1. Learning Goal 3 : Determine the molar proportions of the reactants and products in a balanced chemical reaction.

2.  The Process of using a chemical equation to calculate the relative masses of reactants and products involved in a reaction. › The balanced equation for a chemical reaction describe the stoichiometry of the reaction.

3.  A unit created to describe atoms because the gram and kilogram are too large to use to define an atom.  1amu = 1.66 x 10 -24 g .00000000000000000000000166g

4.  Listed on the periodic table as the atomic mass.  The average mass of all of the isotopes of an atom. Elementamu Hydrogen1.008 Carbon12.01 Nitrogen14.01 Oxygen16.00 Sodium22.99 Aluminum26.98

5.  1 atom = 1 amu  Example › 1 atom of carbon = 12.001 amu of carbon › 3 atoms of carbon = 36.003 amu of carbon › 3.00 X 10 20 amu =2.50 x 10 19 atoms of carbon

6.  The unit all chemists use in describing numbers of atoms.  Defined as the number equal to the number of carbon atoms in 12.01 grams of carbon.  A sample of on element with a mass equal to that element’s average atomic mass expressed in grams contains 1 mol of atoms.

7.  6.022 X 10 23  One mole of something consists of 6.022 X 10 23 units of that substance. › One mole of eggs is 6.022 X 10 23 of eggs. Elements# of atoms Mass of sample (g) Aluminum6.022 X 10 23 26.98 Gold6.022 X 10 23 196.97 Iron6.022 X 10 23 55.85 Sulfur6.022 X 10 23 32.07 Boron6.022 X 10 23 10.81 Xenon6.022 X 10 23 131.3

8. The Mole

9.  The mass of one mole of a substance. › The sum of all the masses of the atoms in a compound. CH 4  C - 1x12.01=12.01  H – 4x1.008=4.032  Total = 16.04g

10.  The “Y” Diagram is a mnemonic aid for remembering the relationship between the mole and some of the basic quantities used in chemistry. Using a play on words, students are told to remember to “mole-tiply” out when converting from moles to other quantities. Mole Volume Particles Mass Multiply by molar mass Divide by molar mass Multiply by 22.4 l/mole Divide by 22.4 l/mole Divide by Avogadro's number 6.022 X 10 23 Multiply by Avogadro's number 6.022 X 10 23

11.  To move from mass to moles or moles to mass, you must use the molar mass of a compound.  Molar mass gives the mass of one mole of the compound.  Moles  Mass › Given molesmolar mass (g) = mass of compound 1 mol  Mass  Moles › Given Mass (g) 1 mol = moles of compound molar mass (g)

12.  Use the “Y” diagram. › Determine where you are starting. › Follow the arrows to where you are trying to go.  Always show your work and write out each step of the equation.  Use a crossbar.  Make sure to include units. › A number alone has no meaning. You must give it meaning; use units!

13.  Mole Relations Mole Relations