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IV Stoichiometry Stoichiometry • The relationship (mole ratio) between elements in a compound and between elements and compounds in a reaction e.g. H2O means: 1 molecule H2O contains 2 atom H and 1 atom O and 1 mole H2O contains 2 moles H atoms and 1 mole O atoms Chemical Compounds • Combination of elements – A compound is made up of specific elements in a specific ratio - Law of Constant Composition • Chemical Formula – Written representation of a chemical compound. So, a specific compound has a specific formula Terms • Formula Unit – Involves the lowest subscript which describes the ratio. – e.g. H2O; CH; CH2; CH4 – May or may not actually exist • Formula Weight – The mass of the formula unit • Empirical Formula – Written representation of the formula unit Terms continued • Molecule – Integral multiple of the formula unit (integer may be 1) that actually exists – e.g. C2H2; C2H4; C2H6 • Molecular Weight (Molar Mass) – Mass of the molecule • Molecular Formula – Written representation of the molecule • • • • Mass to mole conversions Stoichiometry is in mole ratios Most measurements are made in grams So, you need to be able to get from grams to moles and moles to grams • The atomic weight listed on the periodic table is listed without units. Why? • Units depend on what you want. • If you are looking on the atomic scale, atoms or molecules, units are amu. • If you are working on the macroscopic scale, moles of material, units are in grams. Convert 34.0 grams NH3 to moles. 1) Determine M of NH3. M = AW N + 3(AW H) = 14.0 + 3(1.01) = 17.0 2) Determine the moles of NH3 34.0 gNH 3 1molNH 3 2.00 molNH 3 17.0 gNH 3 How many molecules in 32.0 g of oxygen gas. A) B) C) D) E) 2.0 1.0 0.5 6.02 x 1023 1.20 x 1024 How many molecules in 32.0 g of oxygen gas. Oxygen occurs as O2 gas. AW of O = 16.00, so O2 = 32.00 32.0 gO 2 1molO 2 6.022 x10 23 molecules 23 6.02 x10 molecules O 2 1mol 32.00 g • % Composition (% weight/ weight; %w/w) grams ofelement %composition X100 grams ofcompound 1. Calculate the % composition form the Molecular Formula (or the Empirical Formula) What is the % composition by weight of C2H4O2? C = 12.01; H = 1.01; O = 16.00 Step 1: Find the molecular weight of the compound. MW = 2(AW C) + 4(AW H) + 2(AW O) = 2(12.01) + 4(1.01) + 2(16.00) = 60.06 g/ mol Step 2: Find the % of each element. %C grams C x100 MW cpd 2(12.01) 2(AW C) x100 = 39.99% x100 60.06 60.06 grams H 4(1.01) 4(AW H) x100 =6.73% %H x100 x100 MW cpd 60.06 60.06 grams O 2(16.00) 2(AW O) x100 = 53.28% %O x100 x100 MW cpd 60.06 60.06 or : %O 100 %C %H %O 100 39.99 6.73= 53.28% • Find the Chemical Formula from the % composition • Which Chemical formula will you get? – Empirical Formula, to get the molecular formula you would need more information than just the % composition A compound containing only carbon, hydrogen, and oxygen was found to contain 62.02 % C and 10.42% H. What is the formula of the compound? C = 12.01 H = 1.01 O = 16.00 1) Find the amount of O %O = 100 - %C - %H = 100 - 62.02 - 10.42 = 27.57 % O 2) Determine the moles of each element (assume a 100 g sample) grams C 62.02 g C molC 5.163 molC MW C 12.01g/mol grams H 10.42 g H molH 10.3168 molH MW H 1.01g/mol grams O 27.57 g O molO 1.723 molO MW O 16.00 g/mol 3) Divide by the smallest number to get whole number ratio Mol C = 5.163 mol / 1.723 = 2.996 = 3 Mol H = 10.3168 mol / 1.723 = 5.9877 = 6 Mol O = / 1.723 1.723 mol =1 So, the compound has the ratio 3C:6H:1 O And the Empirical formula: C3H6O A further analysis of the compound found that the molar mass was 115.99 g/mol. What is the molecular formula of the compound? Molecular formula = (empirical formula)(# of formula units) You can find the # of formula units from MW/ FW Formula weight = weight of empirical formula: C 3H 6O FW = 3(12.01) + 6(1.01) + 16.00 = 58.048 # formula units = 115.99 g 1.998 2 58.048 g So the molecular formula contains 2 formula units. Multiply the subscripts by 2 2(C3H6O) = C6H12O2 A compound is 84.80% uranium and the balance oxygen. What is the Empirical Formula of the compound? U = 238.0 O = 15.9994 A) U2O5 B) U3O8 C) UO3 D) UO2 E) U3O A compound is 84.80% uranium and the balance oxygen. What is the Empirical Formula of the compound? U = 238.0 O = 15.9994 1. Determine the amount of O 100 - 84.80 = 15.20 2. Determine moles and mole ratio 84.80 g U molU 0.3563 molU / 0.3563 = 1 238.0 g/mol molO 15.20 g O 0.9500 molO 15.9994 g/mol / 0.3563 = 2.666 0.666 = 2/3 so O is 2 2/3 = 8/3 U1O8/3 Clear denominator, U3O8 multiply by 3 3) Determine an unknown element (X) from the % composition A compound XO2 is 78.8% X. What element is X? O = 16.0 A) Ni B) Co C) P D) Sn 3) Determine and unknown element (X) from the % composition A compound XO2 is 78.8% X. What element is X? O = 16.0 % O = 100 - 78.8 = 21.2 21.2 gO 1molO 1.325 molO 16.0 g 1.325 molO 1mol X 0.6625 mol X 2 molO Since Compound is 78.8% X, 0.6625 mol X = 78.8 g X 78.8 g X ? g X so :? 118.9 0.6625 molX 1molX Go to Periodic Table 118.9 = Sn Reactions • Chemical Reaction is represented by a Chemical Equation • General Form: aA + bB cC + dD A/B are ? Reactants C/D are ? a,b,c,d are ? Products Stoichiometric coefficients • Law of Conservation of Mass says? – No mass lost or gained • Total mass of reactants = total mass of products • Elements are re - arranged not changed • This allows us to “Balance” Equations • The number and kinds of atoms in the reactants have to show up as the same number and kind of atoms in the product Ca + H2O 1 Ca + H2O 1Ca + 2H2O 1 Ca + 2 H2O Ca + 2 H2O Ca(OH)2 + H2 1 Ca(OH)2 + H2 1 Ca(OH)2 + H2 1Ca(OH)2 + 1 H2 Ca(OH)2 + H2 C3H7OH + O2 C3H7OH + O2 C3H7OH + O2 C3H7OH + 9/2 O2 2 C3H7OH + 9 O2 CO2 + H2O 3 CO2 + H2O 3 CO2 + 4 H2O 3 CO2 + 4 H2O 6 CO2 + 8 H2O SiF4 + H2O SiF4 + H2O SiF4 + 2 H2O HF + SiO2 4 HF + SiO2 4 HF + SiO2 When the reaction: C 2H 8N 2 + N 2O 4 N2 + H2O + CO2 Is balanced using the smallest whole numbers, what is the coefficient of N2? A) 1 B) 2 C) 3 D) 4 E) 5 When the reaction: C 2H 8N 2 + N 2O 4 N2 + H2O + CO2 Is balanced using the smallest whole numbers, what is the coefficient of N2? 3N2 + 4 H2O + 2CO2 C 2H 8N 2 + 2 N 2O 4 • Hydrates • A compound (solid) that contains intact water molecules as part of the compound. • The water can be removed by heating to leave an anhydrous residue (solid). • Water can then be re - added to the anhydrate to yield the original hydrate CaSO4 • 2H2O Calcium sulfate dihydrate Each mole of compound contains: 1 mole calcium sulfate and 2 mole water or 1 mole Ca 1 mole S 6 mol O 4 mol H CaSO4 • 2H2O(s) 172 CaSO4(s) heat Water vapor CaSO4(s) + 2 H2O(g) 136 + 2(18) CaSO4 • 2H2O(s) 17.0 g of CaSO4 • 2H2O(s) was heated to remove the water. What mass of residue (CaSO4) is left? A) 13.4 B) 13.44 C) 3.56 D) 15.0 E) 15.03 17.0 g of CaSO4 • 2H2O(s) was heated to remove the water. What mass of residue (CaSO4) is left? 17.0 g CaSO4 2H 2 O 1molCaSO 4 2H 2 O 1molCaSO4 136.0 g CaSO4 13.44 g 172.0g 1molCaSO 2H O 1molCaSO 4 2 4 13.4 g What happened to the difference (17.0 - 13.4) = 3.6 grams of material? Lost as water vapor 2. 15.00 grams of the hydrate Na2SO4 • XH2O(s) was heated to remove the water. After heating, 7.95 grams of material remained. What is the formula of the hydrate: (find the value of X) A) Na2SO4 • H2O B) Na2SO4 • 2H2O C) Na2SO4 • 3H2O D) Na2SO4 • 5H2O E) Na2SO4 • 7H2O Na2SO4 • XH2O(s) 15.00 Na2SO4(s) + XH2O 7.95 Find moles of both products and compare 15.00-7.95 7.05 (7.95 g) 1mol molNa 2SO4 0.0559 mol / 0.0559 = 1 142 g (7.05 g) 1mol molH 2O 0.391mol / 0.0559 = 6.99 = 7 18.0 g So, X = 7, formula = Na2SO4 • 7H2O Limiting Reactant • Limiting Reactant is that element or compound that determines the amount of product that you get. • It is the reactant that is used up You are the owner of a bike shop. A shipment came in with 183 frames, 150 seats, 252 pedals, 131 brake assemblies. How many bikes can you sell? (enter the number) Each bike needs, 1 frame, 1 seat, 2 pedals, 1 brake. Given: 183 frames 150 seats 252 pedals 131 brake assemblies Pedals will allow you to make only 126 bikes so that is the limiting reactant NH3 + O2 Balance 2 NH3 + 3/2O2 4 NH3 + 3 O2 N2 + H2O N2 + 2 N2 + 3H2O 6H2O 4 NH3 + 3 O2 2 N 2 + 6 H 2O Remember: Stoichiometry is mole ratios How many moles of N2 can be formed from 4 mol NH3 and 4 mol O2? 1. Determine the LR. 4 molNH 3 3molO 2 3molO 2 required 4 molNH 3 Mol O2 given > mol O2 required. So, NH3 is LR 2. Determine amount of product 4 molNH 3 2 molN 2 2 molN 2 produced 4 molNH 3 4 NH3 + 3 O2 2 N 2 + 6 H 2O How many moles of N2 can be formed from 6 mol NH3 and 4 mol O2? 1. Determine the LR. 6 molNH 3 3molO 2 4.5 molO 2 required 4 molNH 3 Mol O2 given < mol O2 required. So, O2 is LR 2. Determine amount of product 4 molO 2 2 molN 2 2.67 molN 2 produced 3molO 2 4 NH3 + 3 O2 2 N 2 + 6 H 2O How many moles of N2 can be formed from 5 mol NH3 and 4 mol O2? A) B) C) D) E) 5 4 3.75 2.67 2.5 4 NH3 + 3 O2 2 N 2 + 6 H 2O How many moles of N2 can be formed from 5 mol NH3 and 4 mol O2? 1. Determine the LR. 5 molNH 3 3molO 2 3.75 molO 2 required 4 molNH 3 Mol O2 given > mol O2 required. So, NH3 is LR 2. Determine amount of product 5 molNH 3 2 molN 2 2.5 molN 2 produced 4 molNH 3 4 NH3 + 3 O2 2 N2 + 6 H2O If 10.0 grams each of NH3 and O2 are reacted, how many grams of water and N2 are formed? 1. Find moles of each reactant. 10.0 gNH 3 1molNH 3 0.589 molNH 3 17.0 g 10.0 gO 2 1molO 2 0.312 molO 2 32.0 g 2. Determine the Limiting Reactant Determine mole of O2 needed 0.589 molNH 3 3molO 2 0.442 molO 2 required 4 mol NH 3 Compare to what was given: 0.442 mol required > 0.312 mol given So, O2 = LR 3. Determine the amount of product based on the LR 0.312 molO 2 6 molH 2O 18.0 g H 2O 11.2 gH 2O 3 mol O 2 1 mol H 2O 0.312 molO 2 2 molN 2 28.0 g N 2 5.8gN 2 3 mol O 2 1 mol N 2 total mass of products = 17.0 g What happened to conservation of mass? 3.0 g un-reacted NH3 10.0 grams Cr and 10.0 grams S are reacted to give chromic sulfide. How many grams of chromic sulfide are formed? Cr = 52.0 S = 32.0 A) 20.9 g B) 20.0 g C) 19.2 g D) 3.12 g E) 0.80 g 10.0 grams Cr and 10.0 grams S are reacted to give chromic sulfide. How many grams of chromic sulfide are formed? Cr = 52.0 S = 32.0 1. Write and balance the equation: 2Cr + 3S --> Cr2S3 2. Determine moles of each 10.0 gCr 1molCr 0.192 molCr 52.0 g 10.0 gS 1molS 0.313 molS 32.0 g 3. Determine LR 0.192 mol Cr 3 molS 0.288 molS required 2 mol Cr 0.288 molS required < 0.313 molS given so : Cr LR 4. Determine the amount of product based on the LR 0.192 mol Cr 1molCr2S 3 200.0 gCr2S 3 19.2 gCr2S 3 2 mol Cr 1molCr2S 3 • Determination of an unknown element • A metal oxide has the formula XO3 and reacts with H2 to form free metal X and H2O. If 15.99 grams XO3 yields 6.00 grams H2O, what element is X? A) Nd B) Ti C) S D) Mo E) H • Determination of an unknown element • A metal oxide has the formula XO3 and reacts with H2 to form free metal X and H2O. If 15.99 grams XO3 yields 6.00 grams H2O, what element is X? XO3 + 3 H2 --> X + 3 H2O Find moles XO3 from stoichiometry and moles of H2O 6.00 gH 2O 1molH 2O 1molXO 3 0.111molXO 3 18.0 g 3molH O 2 Determine MW XO3 15.99 g XO3 ? g 0.111mol 1mol ? 143.9 AW X = 143.9 - 3(16.0) = 95.9 = MO Theoretical Yield • Theoretical Yield – Maximum amount of product that can obtained if all the reactant is converted to product • Actual Yield – Actual amount of product obtained • % Yield Actual Yield %Yield x100 Theoretical Yield %Yield AY x100 TY One hundred grams of potassium chlorate was heated. What is the final state of affairs? K = 39.1 Cl = 35.45 O = 16.00 One hundred grams of potassium chlorate was heated. What is the final state of affairs? K = 39.1 Cl = 35.45 O = 16.00 1. Determine the formula for potassium chlorate A. KClO B. KClO2 C. KClO3 D. KClO4 One hundred grams of potassium chlorate was heated. What is the final state of affairs? K = 39.1 Cl = 35.45 O = 16.00 1. Determine the formula for potassium chlorate A. KClO B. KClO2 C. KClO3 D. KClO4 2. Write and Balance the reaction 2 KClO3 --> 2 KCl + 3 O2 3. Determine Theoretical Yields of Products 2 KClO3 --> 2 KCl + 3 O2 100. grams Convert to moles-mole ratio-convert to g 100.g KClO 3 1molKClO 3 3molO 2 32.0 gO 2 39.2 gO 2 122.5 g 2 molKClO 3 1molO 2 100.g KClO 3 1molKClO 3 2 molKCl 74.5 g KCl 60.8 g KCl 122.5 g 2 molKClO 3 1molKCl Or: Conservation of mass: 100.0 g - 39.2 g = 60.8 g KCl Part B. If the reaction only produced 50.37 g KCl, 1) What is the %Yield and 2) how many grams of O2 2) were produced. AY 1) %Yield x100 TY %Yield 2) gO 2 x TY 100 50.37 %Yield x100 = 82.8% 60.8 gO 2 82.8% x 39.2 = 32.4 gO 2 100 Consecutive Reactions • Consecutive Reactions – Sequence of reactions (steps) that are required to reach desired products 4 4( 2 8( 4 8 8 FeS2 + 11O2 --> 2 Fe2O3 + SO2 + O2 --> 2 SO3 SO3 + H2O --> H2SO4 8 SO2 FeS2 +11 O2 --> 2 Fe2O3 + SO2 + 4O2 --> 8 SO3 SO3 + 8 H2O --> 8H2SO4 8 SO2 4 FeS2 + 15 O2 + 8 H2O --> 2 Fe2O3 + 8 H2SO4 Now you can have any kind of problem 4 FeS2 + 15 O2 + 8 H2O --> 2 Fe2O3 + 8 H2SO4 How many moles of sulfuric acid can be made from 5 mol FeS2 and 17 mol O2? Have LR problem. 1) Determine the LR 5 molFeS2 15 molO 2 18.75 molO 2 required 4 molFeS2 Mol O2 given < mol O2 required so O2 LR 2) Determine the amount of product 17 molO 2 8 molH 2SO4 9.1molH 2SO4 15 molO 2 Combustion Analysis • Combustion analysis – Reaction of a compound that contains carbon, hydrogen and sometimes oxygen burned in air (O2) to produce CO2 and H2O • Can be used to determine empirical formula and % composition A 0.3000 gram sample containing only carbon, hydrogen and oxygen was burned in air to produce 0.440 grams CO2 and 0.180 grams H2O. Determine the empirical formula of the sample. y CxHyOz + ?O2 --> xCO2 + H2O 2 Determine moles of C and H from moles of CO2 and H2O 0.440 gCO2 1molCO2 1molC molC 0.01molC 44.0 g 1molCO 2 0.180 gH 2O 1molH 2O 2 molH molH 0.02 molH 18.0 g 1molH O 2 Determine moles of O from g of O in sample g O = g sample - g C - g H g O = 0.3000 -(0.01 mol x 12.0 g/mol) - (0.02 mol x 1.01g/mol) g O = 0.16 g mol O = 0.16g/ 16.0 g/mol = 0.01 mol Mol C = 0.01 /0.01 = 1 Mol H = 0.02 /0.01 = 2 Mol O = 0.01 /0.01 = 1 Empirical formula = CH2O How will the exam on Thursday be? • • • • • A) impossible B) Very hard C) able to be done D) un understandable E) hard but workable