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Mass Relationships in Chemical Reactions Chapter 3 4 - 5 Lectures Dr. Ali Bumajdad Chapter 3 Topics Stoichiometry • Average Atomic Masses •The Mole •Molar Mass (M.m.) and Molecular mass (M.w.) •Percent Composition •Calculation Empirical & Molecular Formula •Balancing Chemical Equation •Stoichiometric Calculations & Limiting Reactant •Theoretical yield, actual yield and percentage yield Dr. Ali Bumajdad Atomic mass & Average Atomic Masses Macro World grams Molar mass Micro World atoms & molecules Amu Atomic mass •Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12C “weighs” 12 amu On this scale 1H = 1.008 amu 16O = 16.00 amu (0) Av. At. mass = (m of Isotope 1 x its abundance) + (m of Isotope 2 x its abundance) + ……. 100 Natural lithium is: 7.42% 6Li (6.015 amu) 92.58% 7Li (7.016 amu) Average atomic mass of lithium: 7.42 x 6.015 + 92.58 x 7.016 = 6.941 amu 100 Dr. Ali Bumajdad Average atomic mass (6.941) Sa Ex 3.1: Cu vaporized in a mass spectrometer and It was found to have two isotopes one of mass 62.93 amu and abundance 69.09% and the other of mass 64.93 amu and abundance 30.91% Find its average mass. Av. m = 63.55 amu •The Mole SI unit of amount of substance number number •mole : amount of a substance that contains 6.0221367 x 1023 objects. •mole : number of carbon atoms in 12 g of 12C 1 mol = NA = 6.0221367 x 1023 number Avogadro’s number (NA) Mass of 1 mole Number of atoms in 1 mole 12.00 grams of 12C contains 6.022 x 1023 1.008 grams of H contains 6.022 x 1023 Mass of 1/2 mole Number of atoms in 1/2 mole 6.00 grams of 12C contains 3.011 x 1023 Dr. Ali Bumajdad Molar Mass (M.m.) and Molecular mass (M.w.) eggs Molar mass is the mass of 1 mole of shoes in grams marbles atoms 1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g 1 12C atom = 12.00 amu 1 mole lithium atoms = 6.941 g of Li For any element atomic mass in amu = molar mass in grams/mol •From the periodic table C 12.01 amu 12.01 g mass of 1 atom of C mass of 1mol of C mass of 6.022 x 1023 atoms of C One Mole of atoms S C 12.01 g 32 g 200.59 g Hg 55.85 g 63.55 g Cu Fe Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. 1S SO2 2O SO2 32.07 amu + 2 x 16.00 amu 64.07 amu 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2 For any compound Molecular mass in amu = molar mass in grams/mol M.w. of SO2 = 64.07 amu M.w. of C8H10N4O2 = 194.20 amu One Mole of molecules Q) 1 mol of apple contains 6.022 x 1023 apple Q) 1 mol of CH3OH contains 6.022 x 1023 CH3OH Q) 1 mol of CH3OH contains 6.022 x 1023 O atoms Q) 1 mol of CH3OH contains 4 x 6.022 x 1023 H atoms Q) 1/2 mol of CH3OH contains (6.022 x 1023)/2 CH3OH Q) 1/2 mol of CH3OH contains (4 x 6.022 x 1023)/2 H atoms N = n × NA N = no. of dozen × 12 Dr. Ali Bumajdad n= m (1) M.m. n (2) n= = number of mole M.m. = molar mass in g/mol m = mass NA = Avogadro’s number N = number of objects N NA AaBb ×a No. of A atoms No. of AaBb molecules a No. of AaBb molecules × a 6.022×1023 6.022×1023 × a No. of A moles Q) 1.0 ×103 CH4 molecules contain: 1) How many H atoms. 2) How many moles of H atoms. Q) How many atoms are in 0.551 g of potassium (K) ? (2) I need n and NA n= Find n using: n= N NA m (1) M.m. n = 0.551 g / 39.10 g mol-1 n = 0.0141 mol N = (0.0141) (6.022 x 1023) = 8.49 x 1021 K atoms Expected Dr. Ali Bumajdad Q) How many H atoms are in 72.5 g of C3H8O ? Answer = 5.82 x 1024 atoms H Dr. Ali Bumajdad Because m of 1 apple in g = m of 1 dozen in g 12 m of 1 atom in g = m of 1 mol in g (3) NA m of 1 atom in g = M.m. NA (3) Sa Ex. 3.2: Calculate the mass in gram of 6 atoms of Americium (Am) Method 1: Using Eq. 3 Method 2: 243 g 6.022 x 1023 atoms X 6 atoms Sa Ex. 3.3: Calculate the number of moles and the number of atoms in 10.0 g sample of Al. Use Eq. 1 and 2 n = 0.371 mol Al N = 2.23 × 1023 Al atoms Sa Ex. 3.4: How many Si atoms in 5.68 mg of silicon computer chip? N = 1.22 × 1020 Si atoms Sa Ex. 3.5: Calculate the number of moles and the mass in gram of a sample of Co containing 5.00 ×1020 atoms Method 1: m of 5.00 ×1020 atoms = 4.89 ×10-2g n = 8.30 ×10-4 mol Method 2: n = 8.30 ×10-4 mol m of 5.00 ×1020 atoms = 4.89 ×10-2g Sa Ex. 3.6: (1) Calculate the molar mass of C10H6O3. (2) A sample of 1.56 ×10-2g of C10H6O3, how many moles does this sample represent (1) M.m.= 174.1g (2) No. of moles = 8.96 × 10-5 mol Sa Ex. 3.7: (1) Calculate the molar mass of CaCO3. (2) A sample of 4.86 moles of CaCO3, what is the mass of the CO32- ions present? M.m. = 100.09 g/mol mass of the CO32- = 292 g Sa Ex. 3.8: Bees release Isopentyl acetate (C7H14O2) when sting. The amount release is about 1 g (1) How many molecules of C7H14O2 are released (2) How many atoms of Carbon are present (3) How many atoms are present (1) no. of molecules= 5×1015 molecules (2) no. of C atoms = 4× 1016 C atoms (3) no. of atoms = 1× 1017 atoms We should use Eq.1 and Eq. 2 and … n= m M.m. (1) (2) n= N NA Number of H atoms = 1.03 ×1024 H atoms We should use Eq.1 n= m (1) M.m. Mass of Zn = 23.3 gram We should use Eq.1 and Eq. 2 n= m M.m. (1) (2) n= N NA Number of S atoms = 3.06 × 1023 S atoms Heavy Light Heavy Light •Percent Composition A) If I know M.F. Suppose I have AaBb molecule (M.m.A) × a %A= M.m. AaBb x 100% (4) (12.01 g) x 2 x 100% = 52.14% 46.07 g (1.008 g) x 6 %H = x 100% = 13.13% 46.07 g (16.00 g) x 1 %O = x 100% = 34.73% 46.07 g %C = C2H6O 52.14% + 13.13% + 34.73% = 100.0% Dr. Ali Bumajdad B) If I do not know M.F. but I know the masses Suppose I have AaBb molecule %A= x 100% %C = 24 g 32 g x 100% = 75 % %H = 8g 32 g x 100% = 25 % CyHx m of C =24 g m of H = 8g (5) mA mA +mB Dr. Ali Bumajdad %H= %P= %O= = 3.086% = 31.61% = 65.31% •Calculation Empirical & Molecular Formula A) Empirical formula from combustion of known amount 1) Find the moles of H and C from mass of H2O and CO2 2) Find mass of O (or any other element) using: Mass of O = mtotal – (mH+mC) 3) Find the moles of O from mass of O 4) Divide by the smallest mole value Combust 11.5 g ethanol 22.0 g CO2 13.5 g H2O g CO2 mol CO2 mol C 0.5 mol C g H2O mol H2O mol H 1.5 mol H gO mol O 0.25 mol O Empirical formula C0.5H1.5O0.25 Divide by smallest subscript (0.25) Empirical formula C2H6O Dr. Ali Bumajdad Q) 0.1156g f compound contains C, H and N only. The H2O Absorber increase by 0.1676g and the CO2 absorber increase by 0.1638g what is the empirical formal of the compound? The E.F. is CH5N Dr. Ali Bumajdad B) Empirical Formula from element masses or %mass 1) Find the moles of elements (for %mass assume you have 100 g) 2) Divide by the smallest mole value C) Molecular Formula from element masses or %mass and M.m.M.F. 1) Find the moles of elements (for %mass assume you have 100 g) 2) Divide by the smallest mole value. Now you know E.F. 3) Use: M.F. = E.F. × M.m. M.F. M.m. E.F. (6) Dr. Ali Bumajdad M.F. = N2O4 M.m. = 90.02 g/mol Sa. Ex. 3.11: Compounds consists of 71.65% Cl, 24.27%C And 4.07% H and its molar mass = 98.96 g/mol a) Determine its E.F. b) Determine its M.F. (Assume that we have 100 grams of compound) a) E.F. = CH2Cl b) M.F. = C2H4Cl2 Dr. Ali Bumajdad Sa. Ex 3.12: White powder contain 43.64% P and 56.36% O. The compound molar mass = 283.88 g/mol. a) What is E.F. b) What is M.F. 1) Find the moles of elements (for %mass assume you have 100 g) 2) Divide by the smallest mole value E.F. = P2O5 M.F. = P4O10 Dr. Ali Bumajdad Balancing Chemical Equation Reactant Product 2 Mg + O2 2 MgO Coefficent 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO 2 moles Mg + 1 mole O2 makes 2 moles MgO 1 moles Mg + 1/2 mole O2 makes 1 moles MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO NOT 2 grams Mg + 1 gram O2 makes 2 g MgO Balancing Chemical Equations C2H6 + O2 CO2 + H2O 1.Start by balancing those elements that appear in only one reactant and one product. start with C or H but not O C2H6 + O2 2CO2 + 3H2O 2. Balance those elements that appear in two or more reactants or products. C2H6 + 7 O2 2 2CO2 + 3H2O 2C2H6 + 7O2 4CO2 + 6H2O Dr. Ali Bumajdad Q) Balance: 1) 1C2H5OH (L) + 3O2 (g) 2) Na2CO3 (s) + 2HCl (aq) 3) 1C6H6 (L) + 7.5 O2 (g) 2 15 2CO2 (g) + 3H2O (g) 2NaCl (s) + H2O (L) + CO2 (g) 6 CO2 (g) + 3H2O (g) 12 6 Sa Ex. 3.14: balance (NH4)2Cr2O7 (s) Cr2O3 (s) + N2(g) + 4 H2O (g) Sa Ex. 3.15: balance 2NH3 (g) + 5/2O2 (g) 4 5 2NO (g) + 4 3H2O (g) 6 Dr. Ali Bumajdad 4Al + 3O2 2Al2O3 Stoichiometric Calculations & Limiting Reactant Calculation involving masses ( Coefficient provide moles ratios not exact amount) 1. Balanced the equation 2. Find moles of one reactant or product 3. Use coefficient ratio to calculate the moles of the required substance 4. Convert moles of required substance into masses Dr. Ali Bumajdad Q) How many moles of C atoms are needed to combined with 4.87 mol Cl to form C2Cl6 1) Balanced the equation 2C + 3Cl2 C2Cl6 2) Find moles of one reactant or product 2C ? + 3Cl2 C2Cl6 2.435 mol 3) Use coefficient ratio to calculate the moles of the required substance 2 mol of C X 3 mol Cl2 2.435 mol of Cl2 = 1.62 mol C Dr. Ali Bumajdad Q) 209 g of methanol burns in air according to the equation: CH3OH + O2 CO2 + H2O Find mass of water formed. 1) Balanced the equation 2CH3OH + 3O2 2CO2 + 4H2O 2) Find moles of one reactant or product 2CH3OH + 3O2 2CO2 + 4H2O 209 g = 6.52 mol n= 32.04 g/mol 3) Use coefficient ratio to calculate the moles of the required substance 2 mol of CH3OH 4 mol H2O 6.52 mol of CH3OH X = 13.04 mol H2O 4) Convert moles of required substance into masses m (1) m H O =n ×M.m.=13.04×18.016 = 235 g H O n= 2 2 M.m. Dr. Ali Bumajdad Q) How many grams of Ca must react with 83.0 g of Cl2 to form CaCl2 1) Balanced the equation Ca + Cl2 CaCl2 2) Find moles of one reactant or product Ca + Cl2 CaCl2 83.0 g ? 83.0 g = 1.17 mol n= 70.9 g/mol 3) Use coefficient ratio to calculate the moles of the required substance 1 mol of Ca X 1 mol Cl2 1.17 mol of Cl2 = 1.17 mol Ca 4) Convert moles of required substance into masses m (1) n= m Ca =n ×M.m.= 1.17×40.08 = 46.9 g Ca M.m. Sa. Ex. 3.16: LiOH(s) + CO2 (g) Li2CO3(s) + H2O(L) What is the mass of CO2 require to react with 1.00 Kg of LiOH? 1) Balanced the equation 2LiOH(s) + CO2 (g) Li2CO3(s) + H2O(L) 2) Find moles of one reactant or product 2LiOH(s) + CO2 (g) Li2CO3(s) + H2O(L) ? 1.00 × 103 g 1.00 × 103 g = 41.75 mol (expected) n= 23.95 g/mol 3) Use coefficient ratio to calculate the moles of the required substance 2 mol of LiOH 1 mol CO2 41.75 mol of LiOH X = 20.88 mol CO2 4) Convert moles of required substance into masses m (1) m CO =n ×M.m.= 20.88×44.0 = 920 g CO n= 2 2 M.m. Dr. Ali Bumajdad Limiting Reactant (the reactant the completely consumed in a chemical reaction) 6 green used up 6 red left over Dr. Ali Bumajdad e.g. 2H2 + O2 2 mol 1mol 2H2O 2mol 1 mol 1/2mol 1mol 2.1 mol 1mol limiting 2mol 1.9 mol limiting 1mol 1.9mol Limiting reactant determine the amount of product Dr. Ali Bumajdad • Why limiting reactant is important? Because in the stoichiometric calculation I should only use the coefficient of the limiting reactant • How do I know that a reaction contains a limiting reactant ? If I've been given the masses or number of moles of two reactant then I might have limiting reactant Dr. Ali Bumajdad Calculation involving a limiting reactant 1. Balanced the equation 2. Find moles of the two reactants 3. Identify the limiting reactant (How?) 4. Use the coefficient ratio between the limiting reactant and the required substance to find out the number of mole of the required substance 5. Convert moles of required substance into masses Q) In one process, 124 g of Al are reacted with 601 g of Fe2O3 2Al + Fe O Al O + 2Fe 2 3 2 3 Calculate the mass of Al2O3 formed. 1) Balanced the equation 2Al + Fe2O3 Al2O3 + 2Fe 2) Find moles of the two reactants 2Al + Fe2O3 Al2O3 + 2Fe 124g 601g ? limiting n= 124 g = 4.60 mol 26.98 g/mol n= 601 g = 3.76 mol 159.69 g/mol 3) Identify the limiting reactant (How?) 4.60 = 2.3 2 limiting 3.76 1 = 3.76 Dr. Ali Bumajdad 3) Use the coefficient ratio between the limiting reactant and the required substance to find out the number of mole of the required substance 2 mol of Al 4.60 mol of Al 1 mol AL2O3 X = 2.30 mol Al2O3 4) Convert moles of required substance into masses m (1) n= m Al2O3 =n ×M.m.= 2.30×101.96 = 235 g Al2O3 M.m. Dr. Ali Bumajdad Sa.Ex. 3.18: 18.1 g of NH3 and 90.4g of CuO reacted NH3 + CuO N2 + Cu + H2O 1) Which is the limiting reactant? 2) How many grams of N2 will be formed? 1) Balanced the equation 2NH3 + 3CuO N2 + 3Cu + 3H2O 2) Find moles of the two reactants 2NH3 + 3CuO N2 + 3Cu + 3H2O 18.1g 90.4g ? limiting n= 18.1 g = 1.06 mol 17.03 g/mol n= 90.4 g = 1.14 mol 79.55 g/mol 3) Identify the limiting reactant (How?) 1.06 2 = 0.53 1.14 3 = 0.38 limiting Dr. Ali Bumajdad 3) Use the coefficient ratio between the limiting reactant and the required substance to find out the number of mole of the required substance 3 mol of CuO 1.14 mol of Al 1 mol N2 X = 0.380 mol N2 4) Convert moles of required substance into masses m (1) m N =n ×M.m.= 0.380×28.0 = 10.6 g N n= 2 2 M.m. Dr. Ali Bumajdad •Theoretical yield, actual yield and percentage yield •Mass of product •Maximum yield •Calculated using stiochiometry •Mass of product (7) •Known by experiment Actual Yield x 100 % Yield = •Never more that Theoretical Yield theoretical yield Dr. Ali Bumajdad Sa.Ex. 3.19: 8.60 kg of H2 and 68.5kg of CO reacted H2 + CO CH3OH 1) Theoretical yield? 2) % yield if the actual yield = 3.57 × 104 g CH3OH 1) Balanced the equation 2H2 + CO CH3OH 2) Find moles of the two reactants 2H2 + CO CH3OH 8.60 ×103g 68.5 × 103g ? limiting 8.60 ×103g 68.5 × 103g 3 = 4.27×10 mol n = = 2.44×103 mol n= 2.016 g/mol 28.02 g/mol 3) Identify the limiting reactant (How?) 4.27×103 = 2135 2 limiting 2.44×103 = 2440 1 Dr. Ali Bumajdad 3) Use the coefficient ratio between the limiting reactant and the required substance to find out the number of mole of the required substance 2 mol of H2 4.27×103 mol of H2 1 mol CH3OH X = 2135 mol CH3OH 4) Convert moles of required substance into masses m (1) n= M.m. m CH3OH =n ×M.m.= 2135×32.04 = 6.86×104 g CH3OH % yield = 3.57 × 104 g × 100 = 52.0% 6.86×104 g Dr. Ali Bumajdad