Derived SI Units, continuedDensity

• Density is the ratio of mass to volume, or mass divided by volume.

mass mdensity = D =

volume Vor

Section 2 Units of MeasurementChapter 2

• The derived SI unit is kilograms per cubic meter, kg/m3

• g/cm3 or g/mL are also used • Density is a characteristic physical property of a

substance.

Derived SI Units, continuedDensity

• Density can be used as one property to help identify a substance

Section 2 Units of MeasurementChapter 2

Sample Problem A

A sample of aluminum metal has a mass of

8.4 g. The volume of the sample is 3.1 cm3. Calculate the density of aluminum.

Section 2 Units of MeasurementChapter 2

Derived SI Units, continued

Derived SI Units, continued

Sample Problem A Solution

Given: mass (m) = 8.4 gvolume (V) = 3.1 cm3

Unknown: density (D)

Solution:

mass density =

v

olume 3

3

82.

.4 g

3.7 g

1 cm/ cm

Section 2 Units of MeasurementChapter 2

Re-Arranging the Formula

Here is a method to help you re-arrange the formula to find another variable.

For example the density formula, which is D = m/v, can be rearranged to give the

formula for mass or volume.

Rearranging the formula, continued

Start by drawing a formula circle:

It should be divided into three sections by drawing a “T” in the middle. The horizontal part of the “T” represents the fraction bar.

Since the “M” is above the fraction bar

in the formula, it remains above the fraction

bar in the circle. The “D” and “V” will be written below the

fraction bar on either side of the vertical line.

M

D V

Rearranging formulas.

To find the formula for any of the given variables, cover that variable, and your formula remains. (This technique will work with any three variable equation involving multiplication or division)

When “M” is covered, When “D” is covered When “V” is covered

what remains is “DV” what remains is “M/V” what remains is “M/D”.

Therefore,

M = DV D = M/V V = M/D

M

D V

M

D V

M

D V

Density Practice Problems

Page 40 – Practice Problems (1-3)

Work these problems in your notebooks.

You will have 10 minutes to complete this exercise.

http://www.online-stopwatch.com/full-screen-stopwatch/

M

D V

Let’s go over the problems from page 40.

Problem 1: What is the density of a block of marble that occupies 310 cm³ and has a mass of 853 g?

From our circle, we can determine the formula to be D = M/V.

Therefore, D = 853 g / 310 cm³D = 2.75 g / cm³

Notice that both units remain in the answer because they cannot be cancelled.

Problem 2

Diamond has a density of 3.26 g / cm³. What is the mass of a diamond that has a volume of 0.351 cm³ ?

From our formula circle, we can see that the formula for mass is M = DV.

Therefore, M = 3.26 g / cm³ x 0.351 cm³ M = 1.14 g

(Notice that the cm³ units cancelled because one was under a fraction bar and one after a multiplication sign.)

M

D V

Sample problem 3

What is the volume of a sample of liquid mercury that has a mass of 76.2 grams, given that the density of mercury is 13.6 g/mL?

From the formula circle, we can derive that the formula for volume is M/D. Therefore,

V = M/D

V = 76.2 grams / 13.6 g/mL

V = 5.60 mL

(Grams cancels out of the answer.)

M

D V

•Review this tutorial as needed, then take the quiz labeled “Density”.

•You can also practice this type of problem at the following link:•http://science.widener.edu/svb/tutorial/densitycsn7.html

•

Mini-assessment

Density problems from textbook

Page 59 (17,18,19)

Page 61 (49)

Page 63 (9)

Let’s take fifteen minutes to complete this assessment. Be sure to show your work, beginning with writing the formula.

http://www.online-stopwatch.com/full-screen-stopwatch/(Trade to grade)