1. CH 3 Mass Relations in Chemistry; Stoichiometry

2. Atomic Mass Indicates how heavy an element is compared to another element. Units AMU---Atomic Mass Unit  Defined as 1/12 of the mass of a C-12 atom

3. Atomic Mass from isotope composition Isotopic Abundance: the natural of an isotope. IsotopeAtomic MassPercent abundance Contribution to atomic mass Ne-2020.0090.92= Ne-2121.0000.26= Ne-2222.0008.82=

4. Mass of individual atoms Mass of 1 atom = molar mass/ N A (Avogadro's #) Reacquaint yourself with the mole wheel.

5. The Mole 1 mol= 6.022 x 10 23 items  1mol H = 6.022 x 10 23 H atoms= 1.008g  1mol Cl= 6.022 x 10 23 Cl atoms = 35.45g  1mol Cl 2 = 6.022 x 10 23 Cl 2 molecules = 70.90g

6. Molar mass (MM) Molar mass is numerically equal to the sum of the atomic masses.

7. Mass % from formula % composition of K 2 CrO 4 Use part / whole, assume you have 1 mole of compound. (the math is easier) 1mol = K 2 CrO 4 194.20 g %K = 78.90/194.20 = %Cr= %O=

8. Simplest / empirical formula Simplest whole number ratio of atoms present in a compound.

9. Simplest (empirical) formula from % composition Steps: 1. Find the mass of each element in the sample compound, assume 100g total. 2. Find the numbers of moles of each compound. 3. Divide each by the smallest # of moles and look for obvious ratios. Use the following K= 26.6%, Cr= 35.4%, O = 38.0%

10. Simplest formula (empirical) from analytical data An organic sample containing only C, H, O atoms weighs 1.000g Burning the sample gives 1.466g CO 2, 0.6001g H 2 O Find the simplest formula… All the carbon from the sample is “locked up” in CO 2 The portion of CO 2 that is carbon can be determined by the mass ratio in the formula (12.01/44.01) This multiply this by the mass of CO 2 and you find grams of carbon in the sample.

11. Yield of product in a reaction Ordinarily, reactants are not present in the exact ratio required for reaction. Usually 1 is in excess; some left when reaction is over. 1 is limiting; completely consumed to give the theoretical yield of product.

12. Calculating theoretical yield 1. Calculate the yield expected if the first reactant is limiting. 2. Repeat the calculation for the second reactant 3. The theoretical yield is the SMALLER of these two quantities. The reactant that gave the smaller theoretical yield is the limiting reactant.

13. 2 Ag (s) + I 2 (s)  2 AgI (s) Calculate the theoretical yield of AgI and determine the limiting reactant. There is 1.00g Ag, and 1.00g I 2.

14. % Yield Suppose the actual yield is 1.50g AgI, what was the % yield? Actual/Theoretical (x100) = % Yield