General Chemistry: Atomic Structure, Periodic Table, Stoichiometry, and Dimensional Analysis

Atomic Structure and the Periodic Table

Overview of the Periodic Table

The periodic table organizes all known chemical elements by increasing atomic number and recurring chemical properties. Elements are arranged in rows (periods) and columns (groups or families), which reflect similarities in their electron configurations and chemical behavior.

• Groups: Vertical columns; elements in the same group have similar chemical properties.

• Periods: Horizontal rows; elements in the same period have the same number of electron shells.

• Metals, Nonmetals, and Metalloids: Metals are typically on the left and center, nonmetals on the right, and metalloids form a staircase between them.

Example: Sodium (Na) is an alkali metal in Group 1, Period 3, known for its high reactivity.

Element Classification and Properties

• Nonmetal that is a solid: Examples include sulfur and silicon (metalloid, but often included in such lists).

• Transition metal in Group 10: Nickel (Ni).

• Group 14 metalloid: Silicon (Si).

• Very reactive metal in 3rd period: Sodium (Na).

• Alkali metal in 6th period: Potassium (K).

• Diatomic element: Oxygen (O2).

• Noble gas: Helium (He), Neon (Ne).

• Toxic metal: Mercury (Hg).

• Liquid element: Mercury (Hg) and Bromine (Br2) at room temperature.

• Aqueous element: No element is naturally aqueous; this refers to ions in solution.

Atomic Structure

• Atomic Number (Z): Number of protons in the nucleus.

• Mass Number (A): Total number of protons and neutrons.

• Isotopes: Atoms of the same element with different numbers of neutrons.

Example: Rubidium has two naturally occurring isotopes: Rb-85 and Rb-87. The atomic weight is calculated using their relative abundances.

Calculating Atomic Weight

• Formula: $\text{Atomic weight} = (\text{fractional abundance}_1 \times \text{mass}_1) + (\text{fractional abundance}_2 \times \text{mass}_2)$

• Example: If Rb-85 is 72.17% and Rb-87 is 27.83%, with masses 84.9117 and 86.9091 amu respectively:

$\text{Atomic weight} = 0.7217 \times 84.9117 + 0.2783 \times 86.9091 = 85.47$

Stoichiometry and Chemical Equations

Balancing Chemical Equations

Balancing chemical equations ensures the law of conservation of mass is obeyed. The number of atoms of each element must be the same on both sides of the equation.

• Use the smallest whole number coefficients.

• Example: $2\text{Mg}_3\text{N}_2 + 6\text{H}_2\text{O} \rightarrow 3\text{Mg}(\text{OH})_2 + 2\text{NH}_3$

Law of Definite and Multiple Proportions

• Law of Definite Proportions: A chemical compound always contains the same elements in the same proportion by mass.

• Law of Multiple Proportions: When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in ratios of small whole numbers.

Stoichiometric Calculations

• Use molar mass to convert between grams and moles.

• Use Avogadro's number ($6.022 \times 10^{23}$) to convert between moles and number of particles.

• Example: To find the mass of magnesium hydroxide produced from a given mass of magnesium nitride, use the balanced equation and molar masses.

Dimensional Analysis and Unit Conversions

Dimensional Analysis

Dimensional analysis is a method to convert one unit to another using conversion factors.

• Set up conversion factors so units cancel appropriately.

• Example: To convert the speed of sound from m/s to miles per hour:

$336.43 \frac{\text{m}}{\text{s}} \times \frac{1 \text{ km}}{1000 \text{ m}} \times \frac{1 \text{ mile}}{1.609 \text{ km}} \times \frac{3600 \text{ s}}{1 \text{ hr}} = 752.5 \frac{\text{miles}}{\text{hr}}$

Density and Volume Calculations

• Density: $\text{Density} = \frac{\text{mass}}{\text{volume}}$

• Use density to convert between mass and volume.

• Example: If the density of t-butyl alcohol is 0.810 g/mL, the volume of 5.00 g is:

$\text{Volume} = \frac{5.00 \text{ g}}{0.810 \text{ g/mL}} = 6.17 \text{ mL}$

Empirical and Molecular Formulas

Determining Empirical Formulas

The empirical formula gives the simplest whole-number ratio of atoms in a compound.

• Convert mass percentages to moles.

• Divide by the smallest number of moles to get the ratio.

• Example: For a compound with 71% iron and 29% oxygen by mass:

$\text{Moles Fe} = \frac{71}{55.85} = 1.27$ $\text{Moles O} = \frac{29}{16.00} = 1.81$

Divide by the smallest value to get the ratio and empirical formula.

Determining Molecular Formulas

• Find the empirical formula mass.

• Divide the molar mass by the empirical formula mass to find the multiplier.

• Multiply subscripts in the empirical formula by this number.

Sample Table: Properties of Selected Elements

Element	Symbol	Group	Period	Type	State at Room Temp
Sulfur	S	16	3	Nonmetal	Solid
Nickel	Ni	10	4	Transition Metal	Solid
Oxygen	O	16	2	Nonmetal	Gas
Mercury	Hg	12	6	Metal	Liquid
Helium	He	18	1	Noble Gas	Gas

Additional info:

• Some answers and calculations are inferred from context and standard general chemistry knowledge.

• Some chemical names and formulas are clarified for completeness.