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2. Density & Dimensional Analysis
Chemical math
Density &
Dimensional Analysis

3. Density
Density - the volume which is occupied by a specific mass of a material, usually given in units of grams per cubic centimeters or grams per milliliter.
D = m/v

4. Density of Liquid
The density of a liquid can be determined by massing the liquid and then measuring the exact volume.

5. Displacement/Solid Density
The volume of a solid can be difficult to measure, so its density is determined using a different method. The volume is found by measuring the volume of a liquid that the solid displaces. In the example below, the volume of the compound is 17.2 mL
Volume of water= 15.0 ml Volume of water + compound = 32.2 ml

6. Math problems
To solve all math problems in this class you will use the “u,k,p,s” method.
u: unknown
k: known
p: plan
s: solve
Scientific method

7. Density practice problem 1
An aluminum bar with a volume of 35 cm3 has a mass of 94.5 g. What is the density of aluminum based on this information?
u: density Al
k: m= 94.5 g; v=35cm3
p: d= m v
s: d = 94.5g =
35cm3
2.7 g/cm3

8. Density practice problem 2
A glass tube that holds exactly mL is filled with liquid mercury metal whose mass is shown to be g. From these data, what is the density of mercury metal?
u: density Hg
k: m= g; v= mL
p: d= m v
s: d = 1020g =
75mL
13.6 g/mL

9. Density practice problem 3
The mass of 155 mL of ethyl alcohol is found to be exactly g. What is the density of ethyl alcohol?
u: density of ethyl alcohol
k: m= 122.5g; v=155mL
p: d= m v
s: d = g =
155mL
0.790 g/mL

10. Density practice problem 4
A block of ice with a mass of 519.0g is found to occupy a volume of cm3. What is the density of the ice? u: density H2O(s) k: m= g; v=565.8cm3 p: d= m v s: d = 519.0g = cm3
g/cm3

11. Density practice problem 5
Find the density of a sample of silicon which has a volume of 235.7cm3 and a mass of g. u: density Si k: m= g; v=235.7cm3 p: d= m v s: d = g = cm3
2.42 g/cm3

12. Density practice problem 6
A solid cube of impure zinc metal, 3.50 cm on a side, has a density of 6.95g/cm3. What is the mass of the cube?
u: mass of Zn
k: d= 6.95g/mL; l(per side)=3.50cm
p: d= m volume=l*w*h v
s: volume =3.50cm*3.50cm*3.50cm = 42.88cm3
6.95g = m =
cm cm3
(6.95g)(42.88cm3) = m (cm3)
cm3 cm3
m = 176 g

13. Density practice problem 7
An irregularly shaped piece of gold (density = 19.3g/cm3) has a mass of 428.4g. What is the volume of this piece of gold?
u: volume of Au
k: d= 19.3g/mL; m=428.4g
p: d= m v
s: 19.3g = g =
mL v
  (19.3g)(v) = mL (428.4g)
19.3g g
22.2cm3 Au = v

14. Worksheet 7 Significant Figures Part A
98 -2
– 2
432 – 3
– 3
15.9 – 3
– 4
– 5
30.50 – 4
0.582 – 3
– 7
– 6

15. Worksheet 7 Significant Figures Part B
=14.4
500
44.2
42500
7.6 15
177.36
80.
0.23
0.0327
**6.0**
***10***
3.733
7.9
1E-4

16. worksheet 2 problem 1
What is the density of carbon dioxide gas if g occupies a volume of 100 mL? U: density Carbon dioxide k: mass CO2 = g; vol. CO2 =100mL p: d= m v s: d = 0.196g = g/mL 100mL
2 E -3 g/mL

17. worksheet 3 problem 2 1.0 g/cm3
A block of wood 3.0 cm on each side has a mass of 27g. What is the density of this block? u: dwood k: m = 27g length = 3.0 cm p: d= m volume = l * w * h v s: volume = 3.0cm * 3.0cm * 3.0cm = 27cm3 dwood = 27 g = 27 cm3
1.0 g/cm3

18. worksheet 8 problem 3 5.0 g/mL
An irregularly shaped stone was lowered into a graduated cylinder holding a volume of water equal to 2.0 mL. The height of the water rose to 7.0 mL. If the mass of the stone was 25g. What was its density? u: dstone= k: v1=2.0mL; v2= 7.0mL; m = 25g p: d= m vol = v2 – v1 v S: vol = 7.0mL – 2.0mL = 5.0mL d = 25 g = 5.0mL
5.0 g/mL

19. worksheet 3 problem 4 8.96g/cm3 u: dcopper= k: v = 89.6 cm3; m = 10.0g
A 10.0 cm3 sample of copper has a mass of 89.6g. What is the density of copper?
u: dcopper=
k: v = 89.6 cm3; m = 10.0g
p: d= m v
S: d = g =
10.0 cm3
8.96g/cm3

20. worksheet 3 problem 5 5 g of gold
Silver has a density of 10.5g/cm3 and gold has a density of 19.3 g/cm3. What would have a greater mass, 5 cm3 of silver or 5cm3 of gold?
u: mAg , mAu
k: vAg=5cm3; vAu= 5cm3; d = 10.5g/cm3
p: d= m v
S: g = mAg g = mAu
cm cm cm cm3
10.5g ∙ (5.0cm3) = mAg ∙ (cm3) g ∙ (5.0cm3)= mAu ∙ (cm3)
cm3 cm cm cm3
52.5g = mAg g = mAu
5 g of gold

21. worksheet 3 problem 6 Benzene is denser
Five mL of ethanol has a mass of 3.9g, and 5.0mL of benzene has a mass of 4.4g. Which liquid is denser? u: dethanol , dbenzene k: vethanol=5mL; vbenzene= 5mL; methanol =3.9g mbenzene = 5.0g p: d= m v S: dethanol= 3.9g dbenzene = 4.4 g 5.0mL 5.0mL dethanol = 0.78g/mL dbenzene = 0.88g/mL
Benzene is denser

22. worksheet 3 problem 7 8g/cm3 u: diron k: m = 94g l=3cm, w=2cm, h=2cm
A sample of iron has the dimensions of 2cm x 3cm x 2cm. If the mass of this rectangular-shaped object is 94g, what is the density of iron?
u: diron
k: m = 94g l=3cm, w=2cm, h=2cm
p: d= m volume = l * w * h v
s: volume = 3cm * 2cm * 2cm = 12 cm3
dwood = 94g = 7.8g/cm3
12 cm3
8g/cm3

23. Dimensional Analysis
At times you need to convert from one unit to another, but you do not have a “formula” in which to plug the numbers into. You can use the relationship between any two known units. These are called conversion factors. For example, there are 60 minutes in one hour, so the relationship is 1 hour=60minutes or 1 hour or 60 mins. 60 mins. 1 hour

24. Chemical Math Math formulas conversion Specific heat density
Dimensional
Analysis
conversion

25. Dimensional Analysis Other examples of conversion factors are:
1mile = 5 280ft OR 1mile OR ft
5280 ft 1 mile
11 lbs = 5 kg OR lbs OR kg
5 kg 11 lbs

26. Dimensional Analysis A number ### with a unit √ ### √ √
### √ √
For all problems you MUST use Unknown, Known, Plan, Solve (note how this follows the scientific method.) X
Must equal 1 but not be the same.

27. Dimensional Analysis Ex: 4 m = ______ cm u: cm k: 4 m p: 1m=100cm s:
Begin with KNOWN – what you want to convert from.
Ex: 4 m = ______ cm
u: cm
k: 4 m
p: 1m=100cm s: 4m
Known
100 cm 1 m

28. Dimensional Analysis = 400 cm 4m 100 cm =4*100cm= 400cm 1 1 1 m
Multiply across the top row and divide by each section on the bottom. 4m
100 cm
=4*100cm= 400cm 1 m
= 400 cm

29. Dimensional Analysis = 10 *1 Km = 0.01 Km 1000
Convert 10 m to ____________ Km. U: Km K: 10m P: 1 Km= 1000m S:
10 meters
1 Km
= 10 *1 Km
1000
1000
meters
= 0.01 Km

30. Dimensional Analysis
How much will it cost to take a trip by car. Where? Miles? Type of car? Miles per gallon? $ per gallon? u: how much does it cost to drive to Myrtle Beach k: 1 Myrtle Beach p: 1 Myrtle Beach = 250 miles; 1 gallon = $2.50; 25 miles =1 gallon

31. Dimensional Analysis
u: how much does it cost to drive to Myrtle Beach k:1 Myrtle Beach p :1 Myrtle Beach = 250 miles; 1 gallon = $3.10; 25 miles =1 gallon
1 Myrtle Beach
1 gallon
$3.10
250 miles
1 Myrtle Beach
25 miles
1 gallon
= 1 * 250 * 1 * $3.10 ÷1÷ 25÷ 1
= $31

32. Metric prefix equivalents

33. Prefix >base Base Prefix < base
1 Exe (E) = 1018
1 Penta (P) = 1015
1 Tetra (T) = 1012
1 Giga (G) = 109
1 Mega (M) = 106
1 Kilo (K) = 103 (1000)
1 Hecto(H) = 102 (100)
1 Deka (D) = 10
Only for reference that there are more than the 6 used most often.

34. Prefix >base Base Prefix < base
1 = 10 deci (d)
1 = 100 centi (c)
1 = 1000 milli (m)
1 = 106 micro (μ)
1 = 109 nano (n)
1 = 1012 pico (p)
Only for reference that there are more than the 6 used most often.

35. Prefix >base Base Prefix < base
1 Kilo(K) = 103 (1000)
1 Hecto(H) = 102 (100)
1 Deka (D) = 10
= deci (d)
= 100 centi (c)
1 = 1000 milli (m)
Memorized

36. Sample problem problem 1
500 mL = _____________ L U: _________L K: 500 mL P: 1 L = 1000mL S:
1 Kilo(K) = 103 (1000)
1 Hecto(H) = 102 (100)
1 Deka (D) = 10
= 10 deci (d)
= 100 centi (c)
= 1000 milli (m)
1 L
= 500 * 1 L
1000
500mL
1000 mL
= 0.5L

37. Sample problem problem 2
25 cg = _____________ g U: _________g K: 25 cg P: 1 g= 100cg S:
1 Kilo(K) = 103 (1000)
1 Hecto(H) = 102 (100)
1 Deka (D) = 10
= 10 deci (d)
= 100 centi (c)
= 1000 milli (m)
1 g
= 25 * 1g
100
25 cg cg
100
= 0.25g

38. Sample problem problem 3
400 mg = _____________ kg U: _________kg K: 400 mg P: 1 g= 1000mg; 1000g = 1kg S:
1 Kilo(K) = 103 (1000)
1 Hecto(H) = 102 (100)
1 Deka (D) = 10
= 10 deci (d)
= 100 centi (c)
= 1000 milli (m)
We are not to the unknown
1 g
1kg
400 mg mg g
1000
1000
= 400 * 1 * 1kg÷ 1000 ÷ 1000
= kg

39. Sample problem problem 4
30 cm = _____________ mm U: _________mm K: 30 cm P: 1 m= 100 cm; 1m = 1000mm S:
1 Kilo(K) = 103 (1000)
1 Hecto(H) = 102 (100)
1 Deka (D) = 10
= 10 deci (d)
= 100 centi (c)
= 1000 milli (m)
1 m
1000mm
30cm cm m
100 1
= 30 * 1 * 1000mm÷ 100 ÷ 1
= 300 mm

40. Sample problem problem 5
3500 sec = _____________ hr U: _________hr K: 3500 secs P: 1 min= 60 s; 60 mins=1 hr S:
1 min
3500 s
1 hr s
min 60 60
= 3500 * 1 * 1hr ÷ 60 ÷ 60
=0.97 hr

41. Sample problem problem 6
2 yrs = _____________ secs U: _________secs K: 2 yrs P: 365 days= 1 yr; 1 day=24hr; 1hr = 60 min; 1 min=60sec S:
365 days
60sec
2 yrs
60 min 1 yr
day
1 min 1
= 2 * 365 * 60sec ÷ 1 ÷ 1 ÷ 1
= sec

42. Deep Water Horizon oil spill
barrels 42 gal L mL
1 barrel 1 gallon L
= mL of oil spilled into the Gulf of Mexico
719 trillion 604 billon 400 millon gallons

43. Dimensional Analysis (Factor Label Method) worksheet 9
10000 sec
4.2 moles cg
100 liters
12 lbs
4.46 moles
90. in
1.3 E 24 molecules hr
190 liters
2 E 24 molecules
7.5 liters