preap chemistry chapter 3 chapter 3 outlines due to the box now not later today!
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PreAP Chemistry Chapter 3 Chapter 3 Outlines Due to the box NOW Not later today!. Democritus was the early (around 400BC) Greek philosopher who is credited with the concept of the atom ( atomos ) –which means invisible. - PowerPoint PPT PresentationTRANSCRIPT
PreAP Chemistry Chapter 3
Chapter 3 Outlines Due to the box NOW
Not later today!
Democritus was the early (around 400BC) Greek philosopher who is credited with the concept of the atom (atomos) –which means invisible
Dalton (around 1800AD) is an English school teacher who proposed the law of conservation of mass, the law of definite proportions, and the law of multiple proportions.His many experimentswith gases proved these laws are true, ifatoms exist.
Dalton is also known as the Father of the (Modern) Atomic Theory
Dalton’s atomic theory:1.All matter is composed of very small particles called atoms
Dalton’s atomic theory:1.All matter is composed of very small
particles called atoms2.Atoms of a given element are
identical in size, mass, and other properties; atoms of different elements differ in these properties.
3.Atoms cannot be subdivided, created, or destroyed
3.Atoms cannot be subdivided, created, or destroyed
4.Atoms of different elements combine in simple whole-number ratios to form chemical compounds.
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.
Two aspects of Dalton’s atomic theory proven to be incorrect:
a.We now know atoms are divisible.
Two aspects of Dalton’s atomic theory proven to be incorrect:
a.We now know atoms are divisible.
b. Atoms of the same element can have different masses.
Atom --smallest particle of an element that retains the properties of that element.
J. J. Thomson is the man credited with the discovery of the electrons in the late 1800’s, using cathode ray tubes
Millikan calculated the mass of the electron (very, very small)
Knowledge of electrons led to two inferences about atomic structure:
1.Because atoms are electrically neutral, they must contain positive charge to balance the negative electrons.
Knowledge of electrons led to two inferences about atomic structure:
1.Because atoms are electrically neutral, they must contain positive charge to balance the negative electrons.
2. Because electrons have so little mass, atoms must contain other particles to account for most of their mass
Nucleus of the atom—discovered by Lord Ernest Rutherford
Nucleus of the atom—discovered by Lord Ernest Rutherford
Gold foil experiment—actually done by Hans Geiger and Ernest Marsden
Observations:a.Majority of the alpha (α) particles
penetrated foil undeflected.
b.About 1 in 20,000 were slightly deflected
c.About 1 in 20,000 were deflected back to emitter
Conclusions: 1. Mass of the atom and the positive charge are concentrated in small regions called nucleus
Conclusions: 1. Mass of the atom and the positive charge are concentrated in small regions called nucleus
2. Most of the atom is empty
Conclusions: 1. Mass of the atom and the positive charge are concentrated in small regions called nucleus
2.Most of the atom is empty
3.Magnitude of charge on the nucleus is different for different atoms
4. Number of electrons outside the nucleus = number of units of nuclear charge (to account for the fact that the atom is electrically neutral)
4. Number of electrons outside the nucleus = number of units of nuclear charge (to account for the fact that the atom is electrically neutral)
Atoms are electrically neutral because they contain equal numbers of protons and electrons
A couple years later Rutherford presented evidence for a neutral particle which was also in the nucleus and contained a similar mass to that of a proton – called a neutron
Mass of one proton = mass of neutron = mass of 1837 electrons
Mass of one proton = mass of neutron = mass of 1837 electrons
Thus the total mass of an atom is basically the sum of the protons and neutrons, called the atomic mass or mass number, abbreviated A
Atomic number—the number of protons in the nucleus of the atom.
Atomic number—the number of protons in the nucleus of the atom.
--number of protons identifies the element and is equal to the number of
electrons (of a neutral atom)
--symbol is Z
Isotopes are atoms of the same element that have different masses because they have different numbers of neutrons but they still have similar chemical properties
Mass Number of Isotope
Number of Protons Number of Neutrons
8 6 2
9 6 3
10 6 4
11 6 5
12 6 6
13 6 7
14 6 8
15 6 9
16 6 10
17 6 11
18 6 12
19 6 13
20 6 14
Isotopes of Carbon
Isotopes of CarbonMass Number of Carbon
IsotopesName of Isotopes
8 carbon-8
9 carbon-9
10 carbon-10
11 carbon-11
12 carbon-12
13 carbon-13
14 carbon-14
15 carbon-15
16 carbon-16
17 carbon-17
18 carbon-18
19 carbon-19
20 carbon-20
Nuclide is the general term for any isotope of any element
Nuclide is the general term for any isotope of any element
Atomic Mass Unit (amu) is exactly 1/12 the mass of a carbon-12 atom
Average atomic mass is the weight average of the atomic masses of the naturally occurring isotopes of an element.
Ave. Atomic mass = %abundance(mass of isotope 1) + %abundance(mass of isotope 2) +…..
Example 1:Element Sciencium has two isotopes.
Sciencium-301 has an abundance of 59.5%, and Sciencium-304 is the other. What is the average atomic mass?
301 amu x .595 = 179 amu 304 amu x .405 = 123 amu
302 amu
Example 2:Element Pepsium has an average
atomic mass of 335. Two isotopes of Pepsium exist. If Pepsium-327 is 30.5% abundant, then what is the second isotope?
327 amu x 0.305 = 99.7 amu ? amu x 0.695 = ? amu
335 amu
1 – 0.305 = 0.695
Example 2:Element Pepsium has an average
atomic mass of 335. Two isotopes of Pepsium exist. If Pepsium-327 is 30.5% abundant, then what is the second isotope?
327 amu x 0.305 = 99.7 amu ? amu x 0.695 = ? amu
335 amu
Step 1: Find the missing weighted mass
Example 2:Element Pepsium has an average
atomic mass of 335. Two isotopes of Pepsium exist. If Pepsium-327 is 30.5% abundant, then what is the second isotope?
327 amu x 0.305 = 99.7 amu ? amu x 0.695 = 235 amu
335 amu
335 – 99.7 = 235.3
Example 2:Element Pepsium has an average
atomic mass of 335. Two isotopes of Pepsium exist. If Pepsium-327 is 30.5% abundant, then what is the second isotope?
327 amu x 0.305 = 99.7 amu ? amu x 0.695 = 235 amu
335 amu
Step 2: Find the missing mass
Example 2:Element Pepsium has an average
atomic mass of 335. Two isotopes of Pepsium exist. If Pepsium-327 is 30.5% abundant, then what is the second isotope?
327 amu x 0.305 = 99.7 amu 339 amu x 0.695 = 235 amu
335 amu
235.3 ÷ 0.695 = 339
Unstable Nuclei and Radioactive Decay
1) When referring to nuclear reactions people commonly think of nuclear fission (the splitting of large atoms into smaller pieces)
1) and nuclear fusion (the combining of small atoms into one large one), but on earth these reactions do not occur naturally.
2) Naturally occurring nuclear reactions result from the unusual number of neutrons of an isotope which makes it unstable (unusually high in energy). This often results in the isotope changing from one element into another element in an attempt to become more stable (lower in energy).
A) These reactions are called nuclear reactions, as they involve changes in the nucleus.
A)These reactions are called nuclear reactions, as they involve changes in the nucleus.
B)During these nuclear reactions, rays and particles are given off, which is called radiation.
C) Sometimes an unstable nucleus will change into many different elements as it tries to become more stable. This is called radioactive decay.
3) When radioactive decay occurs, there are three different types of radiation that can be given off. Each type has a different mass, and sometimes a charge.
A) The first type of radiation to be discovered was called alpha radiation and came from alpha particles.
i) Because the mass numbers must be equal, 226 = 222 + x. So the mass of the alpha particle must be
i) Because the mass numbers must be equal, 226 = 222 + x. So the mass of the alpha particle must be 4.
ii) Because the atomic numbers must be equal, 88 = 86 + x. So the atomic number of the alpha particle must be.
ii) Because the atomic numbers must be equal, 88 = 86 + x. So the atomic number of the alpha particle must be 2.
iii) The element Helium has a mass of 4 and an atomic number of 2, so the alpha particle is just like a helium atom without any electrons;
4He or
4 2 2
B) The second type of radiation to be discovered was called beta radiation and came from beta particles.
i) Because the mass numbers must be equal, 14 = 14 + x. So the mass of the beta is
i) Because the mass numbers must be equal, 14 = 14 + x. So the mass of the beta is zero.
ii) Because the atomic numbers must be equal, 6 = 7 + x. So the atomic number must be
ii) Because the atomic numbers must be equal, 6 = 7 + x. So the atomic number must be -1.
iii) The electron has a mass of zero and a charge of -1, so the beta particle is just like an electron;
0
-1 β
C) The last type of radiation to be discovered was called gamma radiation and came from gamma particles.
i) Because the mass numbers must be equal, 238 = 234 + 4 + x. So the mass of the gamma particle must be
i) Because the mass numbers must be equal, 238 = 234 + 4 + x. So the mass of the gamma particle must be zero.
ii) Because the atomic numbers must be equal, 92 = 90 + 2 + x. So the atomic number must also be zero.
iii) The gamma particle was the last to be found because it has no mass and no charge;
00γ
Geiger Counter
Alpha, Beta, and GammaRadiation