Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 4 Section 4.  There are about 118 known types of atoms.  Each element has it’s own type of atom.  All atoms of an element have to have one.

Similar presentations


Presentation on theme: "Chapter 4 Section 4.  There are about 118 known types of atoms.  Each element has it’s own type of atom.  All atoms of an element have to have one."— Presentation transcript:

1 Chapter 4 Section 4

2  There are about 118 known types of atoms.  Each element has it’s own type of atom.  All atoms of an element have to have one common factor, just like all cells of a particular person will have the same DNA.  The identifying factor of an atom is the ATOMIC NUMBER.  The atomic number is the number of p+ in the atom.

3 P+ Proton Positive charge 1amu n Neutron No charge 1amu e- Electron Negative Charge 0amu

4  contains protons and neutrons  The number of protons in all atoms of the same element will be the same  is positively charged  Contains all of the mass of the atom  Each proton and each neutron will add 1amu to the mass of the atom  The atomic number (found on the periodic table) is the number of protons

5  The atomic mass of an atom is equal to the number of protons + the number of neutrons Atomic mass = p+ + n  To find the number of neutrons, subtract the atomic number from the atomic mass  Ex: Phosphorus atomic number = 15 mass # = 31 number of neutron = 31-15 16 neutrons Phosphorus 15 p+ 16n

6  Atoms that have the same number of p+ but different numbers of n Neon-20Neon-21Neon-22 # p+ #n Mass # # e- When an element name or symbol appears as name-x, the x stands for the mass number of that isotope.

7  The number appearing on the periodic table is actually an average atomic mass  This number is the weighted average of the masses of the isotopes of an element  This number accounts for the relative abundance of different isotopes of the element

8 1. Multiply the mass of each isotope by its natural abundance, expressed as a decimal. 2. Add the products. Example: Element X has two natural isotopes. The isotope with a mass of 10.012amu has a relative abundance of 19.91%. The isotope with a mass of 11.009amu has a relative abundance of 80.09%. Calculate the atomic mass of this element. IsotopeMass xAbundance as a Decimal = Product X-1010.012.19911.993amu X-1111.009.80098.817amu Add the Products10.810

9  Multiply your percentage per category by the weight of each category expressed as a decimal.  Add your answers. Homework 10% 60% Lab 25% 75% Quiz 25% 86% Tests 40% 50% Your average is NOT 67.75% CategoryWeight of Category as a Decimal Grade in Category Product HW.10606 Lab.257518.75 Quiz.258621.5 Test.405020 Add the products!66.25%

10  Multiply the mass of the isotope by the percent abundance (expressed as a decimal)  Repeat the first step for each isotope  Add the products of the answers from your first two steps. IsotopeMass of the Isotope % abundance (as a decimal) Product Boron - 1010.013.199 Boron - 1111.009.801 Average Atomic Mass


Download ppt "Chapter 4 Section 4.  There are about 118 known types of atoms.  Each element has it’s own type of atom.  All atoms of an element have to have one."

Similar presentations


Ads by Google