Chemistry Final Exam Review

Calculate the molar mass of: (Ca(C2H3O2)2) Ca = 40.1 g/mol Chapter 10: Molar Mass Calculate the molar mass of: (Ca(C2H3O2)2) Ca = 40.1 g/mol 4 * C = + 48.0 g/mol 6 * H = + 6.0 g/mol 4 * O2 = + 64.0 g/mol molar mass (Ca(C2H3O2)2) = 158.1 g/mol

Molar Mass 3. Pb3(PO4)2 Ans: 811.5 g/mol Cu3(BO3)2 Ans: 308.3 g/mol

2. Number of atoms in a molecule How many atoms of hydrogen are present in four molecules of C3H7O? Answer: 28 How many atoms of hydrogen are present in: 2. Al(OH)3 Answer: 3 (NH4)2HPO4 Answer: 9 C4H10O Answer: 10

Mole conversions Figure 10.12 (pg 303)

Mole conversions: Finding mass in grams Calculate the mass, in grams, of 2 molecules of ribose C5H10O5. a. Calculate molar mass: C5H10O5 molar mass = C: 5 * 12.0 g/mol = 60.0 g/mol H: 10* 1.0 g/mol = +10.0 g/mol O: 5* 16.0 g/mol = +80.0 g/mol C5H10O5 molar mass = 150.0 g/mol Solve using mole conversion: 2 molecules * 1mole * 150.0 g = 4.98 x 10-22 g 6.02 x 1023 mole Problem: Calculate the mass, in grams, of a molecule of aspirin (C9H8O4) Molar mass C9H8O4 = 180.0 g/mole Answer: 2.99 x 10-22 g

Mole conversions: Finding volume in liters 2. Calculate the vol., in liters, of 1.50 mole Cl2 at STP Use molar volume to convert moles to liters: 1.50 mole Cl2 * 22.4 L = 33.6 L mole Problems: Calculate the vol., in liters, of the following gases at STP: i. 7.6 mole Ar ii. 0.44 mole C2H6 iii. 835g SO3 Answer: i. 1.7 x 102 L Ar ii. 9.9 L C2H6 iii. 234 L SO3

Mole conversions: Finding the # of atoms 3. Calculate the number of atoms in 5.78 mol NH4NO3. Answer: 3.13 x 1025 atoms

% Composition 1. Calculate the molar mass 10 * C = 120.0 g/mol Nicotine has a molecular formula of C10H14N2, calculate the % composition of nitrogen (ONLY) in this compound. 1. Calculate the molar mass 10 * C = 120.0 g/mol 14 * H = + 14.0 g/mol 2 * N2 = + 28.0 g/mol molar mass C10H14N2 = 162.0 g/mol

% Composition Divide molar mass of nitrogen by total mass of compound % Composition = 28.0 g/mol * 100 % 162.0 g/mol = 17.3 % C/E: 5/25/10

Empirical Formula from % composition Calculate the empirical formula of a compound that contains 25.9% nitrogen and 74.1% oxygen. unknown: Empirical formula = N?O? Known: % nitrogen = 25.9 % oxygen = 74.1 Change % composition (ratio of the mass) to the ratio of moles by using molar mass 2. Reduce the ratio to the lowest whole-number ratio

Empirical Formula from % Composition and molar mass Solving: 25.9 g N * 1mol N = 1.85 mol N 14.0 g N 74.1 g O * 1mol O = 4.63 mol O 16.0 g O Mole ratio of nitrogen to oxygen is N1.85O4.63. Divide by the smaller number: 1.85 mol N = 1 mol N; 4.63 mol O = 2.50 mol O 1.85 1.85 This becomes: N1O2.5 Multiply by 2: N2O5

Empirical Formula from % Composition and molar mass Calculate empirical formula of each compound: 94.1% O; 5.9% H Answer: HO 67.6% Hg; 10.8% S; 21.6% O Answer: HgSO4

Molecular Formula from % Composition and molar mass The % composition of methyl butanoate is 58.8% C, 9.8% H, and 31.4% O and its molar mass is 102 g/mol. What is the empirical formula? What is the molecular formula? Solve: 1. Compute moles using molar mass C: 58.8 g * 1 mol = 4.9 mol 12.0 g H: 9.8 g * 1 mol = 9.8 mol 1.0 g O: 31.4 g * 1 mol = 1.96 mol 16.0 g

Molecular Formula from % Composition and molar mass Divide by the smallest number of moles 1.96 mol O = 1 mol O; 9.8 mol H = 5 mol H; 4.9 mol C = 2.5 mol 1.96 1.96 1.96 This becomes: C2.5H5O1 Multiply by 2: C5H10O2 (Empirical formula) 3. Compute molar mass of empirical formula

Molecular Formula from % Composition and molar mass molar mass C5H10O2: C = 5 * 12.0 g/mol = 60.0 g/mol H = 10 * 1.0 g/mol = 10.0 g/mol O = 2 * 16.0 g/mol = 32.0 g/mol 102.0 g/mol 3. Compute the ratio of molar mass of compound and empirical formula Molar mass methyl butanoate = 102 g/mol = 1 Empirical formula molar mass 102 g/mol Multiply empirical formula with the ratio in step 3 C5H10O2 * 1 = C5H10O2 So, empirical formula = molecular formula

Molecular Formula from % Composition and molar mass Problem: Caffeine is analyzed and found to have the following % composition: 49.5% carbon, 5.2 % hydrogen, 28.9% nitrogen and 16.4% oxygen. The molar mass of caffeine is 195 g/mol. Find the molecular formula of this compound. 1. C4H5N2O1 2. Molar mass of empirical formula = 97 g/mol 3. Molar mass caffeine = 195 g/mol = 2 Molar mass empirical formula 97 g/mol 4. Molecular formula = C4H5N2O1 * 2 = C8H10N4O2

Balancing equations, types of chemical reactions, and Reactions in aqueous solutions Chapter 11

Balancing Chemical Equations 16 ______C8H18 + _______ O2  ______CO2 +______H2O 2 25 18 4 3 12 _____(NH4)3PO4 + _____ Pb(NO3)4  _____ Pb3(PO4)4 +____(NH4)NO3

Types of Chemical Reactions single replacement synthesis double replacement combustion decomposition

Ba(NO3)2 (aq) + K2CO3 (aq)  BaCO3 (s) + 2KNO3 (aq) Total Ionic: Net ionic equations, formation of a precipitate, and identification of spectator ions Ba(NO3)2 (aq) + K2CO3 (aq)  ? Ba(NO3)2 (aq) + K2CO3 (aq)  BaCO3 (s) + 2KNO3 (aq) Total Ionic: Ba+2 (aq) + NO3-1 (aq) + K +1 (aq) + CO3-2 (aq)  BaCO3 (s) + K+1 (aq) + NO3- (aq) Spectator ions: NO3-1 (aq), K +1 (aq) Net Ionic: Ba+2 (aq) + CO3-2 (aq)  BaCO3 (s) D:5/26/10

Net ionic equations, formation of a precipitate, and identification of spectator ions (continued) AgNO3 (aq) + K3PO4 (aq)  ? AgNO3 (aq) + K3PO4 (aq)  Ag3PO4 (s) + KNO3 (aq) Total Ionic: Ag+ (aq)+ NO3- (aq)+ K+ (aq)+ PO43- (aq) Ag3PO4 (s) + NO3- (aq) + K+ (aq) Spectator Ions: NO3- (aq), K+ (aq) Net Ionic: 3Ag+ (aq) + PO43- (aq)  Ag3PO4 (s)