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1 Chapter 2 Measurements 2.7 Problem Solving Basic Chemistry Copyright © 2011 Pearson Education, Inc. A health professional obtains a measured volume from a vial for an injection.

1 Chapter 2 Measurements 2.7 Problem Solving Basic Chemistry Copyright © 2011 Pearson Education, Inc. A health professional obtains a measured volume from

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Chapter 2 Measurements

2.7Problem Solving

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

A health professional obtains a measured volume from a vial for an injection.

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To solve a problem• identify the given unit• identify the needed unitProblem:

A person has a height of 2.0 m. What is thatheight in inches?

The given unit is the initial unit of height. given unit = meters (m)

The needed unit is the unit for the answer. needed unit = inches (in.)

Given and Needed Units

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Learning Check

An injured person loses 0.30 pints of blood. Howmany milliliters of blood would that be?

Identify the given and needed units in thisproblem.

Given unit = _______

Needed unit = _______

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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Solution

An injured person loses 0.30 pints of blood. Howmany milliliters of blood would that be?

Identify the given and needed units in thisproblem

Given unit = pints Needed unit = milliliters

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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STEP 1 State the given and needed quantities.

STEP 2 Write a plan to convert the given unit to the

needed unit.

STEP 3 State the equalities and conversion factors needed to cancel units.

STEP 4 Set up problem to cancel units and calculate answer.

Unit 1 x Unit 2 = Unit 2Unit 1

Given Conversion Needed unit factor unit

Guide to Problem Solving Using Conversion Factors

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

Guide to Problem Solving

6 Basic Chemistry Copyright © 2011 Pearson Education, Inc.

The steps in the Guides to Problem Solving Using Conversion Factors are useful in setting up a problem with conversion factors.

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Setting up a Problem

How many minutes are 2.5 hours?

STEP 1 Given 2.5 h

Need min

STEP 2 Plan hours minutes

STEP 3 Equalities 1 h = 60 min

STEP 4 Set up problem

2.5 h x 60 min = 150 min

1 h (2 SFs) given conversion needed unit factor unit

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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A rattlesnake is 2.44 m long. How long is the snake

in centimeters?

1) 2440 cm

2) 244 cm

3) 24.4 cm

Learning Check

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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A rattlesnake is 2.44 m long. How long is thesnake in centimeters?STEP 1 Given 2.44 m

Need centimeters

STEP 2 Plan meters centimeters

STEP 3 Equality 1 m = 100 cm

Factors 1 m and 100 cm

100 cm 1 m

STEP 4 Set up problem

2.44 m x 100 cm = 244 cm (answer 2)

1 m

Solution

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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• Often, two or more conversion factors are required to obtain the unit needed for the answer.

Unit 1 Unit 2 Unit 3

• Additional conversion factors are placed in the setup to cancel each preceding unit.Given unit x factor 1 x factor 2 = needed unitUnit 1 x Unit 2 x Unit 3 = Unit 3

Unit 1 Unit 2

Using Two or More Factors

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How many minutes are in 1.4 days?

STEP 1 Given 1.4 days Need minutes

STEP 2 Plan days hours minutes

STEP 3 Equalties: 1 day = 24 h

1 h = 60 minSTEP 4 Set up problem:

1.4 days x 24 h x 60 min = 2.0 x 103 min

1 day 1 h

2 SFs Exact Exact = 2 SFs

Example: Problem Solving

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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• Be sure to check your unit cancellation in the setup.• The units in the conversion factors must cancel to

give the correct unit for the answer.

Example:

What is wrong with the following setup?1.4 day x 1 day x 1 h

24 h 60 min

Units = day2/min, which is not the unit needed.

Units do not cancel properly; the setup is wrong.

Check the Unit Cancellation

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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Using the GPS

What is 165 lb in kilograms?

STEP 1 Given 165 lb Need kg

STEP 2 Plan lb kg

STEP 3 Equalities/conversion factors

1 kg = 2.20 lb 2.20 lb and 1 kg

1 kg 2.20 lb

STEP 4 Set up problem

165 lb x 1 kg = 74.8 kg (3 SFs) 2.20 lb

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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A bucket contains 4.65 L water. How many gallons of water is that?

Learning Check

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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STEP 1 Given 4.65 L Need gallons

STEP 2 Plan liters quarts gallons

STEP 3 Equalities

1 L = 1.057 qt 1 gal = 4 qt

STEP 4 Set Up Problem

4.65 L x 1.057 qt x 1 gal = 1.23 gal

1 L 4 qt 3 SFs 4 SFs Exact 3 SFs

Solution

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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If a ski pole is 3.0 feet in length, how long is the ski pole in mm?

Learning Check

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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STEP 1 Given 3.0 ft Need mm

STEP 2 Plan ft in. cm mm

STEP 3 Equalities 1 ft = 12 in., 2.54 cm = 1 in.,

1 cm = 10 mm

STEP 4 Set up problem

3.0 ft x 12 in. x 2.54 cm x 10 mm = 910 mm (2SFs)

1 ft 1 in. 1 cm

Check initial unit: ft

Check factor setup: units cancel properly

Check needed unit: mm

Solution

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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If your pace on a treadmill is 65 meters per

minute, how many minutes will it take for you to

walk a distance of 7500 feet?

Learning Check

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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STEP 1 Given 7500 ft; 65 m/min Need min

STEP 2 Plan ft in. cm m min

STEP 3 Equalities 1 ft = 12 in. 1 in. = 2.54 cm

1 m = 100 cm

1 min = 65 m (walking pace)

STEP 4 Set Up Problem7500 ft x 12 in. x 2.54 cm x 1m x 1 min

1 ft 1 in. 100 cm 65 m

= 35 min Answer rounded off (2 SFs)

Solution

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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Percent Factor in a Problem

If the thickness of the skin fold at the waist indicates 11% of body fat, how much fat, in kg, does a person have with a body mass of 86 kg?

STEP 1 Given 86 kg; 11% fat Need kg fat

STEP 2 Plan kg body mass kg fat

STEP 3 Equalities/Conversion factors 100 kg of body mass = 11 kg of fat 100 kg body mass and 11 kg fat

11 kg fat 100 kg body massSTEP 4 Set Up Problem

86 kg mass x 11 kg fat = 9.5 kg of fat 100 kg mass

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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How many lb of sugar are in 120 g of candy if the

candy is 25% (by mass) sugar?

Learning Check

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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STEP 1 Given 120 g of candy

Need lb of sugar

STEP 2 Plan g lb of candy lb of sugar

STEP 3 Equalities 1 lb = 453.6 g

100 g of candy = 25 g of sugar

STEP 4 Set Up Problem % factor

120 g candy x 1 lb candy x 25 lb sugar

453.6 g candy 100 lb candy

= 0.066 lb of sugar

Solution

Basic Chemistry Copyright © 2011 Pearson Education, Inc.