CHEMISTRY P
MRS. EDWARDS, MR. HAMM, MR. NELSON, DR. SELGRATH
NAME:
FALL 2013
HANDOUT #3: DIMENSIONAL ANALYSIS
Dimensional analysis (also known as factor-label) is the technique of solving problems easily without actually
cheating. Any situation where a comparison is made between two different quantities (e.g., how long to travel a
certain distance) is a problem ideally suited for dimensional analysis. In such a problem, a relationship is known or
given between the quantities being compared. To solve such a problem, nothing more is needed to know but this
relationship (sometimes called a conversion factor), the given quantity (such as the distance traveled) and what you
are trying to solve (such as the time). You do not need to know or use any formulas, nor do you need special insight
into the problem. Let the dimensions (units) of the measurements guide you through the problem. In the
mathematical solution to the problem, the units may be treated as you would variables.
Addition: 2.0 m + 4.5 m = 6.5 m
Division: 120 mi ÷ 2.0 hr = 60 mi/hr
In order for your answer to be numerically correct, it must be dimensionally correct (it must have the correct units).
Set up conversion factors so that the units combine or cancel to yield the units of the answer—the mathematical
operation will give the correct numerical answer.
Example:
Determine the length of time required to drive 84 miles at a velocity of
58 miles/hour.
In this problem, the given information is the distance traveled (84 miles) and the
unknown quantity that you are solving for is time. As promised you are provided
a relationship between distance and time (58 miles/hour). Here are the steps for
solving the problem:
1) Write down the given quantity:
84 miles
2) Draw a line that will form a fraction:
84 miles __________
3) Set up the units for the fraction in such 84 miles
a way that miles will cancel out of the
problem and the answer will be left in
hours (the units of the desired answer):
Example:
4) Write the number that corresponds to
each unit (58 miles = I hour):
84 miles
1 hour .
58 miles
5) Solve mathematically:
84 miles
1 hour = 1.4 hours
58 miles
Determine the cost of a 325 kilometer automobile trip if gasoline costs $0.389/Liter and your car
gets 11.2 kilometers/Liter.
325 kilometers
325 kilometers
325 kilometers
© BHS Chemistry, 2013
hour .
miles
Given information.
Liter $
.
kilometers
Liter
1 Liter
11.2 kilometers
$ 0.389 = $11.3
1 Liter
Set units to cancel.
Solve.
CHEMISTRY P
MRS. EDWARDS, MR. HAMM, MR. NELSON, DR. SELGRATH
NAME:
FALL 2013
PROBLEMS:
1. If carrots cost $0.39/pound, determine the number of pounds of carrots that may be purchased for $1.50.
2. Determine how long it takes for a fastball thrown with a velocity of 135 feet/second to travel from the
pitcher’s mound to home plate (a distance of 60.5 feet).
3. Determine the weight of 4.5 gallons of water. 1 quart of water weighs 2.1 pounds.
4. A bag of m&m’s costs $0.59 and contains 40 candies. If each one of the 32 students in the class receives
15 m&m’s determine the cost to their teacher.
5. If 3 pencils cost $0.49, determine the cost of 7 pencils.
6. A car can travel 390 miles on a full tank of gasoline (14 gallons). Determine the distance the car can travel
on 9.2 gallons.
7. Determine the volume of 20. pounds of ethanol. The weight of 1 gallon of ethanol is 6.5 pounds
8. Determine the number of three-egg omelets that can be prepared from 4 dozen eggs.
9. If a student requires 6.5 hours to read a 350 page novel, determine the time he would need to read a 500.
page novel.
10. A cyclist peddles a 20. mile race at an average velocity of 22 miles/hour. Determine her velocity in
meters/second. There are 1609 meters in a mile.
© BHS Chemistry, 2013