1. The Mole—Quantifying Equations

2.  The mass of a single atom is far too small in grams to use conveniently.  Chemists use the unit called the unified atomic mass unit (amu) or Dalton (Da).  Definition of amu is exactly 1/12 the mass of an atom of 12 C Amu (Da) = 1.660539 x 10 -24 g

3.  Mass of one 12 C atom = 12 amu (exactly)  1 amu approximates the mass of one proton or neutron.  Mass of electron is neglible in comparison. ParticleMass Charge gramsamucoulombse Electron9.109382 x 10 -28 5.485799 x 10 -4 -1.602176 x 10 -19 Proton1.672622 x 10 -24 1.007276+1.602176 x 10 -19 1 Neutron1.674927 x 10 -24 1.0086650 0

4.  Elements differ in the number of protons in their atoms. The atomic number Z  All atoms of a given element have the same number of protons.  Number of electrons equals protons.  Number of neutrons = N  Mass Number (A) = Z + N  Mass number is the total number of nucleons.

5.  Why do all element not have atomic mass number listed in the periodic table that is not a whole number or very close to it?  Are all atoms of an element the same?

6.  Isotopes are atoms with the same atomic number, but different mass number.  The larger mass size is due to the difference in the number of neutrons that an atom contains. Although mass numbers are whole numbers, the actual masses of individual atoms are never whole numbers (except for carbon-12). This explains how Lithium can have an atomic mass of 6.941 Da.

7.  The atomic masses on the periodic table take these isotopes into account, weighing them based on their abundance in nature, therefore, more weight is given to the isotopes that occur most frequently in nature. Average mass of the element E is defined as:  m(E) = ∑(m(I n ) * p(I n ))  where ∑ represents a n-times summation over all isotopes I n of element E, and p(I) represents the relative abundance of the isotope I.

8. Find the average atomic mass of Boron Mass and abundance of Boron isotopes n isotope I n mass m (Da) isotopic abundance p 1 10 B 10.013 0.199 2 11 B 11.009 0.801 Solution: The average mass of Boron is: m(B) = (10.013 Da)(.199) + (11.009 Da)(.801) = 1.99 Da + 8.82 Da = 10.81 Da

9. Molecular mass: sum of atomic masses of all atoms in a molecule Formula mass: sum of atomic masses of all atoms in a formula unit of any compound, molecular or ionic.

10.  Copper (II) Nitrate Cu(NO 3 ) 2 63.5 + [(14 + {3 x 16}) x 2] = 187.5g  Ca 3 (PO 4 ) 2 3 moles of Ca 2 moles of P 2 x 4 moles of O.  1 mole of Ca is 40.08g, so 3 moles are 120.24 g  1 mole of P is 30.9738g, so 2 moles are 61.9476g  1 mole of O is 15.9994g, so 8 moles are 127.9952g  1 mole of Ca 3 (PO 4 ) 2 is 310.18 g

11.  From a balanced equation, the coefficients define the ratio of reactants needed for the products that result from the reaction.  Counting atoms is impractical.  Use a mass ratio:

12.  Balanced equation:  C 2 H 4 + HCl C 2 H 5 Cl 1 : 1yields 1 for ratio of molecules 28.0 : 35.5yields64.5 for mass ratio Ethylene: Atomic mass of 2C = 2 x 12.0amu = 24.0amu Atomic mass of 4H = 4 x 1.0amu = 4.0amu Molecular mass of C 2 H 4 = 28.0amu Hydrogen chloride: at. mass of H = 1.0amu at. Mass of Cl = 35.5amu Molecular mass of HCl = 36.5amu Ethyl chloride: at. mass of 2C = 2 x 12.0amu = 24.0amu at. mass of 5H = 5 x 1.0amu = 5.0amu at. mass of Cl = 35.5amu = 35.5amu Molecular mass of C 2 H 5 Cl = 64.5amu

13.  1. sodium fluoride  2. potassium hydroxide  3. copper (I) chloride  4. manganese (IV) oxide  5. calcium sulfate  6. magnesium phosphate

14.  Amadeo Avogadro was an Italian physics professor who proposed in 1811 that equal volumes of different gases at the same temperature contain equal numbers of molecules. Amadeo Avogadro

15.  If Avogadro’s hypothesis is true, then atomic weights for gases can be derived by weighing equal volumes of different gases (Cannizzaro).  Johan Loschmidt (HS teacher) took the idea and calculated the size of a molecule of air. He developed an estimate for the number of molecules in a given volume of air.  These three ideas together lead to the number named for Avogadro. Loschmidt was the first to calculate this number.

16.  1 mole of a substance,mole N A = 6.02214179(30)×10 23 is known as the Avogadro constant.Avogadro constant  For calculations please use 6.02 x 10 23  http://www.youtube.com/watch?v=Hj83o RHdezc&safety_mode=true&persist_safet y_mode=1&safe=active http://www.youtube.com/watch?v=Hj83o RHdezc&safety_mode=true&persist_safet y_mode=1&safe=active