Atomic Number, Mass Number, Atomic Mass and Isotopes

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Presentation transcript:

Atomic Number, Mass Number, Atomic Mass and Isotopes

Atomic Number (Z): is the number of protons in the nucleus of the atom. Z=#p+ The number of protons (atomic number) determine the identity of an element.

What about the electrons? How many? Atoms have no overall electrical charge (neutral) so, an atom must have as many electrons as there are protons in its nucleus. # p+ =# e- The atomic number of an element also equals the number of electrons in a neutral atom of that element Z =#e-

Ex: Sodium 11 11 11 What is the atomic number of Sodium? How many protons does sodium have? How many electrons does sodium have? 11 11 11

A=#p+ + #n0 Mass Number (A): The sum of the protons and neutrons in the nucleus. A=#p+ + #n0

Standard atomic Notation

Other ways to write elements: 36 17Cl Cl-36 Mass Number Mass Number Atomic Number

Nucleons: protons and neutrons

Isotopes Atoms of the same element with Same number of protons Same number of electrons Different number of neutrons They have different mass numbers because they have different numbers of neutrons, but they have the same atomic number because they have the same number of protons.

Isotopes Remember: Isotopes have: same # of protons Do isotopes have similar chemical properties? Why? Yes – electrons are the same Do they have similar physical properties? Why? NO – mass is different. Remember: Isotopes have: same # of protons different # of neutrons

Example: Isotopes of Carbon and Hydrogen protium deuterium tritium H H H Isotopes of Hydrogen Isotopes of Carbon

Atomic Mass Unit is a unit used to compare the masses of atoms and has the symbol u or amu.

is approximately equal to the mass of a single proton or neutron. 1 amu or u is approximately equal to the mass of a single proton or neutron.

Carbon-12 Chemists have defined the carbon-12 atom as having a mass of 12 atomic mass units. For carbon: Mass = 12.01 99% 12C AND 1% 13C found in nature 12.01 = average mass based on abundance NOT just 12

Atomic Mass is the weighted average mass of all the naturally occurring isotopes of that element.

How do we deal with the fact that some atoms have several isotopes? The solution is to measure the mass of each individual isotope in the element and then calculate the weighted average. The weighted average is known as the atomic mass of the element

MARBLE ACTIVITY

Types of calculations Finding atomic mass, given mass and abundance of each Finding the mass of one of the isotope given atomic mass and abundance. Finding abundance given atomic mass and mass of each NOTE – if atomic mass is not given for 2 and 3 above use the mass in the periodic table.

Problem 1- finding atomic mass An element consist of two isotopes. Isotope A has an abundance of 75.00 percent, and its mass is 14.000 atomic mass units. Isotope B has an abundance of 25.00 percent, and its mass is 15.000 atomic mass units. What is the atomic mass of the element? Atomic mass = (14.00u x 0.7500) + (15.000 u x 0.2500) = 14.25

Problem 2 – finding mass of one isotope Bromine has two naturally occuring isotopes. Bromine-79 has a mass of 78.918amu and is 50.69% abundant. Using the atomic mass reported in the periodic table, determine the mass of bromine-81, the other isotope of bromine. Atomic mass = (MBr79 x Ra) + (MBr81x Ra) Ra for Br81 = 100 – 50.69 = 49.31 79.90(mass from periodic table) = (78.917 x 0.5069) + (MBr81 x 0.4931) 79.90 = 40.003 + 0.4931Br 79.90 – 40.003 = 0.4931Br 39.897 = 0.4931 Br 80.91 amu = Br

Problem – 3. finding relative abundance Gallium consists of two naturally occuring isotopes with masses of 68.926 and 70.925 amu. The average atomic mass of Ga is 69.72 amu. Calculate the abundance of each isotope. Note – if atomic mass is not given use the mass in the periodic table.

Answer should be:69Ga: 60.3%, 71Ga: 39.7% Average atomic mass = (mass 69Ga)(x) + (mass 71Ga)(y) Where x and y are the relative abundances of 69Ga and 71Ga respectively. So, 69.72 = 68.926(x) + 70.925(y) Cant have 2 unknowns????? Let the abundance of 1 isotope = x, then the abundance of the other will be 1-x . In this way the problem will have 1 unknown. (Substitute y for 1-x) So 69.72 = (68.926) (x) + (70.925)(1-x)                                     69.72 =  68.926x + 70.925 - 70.925x                                 - 1.205 = -1.999x                                           x = 0.60280 = 0.603 x 100 = 60.3%