Unit Conversions and Dimensional Analysis in General Chemistry

Unit Conversions and Dimensional Analysis

Introduction

Unit conversions and dimensional analysis are foundational skills in General Chemistry. They allow chemists to translate measurements between different units, ensuring consistency and accuracy in calculations. Mastery of these concepts is essential for solving problems involving quantities such as mass, volume, length, and amount of substance.

Key Concepts in Unit Conversion

• Unit Conversion: The process of changing a measurement from one unit to another using conversion factors.

• Dimensional Analysis: A systematic method that uses conversion factors to move from one unit to another, ensuring that units cancel appropriately.

• Significant Figures: The digits in a measurement that are known with certainty plus one estimated digit. Answers to conversion problems should reflect the correct number of significant figures.

Helpful Conversion Factors

Quantity	Conversion Factor
Length	1 in = 2.54 cm
Distance	1 mi = 1.609 km
Amount of Substance	$N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$ (Avogadro's number)

Practice Problems and Solutions

a) 102 mmol to mol

• Definition: 1 mmol = $1 \times 10^{-3}$ mol.

• Calculation: $102 \text{ mmol} \times \frac{1 \text{ mol}}{1000 \text{ mmol}} = 0.102 \text{ mol}$

• Example: Converting millimoles to moles is common when dealing with small quantities of substances.

b) 15 km to m

• Definition: 1 km = 1000 m.

• Calculation: $15 \text{ km} \times \frac{1000 \text{ m}}{1 \text{ km}} = 15,000 \text{ m}$

• Example: Useful for expressing distances in SI base units.

c) $2.80 \times 10^4$ pL to mL

• Definition: 1 pL = $1 \times 10^{-12}$ L; 1 mL = $1 \times 10^{-3}$ L.

• Calculation: $2.80 \times 10^4 \text{ pL} \times \frac{1 \text{ mL}}{1 \times 10^9 \text{ pL}} = 2.80 \times 10^4 / 1 \times 10^9 = 2.80 \times 10^{-5} \text{ mL}$

• Example: Converting between very small and more practical liquid volumes.

• Additional info: $1 \text{ mL} = 1 \times 10^9 \text{ pL}$

d) 20 m/s to km/hr

• Definition: 1 km = 1000 m; 1 hr = 3600 s.

• Calculation: $20 \text{ m/s} \times \frac{1 \text{ km}}{1000 \text{ m}} \times \frac{3600 \text{ s}}{1 \text{ hr}} = 20 \times 3.6 = 72 \text{ km/hr}$

• Example: Useful for converting speeds in physics and chemistry problems.

e) 100.0 mol Fe to atoms of Fe

• Definition: 1 mol = $6.022 \times 10^{23}$ entities (Avogadro's number).

• Calculation: $100.0 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} = 6.022 \times 10^{25} \text{ atoms}$

• Example: Used to determine the number of particles in a given amount of substance.

f) 42 miles/hr to miles/min

• Definition: 1 hr = 60 min.

• Calculation: $42 \text{ miles/hr} \times \frac{1 \text{ hr}}{60 \text{ min}} = 0.7 \text{ miles/min}$

• Example: Useful for time-based rate conversions.

g) 232 cm3 to L

• Definition: 1 L = 1000 cm3.

• Calculation: $232 \text{ cm}^3 \times \frac{1 \text{ L}}{1000 \text{ cm}^3} = 0.232 \text{ L}$

• Example: Converting between cubic centimeters and liters is common in laboratory measurements.

Applied Problem: Distance and Speed

Walking Time Calculation

Given: The distance from McAllister Hall to Memorial Library is 0.50 km. The average walking speed is 3.0 miles per hour. Calculate the time (in minutes) to walk from the library to the chemistry building.

• Step 1: Convert 0.50 km to miles using $1 \text{ mi} = 1.609 \text{ km}$.

• Calculation: $0.50 \text{ km} \times \frac{1 \text{ mi}}{1.609 \text{ km}} = 0.311 \text{ mi}$

• Step 2: Calculate time in hours: $\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{0.311 \text{ mi}}{3.0 \text{ mi/hr}} = 0.104 \text{ hr}$

• Step 3: Convert hours to minutes: $0.104 \text{ hr} \times 60 \text{ min/hr} = 6.24 \text{ min}$

• Final Answer: It will take approximately 6.2 minutes to walk from the library to the chemistry building.

Summary Table: Common Conversion Factors

Quantity	SI Unit	Conversion Factor
Length	meter (m)	1 km = 1000 m
Volume	liter (L)	1 L = 1000 cm3
Amount of Substance	mole (mol)	1 mol = $6.022 \times 10^{23}$ entities
Distance	mile (mi)	1 mi = 1.609 km

Conclusion

Unit conversions and dimensional analysis are essential tools for solving quantitative problems in chemistry. By understanding and applying conversion factors, students can confidently approach a wide range of problems involving measurements and calculations.