<<

Warm Up # Cathode Ray Tube

Evacuated tube Anode Cathode

− + 1. What did the cathode ray tube help to discover? Battery 2. Which scientist thought of using the cathode ray tube? How Differ a. Properties of Subatomic Particles

Particle Symbol Location Relative Relative Actual Charge mass (g)

outside 1 e- the 9.11 x nucleus -1 1840 10-28 g

in the 1.673 x p+ nucleus +1 1 10-24 g

in the 1.675 x 0 n nucleus 0 1 10-24 g Elements on the Periodic Table b. • the number of in an • Identifies element c. •represents the total number of protons and in the nucleus

Mass number A ZX atomic number d. • Atoms that have the same number of protons but have a different • Ex: 3 isotopes of :

12 13 14 6C 6C 6C e. Average • the weighted average of the isotopes of that element. • Formula:

% mass % Atomic mass abundance of abundance mass of ( ) ( of ) = of x + of x + … an element Isotope #2 Isotope #1 #1 Isotope #2 Average Atomic Mass • The mass of an atom is so small it is difficult to work with, so have developed an atomic standard to compare all the masses • The standard is the atomic mass unit (amu) • If the mass of an element is not close to a whole number, it is because the atom has several isotopes • The atomic mass is the weighted average of the isotopes of that element Example 1 has two naturally occurring isotopes. Ag-107 has an abundance of 51.82% and mass of 106.9 amu. Ag-109 has a relative abundance of 48.18% and a mass of 108.9 amu. Calculate the average atomic mass of silver. Example 1 • Silver has two naturally occurring isotopes. Ag-107 has an abundance of 51.82% and mass of 106.9 amu. Ag-109 has a relative abundance of 48.18% and a mass of 108.9 amu. Calculate the atomic mass of silver.

.5182(106.9 amu) + .4818(108.9 amu)

(remember to round at the end with more than one operation)

= 107.86 amu **Round to the hundredths for amu values Example 2 is a soft, silvery-white metal that has two 85 87 common isotopes, 37 Rb and 37 Rb. If the abundance of 85Rb is 72.2% with 84.911794 amu and the abundance of 87Rb is 27.8% with 86.909187 amu, what is the average atomic mass of rubidium? Example 2 Rubidium is a soft, silvery-white metal that has 85 87 two common isotopes, 37 Rb and 37Rb. If the abundance of 85Rb is 72.2% with 84.911794 amu and the abundance of 87Rb is 27.8% with 86.909187 amu, what is the average atomic mass of rubidium? .722(84.911794 amu) + .278(86.909187 amu)

(remember to round at the end with more than one operation) **Round to the = 85.467069 amu ≈ 85.47 amu hundredths for amu values Honors Example 3 has two naturally occurring isotopes. If the abundance of 11B is 80.10% with an amu of 11.0093, find the abundance of 10B. Honors Example 3 • Boron has two naturally occurring isotopes. If the abundance of 11B is 80.10% with an amu of 11.0093, find the amu of 10B. Hint: find the abundance first. 0.801(11.0093 amu) + 0.199(X amu) = 10.81 amu 8.8184493 amu + 0.199 (X amu) = 10.81 amu (remember to round at the end with more than one operation) 0.199 X amu = 1.981551 amu = 10.007793 amu for 10B ≈ 10.01 amu for 10B **Round to the hundredths for amu values Vocabulary to Know

• Atomic #- same # of protons & • Mass #-protons + neutrons 14 written 2 ways: Carbon-14 or 6 C • Isotopes-same # of protons, different # of neutrons • Atomic mass-weighted average mass Warm UP #12 1. What should your atomic mass be close to given the percent abundance for each isotope? Use CER to answer this. 2. Calculate the atomic mass of iron given the following data. Show your work. (Use 4 sig. fig.)

Isotope Percent Mass of Isotope Abundance (amu) Iron-54 5.845 % 53.94 Iron-56 91.754 % 55.93 Iron-57 2.119 % 56.94 Iron-58 0.282 % 57.93 The atomic mass should be closest to 55.93 amu, because 55.84 amu it is the most abundant isotope with a percent of 91.754% Procedure for Lab:Candium

1. 2. 3. 4. 5. 6.