Chemistry and Measurements: Problem Solving Using Unit Conversion

Problem Solving Using Unit Conversion

Overview

  • The focus is on using conversion factors to change from one unit to another.

Problem-Solving Process

The problem-solving process for unit conversion involves:

  • Analyzing the problem.
  • Identifying the given unit and the needed unit.
  • Writing a plan to convert the given unit to the needed unit.
  • Identifying one or more conversion factors that cancel units and provide the needed unit.
  • Setting up the calculation.

Example 1: Converting Pounds to Kilograms

  • Problem: Convert 178 lb to kilograms.
  • Step 1: Analyze the Problem
    • Given: 178 lb
    • Need: kilograms
    • Connect: conversion factor (kg/lb)
  • Step 2: Write a Plan
    • lb → kg
  • Step 3: Equalities and Conversion Factors
    • Equality: 1 kg = 2.20 lb
    • Conversion factors: 1 kg2.20 lb\frac{1 \text{ kg}}{2.20 \text{ lb}}, 2.20 lb1 kg\frac{2.20 \text{ lb}}{1 \text{ kg}}
  • Step 4: Set up the Problem
    • 178 lb×1 kg2.20 lb=81 kg178 \text{ lb} \times \frac{1 \text{ kg}}{2.20 \text{ lb}} = 81 \text{ kg}

Learning Check 1: Converting Meters to Centimeters

  • Problem: A rattlesnake is 2.44 m long. How many centimeters long is the snake?
  • Step 1: Analyze the Problem
    • Given: 2.44 m
    • Need: centimeters
    • Connect: conversion factor (cm/m)
  • Step 2: Write a Plan
    • m → cm
  • Step 3: Equalities and Conversion Factors
    • Equality: 1 m = 100 cm
    • Conversion factors: 100 cm1 m\frac{100 \text{ cm}}{1 \text{ m}}, 1 m100 cm\frac{1 \text{ m}}{100 \text{ cm}}
  • Step 4: Set up the Problem
    • 2.44 m×100 cm1 m=244 cm2.44 \text{ m} \times \frac{100 \text{ cm}}{1 \text{ m}} = 244 \text{ cm}

Using Two or More Conversion Factors

  • Often, multiple conversion factors are needed to complete a unit change.
  • Set up the problem so that each factor cancels the preceding unit until the needed unit is obtained.

Example 2: Dosage Calculation

  • Problem: Convert 0.150 mg of Synthroid to the number of tablets, given that each tablet contains 50. mcg.
  • Step 1: Analyze the Problem
    • Given: 0.150 mg of Synthroid
    • Need: number of tablets
    • Connect: 1 tablet = 50. mcg of Synthroid
  • Step 2: Write a Plan
    • mg → mcg → tablets
  • Step 3: Equalities and Conversion Factors
    • Equalities: 1 mg = 1000 mcg, 1 tablet = 50. mcg
    • Conversion factors: 1000 mcg1 mg\frac{1000 \text{ mcg}}{1 \text{ mg}}, 1 mg1000 mcg\frac{1 \text{ mg}}{1000 \text{ mcg}}, 1 tablet50. mcg\frac{1 \text{ tablet}}{50. \text{ mcg}}, 50. mcg1 tablet\frac{50. \text{ mcg}}{1 \text{ tablet}}
  • Step 4: Set up the Problem
    • 0.150 mg×1000 mcg1 mg×1 tablet50. mcg=3 tablets0.150 \text{ mg} \times \frac{1000 \text{ mcg}}{1 \text{ mg}} \times \frac{1 \text{ tablet}}{50. \text{ mcg}} = 3 \text{ tablets}

Learning Check 2: Converting Days to Minutes

  • Problem: How many minutes are in 1.6 days?
  • Step 1: Analyze the Problem
    • Given: 1.6 days
    • Need: minutes
    • Connect: conversion factors (h/day, min/h)
  • Step 2: Write a Plan
    • days → hours → minutes
  • Step 3: Equalities and Conversion Factors
    • Equalities: 1 day = 24 h, 1 h = 60 min
    • Conversion factors: 24 h1 day\frac{24 \text{ h}}{1 \text{ day}}, 1 day24 h\frac{1 \text{ day}}{24 \text{ h}}, 60 min1 h\frac{60 \text{ min}}{1 \text{ h}}, 1 h60 min\frac{1 \text{ h}}{60 \text{ min}}
  • Step 4: Set up the Problem
    • 1.6 days×24 h1 day×60 min1 h=2300 min1.6 \text{ days} \times \frac{24 \text{ h}}{1 \text{ day}} \times \frac{60 \text{ min}}{1 \text{ h}} = 2300 \text{ min}

Learning Check 3: Calculating Time from Distance and Pace

  • Problem: If your pace on a treadmill is 65 m/min, how many minutes will it take for you to walk a distance of 7.5 km?
  • Step 1: Analyze the Problem
    • Given: 7.5 km
    • Need: minutes
    • Connect: conversion factors (m/km, 65 m/min)
  • Step 2: Write a Plan
    • km → m → minutes
  • Step 3: Equalities and Conversion Factors
    • Equalities: 1 km = 1000 m, 65 m = 1 min
    • Conversion factors: 1000 m1 km\frac{1000 \text{ m}}{1 \text{ km}}, 1 km1000 m\frac{1 \text{ km}}{1000 \text{ m}}, 1 min65 m\frac{1 \text{ min}}{65 \text{ m}}, 65 m1 min\frac{65 \text{ m}}{1 \text{ min}}
  • Step 4: Set up the Problem
    • 7.5 km×1000 m1 km×1 min65 m=120 min7.5 \text{ km} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ min}}{65 \text{ m}} = 120 \text{ min}