# How Atoms Differ October 2, 2014. Reading an Element  A – Atomic Number: the number of protons in an atom (this identifies the element)  B – Average.

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How Atoms Differ October 2, 2014

Reading an Element  A – Atomic Number: the number of protons in an atom (this identifies the element)  B – Average Atomic Mass  C – Chemical Symbol  D – Chemical Name

Atomic Number - Z  All atoms are neutral, the charges of the protons and electrons must cancel each other out  Atomic Number = number of protons = number of electrons Fill in the table below. Element Atomic Number ProtonsElectrons Pb82 8 30

Isotopes  Isotopes are atoms with the same number of protons but different number of neutrons.  The atomic number stays the same  Mass will change, since neutrons have a relative mass of 1

Isotope Notation  Mass Number – the sum of the atomic number(or number of protons) and neutrons in the nucleus  These are both saying that the element is carbon, it has a mass number of 14, atomic number of 6  So 6 protons, 6 electrons, and 8 neutrons Carbon -14

Fill in the table below ElementMass #Atomic # Protons & Electrons Neutrons Isotope Symbol Neon2210 Calcium4620 Oxygen178 Iron5726 Zinc6430 Mercury20480

Atomic Mass  An Elements mass is made up of protons (1 amu) and neutrons (1 amu)  However an elements mass is usually never a whole number.  Atomic Mass of an element is the weighted average mass of the isotopes of that element

Calculating Atomic Mass  Chlorine – 35 has a mass of 34.969 amu, a percent abundance of 75.78%  Chlorine – 37 has a mass of 36.966 amu, a percent abundance of 24.22%  Calculate the mass contribution of each isotope  Chlorine – 35: (34.969 amu x 75.78%) = 26.500 amu  Chlorine – 37: (36.966 amu x 24.22%) = 8.9532 amu  Now add the isotopes mass contribution together to get your weighted average atomic mass  (26.500 amu + 8.9532 amu) = 35.453 amu  The weighted average atomic mass should be the same or very close to what is on the periodic table.

Isotope Practice  Boron has two naturally occurring isotopes: boron – 10 (abundance = 19.8%, mass = 10.013 amu) and boron – 11 (abundance = 80.2%, mass = 11.009 amu). Calculate the mass of boron.  Identify the unknown element X, based on the information given in the table.  Nitrogen has two naturally occurring isotopes, N -14 and N- 15. Its atomic mass is 14.007 amu. Which isotope is more abundant? Explain. IsotopeMass (amu)Percent Abundance X – 66.0157.59% X - 77.01692.41%

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