Chapter 3 pages 66 - 871 Modern Chemistry Chapter 3 Atoms: the building block of matter.

Chapter 3 pages 66 - 87 1

Modern ChemistryChapter 3Atoms:

the building block of matter

Section 1Atoms:

From Philosophical Idea to

Scientific Theory

Foundation of Chemical Atomic Theory

• Law of Conservation of Mass– Mass is neither created or destroyed

during ordinary chemical reactions or physical changes

Law of Conservation of Mass Image

• Law of Definite Proportions– A chemical compound contains the

same elements in exactly the same proportions by mass regardless of the size of the sample or the source of the compound.

• Law of Multiple Proportions– If two or more different compounds

are composed of the same two elements then the ratio of the masses of the second element combined with a certain mass of the first element is always a ratio of small whole numbers.

Law of Multiple Proportions Image

Dalton’s Atomic Theory

1. All matter is composed

of extremely small

particles called atoms

2. Atoms of a given

element are identical in

size, mass and other

properties; atoms of

different elements differ

in size, mass and other

properties.

3. Atoms cannot be

subdivided, created or

destroyed.

4. Atoms of different

elements combine in

simple whole-number

ratios to form chemical

compounds.

5. In chemical reactions,

atoms are combined,

separated or

rearranged.

Modern Atomic Theory

Leucippus

Democritus

Atomic Theory Tested by experiment and

modified with new discoveries and experiments

Section 2

The Structure of the Atom

Discovery of the Electron• Cathode Ray Tube Experiment -

Thompson• Observations

– Cathode Rays are deflected a magnetic field.

– Cathode rays are deflected from a negatively charged object.

– Charge to mass ratio is always the same for the cathode rays.

Discovery of the Electron• Cathode Ray Tube

Experiment - Thompson• Conclusion

– Cathode rays are composed of negatively charged particles

– Named “electrons”

Discovery of the Electron• Oil Drop Experiment - Millikan

– Measured the charge of the electron– Calculated the mass of an electron

•9.109 x 10-31 kg

Discovery of the Electron• Inferences

–Atoms are neutral, so there must be a positive charge.

–Electrons are small, so there must be other particles.

Discovery of the Electron• Plum Pudding Model

–Negative electrons were spread evenly throughout the positive charge.

Discovery of the Electron

Discovery of the Atomic Nucleus

• Gold Foil Experiment – Rutherford et. al– Hypothesis: Alpha particles would

pass through with slight deflection.– Observation: 1 in 8000 particles were

deflected back to the source.– Conclusion: The atom contains a

small densely packed bundle of matter with a positive charge

– Named the “nucleus”

Gold Foil Experiment Imagep

Discovery of the Atomic Nucleus

Relative size of the nucleus

Composition of The Atomic Nucleus

• Nuclei contain protons and neutrons• Neutral because number of protons

equal number of electrons• Each element has a different number of

protons in their nucleus – The number of protons determines

the atom’s identity• Nuclear forces hold protons & neutrons

together

Properties of Subatomic Particles

While the number of protons in the nucleus defines an element's

identity, variations on the number of neutrons in the nucleus give

rise to different isotopes of the same element.

Isotopes of hydrogen

Gold Foil Experiment Photo

Thompson and Rutherford Photo

Section 2 Homework

Ch 3 Sec 2 Review Page 76 #1-5

Section 3

Counting Atoms

Atomic Number

• The number of protons of each atom of that element

• Identifies the element

Isotopes• Atoms of the same element that

have different masses equals the isotope

• Isotopes do not differ significantly in their chemical behavior

Mass Numbers

• Mass numbers = # of p+ + # of n0

of a specific isotope

Designating Isotopes• Hyphen notation

– name of element – mass number– Hydrogen – 3

• Nuclear symbol

mass number

atomic number

Number of neutrons in an atom

neutrons = mass number – atomic number

(This is how to calculate isotopes…)

Nuclide – a general term for a specific isotope of an element

**Practice Problems page 87 #2-3

Relative Atomic Mass• One atom, carbon-12, is set as a

standard• All masses are expressed in

relation to this standard• 1 atomic mass unit = 1/12 the

mass of a carbon-12 atom

Relative Atomic Mass• Examples

– Hydrogen – 1 = 1.007825 amu– Oxygen – 16 = 15.994915 amu– Magnesium – 24 = 23.985042 amu

• p+ = 1.007276 amu, n0 = 1.008665 amu, e- = 0.0005486 amu

• Relative mass and mass number are close in value but not the same

Average Atomic Mass• The weighted average of the

atomic masses of the naturally occurring isotopes of an element

• Example– Copper

Cu-63: .6915 x 62.93 amu = 43.52Cu-65: .3085 x 64.93 amu = 20.03 63.55 amu

percentrelative mass

The Mole• An amount of a substance that

contains as many particles as there are atoms in exactly 12 g carbon-12.

• Similar to a dozen or a pair or a gross

• 6.022 x 1023 carbon-12 atoms = 12 grams of carbon-12

• Avogadro’s number = 6.022 x 1023 particles

Molar mass• The mass of one mole of a pure

substance • Unit = g/mol• On the periodic table, use 4 sig.

Gram-Mole Conversions• The conversion factor for gram-

mole conversion is molar mass.

• What is the mass, in grams, of 3.50 moles of Cu?– 222 grams Cu

Practice Problems page 851. What is the mass in grams of 2.25

mol of the element iron?2. What is the mass in grams of 0.357

mol of the element potassium?3. What is the mass in grams of

0.0135 mol of the element sodium?4. What is the mass in grams of 16.3

mol of the element nickel?

126 g Fe

14.7 g K

0.310 g Na

957 g Ni

Gram-Mole Conversions• The conversion factor for gram-

mole conversion is molar mass.

• A Chemist produced 11.9 g of Al. How many moles of Al were produced?– 0.411 moles Al

Practice Problems page 851. How many moles of calcium are in

5.00 g of calcium?2. How many moles of gold are in 3.60

x 10-5 g of gold?3. How many moles of zinc are in

0.535 g of zinc?

0.125 mol Ca

1.83 x 10-7 mol Au

8.18 x 10-3 mol Zn

Conversions with Avogadro’s Number

• The conversion factor for particle-mole conversion is Avogadro’s number.

• How many moles of silver are in 3.01 x 1023 atoms of silver– 0.500 moles Ag

OR6.022x1023atoms

1 mol 6.022x1023atoms

Practice Problems page 861. How many moles of lead are 1.50 x

1012 atoms of lead?2. How many moles of tin are in 2500

atoms of tin?3. How many atoms of aluminum are

in 2.75 mol of aluminum?

2.49 x 10-12 mol Pb

4.2 x 10-21 mol Sn

1.66 x 1024 atoms Al

Conversions with Avogadro’s Number

• The conversion factor for particle-mole conversion is Avogadro’s number.

• What is the mass, in grams, of 1.20x1018 atoms of Cu?– 1.27 x 10-4 g Cu

OR6.022x1023atoms

1 mol 6.022x1023atoms

Practice Problems page 871. What is the mass in grams of 7.5 x

1015 atoms of nickel?2. How many atoms of sulfur are in

4.00 g of sulfur?3. What mass of gold contains the

same number of atoms as 9.0 g of aluminum?

7.3 x 10-7 g Ni

7.51 x 1022 atoms S

66 g Au

Section 1 Homework

Section Review Page 87 #1-7

Chapter 3 pages 66 - 871 Modern Chemistry Chapter 3 Atoms: the building block of matter.

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