Stoichiometry by Kate McKee Stoichiometry Main Ideas: ● Atomic Mass ● Mole ● Stoichiometric Problems o Percent Composition o Determining Formula o Amounts of Reactants and Products o Limiting Reagent o Percent Yield ● Balancing Equations Atomic Mass ● Mass of an atom of a chemical element, approximately equal to the mass number of the element ● Average atomic mass (amu) o percent of element(atomic mass or element) 98.89% of Carbon-12 and 1.11% of 13.0034 (Carbon-13) 0.9889 (12.00 g) + .0111 (13.0034 g) = 12.01 amu for natural Carbon Mole ● ● ● ● ● small mammals adapted to a subterranean lifestyle cylindrical bodies and velvety fur adorable moles can be trapped in almost every season carnivorous, but may scrape at your plant’s roots, ruining your garden Mole - Avogadro's Number 6.022 × 23 10 ● Developed by Avogadro ● measurement of units ● a sample of a natural element with a mass equal to the element’s atomic mass expressed in grams contains 1 mole of atoms ● 1 mole of amu’s (Atomic Mass Units) is exactly 1 gram o (6.022 x 1023 atoms) (12 amu / atom) = 12 g o Molar mass of carbon Molar Mass ● Mass in grams of one mole of the compound Molar Mass of Juglone, C10H6O3 10C: 10 x 12.011 g 6H: 6 x 1.0079 g 3O: 3 x 16.00 g 174.1 grams = molar mass of C10H6O3 Percent Composition ● describes composition of a compound in terms of percentages (by mass) of its elements ● compare the mass of each element present in 1 mole of the compound to the total mass of 1 mole of compound Percent Composition of Water (H2O) Mass of H = 2 mol H x 1.0079 g/mol = 2.0158 g Mass of O = 1 mol O x 15.999 g/mol = 15.999 g Mass of 1 mol H2O = 18.0148 g Mass % of H = 2.0158 g/ 18.0148 g = 11.19% Mass % of O = 15.999 g/18.0148 g = 88.81% Determining Formula ● determine moles of each element present o assume 100 g of substance ● determine smallest whole number ratio by dividing by smallest of mol values ● empirical formula = simplest formula for a compound molecular formula = either empirical or multiple of empirical Molar Mass known = 98.96 g/mol 2.021 mol Cl / 2.021 mol Cl = 1 Molecular formula = 98.96 g/49.48 g = 2 2.021 mol C / 2.021 mol C = 1 Cl2C2H4 4.04 mol H / 2.021 mol H = 2 Empirical = ClCH2 = 49.48 g/mol Chemical Equations ● give reactants and products of a specific equation, state of reactants and products, and relative number of reactants and products ● Balance Chemical reactions for stoichiometry… o determine reactants and products o write unbalanced equation o balance with trial and error ( same # each type of atom appears on both reactant and product sides) C2H3Br + 3O2 2C2H3Br + 3O2 CO + 2H2O + HBr 4CO + 2H2O + 2HBr Chemical Stoichiometry ● ● ● ● Balance equation for reaction convert mass to moles set up appropriate mole ratios use mole ratios to calculate desired reactant or product Find mass of CO2 absorbed by 1.00 kg LiOH LiOH(s)+ CO2 (g) Li2CO3 (s) + H2O(l) Balanced = 2LiOH(s)+ CO2 (g) Li2CO3 (s) + H2O(l) LiOH molar mass = 23.95 g/mol 41.8 mol LiOH = 1.00 kg LiOH/ 23.95 g LiOH 1mol CO2/ 2mol LiOH 41.8 mol LiOH x (1mol CO2/ 2mol LiOH) = 20.9 mol CO2 20.9 mol CO2 x (44.0 g CO2/1mol CO2) = 920. g CO2 absorbed Limiting Reagent ● the reactant that runs out first and thus limits the amounts of products that can form ● deals with moles of molecules instead of individual molecules because of larger quantities of materials ● to determine limiting reagent… o balance equation o determine moles o determine mole ratio o determine moles required to react with each other o compare amount required with amount present to determine limiting reagent o check with mole ratio Limiting Reagent NH3(g)+ CuO(s) N2(g)+ Cu(s)+ H2O(g) Balanced = 2NH3(g)+ 3CuO(s) 18.1 g NH 90.4 g CuO N2(g)+ 3Cu(s)+ 3H2O(g) Moles = 18.1 g NH3 x (1 mol NH3/ 17.03 g NH3) = 1.06 mol NH3 90.4 g CuO x (1 mol CuO/ 79.55 g CuO) = 1.14 mol CuO Mole Ratio = 3 mol CuO/ 2 mol NH3 Moles CuO required to react with 1.06 mol NH3=1.06 mol NH3 x (3 mol/ 2 mol) = 1.59 mol CuO 1.59 mol CuO>1.14 mol CuO CuO is limiting reagent Percent Yield ● actual yield of a product ● theoretical yield is amount of product formed when limiting reactant is completely consumed (Actual Yield/ Theoretical Yield) x 100% = percent yield 6.63 g N2/10.6 g N2 x 100% = 62.5% yield You’re Welcome ● You’re Welcome