1. Jino Park Chris Yang

2.  Average mass=total mass / number of objects  ex) each jelly bean is 5g, you want 100 jelly beans

3. 3.2 Atomic Masses  The modern system of atomic masses is based on the 12 C as the standard.  Carbon’s atomic mass is 12.01 g because it takes the average of the masses of its isotopes.  Basically, the atomic masses of the elements are just averages of their isotopes.

4.  1 mole=6.02 x 10 23 molecules  Dozen  eggs mole  molecules  Use (6.02 x 10 23 molecules / 1 mol) when converting

5. 3.4 Molar Mass  The mass (g) of one mole of the compound.  Found by summing up the masses of each atom in a component.

6. 3.6 Percent Composition of Compounds  Obtained by taking the ratio of the mass of each element in 1 mole of the compound to the total mass of 1 mole of the compound. (mass of element in 1 mole/total mass in 1 mole) x 100 = percent composition

7.  Steps to find empirical formula: 1. Find % mass of each element in the compound 2. Assume there are 100g the compound, so each element would have the same mass (g) as its % mass 3. Convert each mass to moles; divide by the atomic mass 4. Divide all the moles by the lowest to find the ratio between the elements 5. The numbers should be very close to whole numbers; if not, multiply through (usually by 2 or 3) to get whole number ratios *If the molar mass is known, the molecular formula can be found  Ex) If a compound contains 71.65% Cl, 24.27% C, and 4.07% H by mass, and the molar mass is 98.96g/mol, what is the molecular formula of the compound?

8. 3.8 Chemical Equations  A chemical reaction is the rearrangement of the atoms in one or more substances. Represented by a chemical equation with reactants on the left side and products on the right. Atoms are neither created nor destroyed all atoms in reactant must be present in product mass is conserved  Solid (s), liquid (l), gas (g), aqueous (aq)

9.  Change only the coefficient, NOT the subscripts or atoms  The total numbers of an atom must be equal on both reactants and products  ex)C 2 H 5 OH + O 2  CO 2 + H 2 O

10. 3.10 Stoichiometric Calculations: Amounts of Reactants and Products  Coefficients in the equation represent numbers of molecules.  Construct a mole ratio from the balanced equation. Ex. C 3 H 8 + 5O 2  3CO 2 + 4H 2 O; (5 moles of O 2 /1 mol of C 3 H 8 )  Calculating Masses of Reactants & Products 1. Balance the equation 2. Convert mass of reactant/product to moles 3. Set up mole ratios using balanced equation 4. Use mole ratios to calculate number of moles of reactant/product 5. Convert from moles back to grams if required

11.  The reactant that runs out first, limiting the amounts of products  Use the limiting reagent to calculate the products  Determining a limiting reagent 1. Calculate the moles of each reactant 2. Compare ratio to the given equation  Percent Yield=actual/theoretical x 100%