample (if necessary) and multiply the percentage of each element (as a decimal) by the sample mass.

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Percent Composition by Mass – Example Problem

• Example: Calculate the percent composition for Lithium hydroxide, LiOH.

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Percent Composition by Mass – Example Problem (Answer)

• Compute:

  • Li mass fraction: 6.941 g Li in 23.948 g LiOH → 28.98%

  • O mass fraction: 16.00 g O in 23.948 g LiOH → 66.81%

  • H mass fraction: 1.008 g H in 23.948 g LiOH → 4.21%

• For a 68.2 g sample, mass of Li = 0.2898 × 68.2 = 19.76 g; mass of O = 0.6881 × 68.2 = 45.56 g; mass of H = 0.0421 × 68.2 = 2.87 g.

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Percent Composition – Try It Problem

• Problem (a): Calculate the percent composition for Copper(II) Chloride, CuCl₂.

• (b): Calculate how many grams of Copper are in a 452 g sample of CuCl₂.

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Percent Composition – Try It Answer

• (a) % Cu = (63.55 g Cu / 134.45 g CuCl₂) × 100 = 47.27%

• % Cl = (70.90 g Cl / 134.45 g CuCl₂) × 100 = 52.73%

• (b) grams Cu in 452 g CuCl₂: 452 × 0.4727 = 213.66 g Cu

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AP MCQ – % composition Practice #1

• Question: The mass percent of carbon in CH₄ (methane) is 75%. An impure sample contains impurities that reduce this percentage. Which impurity could account for the lower percent carbon? Options: (A) C₃H₈, (B) C₅H₁₂, (C) H₂O, (D) C₂H₆

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AP MCQ Practice #1 Answer

• Answer: (C) H₂O. Water contains no carbon, so introducing it into the sample lowers the overall percent carbon.

• Rationale: Options A, B, D all contain carbon and would raise or maintain the carbon percent, not reduce it.

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AP MCQ – % composition Practice #2

• Question: A 25.0 g sample of AlBr₃ may contain impurities. Which quantities are needed to determine purity of AlBr₃?

• Options: A) mass of Al and Br in the sample; B) mass of Br in sample only; C) number of moles of Br in sample only; D) color and density of the sample.

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AP MCQ Practice #2 Answer

• Answer: A) The mass of Al and the mass of Br in the sample. From these, the moles of each can be determined and compared to the expected 1:3 ratio in AlBr₃ to assess purity.

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Empirical Formulas

• Definition: The lowest (simplest) whole-number ratio of the atoms of the elements in a compound. Example: Glucose molecular formula is C₆H₁₂O₆, empirical formula is CH₂O.

• Try It: Determine the empirical formula of Co₂C₈O₈.

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Steps to determine an empirical formula

1) Assume 100 g from percent composition data (if needed).
2) Convert to moles for each element.
3) Divide by the smallest number of moles (keep at least 4 digits).
4) If whole numbers are not obtained in step 3, multiply by the smallest whole number that yields whole numbers for each element.
5) Write the empirical formula.

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Empirical Formula example problem (from % composition data)

• Determine the empirical formula for a compound with 76.0% Zn and 24.0% P.

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Empirical Formula example problem – Answer (Zn and P)

• 76.0 g Zn → 1.162 mol Zn; 24.0 g P → 0.775 mol P.

• Ratio: Zn 1.162 / 0.775 ≈ 1.5 ≈ 3:2.

• Empirical formula: Zn₃P₂.

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Empirical Formula try it problem (mass data)

• Problem: 33.5 g Fe and 12.8 g O.

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Empirical Formula try it problem – Answers

• Fe: 33.5 g → 0.600 mol Fe.

• O: 12.8 g → 0.800 mol O.

• Ratio: Fe:O = 0.600:0.800 = 3:4.

• Empirical formula: Fe₃O₄.

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Practice Try It AP MCQ

• A 23.0 g sample contains 12.0 g C, 3.0 g H, and 8.0 g O. Which is the empirical formula?

• Options: (A) CHO, (B) C₂H₆O, (C) C₃H₉O₂, (D) C₄H₁₂O₂

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Practice Try It AP MCQ Answer

• Correct option: (B) C₂H₆O, determined by converting masses to moles and finding the simplest whole-number ratio.

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FRQ Try it Problem 2025 Exam Q#2 part a

2. Vitamin C (ascorbic acid) CHO combustion:

• Products: 0.2400 mol CO₂ and 2.883 g H₂O.
  i) Moles of H₂O produced.
  ii) If C:O ratio is 1:1 in ascorbic acid, determine empirical formula from A(i).

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FRQ Try it Problem 2025 Exam Q#2 part a – Answers

• (i) Moles of H₂O: 2.883 g H₂O × (1 mol H₂O / 18.02 g) = 0.1600 mol H₂O.

• (ii) 0.1600 mol H₂O × (2 mol H / 1 mol H₂O) = 0.3200 mol H.

• Empirical formula example: Using C contribution, C = 0.2400 mol, H = 0.3200 mol, O = 0.2400 mol. Simplified to C₃H₄O₃.

• Therefore, empirical formula is C₃H₄O₃.

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Molecular Formulas

• Definition: The actual number of each kind of atom in one molecule.

• Empirical and molecular formulas can be the same; If not, molecular formula is a whole-number multiple of the empirical formula and greater than one.

• Example: CaCl₂ is both empirical and molecular.

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Steps to determine the molecular formula

1) Determine the empirical formula (if needed).
2) Determine the ratio of the molecular mass to the empirical formula mass: $n = \frac{\text{molecular formula mass}}{\text{empirical formula mass}}$
3) Multiply the empirical formula by $n$ to obtain the molecular formula. Note: $n$ must be an integer ≥ 1 and masses should be provided.

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Molecular Formula Try it Problem

• Determine empirical and molecular formula for a compound with 43.65% P and 56.35% O, molar mass 283.88 g/mol.

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Molecular Formula Try it Problem – Answer (Slide #1)

• Empirical formula from percentages: P₂O₅.

• Molar mass of P₂O₅: $2\times30.97 + 5\times16.00 = 141.94$ g/mol.

• Ratio $n = \dfrac{283.88}{141.94} = 2$.

• Molecular formula: $\text{P}2\text{O}5$ multiplied by 2 → $\text{P}4\text{O}{10}$.

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Molecular Formula Try it Problem – Answer (Slide #2)

• Final molecular formula: $\text{P}4\text{O}{10}$.

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Reaction Stoichiometry

• Greek for “measuring elements.” Pronounced “sto-kee-ahh-muh-tree.”

• Defined as calculations of the quantities in chemical reactions, based on a balanced equation.

• There are many ways to interpret a balanced chemical equation.

• KEY Vocab: “A reaction that goes to completion.”

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Stoichiometric flowcharts (Helpful steps)

Given: Grams of substance A.

• Step 1: Use 1 molar mass of A to convert to moles of A.

• Step 2: Use coefficients of A and B from the balanced equation to find moles of B.

• Step 3: Use 1 molar mass of B to convert to grams of B.

• Mole ratio = ratio of coefficients in the balanced equation.

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Stoichiometry Example Problem

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

• How many grams of CO₂ are produced when 20.0 g of propane (C₃H₈) are burned (excess O₂)?

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Stoichiometry Example Problem – Answer

• Reaction: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

• 20.0 g C₃H₈ × (1 mol C₃H₈ / 44.094 g C₃H₈) × (3 mol CO₂ / 1 mol C₃H₈) × (44.01 g CO₂ / 1 mol CO₂) = 59.9 g CO₂

• Note: The answer should be reported with 3 significant figures.

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Stoichiometry Try It Problem

• 2022 FRQ Q#1 Part (a): 0.300 g of methyl salicylate (C₈H₈O₃) reacts with stoichiometric base to form salicylic acid crystals (HC₇H₅O₃).

• (a) For every 1 mole of C₈H₈O₃ (molar mass 152.15 g/mol) that reacts, 1 mole of salicylic acid crystals (HOC₇H₅O₃, molar mass 138.12 g/mol) is produced. Calculate the maximum mass of HC₇H₅O₃ that could be produced.

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Stoichiometry Try It Problem – Answer

• Theoretical yield based on 0.300 g C₈H₈O₃ is:

  • Moles of C₈H₈O₃ = 0.300 / 152.15 ≈ 0.001974 mol.

  • Mass of HC₇H₅O₃ = 0.001974 mol × 138.12 g/mol ≈ 0.2727 g.

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Limiting and Excess Reactants

• The limiting reactant is entirely consumed when a reaction goes to completion.

• The limiting reactant determines the maximum amount of product that can be produced.

• Stoichiometrically proportionate: once one reactant is used up, the reaction stops.

• Any problem with starting amounts for more than one reactant is a limiting reactant problem.

• AP Hint: Be aware of limiting reactant in disguise!

Page 74

Limiting and Excess Reactants – Particle Diagram Example

• Reaction: H₂(g) + I₂(g) → 2HI(g)

• Draw the cylinder after the reaction has gone to completion (constant volume and temperature).

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Limiting and Excess Reactants – Particle Diagram Answer

• Rationale: 2 moles I₂ produce 4 moles HI; 3 moles H₂ produce 6 moles HI.

• Therefore 4 moles HI are produced and I₂ is the limiting reactant and fully consumed; 4 atoms of H₂ used, leaving 2 H₂ atoms left over (one H₂ molecule).

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Limiting and Excess Reactants – Try It Problem

• Reaction: 3H₂(g) + N₂(g) → 2NH₃(g)

• Draw the cylinder after the reaction has gone to completion (constant volume and temperature).

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Limiting and Excess Reactants – Try It Answer

• Rationale: 2 moles of N₂ would produce 4 moles NH₃; 8 moles H₂ would produce 5.33 moles NH₃.

• Therefore 4 moles NH₃ are produced and N₂ is the limiting reactant and is fully consumed; 12 atoms of H₂ are used, leaving 4 atoms left over (two H₂ molecules).

Page 78

Limiting and Excess Reactants – Math Example Problem

• Reaction: 4 Al + 3 O₂ → 2 Al₂O₃

• Given: 12.5 g Al and 22.5 g O₂

• Task: Determine the limiting reactant and calculate moles of Al₂O₃ formed.

Page 79

Limiting and Excess Reactants – Math Example Answer

• Reaction: 4 Al + 3 O₂ → 2 Al₂O₃

• 12.5 g Al → moles Al = 12.5 / 26.98 = 0.463 mol

• 22.5 g O₂ → moles O₂ = 22.5 / 32.00 = 0.703 mol

• Theoretical Al₂O₃ formed from Al: (0.463 mol Al) × (2 mol Al₂O₃ / 4 mol Al) = 0.232 mol Al₂O₃

• From O₂: (0.703 mol O₂) × (2 mol Al₂O₃ / 3 mol O₂) = 0.468 mol Al₂O₃

• Limiting reactant: Al (consumed completely), O₂ is in excess; amount of Al₂O₃ formed = 0.232 mol.

Page 80

Limiting and Excess Reactants – Math Try It Problem

• Reaction: 2 Mg + O₂ → 2 MgO

• Given: 5.00 g Mg and 10.0 g O₂

• Task: Determine the limiting reactant and mass of MgO produced.

Page 81

Limiting and Excess Reactants – Math Try It Answer

• Moles: Mg: 5.00 g / 24.31 g/mol = 0.206 mol

• O₂: 10.00 g / 32.00 g/mol = 0.3125 mol

• Stoichiometry: 2 Mg per 1 O₂; limiting: Mg is limiting because 0.206 mol Mg would need 0.103 mol O₂ to react, leaving excess O₂.

• MgO produced: 0.206 mol Mg × (2 mol MgO / 2 mol Mg) = 0.206 mol MgO

• Mass MgO: 0.206 mol × 40.31 g/mol ≈ 8.29 g MgO

• Conclusion: Limiting reactant is Mg; O₂ is in excess; mass of MgO ≈ 8.29 g.

Page 82

Percentage Yield

• Formula: $\%\text{ yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$

• It measures reaction efficiency. Yields are often < 100% due to impurities, side reactions, or incomplete reactions. (Note: This equation is not on the AP equation sheet.)

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Percentage yield key terms

• Theoretical yield: maximum amount of product from complete conversion of reactants (found by stoichiometric calculation).

• Actual yield: what is obtained experimentally (often less than theoretical).

• Theoretical yield is calculated from the amount of limiting reactant.

Page 84

Percentage yield – Example problem

• Reaction: 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

• 7.50 g NH₃ reacted with excess O₂; If 4.30 g H₂O is obtained, calculate percent yield.

Page 85

Percentage yield – Example problem (Answer)

• Determine theoretical yield of H₂O:

  • 7.50 g NH₃ × (1 mol NH₃ / 17.034 g) × (6 mol H₂O / 4 mol NH₃) × (18.016 g H₂O / 1 mol) = 11.9 g H₂O

• Percent yield: $\%\text{ yield} = \dfrac{4.30}{11.9} \times 100 = 36.1\%$

Page 86

Percentage yield – Try it problem

• Reaction: 3H₂ + N₂ → 2NH₃; 4.5 mol H₂ + excess N₂; If 45.7 g NH₃ is obtained, calculate percent yield.

Page 87

Percentage yield – Try it problem (Answer)

• Theoretical NH₃ yield from 4.5 mol H₂: 4.5 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 3.0 mol NH₃

• Convert to grams: 3.0 mol NH₃ × 17.034 g/mol ≈ 51.1 g NH₃ (rounded appropriately as shown in the slide).

• Percent yield: $\%\text{ yield} = \dfrac{45.7}{51.0\text{ (approx)}} \times 100 ≈ 89.6\%$

Page 88

End of Unit!

• Summary of key themes:

  • Atomic theory and isotopes establish the foundation for understanding mass, composition, and reactions.

  • Average atomic mass combines masses with relative abundances: $\bar{M} = \sum<em>i f</em>i M_i.$

  • Mass spectrometry reveals isotope presence and abundances; spectra show peaks at isotope masses.

  • Avogadro’s number links moles to particles: $N = n \times N<em>A, \quad n = \dfrac{N}{N</em>A}.$

  • Percent composition uses the mass fraction of each element in a compound: $\% \text{Comp}<em>i = \dfrac{\text{mass of part}</em>i}{\text{mass of whole}} \times 100\%.$

  • Empirical formulas reflect simplest whole-number ratios; molecular formulas are whole-number multiples of empirical forms.

  • Stoichiometry uses balanced equations to relate mass, moles, and grams of reactants and products; limiting and excess reactants determine maximum product.

  • Percentage yield assesses reaction efficiency and can indicate practical issues in a lab.

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