Basics of Stoichiometry - Module 1 PDF

A B M DULE 1  macroscopic level microscopic level units = g, L, etc.. units = ??? CHM1311 Basics of Stoichiometry 2 The mole is an amount of substance that contains the same number of elementary entities as there are carbon- 12 atoms in exact...

A B M DULE 1  macroscopic level microscopic level units = g, L, etc.. units = ??? CHM1311 Basics of Stoichiometry 2 The mole is an amount of substance that contains the same number of elementary entities as there are carbon- 12 atoms in exactly 12 g of carbon-12. 12C 12.00 g 6.022 141 29 x 1023 carbon atoms CHM1311 Basics of Stoichiometry 3  NA = 6.022 141 29 x 1023 mol-1 Amedeo Avogadro (1776 – 1856) 12 things = 1 dozen things 6.022 x 1023 things = 1 mole of things CHM1311 Basics of Stoichiometry 4  molecular sum of atomic masses of mass = each atom in the molecule (in atomic mass units, u) molar mass, in grams, of one mass = mole of the substance (units = g/mol) 𝒎 𝒎 𝑴𝑴 = 𝒏= 𝒏 𝑴𝑴 CHM1311 Basics of Stoichiometry 5 For H2O: H H O H O H O H H H O O H H H H H H H O O O H H O H H Molecular Mass = Molar Mass = mass of one molecule mass of one mole 18.015 u 18.015 g/mol CHM1311 Basics of Stoichiometry 6  Concentration à later Percent Composition CHM1311 Basics of Stoichiometry 7  number of parts of a component in 100 parts of the whole – ex. 10% means “10 parts x per 100 parts of the whole” IMPORTANT: must be defined by a unit! – ex. a rock contains 3.5% gold by mass means 3.5 g of gold per 100 g of rock – ex. a bottle of wine contains 10.7% alcohol by volume means 10.7 mL of alcohol per 100 mL of wine – ex. an isotope occurring in nature with 1.8% abundance means 1.8 isotopes per 100 atoms overall CHM1311 Basics of Stoichiometry 8  number of parts of a component in 100 parts of the whole 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝐶𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 = ×100% 𝑡𝑜𝑡𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑚𝑖𝑥𝑡𝑢𝑟𝑒 ex. what % of students earned an A grade (A-/A/A+) last year? 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 = ×100% 269 students total CHM1311 Basics of Stoichiometry 9  when expressed as a conversion factor, the numerator and denominator must have the SAME UNITS – ex. a rock contains 3.5% gold by mass 3.5 g gold 3.5 kg gold 3.5 oz gold = = 100 g rock 100 kg rock 100 oz rock – ex. a bottle of wine contains 10.7% alcohol by volume 10.7 mL EtOH 10.7 L EtOH 10.7 tbsp EtOH = = 100 mL wine 100 L wine 100 tbsp wine CHM1311 Basics of Stoichiometry 10 Calculate the mass percent of hydrogen in ammonium bicarbonate, NH4HCO3. 1. 5.166 % 2. 8.392 % 3. 6.387 % 4. 22.82 % 5. I’m not sure CHM1311 Basics of Stoichiometry 11 Potassium-40 is one of the few naturally occurring radioactive isotopes of elements of low atomic number. Its percent natural abundance among K isotopes is 0.012%. How many 40K atoms do you ingest by drinking one cup of whole milk containing 371 mg of K? 0.012 mol 40K WANT: ? atoms of 40K 100 mol K 0.012 atoms 40K HAVE: abundance of 40K = 0.012% 100 atoms K mass of K = 371 mg 39.098 g K NEED: MM of K = 39.098 g/mol mol K CHM1311 Basics of Stoichiometry 13  Potassium-40 is one of the few naturally occurring radioactive isotopes of elements of low atomic number. Its percent natural abundance among K isotopes is 0.012%. How many 40K atoms do you ingest by drinking one cup of whole milk containing 371 mg of K? ? atoms of 40K 1gK mol K 0.012 mol 40K 6.022x1023 atoms 40K = 371 mg K x x x x 1000 mg K 39.098 g K 100 mol K mol 40K = 6.9x1017 atoms of 40K Sept 5 DGD: covers Dimensional Analysis CHM1311 Basics of Stoichiometry 14  Potassium-40 is one of the few naturally occurring radioactive isotopes of elements of low atomic number. Its percent natural abundance among K isotopes is 0.012%. How many 40K atoms do you ingest by drinking one cup of whole milk containing 371 mg of K? STEPWISE SOLUTION: 𝟒𝟎 𝒎 𝟒𝟎 𝒂𝒕𝒐𝒎𝒔 𝑲 𝒏= 𝒏( 𝑲) = 𝑴𝑴 𝑵𝑨 atoms of mass of K mol of K mol of 40K 40K 𝟒𝟎 𝟒𝟎 𝒏( 𝑲) % 𝑲= ×𝟏𝟎𝟎% 𝒏(𝑲) CHM1311 Basics of Stoichiometry 15  chemical equations depict the kind of reactants and products and their relative amounts in a reaction the physical state of the reactants and products may be indicated: 2 H2 (g) + O2 (g) ® 2 H2O (l) for the dissolution of compound in water, the following notation may be used: H2O NaCl (s) NaCl (aq) CHM1311 Basics of Stoichiometry 16  Because of the principle of the conservation of matter, an equation must be balanced. It must have the same number of atoms of the same kind on both sides. Antoine Lavoisier 1743 – 1794 CHM1311 Basics of Stoichiometry 17  NO(g) + O2(g) ® NO2(g) 1. Never introduce extraneous atoms to balance. NO(g) + O2(g) ® NO2(g) + O(g) X 2. Never change a formula for the purpose of balancing an equation. NO(g) + O2(g) ® NO3(g) X CHM1311 Basics of Stoichiometry 18 Empirical formula: simplest whole number RATIO of elements HO Molecular formula: H2O2 exact number of atoms of each element Structural formula: H–O–O–H shows relative placement and connectivity of atoms CHM1311 Basics of Stoichiometry 19  the data obtained from elemental analysis of a sample: – elements present – relative proportions CHM1311 Basics of Stoichiometry 21  Elemental Analysis Mass Elements present Spectrum and their relative (or other proportions techniques) EMPIRICAL Molecular MOLECULAR FORMULA Mass FORMULA CHM1311 Basics of Stoichiometry 22  compound + excess O2 (g) ® oxidation products (g) carbon CO2 hydrogen H2O nitrogen NO2 sulfur SO2 CHM1311 Basics of Stoichiometry 23 Complete combustion of a 1.505 g sample of an unknown compound consisting of C, H and S yields 3.149 g CO2, 0.645 g H2O and some SO2. a) What is the empirical formula for the unknown? b) We experimentally determine the molecular mass to be 168.1 u. Determine the molecular formula of the unknown. CHM1311 Basics of Stoichiometry 24  the study of the quantitative aspects of chemical reactions if the quantity of a reactant is known, it is possible to calculate the quantity of product that will be formed (or vice-versa) to simplify the process, we use the mole method: we use units of moles instead of grams or millilitres CHM1311 Basics of Stoichiometry 25 Step 1: Identify all reactants and products Step 2: Balance the chemical equation Step 3: Follow the mole method mass of mass of reactant product MOLAR MOLAR MASS MASS STOICHIOMETRIC moles of FACTOR moles of reactant product CHM1311 Basics of Stoichiometry 26 Lithium, like all alkali metals, reacts readily with water to produce hydrogen gas and aqueous metal hydroxide: 2 Li (s) + 2 H2O (l) à 2 LiOH (aq) + H2 (g) How many grams of solid lithium metal is needed to produce 9.89 g of hydrogen gas? 2 mol Li WANT: ? g Li 1 mol H2 HAVE: balanced equation mass of H2 = 9.89 g 6.941 g Li mol Li NEED: MM of Li = 6.941 g/mol MM of H2 = 2.016 g/mol 2.016 g H2 mol H2 CHM1311 Basics of Stoichiometry 27  Lithium, like all alkali metals, reacts readily with water to produce hydrogen gas and aqueous metal hydroxide: 2 Li (s) + 2 H2O (l) à 2 LiOH (aq) + H2 (g) How many grams of solid lithium metal is needed to produce 9.89 g of hydrogen gas? mol H2 2 mol Li 6.941 g Li ? g of Li = 9.89 g H2 x x x 2.016 g H2 1 mol H2 mol Li = 68.1 g Li MOLAR STOI’C MOLAR MASS RATIO MASS mass of moles of moles of mass of H2 H2 Li Li CHM1311 Basics of Stoichiometry 28  stoichiometry and the mole method is an extremely powerful tool it can be used with other conversion factors, such as: – density – concentration – percent composition CHM1311 Basics of Stoichiometry 29 An alloy used in aircraft structures consists of 93.7% Al and 6.3% Cu by mass. The alloy has a density of 2.85 g/cm3. A 0.691 cm3 piece of the alloy reacts with an excess of HCl(aq) to produce H2(g) and AlCl3(aq). If we assume that all the Al but none of the Cu reacts with HCl(aq), what is the mass of H2 obtained? CHM1311 Basics of Stoichiometry 30  the theoretical yield is the amount of product expected if the reactants react to completion the actual yield is always smaller than this value! – the inverse reaction occurs – other side-products are products – difficult to collect all of the product actual yield % yield = × 100% theoretical yield CHM1311 Basics of Stoichiometry 31  2 Li (s) + 2 H2O (l) à 2 LiOH (aq) + H2 (g) For the earlier example, if only 7.82 g of H2 were formed, what is the yield of the reaction? 7.82 g 9.89 g = 79.1% CHM1311 Basics of Stoichiometry 32 Two sets of successive reactions, A à B and B à C, have respective yields of 48% and 73%. What is the overall percent yield for conversion of A to C? 1. 66% 2. 48% 3. 73% 4. 35% 5. I’m not sure CHM1311 Basics of Stoichiometry 33  the mole and Avogadro’s number percent composition empirical vs. molecular formulae balancing chemical equations stoichiometry and the mole method percent yields CHM1311 Basics of Stoichiometry 35 Complete combustion of 200 g of a CH4/C3H8 mixture that is 25.0% methane and 75.0% propane by mass. What mass (in g) of carbon dioxide is produced? Try this on your own – the solution is posted on Brightspace CHM1311 Basics of Stoichiometry 36