Mass Relationships in Chemical Reactions PDF

Mass Relationships in Chemical Reactions Chapter 3 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.  Introduction Micro World Macro World atoms & molecules moles atomic mass...

Mass Relationships in Chemical Reactions Chapter 3 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.  Introduction Micro World Macro World atoms & molecules moles atomic mass molar mass amu grams  Atomic Mass or Atomic Weight It is the mass of an atom in atomic mass units (atomic mass = mass number in amu) Examples: 1H = 1.001 amu 12C = 12.00 amu 16O = 16.00 amu By definition: 1 atom 12C “weighs” 12 amu 1 amu = exactly 1/12 of the mass of one 12C atom  Average Atomic Mass 6 C 12.01 Atomic mass is not 12.00 Isotope Atomic mass Natural abundance Carbon-12 12.00000 amu 98.90 % Carbon-13 13.00335 amu 1.1 %  Average Atomic Mass is a weighted average of the masses of all isotopes present in the sample Natural lithium exists in two isotopes 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 Average atomic mass (6.941)  Avogadro’s Number The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12 grams of 12C Chemists measure atoms and molecules in moles 1 mol = NA = 6.022 x 1023 Avogadro’s number (NA)  Molar Mass Molar mass (g/mol) is the mass of 1 mole of units (such as atoms or molecules) in grams 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 The molar mass (in grams) contains Avogadro’s number of atoms and is numerically equal to the atomic mass in amu For any element atomic mass (amu) = molar mass (grams) (Numerically equal)  One Mole of: C S Hg Cu Fe  Relationship between atomic mass units (amu) and grams (g) Mass of 1 mole of atoms 12C = 12 g Mass of 6.022 x 1023 atoms 12C = 12 g mass of 1 atom 12C = 12 g / 6.022 x 1023 = 12 amu 1 g = 6.022 x 1023 amu 1 amu = 1.66 x 10-24 g  Relationship between mass (g); number of moles and Avogadro’s number M = Molar mass in (g/mol) NA = Avogadro’s number n = Number of moles N = Number of atoms mass (g) n= molar mass (g/mol) N = n (mol) x Avogadro’s number (atom/mol)  Do You Understand Molar Mass? How many atoms are in 0.551 g of potassium (K) ? g of K mol of K atoms of K 1 mol K = 39.10 g K 1 mol K = 6.022 x 1023 atoms K 0.551 g K x 1 mol K x 6.022 x 1023 atoms K = 39.10 g K 1 mol K 8.49 x 1021 atoms K  Molecular Mass Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. 1S 32.07 amu 2O + 2 x 16.00 amu SO2 SO2 64.07 amu For any molecule molecular mass (amu) = molar mass (grams) (same numerical value) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2 Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. 1Na 22.99 amu NaCl 1Cl + 35.45 amu NaCl 58.44 amu For any ionic compound formula mass (amu) = molar mass (grams) 1 formula unit NaCl = 58.44 amu 1 mole NaCl = 58.44 g NaCl What is the formula mass of Ca3(PO4)2 ? 1 formula unit of Ca3(PO4)2 3 Ca 3 x 40.08 2P 2 x 30.97 8O + 8 x 16.00 310.18 amu  How many H atoms are in 72.5 g of C3H8O ? g of C3H8O mol of C3H8O mol of H atoms of H 1 mol C3H8O = (3 x 12.01) + (8 x 1.008) + 16.00 = 60.10 g C3H8O 1 mol C3H8O molecules = 8 mol H atoms 1 mol H = 6.022 x 1023 atoms H 1 mol C3H8O 8 mol H atoms 6.022 x 1023 H atoms 72.5 g C3H8O x x x = 60.10 g C3H8O 1 mol C3H8O 1 mol H atoms 5.82 x 1024 atoms H Mass Spectrometer The Mass spectrometer is the most direct and accurate method for determining the atomic and molecular masses.  Percent Composition of Compounds Percent composition by mass is the percent by mass of each element in a compound. Percent composition of an element in a compound = n x molar mass of element x 100% molar mass of compound n is the number of moles of the element in 1 mole of the compound  What is the percent composition by mass of C2H6O? Molar mass = (2 x 12.01g) + (6 x 1.008 g) + (1 x 16.00 g) = 46.07g 2 x (12.01 g) %C = x 100% = 52.14% 46.07 g 6 x (1.008 g) %H = x 100% = 13.13% 46.07 g 1 x (16.00 g) %O = x 100% = 34.73% 46.07 g 52.14% + 13.13% + 34.73% = 100.0% C2H6O  Determination of Empirical Formula Step 1. Assume a definite starting quantity usually 100.0g of the compound, express the mass of each element in grams. if percentages are given convert to grams. Step 2. Convert the mass of each element into moles Mass (g) # of moles = Molar mass (g/mol) Step 3 Divide each value by the smallest of these values if numbers obtained are whole numbers, use them as subscripts in empirical formula If not, go to step 4 Step 4 Multiply the values obtained in step 3 by the smallest numbers that will convert them to whole numbers FeO1.5 Fe1 x 2O1.5 x 2 Fe2O3 Use these whole numbers as the subscripts in the empirical formula. Rounding of numbers obtained in calculations The results of calculations may differ from a whole number. – If they differ ± 0.1 round off to the next nearest whole number. 2.9 → 3 – Deviations greater than 0.1 unit from a whole number usually mean that the calculated ratios have to be multiplied by a whole number.  What is the empirical formula of a compound with the following percent composition? C (40.92%), H (4.58%) and O (54.50%) Step 1 Assume 100 g convert percentages to grams: 40.92 c of C; 4.58 g of H , 54.50 g of O Step 2. Convert the mass of each element into moles 1 mol C nC = 40.92 g C × = 3.407 mol C 12.01 g C 1 mol H = 4.54 mol H nH = 4.58 g H × 1.008 g H 1 mol O = 3.406 mol O nO = 54.50 g O × 16.00 g O Step 3 Divide each value by the smallest of these values 3.407 4.54 3.406 C: ≈1 H: = 1.33 O: =1 3.406 3.406 3.406 Empirical formula CH1.33O Step 4 Multiply the values obtained in step 3 by the smallest numbers that will convert them to whole numbers C1 x 3H 1.33 x 3O1 x 3 Empirical formula C3H4O3  Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Experimental Determination of Empirical Formula Example : When ethanol is burned in an apparatus, carbon dioxide and water vapor are given off. O2 Ethanol Unused O2 Heat H2O CO2 absorber absorber The masses of CO2 and H2O produced can be determined from the increase in mass of the CO2 and H2O absorbers  The combustion of 11.5 g ethanol produced 22.0 g CO2 and 13.5 g of H2O. Determine its empirical formula g CO2 mol CO2 mol C gC 1 mol CO2 1 mol C 12.01 g C mass of C = 22.0 g CO2 × × × 44.01 g CO2 1 mol CO2 1 mol C = 6.00 g C g H2O mol H2O mol H gH 1 mol H2O 2 mol H 1.008 g H mass of H = 13.5 g H2O × × × 18.02 g H2O 1 mol H2O 1 mol H = 1.51 g H mass of O = mass of sample – (mass of C + mass of H) = 11.5 g – (6.00 g + 1.51 g) = 4.0 g The number of moles of C, H and O can now be calculated 1 mol C nC = 6.00 g C × = 0.500 mol C 12.01 g C 1 mol H nH = 1.51 g H × = 1.50 mol H 1.008 g H 1 mol O = 0.25 mol O nO = 4.0 g O × 16.00 g O Divide by smallest number (0.25) Empirical formula C2H6O  Determination of Molecular Formulas The molecular formula can be calculated from the empirical formula if the molar mass is known. The molecular formula will be equal to the empirical formula or some multiple x of it. ( mass of empirical formula) x = molar mass To determine the molecular formula evaluate x. molar mass x= empirical molar mass x is the number of units of the empirical formula contained in the molecular formula.  A sample of a compound contains 1.52 g of nitrogen (N) and 3.47 g of Oxygen (O). The molar mass of this compound is 92 g. Determine the molecular formula. 1. Determine the empirical formula of the compound 1 mol N nN = 1.52 g N × = 0.108 mol N 14.01 g N 1 mol O = 0.217 mol O nO = 3.47 g O × 16.00 g O Divide by smallest number (0.108) Empirical formula NO2 Empirical molar mass = 14.01 g + 2(16.00 g) = 46.01 g molar mass x= empirical molar mass Molecular formula (NO2)x 92 =2 Molecular formula N2O4 x= 46.01  Chemical Reactions and Chemical Equations A chemical reaction is the process in which one or more substances is changed into one or more new substances A chemical equation uses chemical symbols to show what happens during a chemical reaction reactants products  How to “Read” Chemical Equations 2 Mg + O2 2 MgO 2 atoms Mg + 1 molecule O2 2 formula units MgO 2 moles Mg + 1 mole O2 2 moles MgO 48.6 grams Mg + 32.0 grams O2 80.6 g MgO IS NOT 2 grams Mg + 1 gram O2 2 g MgO  Balancing Chemical Equations 1. Write the unbalanced equation using the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C2H6 + O2 CO2 + H2O 2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2C2H6 NOT C4H12  Balancing Chemical Equations 3. Start by balancing those elements that appear in only one reactant and one product. C2H6 + O2 CO2 + H2O start with C or H but not O 2 carbon 1 carbon multiply CO2 by 2 on left on right C2H6 + O2 2CO2 + H2O 6 hydrogen 2 hydrogen multiply H2O by 3 on left on right C2H6 + O2 2CO2 + 3H2O  Balancing Chemical Equations 4. Balance those elements that appear in two or more reactants or products. C2H6 + O2 2CO2 + 3H2O multiply O2 by 7 2 2 oxygen 4 oxygen + 3 oxygen = 7 oxygen on left (2x2) (3x1) on right C2H6 + 7 O2 remove fraction 2CO2 + 3H2O 2 multiply both sides by 2 2C2H6 + 7O2 4CO2 + 6H2O  Balancing Chemical Equations 5. Check to make sure that you have the same number of each type of atom on both sides of the equation. 2C2H6 + 7O2 4CO2 + 6H2O Reactants Products 4 C (2 x 2) 4C 4C 4C 12 H 12 H 12 H (2 x 6) 12 H (6 x 2) 14 O 14 O 14 O (7 x 2) 14 O (4 x 2 + 6)  Amounts of Reactants and Products Stoichiometry is the quantitative study of reactants and products in a chemical reaction. The Mole Method The stoichiometric coefficients in a balanced equation indicates the number of moles of each substance If the quantity of a reactant or product is given , it can be used to calculate the quantity of any other substance in the reaction. Use moles to calculate the amount of reactants or products in a reaction.  Mass Changes in Chemical Reactions 1. Write balanced chemical equation 2. Convert quantities of known substances into moles 3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4. Convert moles of sought quantity into desired units  Methanol burns in air according to the equation 2CH3OH + 3O2 2CO2 + 4H2O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH3OH moles CH3OH moles H2O grams H2O molar mass coefficients molar mass CH3OH chemical equation H2O 1 mol CH3OH 4 mol H2O 18.0 g H2O 209 g CH3OH x x x = 32.0 g CH3OH 2 mol CH3OH 1 mol H2O 235 g H2O Limiting Reagents 2NO + O2 2NO2 NO is the limiting reagent O2 is the excess reagent  Limiting Reagents Limiting Reagent is the reactant that is used up first in a reaction. It is called the limiting reactant because the amount of it present is insufficient to react with the amounts of other reactants that are present. The limiting reactant limits the amount of product that can be formed. Excess reactant is present in quantities greater than necessary to react with the quantity of the limiting reactant  Limiting Reactant Calculations: Method 1 Step 1 Calculate the amount of product (moles or grams, as needed) formed from each reactant. Step 2 Determine which reactant is limiting.(The reactant that gives the least amount of product is the limiting reactant; the other reactant is in excess. The limiting reactant is used in all calculations in a problem  Excess Reactant To calculate the amount of the substance that remains after the reaction ( in excess) Step 3 Calculate the amount of the other reactant required to react with the limiting reactant, then subtract this amount from the starting quantity of the reactant. Excess mass = given mass – mass (reacted) Excess moles = given moles – moles (reacted)  Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe2O3 2Al + Fe2O3 Al2O3 + 2Fe a) Calculate the mass of Al2O3 formed. g Al mol Al mol Al2O3 g Al2O3 g Fe2O3 mol Fe2O3 mol Al2O3 g Al2O3 1 mol Al 1 mol Al2O3 102 g Al2O3 124 g Al x x x = 234 g Al2O3 27.0 g Al 2 mol Al 1 mol Al2O3 1 mol Fe2O3 1 mol Al2O3 102 g Al2O3 601 g Fe2O3 x x x = 383 g Al2O3 160 g Fe2O3 1mol Fe2O3 1 mol Al2O3 124 g of Al gives the least amount of product Al2O3 Al is the limiting reagent Alternative method In one process, 124 g of Al are reacted with 601 g of Fe2O3 2Al + Fe2O3 Al2O3 + 2Fe a) Calculate the mass of Al2O3 formed. g Al mol Al mol Fe2O3 needed g Fe2O3 needed OR g Fe2O3 mol Fe2O3 mol Al needed g Al needed 1 mol Al 1 mol Fe2O3 160. g Fe2O3 124 g Al x x x = 367 g Fe2O3 27.0 g Al 2 mol Al 1 mol Fe2O3 Start with 124 g Al need 367 g Fe2O3 Have more Fe2O3 (601 g) so Al is limiting reagent  Use limiting reagent (Al) to calculate amount of product that can be formed. g Al mol Al mol Al2O3 g Al2O3 2Al + Fe2O3 Al2O3 + 2Fe 1 mol Al 1 mol Al2O3 102 g Al2O3 124 g Al x x x = 234 g Al2O3 27.0 g Al 2 mol Al 1 mol Al2O3 b) How much of the excess reagent is left at the end of the reaction? The excess reagent is Fe2O3 Excess mass = given mass – mass (reacted) = 601 g - 367 g = 234 g Fe2O3  Limiting Reactant Calculations: Method 2 aA + bB cC + dD a moles b moles c moles d moles Step 1 Calculate the number of moles of each given reactant Step 2 Divide the given number of moles of each reactant by the number of moles in the balanced equation If nA nB A is in excess, a b B is the limiting reactant n A nB B is in excess, If A is the limiting reactant a b  Reaction Yield The quantities of products calculated from equations represent the maximum yield (100%) of product according to the reaction represented by the equation. Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. Actual Yield % Yield = x 100 Theoretical Yield  Consider the reaction MnO2 + 4HCl MnCl2 + Cl2 + 2H2O If 0.86 mol of MnO2 and 48.2 g of HCl react, 1. Which reagent will be used up first? 2. How many grams of Cl2 will be produced? 3. Calculate the number of moles of the excess reagent remaining at the end of the reaction 4. Calculate the percent yield, if the actual yield of Cl2 is 18.6 g