1.2 Atomic Mass vs Atomic Weight **

Question 1

atomic mass

Answer 1

• atomic mass: the mass of an atom expressed in atomic mass units (amu)
  
  • atomic mass ≈ mass number (A)
    
    • 1 amu = ½ the mass of a carbon-12 atom
      
      • 6 protons + 6 neutrons /12 = 1 amu
        
        • 1 amu = 1 proton
        • 1 amu = 1 neutron
          
          • *some mass is lost as binding energy (≈)
  • 1 amu = 1.66 x 10^-24 g
  • atomic mass varies between isotopes

Question 2

Example of varying atomic masses of isotopes:

Answer 2

Question 3

atomic weight

Answer 3

• atomic weight: the weighted average of the naturally occurring isotopes of an element
• listed on the periodic table of elements
• example: 2 naturally occurring isotopes of chlorine: Cl-35 and Cl-37
  
  • Cl-35 is 3x more abundant than Cl-37 -> the atomic weight of Cl is closer to Cl-35 than Cl-37
• atomic weight (in amu) of an element = mass of 1 mole of that element (in grams)
  
  • example: atomic weight of carbon = 12amu
  • 6.022 x 10²³ carbon atoms have a mass = 12g

Question 4

atomic weight (continued…)

Answer 4

• atomic weight represents:
  
  • the mass of the “average” atom in an element (in amu)
  • ***AND the mass of one mole of an element (in grams)
    
    • example:
      
      • atomic weight of carbon = mass of 12 amu
      • 1 mole of carbon (6.022 x 10²³ carbon atoms) = mass of 12g

Question 5

atomic mass vs atomic weight

Answer 5

• atomic weight represents:
  
  • the mass of the “average” atom in an element (in amu)
  • ***AND the mass of one mole of an element (in grams)
    
    • example:
      
      • atomic weight of carbon = mass of 12 amu
      • 1 mole of carbon (6.022 x 10²³ carbon atoms) = mass of 12g

Question 6

mole

Answer 6

• the # of “things” (atoms, ions, molecules, basketballs) equal to 6.022 x 10²³
  
  • example: 1 mole of atoms → 6.022 X 10²³ atoms

Question 7

Avogadro’s number (N_A)

Answer 7

N_A = 6.022 x 10²³

example: 1 mole of basketballs = 6.022 x 10²³

1 mole of carbon atoms = 6.022 x 10²³ carbon atoms

Question 8

1.2 Practice Problem

Answer 8

Question 9

1.2 Concept Check

Answer 9