General Chemistry Study Guide: Fundamental Constants, Units, and Chemical Calculations

Fundamental Constants and Units

Key Physical Constants

Fundamental physical constants are essential for calculations in chemistry, especially when dealing with atomic and molecular properties. These constants are used in equations for energy, mass, charge, and other properties.

• Avogadro's number (NA): $6.0221367 \times 10^{23}$ mol-1

• Electron charge (e): $1.6022 \times 10^{-19}$ C

• Electron mass: $9.1093897 \times 10^{-28}$ g

• Faraday constant (F): $96,485.3$ C/mol e-

• Gas constant (R): $0.082057$ L·atm/(mol·K)

• Planck's constant (h): $6.626 \times 10^{-34}$ J·s

• Proton mass: $1.6726 \times 10^{-24}$ g

• Neutron mass: $1.6749 \times 10^{-24}$ g

• Speed of light in vacuum (c): $2.99792458 \times 10^{8}$ m/s

Useful Conversion Factors and Relationships

Conversion factors allow for the transformation of units in chemical calculations. Mastery of these is crucial for solving problems involving mass, length, volume, and energy.

Quantity	Conversion
Mass	1 lb = 453.6 g
Length	1 in = 2.54 cm (exactly)
Distance	1 mi = 1.609 km
Energy	1 cal = 4.184 J (exactly)
Volume	1 L = 1.01 qt; 1 km = 0.6215 mi
Pressure	1 atm = 760 mmHg = 101.325 kPa
Other	1 J = 1 C × 1 V; 1 pm = 1 × 10-12 m; 1 pm = 1 × 10-10 cm

Pressure Units

Pressure is commonly measured in several units. Understanding their equivalence is important for gas law calculations.

Unit	Equivalent
1 atm	760 mmHg = 760 torr = 101,325 Pa = 101.325 kPa
1 bar	100 kPa = 0.9869 atm
1000 Pa	1 kPa

Equations and Formulas

Common equations used in general chemistry for temperature conversion, density, and molarity:

• Celsius to Fahrenheit: $^{\circ}C = \frac{5}{9}(^{\circ}F - 32^{\circ})$

• Fahrenheit to Celsius: $^{\circ}F = \frac{9}{5}(^{\circ}C) + 32^{\circ}$

• Density: $d = \frac{m}{V}$

• Molarity: $M_1 V_1 = M_2 V_2$

Significant Figures and Measurement

Significant Figures

Significant figures reflect the precision of a measured quantity. The number of significant figures in a value depends on the digits that are known with certainty plus one estimated digit.

• Example: In 0.05070 × 103, there are 4 significant figures.

• Rule: Leading zeros are not significant; zeros between nonzero digits and trailing zeros in a decimal number are significant.

Measurement and Density

Density is calculated as mass divided by volume. The number of significant figures in the result should match the least precise measurement.

• Formula: $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$

• Example: If mass = 290.4 g and volume = 63.498 mL, the density should be reported with three significant figures.

Atoms, Elements, and Compounds

Atomic Structure and Isotopes

Atoms consist of protons, neutrons, and electrons. Isotopes are atoms of the same element with different numbers of neutrons.

• Atomic mass: Calculated as the weighted average of isotopic masses.

• Example: If an element has two isotopes with masses 106.90509 amu (51.84%) and 108.90476 amu (48.46%), the atomic mass is: $\text{Atomic mass} = (106.90509 \times 0.5184) + (108.90476 \times 0.4846) = 107.87$ amu$

Empirical and Molecular Formulas

The empirical formula shows the simplest whole-number ratio of atoms in a compound, while the molecular formula shows the actual number of atoms.

• Example: The empirical formula of tetanitrain is C2H2N2O. If the molar mass is 300 g/mol, the molecular formula is C4H4N4O2.

Chemical Reactions and Stoichiometry

Balancing Chemical Equations

Balancing equations ensures the same number of atoms of each element on both sides. Coefficients are used to achieve balance.

• Example: For the reaction H2S + HCl → H2S + S + HCl, the smallest whole-number coefficients are 1, 2, 1, 1, 1, 3.

Stoichiometric Calculations

Stoichiometry involves calculating the amounts of reactants and products in chemical reactions using balanced equations and molar relationships.

• Example: To prepare 100 mL of 1.00 M H2SO4 from 4.00 M H2SO4: $M_1 V_1 = M_2 V_2$ $4.00 \times V_1 = 1.00 \times 100$ $V_1 = 25.0$ mL

Limiting Reactant and Percent Yield

The limiting reactant is the reactant that is completely consumed first, limiting the amount of product formed. Percent yield compares actual yield to theoretical yield.

• Percent yield formula: $\text{Percent yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\%$

Solutions and Concentrations

Solubility of Ionic Compounds

Solubility rules determine which ionic compounds dissolve in water. For example, NH4Cl is soluble, while PbSO4 is not.

• Example: Of NH4Cl, PbSO4, Li2CO3, KI, Na2S, only NH4Cl, KI, and Na2S are soluble.

Molarity and Solution Preparation

Molarity (M) is the number of moles of solute per liter of solution. It is used to calculate the volume or mass of substances needed for reactions.

• Formula: $M = \frac{\text{moles of solute}}{\text{volume of solution (L)}}$

• Example: To prepare 100 mL of 1.00 M H2SO4 from 4.00 M stock, use 25.0 mL of stock solution.

Properties of Elements and Compounds

Atomic Mass and Periodic Trends

Atomic mass increases across a period and down a group. Elements with higher atomic numbers generally have greater atomic mass.

• Example: Bromine has a higher atomic mass than potassium, phosphorus, or magnesium.

Malleability and Physical Properties

Malleability is the ability of a substance to be hammered or rolled into sheets. Metals like silver (Ag) are malleable, while nonmetals are not.

Sample Calculations and Applications

Calculating Number of Atoms

To find the number of atoms in a given mass, use Avogadro's number and the molar mass.

• Formula: $\text{Number of atoms} = \frac{\text{mass}}{\text{molar mass}} \times N_A$

• Example: 0.67 moles of hydrogen contains $0.67 \times 6.022 \times 10^{23} = 4.0 \times 10^{23}$ atoms.

Percent Composition

Percent composition is the percentage by mass of each element in a compound.

• Formula: $\text{Percent composition} = \frac{\text{mass of element}}{\text{total mass of compound}} \times 100\%$

Practice Problems and Applications

Sample Multiple Choice and Calculation Questions

Practice problems cover significant figures, solubility, solution preparation, empirical formulas, stoichiometry, atomic mass, and more. These are essential for mastering general chemistry concepts and preparing for exams.

• Example: Calculate the mass of product generated by reacting 104 g of arsenic with excess oxygen: $4As + 5O_2 \rightarrow 2As_2O_5$

• Example: Determine the empirical formula of cinnamic alcohol from combustion data.

Additional info:

• Some context and explanations have been expanded for clarity and completeness.

• Tables have been recreated from the original images and text.