Quantifying Atoms and Compounds

Key Knowledge and Study Objectives

• Relative Isotopic Masses: Investigation of the mass of isotopes relative to the scale where the carbon-12 ($^{12}C$) isotope is assigned a value of exactly 12.
• Mass Spectrometry: Determination of the relative atomic mass of an element using a mass spectrometer (specific instrument details are not required).
• Avogadro’s Constant ($N_A$): Defined as current value $6.02 \times 10^{23}$, representing the number of atoms or molecules in one mole of any substance.
• Moles and Mass: Determination of the amount in moles ($n$) of atoms or molecules in a pure sample of known mass ($m$).
• Compound Analysis: Determination of the molar mass ($M$) of compounds, the percentage composition by mass of covalent compounds, and the empirical and molecular formulas based on percentage composition by mass.

Relative Isotopic Mass and the Carbon-12 Scale

• Definition: Relative Isotopic Mass (RIM) is the mass of an atom of an isotope relative to the mass of an atom of the carbon-12 isotope.
• Standard Scale: The Carbon-12 isotope ($^{12}C$) is assigned a mass of exactly $12$ unified atomic mass units (symbol $u$).
• Unified Atomic Mass Unit (u): Defined as $1\,u = \frac{1}{12} \times \text{mass of an atom of } {}^{12}C$.
• Examples of RIM Calculations:     * For Hydrogen: ${}^1H = \frac{1}{12} \times 12 = 1\,u$     * For Iron: ${}^{26}Fe = \frac{26}{12} \times 12 = 26\,u$

Measuring Atomic Mass via Mass Spectrometry

• Instrument: A Mass Spectrometer is used to determine the isotopic composition of elements.
• Process Overview:     * A sample ($M$) is introduced.     * The sample is ionized to form positive ions ($M^+$).     * The ionized sample passes through a magnet which deflects ions based on their mass-to-charge ($m/z$) ratio.     * Low mass ions: Experience high deflection.     * High mass ions: Experience low deflection.     * A collector detects the ions, and the system is connected to a vacuum pump.
• Information Obtained from Mass Spectrum:     * Number of peaks: Indicates the number of isotopes present.     * Position of each peak (m/z): Indicates the relative mass of the isotopes.     * Height of the peaks: Indicates the relative abundance of the isotopes.
• Mass Spectrum Graph: A plot of the relative abundance of isotopes ($\%$) versus the charge-to-mass ratio ($m/z$).

Relative Atomic Mass ($A_r$)

• Definition: The weighted average of all the relative isotopic masses for the isotopes that make up the naturally occurring isotopic mixture of an element.
• Formula:     * $A_r = \frac{(\text{RIM}_1 \times \%\text{ abundance}_1) + (\text{RIM}_2 \times \%\text{ abundance}_2) + \dots}{100}$
• Sample Problem 1 (Lithium):     * Isotope 1: ${}^6Li$; RIM = $6.02$; Abundance = $7.42\%$     * Isotope 2: ${}^7Li$; RIM = $7.02$; Abundance = $92.58\%$     * Calculation: $A_r = \frac{(6.02 \times 7.42) + (7.02 \times 92.58)}{100}$
• Sample Problem 2 (Copper - Composition):     * Lighter isotope: RIM = $62.95\,u$; Abundance = $69.2\%$     * Heavier isotope: RIM = $64.95\,u$     * Calculation: Determine the abundance of the heavier isotope ($100\% - 69.2\% = 30.8\%$) and apply the $A_r$ formula.
• Sample Problem 3 (Copper - Percentage Abundance):     * ${}^{63}Cu$ RIM = $62.95$     * ${}^{65}Cu$ RIM = $64.95$     * $A_r$ of copper = $63.54$     * Goal: Solve for the unknown percentages using the $A_r$ formula.

Relative Masses for Molecules and Ionic Compounds

• Relative Formula Mass: The sum total of relative atomic masses of all atoms in the formula of an ionic compound.
• Relative Molecular Mass ($M_r$): The sum total of relative atomic masses of all atoms in the formula of a covalent (molecular) compound.

Avogadro's Constant and the Mole ($n$)

• Definition of the Mole: A counting unit similar to a dozen, but specialized for chemistry.
• Avogadro's Number ($N_A$): $6.02 \times 10^{23}$ (602 billion trillion particles).     * Standard scientific notation: $6.02 \times 10^{23}$.
• Unit: The unit for the mole is the "mol".
• Scale and Magnitude Metaphors:     * A mole of soft drink cans would cover the Earth's surface to a depth of over $200\,miles$.     * A mole of unpopped popcorn kernels spread across the USA would cover the country to a depth of over $9\,miles$.     * Counting atoms at a rate of $10\text{ million/second}$ would take approximately $2\text{ billion years}$ to count one mole.
• Conceptual Understanding:     * $1\text{ dozen cookies} = 12\text{ cookies}$     * $1\text{ mole of cookies} = 6.02 \times 10^{23}\text{ cookies}$     * Note: While the number of particles in a mole is always constant ($N_A$), the mass varies depending on the substance.
• Substance Examples:     * $1\,mol$ of $C = 6.02 \times 10^{23}\text{ C atoms}$     * $1\,mol$ of $Ag = 6.02 \times 10^{23}\text{ Ag atoms}$     * $1\,mol$ of $CO_2 = 6.02 \times 10^{23}\text{ } CO_2 \text{ molecules}$
• Molar Ratios in Compounds:     * $1\,mol$ of $CO_2$ contains $1\,mol$ of carbon ($C$) atoms and $2\,mol$ of oxygen ($O$) atoms.     * Equating to particles: $1 \times 6.02 \times 10^{23}\text{ C atoms}$ and $2 \times 6.02 \times 10^{23}\text{ O atoms}$.

Molar Calculations: Number of Particles

• Formula for Moles from Particles:     * $n = \frac{N}{N_A}$     * Where $n$ is the amount in moles, $N$ is the known number of particles, and $N_A$ is Avogadro's number ($6.02 \times 10^{23}$).
• Sample Problems:     * Number of atoms in $0.500\,mol$ of Al: $3.01 \times 10^{23}\text{ Al atoms}$.     * Amount in mol of $1.8 \times 10^{24}\text{ S atoms}$: $n = \frac{1.8 \times 10^{24}}{6.02 \times 10^{23}} = 3.0\,mol\,S\text{ atoms}$.     * Oxygen molecules in $2.5\,mol$ of $O_2$ gas: Calculate molecules using $N = n \times N_A$.     * Amount in mol of $3.01 \times 10^{23}$ copper atoms: Calculate moles using $n = \frac{N}{N_A}$.     * Amount in mol of oxygen atoms in $5.0\,mol$ of oxygen molecules ($O_2$): Since $1\,mol\,O_2$ contains $2\,mol\,O$, $5.0\,mol\,O_2$ contains $10\,mol\,O\text{ atoms}$.

Molar Mass ($M$)

• Definition: The mass of one mole of an element or molecule.
• Equivalencies:     * Mass of $1\,mol$ of an element = Relative Atomic Mass ($A_r$) expressed in grams.     * Mass of $1\,mol$ of a molecule = Relative Molecular Mass ($M_r$) expressed in grams.
• Unit: $g/mol$ or $g\,mol^{-1}$.
• Examples:     * Molar Mass of Sodium ($Na$): $A_r = 23.0$, therefore $M = 23.0\,g\,mol^{-1}$.     * Molar Mass of Water ($H_2O$): $(2 \times 1.0) + (1 \times 16.0) = 18.0\,g$. Therefore $M = 18.0\,g\,mol^{-1}$.     * Prozac ($C_{17}H_{18}F_3NO$): Used as an antidepressant; calculations require summing the relative atomic masses of all constituent atoms to find the mass of $1.0\,mol$.
• Practice Calculations for $M_r$:     * a. $SO_2$     * b. $Mg(OH)_2$     * c. $NH_4NO_3$

Molar Calculations: Mass and Moles

• Formula for Moles from Mass:     * $n = \frac{m}{M}$     * Where $n$ is moles, $m$ is mass in grams ($g$), and $M$ is molar mass in $g/mol$.
• Sample Problem 1 (Aluminum Bicycle Frame):     * Determine grams of $Al$ in $3.00\,moles$ of $Al$.     * Rearrange formula: $m = n \times M$.
• Sample Problem 2 (Aspartame):     * Formula: $C_{14}H_{18}N_2O_5$     * Calculate moles present in $225\,g$ of aspartame using $n = \frac{m}{M}$.

Conversion Between Mass, Moles, and Particles

• Unified Relationships:     * $n = \frac{m}{M}$     * $n = \frac{N}{N_A}$     * Combined equality: $\frac{m}{M} = \frac{N}{N_A}$
• Conversion Logic:     * To go from Mass (Grams) to Moles: Divide by Molar Mass ($M$).     * To go from Moles to Mass (Grams): Multiply by Molar Mass ($M$).     * To go from Moles to Particles (Atoms/Molecules): Multiply by $6.02 \times 10^{23}$.     * To go from Particles to Moles: Divide by $6.02 \times 10^{23}$.
• Sample Problems:     * Atoms of $Cu$ in $35.4\,g$ of $Cu$: First calculate moles ($n = \frac{35.4}{A_r(Cu)}$), then multiply by $N_A$.     * Mass of $1.20 \times 10^{24}$ molecules of Glucose ($C_6H_{12}O_6$): First calculate moles ($n = \frac{N}{N_A}$), then multiply by $M(C_6H_{12}O_6)$.     * Atoms of oxygen in $66.05\,g$ of ammonium sulfate $(NH_4)_2SO_4$:         1. Calculate moles of $(NH_4)_2SO_4$ using $n = \frac{m}{M}$.         2. Identify there are $4\,mol$ of $O$ atoms per mole of compound.         3. Multiply moles of compound by $4$ to find moles of $O$.         4. Multiply moles of $O$ by $N_A$ to find the number of oxygen atoms.

Percentage Composition

• Definition: The percentage by mass of each element in a compound.
• Formula:     * $\% \text{ of } x \text{ in compound} = \frac{\text{mass of } x \text{ in 1 mol of compound}}{\text{molar mass of compound}} \times 100$
• Sample Problem (Glutamic Acid):     * Formula: $C_5H_8NO_4$ (used to make Monosodium Glutamate - MSG).     * Calculate the percent carbon ($C$) by taking the mass of $5$ moles of carbon and dividing it by the total molar mass of the molecule, then multiplying by $100$.