General Chemistry Practice Questions: Atomic Mass, Nomenclature, Stoichiometry, and Gas Laws

Atomic Mass and Isotopes

Isotopic Abundance and Atomic Mass Calculations

Isotopes are atoms of the same element with different numbers of neutrons, resulting in different mass numbers. The atomic mass of an element is the weighted average of the masses of its naturally occurring isotopes.

• Atomic Mass: The average mass of atoms of an element, weighted by the relative abundance of each isotope.

• Isotopic Abundance: The percentage of a particular isotope in a natural sample of the element.

• Calculation Example: If Gallium has an atomic mass of 69.723 u, and the abundance of 69Ga (68.926 u) is 60.11%, the atomic mass of the other isotope can be calculated using the formula: $\text{Atomic mass} = (\text{fraction}_1 \times \text{mass}_1) + (\text{fraction}_2 \times \text{mass}_2)$

Chemical Nomenclature and Structure

Naming Inorganic and Organic Compounds

Chemical nomenclature is the systematic naming of chemical compounds. It follows specific rules set by IUPAC for both inorganic and organic compounds.

• Inorganic Compounds: For example, Ba(NO3)2 is named barium nitrate.

• Organic Compounds: Naming follows rules based on the structure and functional groups present. For example, 3,3-dimethylethynyl-1-yne is an alkyne with two methyl groups at the third carbon.

• Structural Formulas: Show the arrangement of atoms in a molecule. For 3,3-dimethylethynyl-1-yne, the structure would include a triple bond and methyl substituents.

• Functional Groups: Groups of atoms responsible for characteristic reactions. For example, the ethynyl group (–C≡CH) is characteristic of alkynes.

Combustion Reactions and Empirical/Molecular Formulas

Determining Empirical and Molecular Formulas from Combustion Data

Combustion analysis is used to determine the empirical and molecular formulas of organic compounds by measuring the amounts of CO2 and H2O produced.

• Empirical Formula: The simplest whole-number ratio of atoms in a compound.

• Molecular Formula: The actual number of atoms of each element in a molecule.

• Combustion Reaction Example: For a hydrocarbon CxHyOz combusted to produce CO2 and H2O, use the masses of products to back-calculate the moles of C, H, and O in the original sample.

Balancing Redox Reactions

Balancing Equations in Acidic Media

Redox (reduction-oxidation) reactions involve the transfer of electrons between species. In acidic media, balancing requires accounting for H+ ions and H2O molecules.

• Steps to Balance:

  1. Write the half-reactions for oxidation and reduction.4

  2. Balance all elements except H and O.

  3. Balance O by adding H2O, and H by adding H+.

  4. Balance charges by adding electrons.

  5. Combine the half-reactions and simplify.

• Example Equation: $\mathrm{SO_3^{2-} (aq) + MnO_4^- (aq) \rightarrow SO_4^{2-} (aq) + Mn^{2+} (aq)}$

Precipitation Reactions and Stoichiometry

Predicting and Calculating Precipitate Formation

Precipitation reactions occur when two soluble salts react to form an insoluble product (precipitate). Stoichiometry is used to calculate the theoretical yield and percent yield of the precipitate.

• Net Ionic Equation: Shows only the species that change during the reaction. $\mathrm{Pb^{2+} (aq) + 2Cl^- (aq) \rightarrow PbCl_2 (s)}$

• Theoretical Yield: The maximum amount of product that can be formed from the given reactants.

• Percent Yield: The ratio of actual yield to theoretical yield, expressed as a percentage. $\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%$

• Example: When 28.5 g KCl is added to a solution containing 7.5 g Pb(NO3)2, PbCl2 precipitates. If 2.49 g of PbCl2 is obtained, calculate the theoretical yield and percent yield.

Gas Laws and Calculations

Volume, Pressure, and Temperature Relationships

Gas laws describe the relationships between pressure, volume, temperature, and amount of gas. The ideal gas law is commonly used for calculations.

• Boyle's Law: At constant temperature and moles, pressure and volume are inversely related. $P_1 V_1 = P_2 V_2$

• Ideal Gas Law: $PV = nRT$ Where:

  • P = pressure (in kPa or bar)

  • V = volume (in L)

  • n = moles of gas

  • R = gas constant (8.314 J·mol−1·K−1)

  • T = temperature (in K)

• Example: A gas occupies 3.33 L at 2.23 bar. What is the volume at 2.50 bar if temperature and moles are constant? Use Boyle's Law.

• Example: 0.35 g of Ar(g) in a 300 mL vessel at 1.3 bar. Calculate the temperature using the ideal gas law.

Summary Table: Key Concepts and Formulas

Concept	Key Formula	Example Application
Atomic Mass Calculation	$\text{Atomic mass} = (\text{fraction}_1 \times \text{mass}_1) + (\text{fraction}_2 \times \text{mass}_2)$	Finding unknown isotope mass from average atomic mass and abundance
Empirical Formula	Convert mass to moles, divide by smallest number of moles	Determining formula from combustion data
Percent Yield	$\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%$	Comparing actual and theoretical mass of precipitate
Boyle's Law	$P_1 V_1 = P_2 V_2$	Calculating new volume when pressure changes
Ideal Gas Law	$PV = nRT$	Finding temperature, pressure, or volume of a gas sample