General Chemistry Study Notes: Units, Density, and Moles

Units, Conversions, and Dimensional Analysis

Unit Conversions

Unit conversion is a fundamental skill in chemistry, allowing scientists to express measurements in different units for clarity and calculation. It is essential to use conversion factors correctly and maintain significant figures throughout calculations.

• Conversion Factors: Ratios that express how many of one unit are equal to another unit. For example, 1 m = 2.54 cm, 1 mi = 1.609 km.

• Dimensional Analysis: A method to convert units using multiplication by conversion factors.

• Example: To convert 8.45 kg/m3 to g/cm3:

  • 1 kg = 1000 g

  • 1 m3 = 1,000,000 cm3

  • Calculation: $8.45 \frac{\text{kg}}{\text{m}^3} \times \frac{1000\ \text{g}}{1\ \text{kg}} \times \frac{1\ \text{m}^3}{1,000,000\ \text{cm}^3} = 0.00845\ \frac{\text{g}}{\text{cm}^3}$

Volume Calculations

Calculating the volume of containers is a common task in chemistry, especially when preparing solutions or measuring substances.

• Volume of a Rectangular Prism: $V = l \times w \times h$

• Example: For a cup with dimensions 1.2 in × 1.3 in × 2.1 in:

  • Calculate volume in cubic inches: $V = 1.2 \times 1.3 \times 2.1 = 3.276\ \text{in}^3$

  • Convert to milliliters (1 in3 = 16.387 mL): $3.276 \times 16.387 = 53.7\ \text{mL}$

Moles and Atomic Mass

Calculating Moles from Mass

The mole is a central concept in chemistry, representing a specific number of particles (atoms, molecules, etc.). The mass of a mole of atoms is determined by the atomic mass and Avogadro's number.

• Avogadro's Number: $N_A = 6.022 \times 10^{23}$ mol-1

• Atomic Mass: The mass of one atom, usually expressed in grams or kilograms.

• Example: If a beaker contains 0.056 kg of oxygen and each oxygen atom has a mass of $2.66 \times 10^{-26}$ kg:

  • Number of atoms: $\frac{0.056\ \text{kg}}{2.66 \times 10^{-26}\ \text{kg/atom}} = 2.11 \times 10^{24}$ atoms

  • Moles of atoms: $\frac{2.11 \times 10^{24}}{6.022 \times 10^{23}} = 3.50$ mol

Density and Identification of Substances

Definition and Calculation of Density

Density is a physical property defined as mass per unit volume. It is used to identify substances and predict their behavior in mixtures.

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

• Units: Commonly expressed in g/cm3 or kg/m3

• Example: If a liquid has a mass of 500 g and a volume of 0.584 L:

  • Convert volume to cm3: 1 L = 1000 cm3, so 0.584 L = 584 cm3

  • Calculate density: $\frac{500\ \text{g}}{584\ \text{cm}^3} = 0.856\ \text{g/cm}^3$

Using Density to Identify Liquids

By comparing the measured density of an unknown liquid to known values, chemists can identify the substance.

Liquid	Density (g/cm3)
Diethylamine	0.71
Glycerol	1.26
Ethanolamine	1.03
Carbon tetrachloride	1.59
Octane	0.70

• Example: A measured density of 0.856 g/cm3 does not match any listed liquid exactly, but is closest to diethylamine or octane.

Significant Figures in Calculations

Importance of Significant Figures

Significant figures reflect the precision of measurements and must be maintained throughout calculations to ensure accuracy.

• Rules:

  • Nonzero digits are always significant.

  • Zeros between nonzero digits are significant.

  • Leading zeros are not significant.

  • Trailing zeros in a decimal number are significant.

• Example: If a student needs 6.6 g of pentane and the density is 0.626 g/cm3:

  • Volume needed: $\frac{6.6\ \text{g}}{0.626\ \text{g/cm}^3} = 10.5\ \text{cm}^3$ (rounded to 2 significant figures)

Summary Table: Key Conversion Factors and Constants

Quantity	Value
1 m	2.54 cm
1 mi	1.609 km
Avogadro's Number ($N_A$)	$6.022 \times 10^{23}$ mol-1

Additional info: These notes expand on the original questions by providing full academic context, definitions, and step-by-step examples for each concept.