AP Chemistry. Mole, Molecular Formula Review One Mole = o This is Avogadro s number. Molar Mass: The mass of one mole of a substance in grams.

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AP Chemistry Ms Ye Name Date Block Mole, Review One Mole = o This is Avogadro s number Molar Mass: The mass of one mole of a substance in grams Percent Composition: Identifies the elements present in a compound as a mass % of the total compound mass Practice: Molar Mass of Compound % Composition of Each Element PbO 2 Pb: 1 x 20722 = 20722 O: 2 x 1600 = 3200 1 mole PbO2 = 23922 grams PbO2 CrF 3 %Pb = %O= 100 = % Pb 100 = % O Al 2 (SO 4 ) 3 KNO 3 *Remember: multiply the subscripts when there is a parentheses

Molar Conversions: Show work using Dimensional Analysis! 1 How many grams are in 45 moles of sodium fluoride, NaF? 2 How many moles are in 983 grams of aluminum hydroxide, Al(OH)3? 3 How many molecules are there in 450 grams of sodium sulfate, Na2SO4? 4 How many grams are there in 75 x 10 23 molecules of sulfuric acid, H2SO4?

Empirical/ Review Empirical = = Remember: the empirical and molecular formula can be the same! 1 Determine the empirical formula of the given molecular formulas below Empirical Empirical HgBr2 C2H6O C8H14 C3H6 2 Determine the molecular formula based on the information given a 8418 g/mol b c Empirical Empirical Empirical CH2 CaCl2 CH2O 11098 g/mol 12012 g/mol 3 A compound has an empirical formula of CF3 and a molar mass of 13802 g/mol What is the molecular formula of the compound?

Determining Empirical s from Mass Data Ex: Assume you have a sample that is 0504 g hydrogen and 400 g oxygen 1 Convert grams moles for each element 2 Simplify the mole ratio for each element by dividing each number (of moles) by the smallest # of moles 3 Use the smallest whole-number mole ratio as the subscripts for the empirical formula Practice: 1 A 5325 g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain 3758 g of carbon, 0316 g of hydrogen and 1251 g of oxygen What is the empirical formula of methyl benzoate? 2 A 2498g sample of an unknown compound was analyzed and found to contain 135 g Ca, 108 g O, and 0675 g H What is the empirical formula of the compound?

Determining Empirical s from Percent Composition Data: Ex: Determine the empirical formula for a compound containing 740% carbon (C), 865% hydrogen (H), and 173% nitrogen (N) by mass 1 Assume a 100 gram sample for the compound The percent composition of each element will then be equal to the mass of the element found in the 100 g sample 2 Convert grams moles for each element C: 74 = Assume a 100 g sample: C: 740 % 74 g C in 100 g sample H: 865 % 865 g H in 100 g sample N: 173% 173 g N in 100 g sample 3 Simplify the mole ratio for each element by dividing each number (of moles) by the smallest # of moles C: H: N: = 501 5 = 696 7 = 100 H: 865 = N: 173 = 4 Use the smallest whole-number mole ratio as the subscripts for the empirical formula C: 5 H: 7 N: 1 C 5 H 7 N Note: If the simplified mole ratios are not all whole numbers, multiply by the smallest factor that will produce whole number subscripts Practice: 1 A compound is found to contain 6352% iron and 3648% sulfur Determine the empirical formula 2 A combustion analysis determined that the molecule consisted of 545% carbon, 915% hydrogen, and 363% oxygen a Determine the empirical formula b Determine the molecular formula knowing that the molecular mass = 13216g/mol

Finding Empirical from Percent Composition Data what happens when your mole ratios aren t whole numbers? If when you divide all the # of moles by the smallest # of moles and your mole ratios end in: ~05 (1/2) ~33 (1/3), ~66 (2/3) ~025 (1/4), ~075 (3/4) Multiply all mole ratios by to get all whole numbers for your mole ratio Examples 3 A compound contains 2594% Nitrogen and 7406% oxygen by mass What is the empirical formula? 4 A 100 gram sample of a compound contains 722 grams of magnesium and 278 grams of nitrogen Determine the empirical formula

Balancing & Types of Reactions Review Types of Reactions: Synthesis: 2 or more elements combine to make a compound; Ex: 8 Fe + S8 8 FeS Decomposition: A compound breaks down into 2 or more simpler substances; Ex: 2 H2O 2 H2 + O2 Combustion: A compound containing carbon and hydrogen is burned in the presence of oxygen, and the products are carbon dioxide & water vapor Ex: CH4 + 2 O2 CO2 + 2 H2O Single Replacement: 1 element replaces another element in a compound; Ex: Zn + 2 HCl ZnCl2 + H2 Double Replacement: 2 elements of 2 different compounds switch places with one another; Ex: NaCl(aq) + AgNO3(aq) NaNO3(aq) + AgCl(s) Neutralization: A double replacement reaction that takes place when an acid and a base react to form a metal salt and water (This is also known as an Acid-Base Reaction) Ex: HCl + NaOH NaCl + H2O Chemical Reactions and Law of Conservation of Matter: Matter cannot be created or destroyed This means that you end up with the same number of atoms of each element that you started with in a chemical reaction We balance chemical reactions by adding coefficients in front of the reactants or products so that there are equal numbers of each element on both sides In balanced chemical reaction equations: the coefficients represent the relative number of moles of substance that is being reacted or produced Ex: Consider the balanced equation 2 H2 + 1 O2 2 H2O This equation can be read as 2 moles of H2 react with 1 mole of O2 to make 2 moles of H2O

PRACTICE: Balance the following reactions AND identify the reaction type Balanced Reaction Reaction Type 1) Cl2 (aq) + KBr (aq) KCl (aq) + Br2 (aq) 2) HgO (s) Hg (l) + O2 (g) 3) AlBr3 + K2SO4 Al2(SO4)3 + KBr 4) Al (s) + O2 (g) Al2O3 (s) 5) FeCl3 + NaOH Fe(OH)3 + NaCl 6) C3H8 (g) + O2 (g) CO2 (g) + H2O (g) 7) NaOH (aq) + HCl (aq) NaCl (aq) + H2O (aq) 8) H2O (l) + Fe (s) Fe2O3 (s) + H2 (g) 9) AgNO3 (aq) + Cu (s) Cu(NO3)2 (aq) + Ag (s) 10) KOH(aq) + H2SO4 (aq) H2O (l) + K2SO4(aq) 11) (NH4)2CO3 NH3 + CO2 + H2O 12) C7H16 (l) + O2 (g) CO2 (g) + H2O (l)