Agenda 10/6/2017 Per 2. Isotope - One More Time

Agenda 10/6/2017 Per 2. Isotope - one more time https://drive.google.com/drive/folders/0BzGRyLjDEz_XdFJrN2ZBNkM4TDg

• Slip Quiz • Modern Model of the Atom • Limitations of Model 1 in Isotopes Pogil • Average Atomic Mass • Looking for Patterns - Analysis Activity • Homework Slip Quiz

1.How many neutrons in an atom of neon-21 (21Ne). State clearly how you figure it out.

2. How many electrons in a neon-21 atom? How do you know? Slip Quiz 1.How many neutrons in an atom of neon-21 (21Ne). State how you know. I find neon’s atomic number on periodic table, neon has atomic number 10 which tells me it has 10 protons in its nucleus. The remainder of the particles in the nucleus are neutrons. Mass number - atomic number = number of neutrons 21 - 10 = 11 So Ne-21 has 11 neutrons. Slip Quiz 2. How many electrons in a neon-21 atom? How do you know? I know that for an electrically neutral atom, the number of protons in the nucleus is equal to the number of electrons in the electron cloud around the nucleus. For question 1 I found the atomic number of neon-21 is 10, hence the atoms have 10 protons and must also have 10 electrons. Isotopes Back to last Extension Questions 17. What characteristics of Model 1 are inconsistent with our understanding of what atoms look like?

Note, limitations of models always exist, and we should be aware of them so we don’t try to use the model in situations where it does not really apply. Isotopes Back to last Extension Questions 17. What characteristics of Model 1 are inconsistent with our understanding of what atoms look like? A scientific model is a representation (mathematical, conceptual or even physical) to help us understand, explain and predict phenomena that cannot be easily observed directly. Limits of the Atom Model in our Pogil packet. 17. What characteristics of Model 1 are inconsistent with our understanding of what atoms look like?

??? Limits of the Atom Model in our Pogil packet. 17. What characteristics of Model 1 are inconsistent with our understanding of what atoms look like? •Electrons are much smaller than protons and neutrons yet in Model 1 they are all represented by similar sized symbols. •Electrons are much further from the nucleus than represented in Model 1. •Atoms and all the subatomic particles are 3Dimensional and not flat • Protons are not black Average Atomic Mass How are the masses on the periodic table determined?

Why? Most elements have more than one naturally occurring isotope. As you learned previously, the atoms of those isotopes have the same atomic number (number of protons), making them belong to the same element, but they have different mass numbers (total number of protons and neutrons) giving them different atomic masses. Skip to page 4 Average Atomic Mass How are the masses on the periodic table determined? So which mass is put on the periodic table for each element? Is it the most common isotope’s mass? The heaviest mass? This activity will help you answer that question. Average Atomic Mass How are the masses on the periodic table determined? Model 1 - A Strip of Magnesium Metal

1. Write in the atomic number for each Mg atom in Model 1.

24Mg 25Mg 26Mg

What, where? Average Atomic Mass How are the masses on the periodic table determined? Model 1 - A Strip of Magnesium Metal

1. Write in the atomic number for each Mg atom in Model 1. 24Mg 25Mg 26Mg 12 12 12

2. What are the mass numbers of the naturally occurring isotopes of magnesium? Average Atomic Mass How are the masses on the periodic table determined? Model 1 - A Strip of Magnesium Metal

1. Write in the atomic number for each Mg atom in Model 1. 24Mg 25Mg 26Mg 12 12 12

2. What are the mass numbers of the naturally occurring isotopes of magnesium? 24, 25 and 26 3. Do all of the atoms in magnesium in Model 1 have the same atomic mass? Explain. 3. Do all of the atoms in magnesium in Model 1 have the same atomic mass? Explain. All of the atoms of magnesium in the model do not have the same atomic mass. There are three different isotopes of magnesium shown in the model. The three isotopes have different mass numbers and so the atomic mass of the atoms of each isotope will also be different. 4. For the sample of 20 atoms, draw a table indicating the mass numbers of the three isotopes and the number of atoms of each isotope present. 4. For the sample of 20 atoms, draw a table indicating the mass numbers of the three isotopes and the number of atoms of each isotope present. 24Mg 25Mg 26Mg

Number 16 2 2 of atoms 5. Which isotope of magnesium is most common in Model 1?

24Mg 25Mg 26Mg

Number 16 2 2 of atoms 6. ...for every 10 atoms of magnesium, approximately how many atoms of each isotope will be found?

24Mg 25Mg 26Mg

Number 16 2 2 of atoms 8 1 1 6. ...for every 10 atoms of magnesium, approximately how many atoms of each isotope will be found? approx. one Mg-25, one Mg-26 and eight Mg-24 atoms 24Mg 25Mg 26Mg

Number 16 2 2 of atoms 8 1 1 Model 2 - Natural Abundance Information for Magnesium a. 78.99/100 x 20 = 15.8

=16atoms

10/100 x 20 = 2 atoms

11.01/100 x 20 = 2.2 = 2 atoms b) Is Model 1 accurate…? Model 2 - Natural Abundance Information for Magnesium b) Is Model 1 accurate…? Yes, for the limited number of atoms shown in Model 1, the percentage is close to the actual values for natural abundance of magnesium isotopes (accurate). 8. If you could pick up an atom and put it on a balance, the mass of that atom would most likely be ______amu because 8. If you could pick up an atom and put it on a balance, the mass of that atom would most likely be 23.9850 amu because there are more Mg-24 in the sample than any other isotope.

This isotope is most common (most abundant), so chances are highest that I would pick an atom of that mass. 9. From periodic table mass of Mg is 24.305 amu Does the decimal number shown on the periodic table for magnesium match any of the atomic masses listed in Model 2? No 10. The periodic table does not show the atomic mass of every isotope for an element. a. There is not enough room in the boxes on the periodic table to put information about 2, 3, or even more isotopes. 10. The periodic table does not show the atomic mass of every isotope for an element. b. In most cases a scientist will be working with a mixture of all the (naturally occurring) isotopes of an element, so information about individual isotopes in not necessary. 11. What would be a practical way of showing the mass of magnesium atoms in the periodic table given that most elements occur as a mixture of isotopes? 11. What would be a practical way of showing the mass of magnesium atoms in the periodic table given that most elements occur as a mixture of isotopes?

Calculate an average value. 12. Propose a possible way to calculate the average atomic mass of 100 magnesium atoms. 12. Add up all the masses of all the 100 atoms in the sample and divide by the number of atoms (100) in the sample.

Find the weighted average mass of atoms in the sample. Model 3 – Proposed Average Atomic Mass Calculations 13. Complete the proposed calculations for the average atomic mass of magnesium in Model 3. Model 3 – Proposed Average Atomic Mass Calculations

Mary’s method (78.99)(23.9850 amu) + (10.00)(24.9858 amu) +(11.01)(25.9826amu) 100 = 24.305 amu

Jack’s Method (0.7899)(23.9850 amu) + (0.1000)(24.9858 amu) + (0.1101)(25.9826amu) = 24.305 amu

Alan’s method (23.9850 amu) + (24.9858 amu) + (25.9826 amu) = 24.984 amu 3 Model 3 – Proposed Average Atomic Mass Calculations

Mary’s method (78.99)(23.9850 amu) + (10.00)(24.9858 amu) +(11.01)(25.9826amu) 100 = 24.305 amu

Jack’s Method (0.7899)(23.9850 amu) + (0.1000)(24.9858 amu) + (0.1101)(25.9826amu) = 24.305 amu

Alan’s method ONLY found average for 3 atoms (23.9850 amu) + (24.9858 amu) + (25.9826 amu) = 24.984 amu 3 assumes equal representation of all isotopes 14.c. show Mary and Jack’s methods are mathematically equivalent models.

Mary’s method (78.99)(23.9850 amu) + (10.00)(24.9858 amu) + (11.01)(25.9826 amu) 100 =78.99)(23.9850 amu) + (10.00)(24.9858 amu) + (11.01)(25.9826 amu) 100 100 100

=(0.7899)(23.9850 amu)+(0.1000)(24.9858 amu)+(0.1101)(25.9826amu) =

=(0.7899)(23.9850 amu) + (0.1000)(24.9858 amu)+(0.1101)(25.9826amu)

= 24.305 amu Jack’s Method 15. Use one of the methods in Model 3 (Mary or Jack) to calculate the average atomic mass for oxygen. Isotope information is provided below. Show all of your work and check your answer against the mass listed on the periodic table.

Use separate paper - or back of one of the sheets. 15. Calculation of Average atomic mass for oxygen

Mary’s method (99.76)(15.9949 amu) + (0.04)(16.9991 amu) + (0.20)(17.9992 amu) 100 =

=(0..9976)(15.9949 amu) + (0.0004)(16.9991 amu)+(0.1101)(17.9992amu)

= 24.305 amu Jack’s Method 15.9565122 amu 0.00679964 amu + 0.0359984 amu

15.99931024 amu

Given 6 sig digs with original measurements

15.9993 amu vs. PT value 15.999 amu Read This! PAGE 4

Recall that all isotopes of an element have the same physical and chemical properties, with the exception of atomic mass (and for unstable isotopes, radioactivity). Therefore, the periodic table lists a weighted average atomic mass for each element. In order to calculate this quantity, the natural abundance and atomic mass of each isotope must be provided. 16. Consider the individual atomic masses for magnesium isotopes give in Model 2. a) Which isotope has an atomic mass closest to the average atomic mass

listed on the periodic table? 16. Consider the individual atomic masses for magnesium isotopes give in Model 2. a) Which isotope has an atomic mass closest to the average atomic mass

listed on the periodic table?

24 Mg b) Give a mathematical reason for answer to part a. Weighted average atomic mass cont. 24Mg is the most common isotope and is thus most heavily “weighted” in the equation for average atomic mass. There are more 24Mg atoms in a sample than atoms of the other isotopes since it is the most abundant naturally occurring isotope. The mass of 24Mg atoms contribute the most to the average mass of the sample. Weighted average atomic mass cont.

17. Boron has two naturally occurring isotopes: boron-10 and boron-11. Which isotope is more abundant on Earth? Use grammatically correct sentences to explain the

answer. Weighted average atomic mass cont. The expected mass of boron-10 would be 10 amu and the expected mass of boron

-11 would be 11 amu. On the periodic table the average atomic mass for boron is listed as 10.811amu, which is numerically closer to 11 than it is to 10. Therefore we can conclude that boron-11 is more abundant on Earth than boron-10. Write in space on Atomic Mass Units back sheet. (amu) The masses of protons, neutrons and electrons in g are very small and difficult to work with.

Chemists developed a method of measuring the mass of an atom relative to the mass of a specifically chosen atomic standard. Atomic Mass Units (amu)

…atomic standard

Carbon-12 atom One carbon-12 atom was assigned a mass of exactly 12 atomic mass units (amu). One atomic mass unit (amu) is defined as 1/12th (one twelfth) of the mass of a carbon-12 atom. 1 amu is close to the mass of a single proton or single neutron. See p 102 Table 4-2 Finding average atomic mass and using it to identify an element Given the data in the table below, calculate the atomic mass of unknown element x. Then identify the unknown element, which is used medically to treat some mental disorders. Isotope Abundance for Element X

Isotope Mass Percent (amu) abundance 6X 6.015 7.5%

7X 7.016 92.5% Finding average atomic mass and using it to identify an element For isotope 6X Mass contribution = (6.015 amu)(0.0750) = ______amu

For isotope 7X Mass contribution = (7.016 amu)(0.925) = _____amu Finding average atomic mass and using it to identify an element For isotope 6X Mass contribution = (6.015 amu)(0.0750) = 0.4511 amu

For isotope 7X Mass contribution = (7.016 amu)(0.925) = 6.490 amu

Sum the mass contributions to find the average atomic mass Av. Atomic mass of X = ______amu = ______amu Finding average atomic mass and using it to identify an element For isotope 6X Mass contribution = (6.015 amu)(0.0750) = 0.4511 amu

For isotope 7X Mass contribution = (7.016 amu)(0.925) = 6.490 amu

Average Atomic mass of X = (0.4511+6.490)amu= ______amu Finding average atomic mass and using it to identify an element For isotope 6X Mass contribution = (6.015 amu)(0.0750) = 4.511 amu

For isotope 7X Mass contribution = (7.016 amu)(0.925) = 6.491 amu Average Atomic mass of X =(4.511 + 6.491)amu= 6.941 amu Look it up in the periodic table. Element with av. atomic mass of 6.941amu is ______Finding average atomic mass and using it to identify an element For isotope 6X Mass contribution = (6.015 amu)(0.0750) = 4.511 amu

For isotope 7X Mass contribution = (7.016 amu)(0.925) = 6.491 amu Average Atomic mass of X =(4.511 + 6.491)amu= 6.941 amu Look it up in the periodic table. Element with av. atomic mass of 6.941amu is Lithium. Properties of Subatomic Particles ( 4-1 p 97, 4-2 p 102)

Electron Proton Neutron 0 - + n Symbol e p

in cloud In the Nucleus of In the Nucleus of

Location surrounding the atom the atom the nucleus

Relative electrical 1- 1+ 0 charge

Relative mass (amu) 1 1.007 276 1.008 665 1840 0.000 549 Actual mass (g) 9.11 x 10 –28 1.673 x 10 – 24 1.675 x 10 – 24 Add to your “Periodic Table of The Elements”

Average Add to your “Periodic Table of The Elements” • Atomic number defines the element name • Average is the number of protons in the nucleus of each atom of the element • Atomic number defines the element name

Average • is the number of protons in the nucleus of each atom of this

All atoms are neutral element have no net charge

Number of protons (+) = number of electrons (-) np + (ne-) = 0 Periodic Table of the Elements Why are some average atomic masses in ( ) - everything after 92? (U) ?

50 a Isotopes and Atomic Structure Complete the table - what patterns do you notice? Isotope Atomi Mass Numbe Numbe numbe Neutro Av. Most c numbe r r of r of n/ atomic abundan Numb r proton neutro electro proton mass t isotope er s ns ns ratio elemen t (amu) Hydroge 1 1 n-1 carbon-1 6 12 6 6 6 6/6 = 1 12.011 C-12 2 Oxygen- 8/8 = 1 O-16 16 fluorine- 9 19 19 neon-20 10 10 sodium- 23 aluminu 13 m-27 cobalt-_ 27 32/27 = __ 1.185 krypton- 36 48 __ uranium -__ Isotopes and Atomic Structure Complete the table - what patterns do you notice? Isotope Atomi Mass Numbe Numbe numbe Neutro Av. Most c numbe r r of r of n/ atomic abundan Numb r proton neutro electro proton mass t isotope er s ns ns ratio elemen t (amu) Hydroge 1 1 1 0 1 ------1.0079 H-1 n-1 carbon-1 6 12 6 6 6 6/6 = 1 12.011 C-12 2 Oxygen- 8/8 = 1 O-16 16 fluorine- 9 19 19 neon-20 10 10 sodium- 23 aluminu 13 m-27 cobalt-_ 27 32/27 = __ 1.185 krypton- 36 48 __ uranium -__ Isotopes and Atomic Structure Complete the table - what patterns do you notice? Isotope Atomi Mass Numbe Numbe numbe Neutro Av. Most c numbe r r of r of n/ atomic abundan Numb r proton neutro electro proton mass t isotope er s ns ns ratio elemen t (amu) Hydroge 1 1 1 0 1 ------1.0079 H-1 n-1 carbon-1 6 12 6 6 6 6/6 = 1 12.011 C-12 2 Oxygen- 8 16 8 8 8 8/8 = 1 15.999 O-16 16 fluorine- 9 19 9 10 9 10/9 18.998 F-19 19 =1.11 neon-20 10 20 10 10 10 10/10 20.180 Ne-20 =1 sodium- 11 23 11 12 11 12/11= 22.990 Na-23 23 1.09 aluminu 13 27 13 14 13 14/13 26.982 Al-27 m-27 =1.077 cobalt-_ 27 32/27 = __ 1.185 krypton- 36 48 __ uranium -__ Isotopes and Atomic Structure Complete the table - what patterns do you notice? Isotope Atom Mass Numbe Numbe numbe Neutro Av. Most ic numbe r r of r of n/ atomic abundan Num r proton neutro electro proton mass t isotope ber s ns ns ratio elemen t (amu) aluminu 13 27 13 14 13 14/13 26.982 Al-27 m-27 =1.077 cobalt-59 27 59 27 32 27 32/27 = 58.933 Co-59 1.185 krypton- 36 36 36 48 36 48/36 = 83.80 Kr-84 84 +48= 1.333 84 niobium- 41 41 + 52 41 52 41 52/41 92.906 Nb-93 93 = 93 =1.268

uranium- 92 238 92 238-92 92 146/92 238.03 U-238 238 = 146 = 1.587 Isotopes and Atomic Structure Complete the table - what patterns do you notice? Isotope Atom Mass Numbe Numbe numbe Neutro Av. Most ic numbe r r of r of n/ atomic abundan Num r proton neutro electro proton mass t isotope ber s ns ns ratio elemen t (amu) xenon-13 54 134 54 134 -54 54 80/54 = 131.29 Xe-131 4 = 80 1.482 Lead-207 82 207 82 207-82 82 125/82 207.2 Pb-207 = 125 = 1.524 thulim-1 69 169 69 100 69 100/69 168.93 Tm-169 69 =1.449 4 uranium- 92 238 92 238-92 92 146/92 238.03 U-238 238 = 146 = 1.587 bohrium- 107 264 107 264-10 107 167/10 ?? ?? 264 7 = 167 7 =1.561 Section 25.2 Radioactive Decay copy for Nuclear Stability homework

Neutron-to-proton (n/p) ratio and

nuclear stability. To some degree, the stability of a nucleus can be correlated with its neutron-to-proton ratio. For light atoms 1/1 ratio, for heavier atoms closer to 1.5/1 Section 25.2 Radioactive Decay copy for Nuclear Stability homework The strong nuclear force acts on protons and neutrons that are very close together in a nucleus to overcome electrostatic repulsion

between the protons. copy for Section 25.2 Radioactive Decayhomework Nuclear Stability ΔE = Δmc2 As atomic number increases, more neutrons are needed to produce the strong nuclear force to balance the electrostatic repulsion forces between the protons. Plotting the number of neutrons versus the number of protons for all stable

nuclei illustrates the band of stability.

The Band of Stability copy for Ratio gradually homework increases to maximum of 1.5: 1 for largest atoms

For atoms with low atomic number (<20) 1:1 n/p ratio page 811 textbook Section 25.2 Radioactive Decay copy for Nuclear Stability homework

Radioactive nuclei are found outside the band of stability - either above or below - and undergo radioactive decay in order to gain stability. After decay, the new atom is positioned more closely to, if not within, the band of stability. Section 25.2 Radioactive Decay copy for Nuclear Stability homework

The band of stability ends at bismuth-209 ; all the elements with atomic numbers greater than 83 are radioactive. Section 25.2 Types of Radioactive copy for homework - see page 811 in textbook for correct Decay appearance of equation which is not working with my google slides conversion to a pdf. Alpha decay 210 206 4 Po Pb + He 84 82 2

alpha (helium nucleus) particle emitted from nucleus of Po-210. What happens to neutron/proton ratio? Section 25.2 Types of Radioactive copy for homework - see page 811 in textbook for correct Decay appearance of equation which is not working with my google slides conversion to a pdf. Beta decay 14 14 0 C N + β 6 7 -1

beta particle emitted from nucleus of C-14 (a neutron splits into a proton and the beta particle that is emitted and we

detect as radioactive decay.) copy for homework Properties of Alpha, Beta and Gamma Radiation Radiation type Charge Mass Relative penetrating power -24 Alpha (α ) 2+ 6.64 x 10 kg Blocked by paper (least penetrating) -28 Beta (β) 1- 9.11 x 10 kg Blocked by thin metal sheets (foil) Gamma (γ) 0 0 Not completely (photons) blocked by lead or concrete (most penetrating) Types of Radiation – Penetrating copy for Power homework

Gamma rays are very dangerous because they penetrate living tissue Skin can stop alpha radiation Beta radiation – usually only penetrates 1-2cm beneath the skin Alpha, beta and gamma are “ionizing radiation” – they have enough energy to break bonds in molecules which ionizes them, which makes them unstable, and very reactive inside the organism Find out more at Ionizing vs. nonionizing radiation at the EPA website. The Discovery of Radioactivity copy for homework • 1895 Roentgen (X-rays) • 1895 Becquerel • 1898 The Curies (Marie and Pierre) • 1903 The Curies and Becquerel • 1911 Marie Curie

Today we call isotopes that are radioactive, radioisotopes. Section 25.3 Transmutation copy for Vocabulary homework Half-life the time required for one half of a radio-isotope’s nuclei to decay into

its products Radiochemical dating – process of determining the age of an object by measuring the amount of certain radioisotopes remaining in the object Homework Copy the notes from the slides marked above. On the Study Guide packet for the test: Draw “Paper chromatography” equipment set up on Uses of Chromatography paper. Label the distillation diagram. Compete the rest of the packet using your textbook. (Chapter 4) Heads up - Unit test - Classification of Matter and Atomic Structure - Thursday October 12th. (Sections 3.3, 3.4, and Chapter 4 in textbook)