1. Chapter 1: Chemistry and Measurement
Vanessa Prasad-Permaul
Valencia College
CHM 1045

2. Matter: Anything that has both mass & volume
Properties of Matter
Chemistry: The study of composition, properties, and transformations of matter
Matter: Anything that has both mass & volume
Hypothesis: Interpretation of results
Theory: Consistent explanation of observations

3. Conservation of Mass
Law of Mass Conservation: Mass is neither created nor destroyed in chemical reactions.

4. Example 1: Conservation of Mass
C(s) + O2(g)  CO2(g)
12.3g C reacts with 32.8g O2, ?g CO2
12.3g g = 45.1g
0.238g C reacts with ?g O2 to make .873g CO2
0.238g + x = 0.873g = 0.873g-0.238g = 0.635g of O2
?g C reacts with 1.63g O2 to make 2.24g CO2
x g = 2.24g = 2.24g g = 0.61g C

5. Example 1: Conservation of Mass
Exercise 1.1
1.85g of wood is placed with 9.45g of air in a sealed vessel. It is
heated and the wood burns to produce ash and gases. The ash
is weighed to yield 0.28g. What is the mass of the gases in the
vessel?
1.85g Wood g Air heat g Ash + ? g gases
= 11.02g of gases
What is the mass of wood that is converted to gas by the end of
the experiment?
1.85g of Wood – 0.28g of ash = 1.57g

6. Matter is any substance that has mass and occupies volume.
Matter exists in one of three physical states:
solid
liquid
gas

7. In a solid, the particles of matter are tightly packed together.
Solids have a definite, fixed shape.
Solids cannot be compressed and have a definite volume.
Solids have the least energy of the three states of matter.

8. Liquids cannot be compressed and have a definite volume.
In a liquid, the particles of matter are loosely packed and are free to move past one another.
Liquids have an indefinite shape and assume the shape of their container.
Liquids cannot be compressed and have a definite volume.
Liquids have less energy than gases but more energy than solids.

9. Gases can be compressed and have an indefinite volume.
In a gas, the particles of matter are far apart and uniformly distributed throughout the container.
Gases have an indefinite shape and assume the shape of their container.
Gases can be compressed and have an indefinite volume.
Gases have the most energy of the three states of matter.

10. Phases

11. Properties of Matter
A physical change is a change in the form of matter but not in its chemical identity
A chemical change or a chemical reaction is a change in which one of more kinds of matter are transformed into a new kind of matter or several new kinds of matter

12. Properties of Matter
Physical Properties can be determined without changing the chemical makeup of the sample.
Some typical physical properties are:
Melting Point, Boiling Point, Density, Mass, Touch, Taste, Temperature, Size, Color, Hardness, Conductivity.
Some typical physical changes are:
Melting, Freezing, Boiling, Condensation, Evaporation, Dissolving, Stretching, Bending, Breaking.

13. Some typical chemical properties are:
Properties of Matter
Chemical Properties are those that do change the chemical makeup of the sample.
Some typical chemical properties are:
Burning, Cooking, Rusting, Color change, Souring of milk, Ripening of fruit, Browning of apples, Taking a photograph, Digesting food.
Note: Chemical properties are actually chemical changes

14. Properties of Matter
Exercise 1.2
Potassium (K) is a soft, silvery-colored metal that 64oC. It reacts vigorously with water (H2O), Oxygen (O2) and Chlorine (Cl2).
Identify all physical properties:
Soft
Silvery-colored
Melting point of 64oC
Identify all chemical properties:
Metal (its chemical identity)
K reacts vigorously with H2O
K reacts vigorously with O2
K reacts vigorously with Cl2

15. Classifications of Matter
Matter can be divided into two classes:
mixtures
pure substances
Mixtures are composed of more than one substance and can be physically separated into its component substances.
Pure substances are composed of only one substance and cannot be physically separated.

16. There are two types of pure substances:
Compounds
Elements
A compound is a substance composed of two or more elements chemically combined
Compounds can be chemically separated into individual elements.
Water is a compound that can be separated into hydrogen and oxygen.
An element cannot be broken down further by chemical reactions.

17. Dalton’s Atomic Theory
Law of Definite Proportions: Different samples of a pure chemical substance always contain the same proportion of elements by mass.
Any sample of H2O contains 2 hydrogen atoms for every oxygen atom

18. There are two types of mixtures:
homogeneous mixtures
heterogeneous mixtures
Homogeneous mixtures have uniform properties throughout.
Salt water is a homogeneous mixture.
Heterogeneous mixtures do not have uniform
properties throughout.
Sand and water is a heterogeneous mixture.

19. Which of the following represents a mixture?
Example 2: Matter
Which of the following represents a mixture?

21. Accuracy, Precision, and Significant Figures in Measurement
Accuracy is how close to the true value a given measurement is.
Precision is how well a number of independent measurements agree with one another.

22. Example 8: Accuracy & Precision
Which of the following is precise but not accurate?

23. Accuracy, Precision, and Significant Figures in Measurement
Significant Figures are the total number of digits in the measurement.
The results of calculations are only as reliable as the least precise measurement!!
Rules exist to govern the use of significant figures after the measurements have been made.

24. Accuracy, Precision, and Significant Figures in Measurement
Rules for Significant Figures:
Zeros in the middle of a number are significant
Zeros at the beginning of a number are not significant
Zeros at the end of a number and following a period are significant
Zeros at the end of a number and before a period may or may not be significant.

25. Example 4: Significant Figures
How many Significant Figures ?
a) = 3
b) = 4
c) 36,450 = 4
d) = 4
e) = 5

26. Accuracy, Precision, and Significant Figures in Measurement
Rules for Calculating Numbers:
During multiplication or division, the answer can’t have more significant figures than any of the original numbers.

27. Example 5: Significant Figures
218.2 x 79 = = 1.7 x 104
12.5 / = = 94.3
x 0.29 = = 0.084
/ = = 1.417

28. Accuracy, Precision, and Significant Figures in Measurement
-During addition or subtraction, the answer can’t have more digits to the right of the decimal point than any of the original numbers.

29. Example 6: Significant Figures
= = 297
= = 12.4
= = 0.58
= = 55.55
= = 187.7

30. Accuracy, Precision, and Significant Figures in Measurement
Rules for Rounding Numbers:
If the first digit removed is less than 5
round down (leave # same)
If the first digit removed is 5 or greater
round up
Only final answers are rounded off, do not round intermediate calculations

31. Example 7: Rounding and Significant Figures
Round off each of the following measurements a) L to 4 sig. figs. = 3.774L b) K to 3 sig. figs. = 255K c) kg to 4 sig. figs. = 55.27kg d) ml to 3 sig. figs. = 1.22ml e) g to 3 sig. figs. = 1.21g

32. Significant Figures
Exercise 1.3
Give answers to the following arithmetic setups. Round to the correct number of significant figures:
a) 5.61 x = =
9.1
b) = =
c) 6.81 – = =
d) x ( ) = x = = 3

33. How to put into calculator
Scientific Notation
Changing numbers into scientific notation
Large # to small #
Moving decimal place to left, positive exponent
123,987 = x 105
Small # to large #
Moving decimal place to right, negative exponent
= 2.39 x 10-4
How to put into calculator

34. Example 3: Scientific Notation
Put into or take out of scientific notation
1973 = x 103
x 10-4 =
0.775 = 7.75 x 10-1
3.55 x 107 = 35,500,000
8500 = 8.5 x 103

35. Measurement and Units
SI Units

36. Some prefixes for multiples of SI units
Measurement and Units
Some prefixes for multiples of SI units * * * * * * *
* Important

37. Exercise 1.4 Express the following quantities using an SI prefix and a
Measurement and Units
Exercise 1.4
Express the following quantities using an SI prefix and a
base unit. For instance, 1.6 x 10-6m = 1.6mm. A quantity such
as g could be written 0.168mg or 168mg.
a) x 10-9 m = 1.84 nm (nanometer)
b) x s = 5.67 ps (picosecond)
c) x 10-3 g = 7.85 mg (milligram)
d) 9.7 x 103 m = 9.7 km (kilometer)
e) s = ms (millisecond)
= 732us (microsecond)
f) m = 0.154nm (nanometer)
= 154pm (picometer)

38. Changes in Physical State
Most substances can exist as either a solid, liquid, or gas.
Water exists as a solid below 0 °C; as a liquid between 0 °C and 100 °C; and as a gas above 100°C.
A substance can change physical states as the temperature changes.

39. When a solid changes to a liquid, the phase change is called melting.
Solid  Liquid
When a solid changes to a liquid, the phase change is called melting.
A substance melts as the temperature increases.
When a liquid changes to a solid, the phase change is called freezing.
A substance freezes as the temperature decreases.

40. A substance vaporizes as the temperature increases.
Liquid  Gas
When a liquid changes to a gas, the phase change is called vaporization.
A substance vaporizes as the temperature increases.
When a gas changes to a liquid, the phase change is called condensation.
A substance condenses as the temperature decreases.

41. When a solid changes directly to a gas,
Solid  Gas
When a solid changes directly to a gas,
the phase change is called sublimation.
A substance sublimes as the temperature
increases.
When a gas changes
directly to a solid, the phase
change is called deposition.
A substance undergoes
deposition as the
temperature decreases.

42. Diagram of the various phases of temperature change

43. Temperature Conversions:
The Kelvin and Celsius scales have equal size units (a change of 1oC is equivalent to a change of 1K)
180oF
100 oC
100 K

44. Temperature Conversions:
Celsius (°C) — Kelvin (K) temperature conversion:
Kelvin (K) = t°C x 1K K
1oC
Fahrenheit (°F) — Celsius (C) temperature conversions: there are exactly 9oF for every 5oC. Knowing that 0oC = 32oF
tF = tC x 9oF
5oC
tC = 5oC x (toF – 32)
9oF

45. Example 9: Temp. Conversions
Carry out the indicated temperature conversions:
a) –78°C = ? K = (-78oC x 1K/1oC) K = = 195K
b) 158°C = ? °F = (158oC x 9oF/5oC)+32oF = = 316oF
c) K = ? °C = (373.15K x 1oC/1K)– K = 100K
d) 98.6°F = ? °C = 5oC/9oF x (98.6oF – 32oF) = 37oC
e) 98.6°F = ? K = (37oC x 1K/1oC) K = = 310K

46. A person with a fever has temperature of 102.5oF.
Measurement and Units
Exercise 1.5
A person with a fever has temperature of 102.5oF.
What is this temperature in oC? A cooling mixture
of dry ice and isopropyl alcohol has a temperature
of -78 oC. What is the temperature in kelvins?
a) oC = 5oC x (oF – 32 ) = x (102.5 – 32) = oC
9oF
b) K = oC = = 195 K

47. Volume
Volume: how much three-dimensional space a substance (solid, liquid, gas) or shape occupies or contains often quantified numerically using the SI derived unit (m3) the cubic meter.
The volume of a container is generally understood to be the capacity of the container, i. e. the amount of fluid (gas or liquid) that the container could hold, rather than the amount of space the container itself displaces.

48. Traditionally chemists use liter (L)
Volume
units of Volume:
m3 or cm3 (cc)
Traditionally chemists use liter (L)
1cm3 = 1cc = 1mL

49. Density: relates the mass of an object to its volume.
Measurement and Units
Density: relates the mass of an object to its volume.
Density = mass / Volume D = m / V
V = m / D
m = V  D
Density decreases as a substance is heated because the substance’s volume increases.

50. Density
What is the density of glass (in mL) if a sample weighing g has a volume of cm3? d = ? m = g V = cm3 = mL d = m = g = = g/mL V mL

51. Density
What is the volume of an unknown solution if the mass is g and the density is g/mL ? d = m/V V x d = m V = m/d V = g / g/mL = mL g x 1mL = mL g

52. Density
What is the mass of an unknown solution if the volume is 20.2 mL and the density is g/mL? d = m/V m = d x V m = 2.613g x 20.2 mL = = 52.8 g mL

53. Exercise 1.6 A piece of metal wire has a volume of 20.2 cm3 and a
Density
Exercise 1.6
A piece of metal wire has a volume of 20.2 cm3 and a
mass of 159 g. What is the density of the metal?
D = m = g = = g /cm3
V cm3
We know that the following metals have the following
densities. Which metal is the wire made of?
Mn = 7.21 g/cm3
Fe = 7.87 g/cm3
Ni = 8.90 g/cm3

54. Ethanol (grain alcohol) has a density of 0.789 g/cm3.
Exercise 1.7
Ethanol (grain alcohol) has a density of g/cm3.
What volume (mL) of ethanol must be poured into
a graduated cylinder to equal 30.3 g?
d = m/V V x d = m V = m / d
V = 30.3 g x 1 cm = cm3
0.789 g

55. Dimensional Analysis & Units
Dimensional-Analysis method uses a conversion factor to express the relationship between units.
Original quantity x conversion factor = equivalent
quantity
Example: express 2.50 kg  lb.
Conversion factor: kg = lb
2.50 kg x lb = 6.00 lb
1.00 kg

56. Dimensional Analysis & Units

57. The oxygen molecule (O2) consists of two oxygen
Dimensional Analysis & Units
Exercise 1.8
The oxygen molecule (O2) consists of two oxygen
atoms a distance of 121 pm apart. How many
millimeters (mm) is this distance?
121 pm x m x 1mm = x 10-7 mm
1 pm

58. Dimensional Analysis & Units
Exercise 1.9
A large crystal is constructed by stacking small identical
pieces of crystal. A unit cell is the smallest piece from
which a crystal can be made. A unit cell of a crystal of
gold metal has a volume of 67.6 A3. What is the volume
in dm3?
67.6 A3 x m x 10 dm = x dm3
1 A m 3

59. 1 in = 2.54cm (exactly) 1 yd = 36in (exactly)
Dimensional Analysis & Units
Exercise 1.10
Using the following definitions, obtain the
conversion factor for yards to meters. How many
meters are there in 3.54 yd?
1 in = 2.54cm (exactly) yd = 36in (exactly)
1 yd x 1 in x 1 cm = = yd/m
36 in cm m
3.54 yd x m = m
1.094 yd

60. a) 1.267 km  m  cm b) 0.784 L  mL c) 3.67 x 105 cm  in Conversions
1.267km x 1000m x 100cm = cm = x 105
1km m
b) L  mL
0.784L x 1000mL = 784L 1L
c) 3.67 x 105 cm  in
3.67 x 105cm x 1in = in = 1.44 x 105in
2.54cm

61. d) 79 oz  g e) 9.63 x 10-3 yd  ft f) 23.5 cm2  m2 Conversions
79oz x g = g = 2.2 x 103g
1oz
e) 9.63 x 10-3 yd  ft
9.63 x 10-3yd x 1m x 1km x mile x 5280ft
1.0936yd m km mile
= ft
f) 23.5 cm2  m2
23.5cm2 x 1m2 = m2
100cm2

62. g) 1.34 x 1012 pm  m 10-12pm h) 4.67 x 10-7 nm  pm 10-9nm 1m
Conversions
g) 1.34 x 1012 pm  m
1.34 x 1012pm x m = x 1024m
10-12pm
h) 4.67 x 10-7 nm  pm
4.67 x10-7nm x 1m x pm = x 10-12pm
10-9nm m