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Chapter 1 (Lesson 5)

Dimensional Analysis (1 of 4)

  • Key idea: Many chemistry problems are unit conversion problems, solved using dimensional analysis.
  • Definition: A unit equation is a statement of two equivalent quantities, e.g. 2.54\ \,\text{cm} = 1\ \,\text{in}.
  • Conversion factor: A fractional quantity of a unit equation with the units we are converting from on the bottom (given unit) and the units we are converting to on the top (desired unit).
    • Purpose: Enables information given to cancel with the given unit and leave the desired unit.
  • Method: Information given × conversion factor(s) = information required.
  • Practical rule: Include units in calculations; they should be multiplied, divided, and canceled just like algebraic quantities.

Dimensional Analysis (2 of 4)

  • Units Raised to a Power:
    • When building conversion factors for units raised to a power, raise both the number and the unit to that power.
    • Example: Converting from 2.54\ \text{cm} = 1\ \text{in} to a relation for cm^2 and in^2.
    • Steps:
    • Begin with the known conversion factor: 2.54\ \text{cm} = 1\ \text{in}
    • Square both sides: (2.54\ \text{cm})^2 = (1\ \text{in})^2
    • Separate numbers and units: (2.54)^2\ \text{cm}^2 = 1^2\ \text{in}^2
    • Evaluate numeric square: 6.4516\ \text{cm}^2 = 1\ \text{in}^2
    • Therefore:
    • 1\ \text{in}^2 = 6.4516\ \text{cm}^2
    • And conversely 1\ \text{cm}^2 = \frac{1}{6.4516}\ \text{in}^2 \approx 0.1550\ \text{in}^2
    • Practical simplifications (to three or four significant figures, as appropriate):
    • 1\ \text{in}^2 \approx 6.45\ \text{cm}^2
    • 1\ \text{cm}^2 \approx 0.155\ \text{in}^2
  • Important note: When converting squared or otherwise powered units, carry the power through each conversion so the resulting units retain the intended exponent.

Dimensional Analysis (3 of 4)

  • Example 1: Convert 5\ \text{in}^2 to \text{cm}^2.
    • Use the conversion 1\ \text{in}^2 = 6.4516\ \text{cm}^2.
    • Calculation: 5\ \text{in}^2 \times 6.4516\ \text{cm}^2/\text{in}^2 = 32.258\ \text{cm}^2.
    • Rounding note: Depending on significant figures, this may be reported as approximately 32.3\ \text{cm}^2 (if rounding to 3 significant figures) or 32\ \text{cm}^2 (depending on the given data's precision).
  • Example 2: A rectangular piece of paper measures 3 in by 4 in. Calculate the area in square centimeters.
    • Area in in^2: 3\ \text{in} \times 4\ \text{in} = 12\ \text{in}^2.
    • Convert: 12\ \text{in}^2 \times 6.4516\ \text{cm}^2/\text{in}^2 = 77.4192\ \text{cm}^2.
    • Rounding note: Depending on significant figures, report as about 77\ \text{cm}^2 (2 significant figures) or 77.4\ \text{cm}^2 (if allowed by data precision).

General Problem-Solving Strategy

  • There are many ways to solve a problem; understanding the problem and the associated concepts is essential.
  • Know where to begin: identify the given information and the quantity you need to find.
  • Ensure units are consistent: use like units so they can be added, subtracted, multiplied, divided, or canceled.
  • Check your work: mistakes can happen, so always verify the answer.

Practice Problem (1)

  • Scenario: A chemist is preparing sodium stearate (soap) in two rounds, producing 1.7 kilograms and 552.5 grams, respectively.
  • Task: Find the total mass of the compound produced in kilograms, with the correct number of significant figures.
  • Solution steps:
    • Convert to a common unit:
    • 1.7 kg = 1.7 kg
    • 552.5 g = 0.5525 kg
    • Add: 1.7\ \text{kg} + 0.5525\ \text{kg} = 2.2525\ \text{kg}
    • Rounding to match the least precise decimal place among the terms (1 decimal place in 1.7 kg): final result should be to 1 decimal place => 2.3\ \text{kg}.
  • Answer: 2.3\ \text{kg} (correct number of decimal places/significant figures given the data).

Practice Problem (2)

  • Scenario: A sample of aluminum has mass 27.891\ g and volume 10.5\ mL.
  • Task: Find the density with the correct number of significant figures.
  • Calculation:
    • Density = mass/volume = \frac{27.891\ g}{10.5\ mL} = 2.657714…\ \frac{g}{mL}
    • Significant figures: mass has 5 SF, volume has 3 SF; result should have 3 SF.
    • Reported density: 2.66\ \frac{g}{mL}
  • Answer: 2.66\ \text{g/mL}.

1.9 Interpreting Data and Graphs

  • Purpose: Interpreting data and graphs allows drawing meaningful conclusions, identifying patterns, and making informed decisions from empirical evidence.
  • Applications across fields:
    • Health Science: Analyzing patient data (weight, BMI, BP) to track health trends and outcomes.
    • Biomedical Science: Understanding experimental results and biological processes (drug efficacy, side effects, patient outcomes).
    • Engineering: Evaluating performance metrics and optimizing designs.
    • Forensic Science: Interpreting crime scene data and laboratory results.
    • Arts and Music: Analyzing sound wave patterns and frequency distributions to improve music production and sound quality.

Interpreting Data (1 of 3)

  • Importance: Analyzing data is a core scientific skill.
  • Example context: As an early chemist, you study the composition of water by performing several experiments with different water samples.
  • Core question: Do you notice any patterns in these data?

Interpreting Data (2 of 3)

  • Two notable patterns observed:
    • The sum of the masses of oxygen and hydrogen equals the mass of the water sample.
    • The ratio of the masses of oxygen to hydrogen is the same across samples.
  • Data table (example):
    • Sample A: Mass Hydrogen Formed = 2.2\ g, Mass Oxygen Formed = 17.8\ g; Mass O/H = \frac{17.8}{2.2} = 8.090909… \approx 8.1
    • Sample B: Mass Hydrogen Formed = 5.6\ g, Mass Oxygen Formed = 44.4\ g; Mass O/H = \frac{44.4}{5.6} = 7.928571… \approx 7.9
    • Sample C: Mass Hydrogen Formed = 11.1\ g, Mass Oxygen Formed = 88.9\ g; Mass O/H = \frac{88.9}{11.1} = 8.009009… \approx 8.01
  • Conclusion: The ratio is about 8\,, with small experimental error.

Interpreting Data (3 of 3)

  • When identity of the substance/sample is not known, use patterns to infer composition.
  • Using the same data (Hydrogen mass, Oxygen mass, and their ratio), you can infer the substance is water based on the observed constant ratio and mass-sum pattern.
  • Example evaluation from table:
    • Mass of Constituent I and II across samples align with the pattern that suggests water.
  • Final insight: The consistent mass-sum and the consistent ratio support identifying the substance as water.

Interpreting Graphs (1 of 2)

  • Data visualization: Graphs/images are used to convey trends; scientists must analyze them critically.
  • Example to practice interpretation: Carbon dioxide (CO₂) as a greenhouse gas rising due to burning fossil fuels.
  • Key steps when analyzing a graph:
    • Examine the x-axis and y-axis to understand what each represents.
    • Check the numerical range of the axes.
    • Determine the concentration of CO₂ at a specific year and the extent of its increase.

Interpreting Graphs (2 of 2)

  • Observations about CO₂ data:
    • The increase in CO₂ is not constant over time.
    • The slope of the line, representing the rate of increase, has intensified since about 1960.
  • Implication: The rate of CO₂ accumulation in the atmosphere has accelerated in recent decades, indicating changing dynamics in emissions and atmospheric chemistry.