Chemistry

LessonAbout 1 h2 of 2 in Chemistry

Chemistry is the study of matter, its properties, composition, structure, and the changes it undergoes.

1. Matter and Thermometric Calculations

Matter and thermometric calculations involve physical measurements, units of volume and mass, and temperature conversions.

1.1 Properties of Matter and Density

Density: mass per unit volume of a substance.
Mass: quantitative measure of matter in a sample.
Volume: three-dimensional space occupied by matter.

Density Formula

\[ d = \dfrac{m}{V} \]

Example. A piece of platinum metal with a density of \(21.5\ \text{g/cm}^3\) has a volume of \(4.49\ \text{cm}^3\). Calculate its mass in grams, given the density \(d = 21.5\ \text{g/cm}^3\).

Solution. Rearrange the density formula to solve for mass:

\[ m = d \times V = (21.5\ \text{g/cm}^3)(4.49\ \text{cm}^3) = 96.535\ \text{g} \approx 96.5\ \text{g} \]

Answer: \(96.5\ \text{g}\)

1.2 Temperature Scales and Conversions

Celsius scale: scale based on water freezing at \(0^\circ\text{C}\).
Fahrenheit scale: scale based on water freezing at \(32^\circ\text{F}\).
Kelvin scale: absolute thermodynamic temperature scale starting at absolute zero.

Temperature Conversions

\[ T_{^\circ\text{F}} = \dfrac{9^\circ\text{F}}{5^\circ\text{C}} (T_{^\circ\text{C}}) + 32^\circ\text{F} \]

\[ T_{^\circ\text{C}} = (T_{^\circ\text{F}} - 32^\circ\text{F}) \times \dfrac{5^\circ\text{C}}{9^\circ\text{F}} \]

\[ T_{\text{K}} = T_{^\circ\text{C}} + 273.15 \]

Example. The melting point of lead is \(327.5^\circ\text{C}\). Convert this temperature to degrees Fahrenheit, given \(T_{^\circ\text{C}} = 327.5^\circ\text{C}\).

Solution. Apply the Celsius-to-Fahrenheit conversion equation:

\[ T_{^\circ\text{F}} = \dfrac{9^\circ\text{F}}{5^\circ\text{C}}(327.5^\circ\text{C}) + 32^\circ\text{F} = 589.5^\circ\text{F} + 32^\circ\text{F} = 621.5^\circ\text{F} \]

Answer: \(621.5^\circ\text{F}\)

2. Atomic Structure and Composition Statistics

Atomic structure describes the arrangement and subatomic particle composition of atoms and ions.

2.1 Atomic Theory and Subatomic Particles

Proton: positively charged subatomic particle located in the nucleus.
Electron: negatively charged subatomic particle orbiting the central nucleus.
Neutron: neutral subatomic particle present inside the atomic nucleus.

Particle	Mass (g)	Charge (C)	Charge Unit
Electron	\(9.10938 \times 10^{-28}\ \text{g}\)	\(-1.6022 \times 10^{-19}\ \text{C}\)	\(-1\)
Proton	\(1.67262 \times 10^{-24}\ \text{g}\)	\(+1.6022 \times 10^{-19}\ \text{C}\)	\(+1\)
Neutron	\(1.67493 \times 10^{-24}\ \text{g}\)	\(0\ \text{C}\)	\(0\)

Example. Calculate the total positive electric charge in coulombs of the protons in an atomic nucleus containing \(6\) protons, given the elementary proton charge constant \(e = 1.6022 \times 10^{-19}\ \text{C}\).

Solution. Multiply the number of protons by the charge of a single proton:

\[ Q = 6 \times (1.6022 \times 10^{-19}\ \text{C}) = 9.6132 \times 10^{-19}\ \text{C} \]

Answer: \(9.6132 \times 10^{-19}\ \text{C}\)

2.2 Atomic Number and Mass Number

Atomic number: total number of protons in an atomic nucleus.
Mass number: combined sum of protons and neutrons in an atom.
Isotopes: atoms of one element with differing neutron counts.

Mass Number Relation

\[ A = Z + N \]

Where: \(A\) = mass number, \(Z\) = atomic number, \(N\) = number of neutrons

Example. Give the number of protons, neutrons, and electrons in an atom of oxygen-17 (\(^{17}_8\text{O}\)), given atomic number \(Z = 8\) and mass number \(A = 17\).

Solution. The atomic number gives protons and electrons; subtract \(Z\) from \(A\) for neutrons:

\[ \text{Protons} = Z = 8 \qquad\qquad \text{Electrons} = 8 \]

\[ N = A - Z = 17 - 8 = 9 \]

Answer: \(8\ \text{protons},\ 9\ \text{neutrons},\ 8\ \text{electrons}\)

2.3 Nomenclature: Common Inorganic Cations and Anions

Cation: positively charged ion formed by losing electrons.
Anion: negatively charged ion formed by gaining electrons.
Polyatomic ion: ion composed of two or more covalently bonded atoms.
Stock system: naming system specifying cation charge with Roman numerals.

Ionic Charge Balance

\[ \sum (\text{Cation Charge} \times \text{Subscript}) + \sum (\text{Anion Charge} \times \text{Subscript}) = 0 \]

Example. Write the chemical formula for calcium phosphate, given that the calcium cation is \(\text{Ca}^{2+}\) (charge \(+2\)) and the phosphate anion is \(\text{PO}_4^{3-}\) (charge \(-3\)).

Solution. Balance total positive and negative charges to zero:

\[ 3(+2) + 2(-3) = +6 - 6 = 0 \qquad\qquad \Rightarrow \qquad\qquad \text{Formula} = \text{Ca}_3(\text{PO}_4)_2 \]

Answer: \(\text{Ca}_3(\text{PO}_4)_2\)

3. Stoichiometry and Chemical Quantities

Stoichiometry is the quantitative study of reactants, products, and mass-to-mole relationships in chemistry.

3.1 Atomic Mass, Molar Mass, and Molecular Mass

Atomic mass unit: mass unit equal to one-twelfth carbon-12 atom mass.
Average atomic mass: weighted average mass of natural isotopic mixture.
Molecular mass: sum of atomic masses of all atoms in a molecule.

Average Atomic Mass

\[ \text{Average Atomic Mass} = \sum (\text{Isotope Mass}_i \times \text{Fractional Abundance}_i) \]

Example. Copper has two stable isotopes: \(^{63}_{29}\text{Cu}\) (abundance \(69.09\%\), mass \(62.93\ \text{amu}\)) and \(^{65}_{29}\text{Cu}\) (abundance \(30.91\%\), mass \(64.9278\ \text{amu}\)). Calculate the average atomic mass of copper.

Solution. Multiply isotopic masses by fractional abundances and sum:

\[ \text{Average Mass} = (0.6909)(62.93\ \text{amu}) + (0.3091)(64.9278\ \text{amu}) = 43.478\ \text{amu} + 20.069\ \text{amu} = 63.55\ \text{amu} \]

Answer: \(63.55\ \text{amu}\)

3.2 Avogadro’s Number and the Mole Concept

Mole: SI unit for amount containing Avogadro’s number of entities.
Avogadro’s number: constant representing \(6.022 \times 10^{23}\ \text{entities/mol}\).
Molar mass: mass in grams of one mole of units.

Mole Mass Formula

\[ n = \dfrac{m}{MW} \]

Example. Calculate the mass in grams of a single silver (\(\text{Ag}\)) atom, given the molar mass of silver \(MW = 107.9\ \text{g/mol}\) and Avogadro’s number \(N_A = 6.022 \times 10^{23}\ \text{atoms/mol}\).

Solution. Divide the molar mass of silver by Avogadro’s number:

\[ m_{\text{atom}} = \dfrac{107.9\ \text{g/mol}}{6.022 \times 10^{23}\ \text{atoms/mol}} = 1.792 \times 10^{-22}\ \text{g} \]

Answer: \(1.792 \times 10^{-22}\ \text{g}\)

3.3 Percent Composition of Compounds

Percent composition: percent by mass of each element in a compound.
Empirical formula: formula showing simplest whole-number ratio of constituent atoms.

Percent Composition by Mass

\[ \text{Percent Composition} = \dfrac{n \times \text{Molar Mass of Element}}{\text{Molar Mass of Compound}} \times 100\% \]

Example. Calculate the percent composition by mass of \(\text{H}\) and \(\text{O}\) in hydrogen peroxide (\(\text{H}_2\text{O}_2\)), given atomic mass of \(\text{H} = 1.008\ \text{g/mol}\), \(\text{O} = 16.00\ \text{g/mol}\), and molar mass of \(\text{H}_2\text{O}_2 = 34.02\ \text{g/mol}\).

Solution. Compute mass fraction for each element multiplied by 100%:

\[ \% \text{H} = \dfrac{2(1.008\ \text{g/mol})}{34.02\ \text{g/mol}} \times 100\% = 5.926\% \]

\[ \% \text{O} = \dfrac{2(16.00\ \text{g/mol})}{34.02\ \text{g/mol}} \times 100\% = 94.06\% \]

Answer: \(5.926\%\ \text{H},\ 94.06\%\ \text{O}\)

4. Gas Laws and Kinetic Molecular Theory

Gases are fluids that expand spontaneously to fill the volume and shape of their containers.

4.1 Empirical Gas Laws

Boyle’s law: volume varies inversely with pressure at constant temperature.
Charles’s law: volume varies directly with absolute temperature at constant pressure.
Gay-Lussac’s law: pressure varies directly with absolute temperature at constant volume.
Avogadro’s law: volume varies directly with gas moles at constant pressure and temperature.

Empirical Gas Laws

\[ P_1 V_1 = P_2 V_2 \]

\[ \dfrac{V_1}{T_1} = \dfrac{V_2}{T_2} \]

\[ \dfrac{P_1}{T_1} = \dfrac{P_2}{T_2} \]

Example. A sample of chlorine gas occupies \(0.69\ \text{L}\) at \(0.986\ \text{atm}\). What is the volume if the pressure is increased to \(1.2\ \text{atm}\) at constant temperature, given \(P_1 = 0.986\ \text{atm}\), \(V_1 = 0.69\ \text{L}\), and \(P_2 = 1.2\ \text{atm}\)?

Solution. Apply Boyle’s law to solve for final volume \(V_2\):

\[ V_2 = \dfrac{P_1 V_1}{P_2} = \dfrac{(0.986\ \text{atm})(0.69\ \text{L})}{1.2\ \text{atm}} = 0.56695\ \text{L} \approx 0.57\ \text{L} \]

Answer: \(0.57\ \text{L}\)

4.2 The Ideal Gas Equation and Applications

Ideal gas law: state equation relating pressure, volume, temperature, and moles.
Gas density: mass of gas per unit volume under given pressure and temperature.

Ideal Gas Law and Molar Mass

\[ P V = n R T \]

\[ MW = \dfrac{d R T}{P} \]

Where: \(R\) = gas constant (\(0.0821\ \text{L}\cdot\text{atm/(K}\cdot\text{mol)}\)), \(d\) = gas density (\(\text{g/L}\))

Example. Sulfur hexafluoride (\(\text{SF}_6\)) is a gas. Calculate the pressure in \(\text{atm}\) exerted by \(1.82\ \text{mol}\) of the gas in a vessel of volume \(5.43\ \text{L}\) at \(45^\circ\text{C}\) (\(318\ \text{K}\)), given gas constant \(R = 0.0821\ \text{L}\cdot\text{atm/(K}\cdot\text{mol)}\).

Solution. Solve ideal gas equation for pressure \(P\):

\[ P = \dfrac{n R T}{V} = \dfrac{(1.82\ \text{mol})(0.0821\ \text{L}\cdot\text{atm/(K}\cdot\text{mol)})(318\ \text{K})}{5.43\ \text{L}} = 8.751\ \text{atm} \approx 8.75\ \text{atm} \]

Answer: \(8.75\ \text{atm}\)

4.3 Gas Mixtures and Transport Phenomena

Dalton’s law: total pressure equals sum of component partial pressures.
Root-mean-square speed: molecular velocity proportional to square root of temperature.
Graham’s law: rate of effusion varies inversely as square root of molar mass.

Dalton, RMS Speed, and Graham Formulas

\[ P_i = X_i P_{\text{total}} \]

\[ u_{\text{rms}} = \sqrt{\dfrac{3 R T}{MW}} \]

\[ \dfrac{r_1}{r_2} = \sqrt{\dfrac{MW_2}{MW_1}} \]

Where: \(R\) = gas constant (\(8.314\ \text{J/(K}\cdot\text{mol)}\)), \(MW\) = molar mass in \(\text{kg/mol}\)

Example. Calculate the root-mean-square speed of helium (\(\text{He}\)) atoms in \(\text{m/s}\) at \(25^\circ\text{C}\) (\(298\ \text{K}\)), given gas constant \(R = 8.314\ \text{J/(K}\cdot\text{mol)}\) and molar mass of helium \(MW = 4.003 \times 10^{-3}\ \text{kg/mol}\) (\(1\ \text{J} = 1\ \text{kg}\cdot\text{m}^2/\text{s}^2\)).

Solution. Substitute values into the root-mean-square speed equation:

\[ u_{\text{rms}} = \sqrt{\dfrac{3(8.314\ \text{J/(K}\cdot\text{mol)})(298\ \text{K})}{4.003 \times 10^{-3}\ \text{kg/mol}}} = \sqrt{\dfrac{7432.716}{4.003 \times 10^{-3}}} = \sqrt{1.8568 \times 10^6} = 1.36 \times 10^3\ \text{m/s} \]

Answer: \(1.36 \times 10^3\ \text{m/s}\)

4.4 Real Gases (Non-Ideal Behavior)

Van der Waals equation: state equation accounting for real gas volume and intermolecular forces.

Van der Waals Equation

\[ \left( P + \dfrac{a n^2}{V^2} \right) (V - n b) = n R T \]

Where: \(a\) = molecular attraction constant, \(b\) = finite molecular volume constant

Example. Given \(3.50\ \text{mol}\) of \(\text{NH}_3\) in \(5.20\ \text{L}\) at \(47^\circ\text{C}\) (\(320\ \text{K}\)), with van der Waals constants \(a = 4.17\ \text{atm}\cdot\text{L}^2/\text{mol}^2\) and \(b = 0.0371\ \text{L/mol}\), and gas constant \(R = 0.0821\ \text{L}\cdot\text{atm/(K}\cdot\text{mol)}\), calculate the pressure in \(\text{atm}\).

Solution. Compute pressure correction term and volume correction term first:

\[ \dfrac{a n^2}{V^2} = \dfrac{(4.17\ \text{atm}\cdot\text{L}^2/\text{mol}^2)(3.50\ \text{mol})^2}{(5.20\ \text{L})^2} = \dfrac{51.0825}{27.04} = 1.889\ \text{atm} \]

\[ n b = (3.50\ \text{mol})(0.0371\ \text{L/mol}) = 0.12985\ \text{L} \]

\[ (P + 1.889\ \text{atm})(5.20\ \text{L} - 0.12985\ \text{L}) = (3.50\ \text{mol})(0.0821\ \text{L}\cdot\text{atm/(K}\cdot\text{mol)})(320\ \text{K}) \]

\[ (P + 1.889)(5.07015) = 91.952 \qquad\qquad \Rightarrow \qquad\qquad P + 1.889 = 18.136 \qquad\qquad \Rightarrow \qquad\qquad P = 16.247\ \text{atm} \approx 16.2\ \text{atm} \]

Answer: \(16.2\ \text{atm}\)

5. Solutions and Mixture Volumetrics

Solutions are homogeneous mixtures composed of a solute dissolved in a liquid solvent.

5.1 Molarity

Molarity: number of moles of solute per liter of solution.
Solute: substance dissolved in solvent to form solution.
Solvent: dissolving medium present in larger quantity in solution.

Molarity Formula

\[ M = \dfrac{n}{V} \]

Where: \(n\) = moles of solute, \(V\) = liters of solution

Example. How many grams of potassium dichromate (\(\text{K}_2\text{Cr}_2\text{O}_7\)) are required to prepare a \(250\ \text{mL}\) (\(0.250\ \text{L}\)) solution whose concentration is \(2.16\ \text{M}\), given molar mass of \(\text{K}_2\text{Cr}_2\text{O}_7 = 294.2\ \text{g/mol}\)?

Solution. Find moles required and convert to mass in grams:

\[ n = M \times V = (2.16\ \text{mol/L})(0.250\ \text{L}) = 0.540\ \text{mol} \]

\[ m = n \times MW = (0.540\ \text{mol})(294.2\ \text{g/mol}) = 158.868\ \text{g} \approx 159\ \text{g} \]

Answer: \(159\ \text{g}\)

5.2 Dilution Engineering

Dilution: process of reducing solution concentration by adding solvent.

Dilution Equation

\[ M_i V_i = M_f V_f \]

Example. Describe how you would prepare \(500\ \text{mL}\) (\(0.500\ \text{L}\)) of a \(1.75\ \text{M}\ \text{H}_2\text{SO}_4\) solution, starting with an \(8.61\ \text{M}\) stock solution of \(\text{H}_2\text{SO}_4\), given \(M_f = 1.75\ \text{M}\), \(V_f = 0.500\ \text{L}\), and \(M_i = 8.61\ \text{M}\).

Solution. Apply the conservation of moles dilution formula:

\[ M_i V_i = M_f V_f \qquad\qquad \Rightarrow \qquad\qquad (8.61\ \text{M}) V_i = (1.75\ \text{M})(0.500\ \text{L}) \]

\[ V_i = \dfrac{(1.75\ \text{M})(0.500\ \text{L})}{8.61\ \text{M}} = 0.1016\ \text{L} \approx 102\ \text{mL} \]

Answer: \(102\ \text{mL}\)