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History Dmitri Mendeleev 1896 1951 - Seaborg Developed the modern periodic table.

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Presentation on theme: "History Dmitri Mendeleev 1896 1951 - Seaborg Developed the modern periodic table."— Presentation transcript:

1

2 History Dmitri Mendeleev 1896

3 1951 - Seaborg Developed the modern periodic table

4 The Whole Periodic table

5 Groups vs. Periods Need to Know: Halogens Noble Gas Alkali Metals Alkaline Earth Transition Metals Lanthanides Actanides Groups: Up and Down Periods: Left to Right

6 Periodic Table Atomic Number Atomic Mass Name Symbol **Mass Number can not be found on the periodic table

7 Element Symbol Symbol Atomic Number Mass Number

8

9 Atoms What do we know MUST be in an atom? (3 particles)

10 Ions How is an ion formed?

11 If an ion has a charge of -2, what has happened to the atom? Ions If an ion has a charge of +2, what has happened to the atom? The atom has gained two electrons The atom has lost two electrons

12 http://www.green-planet-solar-energy.com/images/helium-isotopes-5-to-8.gif Isotopes Count the neutrons and protons Proton Neutron

13 Natural State of an Element

14 Questions?

15 Quiz Time Acceptable Answers Correct Answer

16 Good Luck You have 7 minutes to complete your quiz. When you are done, hold on to it and wait patiently.

17 1 copper atom = 1.0552 x 10 -25 kg on average 69.2% of Cu atoms are 1.04497 x 10 -25 kg 30.8% of Cu atoms are 1.07815 x 10 -25 kg 1 copper penny = 3.1276 x 10 -3 kg 3.1276 x 10 -3 kg 1 Cu atom = 2.9640 x 10 22 atoms 1.0552 x 10 -25 kg Relative Size

18 Average Atomic Mass Calcs 1 atomic mass unit (amu) = 1.66054x10 -27 kg, Mass of an electron = 9.10939x10 -31 kg Mass of a proton = 1.67262x10 -27 kg Mass of a neutron = 1.67493x10 -27 kg **You will NOT need to memorize these numbers. Always given to you. Relative Size

19 Carbon has 2 common isotopes. 12 6 C and 13 6 C. If the abundance of the two isotopes is 98.89% and 1.11 % respectively, what is the average atomic mass of carbon in amu? Average Atomic Mass Calcs Carbon 12Carbon 13 Neutrons Protons Electrons

20 12 6 C 6 neutrons1.67493x10 -27 kg1 amu =6.05200 amu 1 neutron1.66054x10 -27 kg 6 electrons9.10939x10 -31 kg1 amu =0.0032915 amu 1 electron1.66054x10 -27 kg 6 protons1.67262x10 -27 kg1 amu =6.04364 amu 1 proton1.66054x10 -27 kg 6.04364 amu + 6.05200 amu + 0.0032915 amu =12.09893 amu

21 6 protons1.67262x10 -27 kg1 amu =6.04364 amu 1 proton1.66054x10 -27 kg 7 neutrons1.67493x10 -27 kg1 amu =7.06066 amu 1 neutron1.66054x10 -27 kg 6 electrons9.10939x10 -31 kg1 amu =0.0032915 amu 1 electron1.66054x10 -27 kg 6.04364 amu + 7.06066 amu + 0.0032915 amu = 13.10755 amu 13 6 C

22 Putting It All Together What we know: Carbon 12: 12.09893 amu (98.89% abundance) Carbon 13: 13.10755 amu (1.11% abundance) Therefore: (12.09893 amu x 0.9889) + (13.10755 amu x 0.0111) 12.01101 amu

23 Homework *Attempt problem #1 on Average Atomic Mass Calculations WS Read and notes 3.5, 3.7, 3.8


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