1. Ch. 3 Molar Mass; The Mole; Empirical and Molecular Formulas; % Composition; Isotopic Masses and abundance; Balancing Equations; Mass to Mass; Limiting Reactants; % Yield;

2. The Mole: 1 mol of anything = 6.022 x 1023 units of that thing; Average Atomic Masses have units: g/mol • A sample of silver has a mass of 21.46 g • How many mols? Remember 1mol Ag= 107.9g Ag • 21.46g Ag 1mol Ag 107.9 g Ag 0.1989 mols Ag

3. What about Atoms? • If 21.46 g Ag has 0.1989 mol of Ag then how many atoms does it have? • 0.1989mol Ag x (6.022 x 1023) atoms/mol • 1.198 x 1023 atoms Ag

4. Molar Mass: the mass in grams of one mole of a compound • Calculate the molar masses of: • AL2O3 and Na3AlF6 • AL2O3 • (2 x 26.98) + (3 x 16.00)= 101.96 amu • Na3AlF6: • (3 x 22.99) + 26.98 + (6 x 19.00)= 209.95 amu

5. % Composition: there are two ways of describing the composition of a compound: by its atoms and % by mass of its elements. • We can obtain the % composition by mass by comparing the mass of the elements in the formula to the total molecular mass • Example: What is the % by mass of the elements in ethanol (C2H5OH)? • Molecular Mass: (2 x 12.01) + (5 x 1.008) + 16.00+ 1.008 • Mass of 1 mol of ethanol = 46.068 g • % C= (2 x 12.01)/46.068 = 0.5212 x 100= 52.12% • % O= 16.00/46.068= .3473 x 100= 34.73% • %H= (6 x 1.008)/46.068= .1313 x 100= 13.13% • % add to 100%

6. Empirical Formulas: the formula with the smallest whole number ratio which cannot be simplified any more • Given % composition what is the empirical formula? • Assume the compound has a mass of 100g , convert your % to grams. • Calculate mols of each kind of atom present • Determine simplest whole number ratio by dividing by smallest # mols. Example: Rocket fuel contains 87.4% N and 12.6% H • 87.4 g N and 12.6 g H • 87.4g/(14.01g/mol)= 6.24 mols N and 12.6g/(1.008g H) = 12.5 mol H • Ratios: N= 6.24/6.24=1 H= 12.5/6.24= 2.0 these are the subscripts • Empirical formula is NH2

7. Molecular Formula: actual ratio of atoms in a compound • To determine the molecular formula: divide the molecular mass by the mass of the empirical formula, this result will give you the number of empirical formula units. • If the fuel had a molecular mass of 32.05g what is it’s Molecular formula? Answer: • Mass of empirical formula (NH2)= 16.026g • Mass of compound= 32.05 • Units of empirical formula = 32.05/16.026= 2.00units • 2(NH2) = N2H4

8. Isotopic Masses: average mass of an element (also known as weighted average) • A sample of metal “M” is vaporized and injected into a mass spectrometer. The mass spectrum tells us that 60.10% of the metal is present as 69M and the rest, or 39.9% is 71M. The mass values are 68.93amu and 70.92amu respectively. What is the average atomic mass?

9. Answer: • (.6010 x 68.93)+ (.3990 x 70.92)= 69.72 amu (it makes sense that more of the isotopes are 69M than 71M, the answer is closer to 69 than 71) • What element is this? Check the P.T.

10. Isotope abundances: What is the % of each isotope in an element? • The element Indium has two naturally occurring isotopes, 113In (112.9043 amu) &115In (114.9041 amu). Average atomic mass of 114.82 amu. Calculate the % relative abundance of the two isotopes. • You must use two equations to solve a problem like this: (Atomic mass 113In)(% of 113In)+(Atomic Mass 115In)(%115In)= average atomic mass and (% of 113In) + (% of 115In) = 1

11. So… • 113In = x and 115In = y • 112.9043x + 114.9041y = 114.82 • X + Y = 1… so y= 1-x • Substituting: 112.9043x + 114.9041(1-x)= 114.82 and x = .042 and y= 1-.042 • % abundance of 113In = 4.2% and % of 115In= 95.8%

12. Chemical Equations: Chemical equations describe chemical changes that occurs in a reaction. • Using this equation explain what is occurring Na2CO3(aq) + 2HCl(aq)CO2(g) + H2O(l) + 2NaCl(aq) • 1mol of aqueous sodium carbonate is reacting with 2 mol of aqueous hydrochloric acid to form 1 mol of carbon dioxide gas, 1 mol water and 2 mols aqueous sodium chloride

13. Stoichiometry: to calculate the amount of product from a given amount of reactant • Key relationship in any balanced equation will be the mole ratios which can be used as conversion factors • How many grams of Silver Sulfide can be made from the reaction of 3.94 g of Silver Nitrate? • Na2S (aq) + AgNO3 (aq)  Ag2S (s) + NaNO3 (aq) • Balance: Na2S (aq) + 2 AgNO3 (aq)  Ag2S (s) + 2 NaNO3 (aq)

14. The mole ratio between AgNO3 and Ag2S is the key:

15. Answer = 2.87 g Ag2S 3.94 AgNO3 x 1 mol AgNO3 x 1mol Ag2Sx 247.8 g Ag2S = 169.88g AgNO3 2 mol AgNO3 1mol Ag2S

16. Limiting Reactant: These are the reactants that produce the least amount of product thus, “limiting” the product. • Solutions of Barium Nitrate and Sodium Chloride are reacted to form aqueous Sodium nitrate and solid Barium Chloride. If 5.0 grams of Barium Nitrate and 3.3 grams of Sodium Chloride are used which is the limiting reactant? How much Sodium Nitrate is produced? If only 1.5 grams are produced, what is the % yield?

17. Write and Balance Equation: • Ba(NO3)2 + 2 NaCl  2 NaNO3 + BaCl2 Mole • Ratio: 1:2:2:1 This means it takes 1 mol of barium nitrate and 2 mols of sodium chloride; but if the ratio is not 1:2 then there is a limiting reactant.

18. Continued: These are the reactants that produce the least amount of product thus, “limiting” the product. • Calculate mols of reactants: • 5.0 grams of Ba(NO3)2 = 5.0g / 261.347g/mol = .019 mols Ba(NO3)2 • 3.3 g of NaCl = 3.3g/ 58.44 g/mol = .056 mols NaCl • Ratio: • .019: .056 or 1: 3 the Barium Nitrate is not 1.5 mols as it should be for a 1:2 ratio, so it is the limiting reactant

19. Calculate amount of NaNO3 produced from the limiting reactant: 5 g Ba(NO3)2 x1 mol Ba(NO3)2 x2 mol NaNO3x85g NaNO3 = 3.3 g NaNO3 261.347 g 1 mol Ba(NO3)2 1 mol NaNO3

20. % yield= • 1.5g x 100 = 45% 3.3