Matter, Measurement, and Problem Solving (Ch. 1.8–1.9): Dimensional Analysis and Chemical Problem Solving

Matter, Measurement, and Problem Solving

Introduction

This section covers essential strategies for solving chemical problems, focusing on dimensional analysis, unit conversions, and the application of these skills to real-world and laboratory scenarios. Mastery of these concepts is foundational for success in general chemistry.

Chemical Problems and Dimensional Analysis

Understanding Chemical Problem Solving

• Problem solving is a core skill in chemistry, involving the application of logical steps to reach a solution.

• One of the most effective methods for solving quantitative problems in chemistry is dimensional analysis (also called the factor-label method or unit factor method).

Dimensional Analysis: The Method

• Dimensional analysis uses conversion factors to convert one unit to another.

• A conversion factor is a ratio (fraction) that expresses how many of one unit are equal to another unit.

• The general formula for converting units is:

$\text{Given unit} \times \frac{\text{Desired unit}}{\text{Given unit}} = \text{Desired unit}$

• Units you want to cancel should appear in opposite positions (numerator/denominator) in the conversion factor.

Problem Solving Strategy

• Identify the starting point (the information given in the problem).

• Identify the end point (what you are asked to find).

• Devise a pathway from the starting point to the end point using known relationships and conversion factors.

• Gather any additional information needed (from memory, tables, or reference materials).

Worked Example: Calculating Mass of Jet Fuel

Example Problem

Problem: A 747 is fueled with 173,231 L of jet fuel. If the density of the fuel is 0.768 g/mL, what is the mass of the fuel in kilograms?

• Given: Volume = 173,231 L; Density = 0.768 g/mL

• Find: Mass in kilograms

Stepwise Solution:

1. Convert liters to milliliters: $1\ \text{L} = 1000\ \text{mL}$

2. Use density to convert milliliters to grams: $0.768\ \text{g}/\text{mL}$

3. Convert grams to kilograms: $1\ \text{kg} = 1000\ \text{g}$

$173,231\ \text{L} \times \frac{1000\ \text{mL}}{1\ \text{L}} \times \frac{0.768\ \text{g}}{1\ \text{mL}} \times \frac{1\ \text{kg}}{1000\ \text{g}} = 1.33 \times 10^5\ \text{kg}$

• Answer: $1.33 \times 10^5$ kg of fuel

Practice Problem: Dosage Calculation

Example Problem

Problem: An acetaminophen suspension for infants contains 80 mg/mL. The recommended dose is 15 mg/kg body weight. How many mL of this suspension should be given to an infant weighing 0.90 kg?

• Given: Concentration = 80 mg/mL; Dose = 15 mg/kg; Infant mass = 0.90 kg

• Find: Volume in mL

Stepwise Solution:

1. Calculate required dose: $0.90\ \text{kg} \times 15\ \text{mg}/\text{kg} = 13.5\ \text{mg}$

2. Convert mg to mL using the concentration: $\frac{13.5\ \text{mg}}{80\ \text{mg}/\text{mL}} = 0.16875\ \text{mL}$

3. Alternatively, set up the dimensional analysis as shown in the image:

$0.90\ \text{kg} \times \frac{15\ \text{mg}}{1\ \text{kg}} \times \frac{1\ \text{mL}}{80\ \text{mg}} = 0.95\ \text{mL}$

• Answer: 0.95 mL

Unit Conversions and Useful Relationships

• $1\ \text{cm}^3 = 1\ \text{mL}$

• $1\ \text{kg} = 1000\ \text{g}$

• $1\ \text{kg} = 2.205\ \text{lbs}$

Data Analysis: Atmospheric Carbon Dioxide Concentration

Interpreting Graphical Data

Graphs are often used in chemistry to represent changes in concentration, temperature, or other variables over time. The provided graph shows atmospheric carbon dioxide concentration (in parts per million, ppm) from 1860 to 2010.

Example Data Table: Atmospheric CO2 Concentration

Year	CO2 Concentration (ppm)
1960	317
2010	389

Calculating Rate of Increase

• Average rate of increase is calculated as:

$\text{Average rate} = \frac{\text{Change in concentration}}{\text{Number of years}}$

• For 1960 to 2010: $\frac{389\ \text{ppm} - 317\ \text{ppm}}{2010 - 1960} = \frac{72\ \text{ppm}}{50\ \text{years}} = 1.44\ \text{ppm/year}$

Estimating Future Concentrations

• To estimate future CO2 concentration, add the average rate of increase multiplied by the number of years to the most recent value.

$\text{Estimated CO}_2\ \text{in 2050} = 389\ \text{ppm} + (1.44\ \text{ppm/year} \times 40\ \text{years}) = 446.6\ \text{ppm}$

Additional info: This method assumes the rate of increase remains constant, which may not reflect real-world changes due to policy or environmental factors.

Summary Table: Key Conversion Factors and Relationships

Quantity	Conversion Factor
Length	1 cm = 0.01 m
Volume	1 cm3 = 1 mL
Mass	1 kg = 1000 g
Mass	1 kg = 2.205 lbs

Key Takeaways

• Dimensional analysis is a powerful tool for solving quantitative problems in chemistry.

• Always keep track of units and use conversion factors to guide your calculations.

• Interpreting and analyzing data, including graphical data, is essential for understanding chemical trends and making predictions.