1. Stoichiometry by Kate McKee

2. 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

3. 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

4. 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

5. Mole - Avogadro's Number 6.022 × 10 23 ●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 10 23 atoms) (12 amu / atom) = 12 g o Molar mass of carbon

6. Molar Mass ●Mass in grams of one mole of the compound Molar Mass of Juglone, C 10 H 6 O 3 10C: 10 x 12.011 g 6H: 6 x 1.0079 g 3O: 3 x 16.00 g 174.1 grams = molar mass of C 10 H 6 O 3

7. 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 (H 2 O) 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 H 2 O = 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%

8. 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 = 1Cl 2 C 2 H 4 4.04 mol H / 2.021 mol H = 2 Empirical = ClCH 2 = 49.48 g/mol

9. 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) C 2 H 3 Br + 3O 2 CO + 2H 2 O + HBr 2C 2 H 3 Br + 3O 2 4CO + 2H 2 O + 2HBr

10. 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 CO 2 absorbed by 1.00 kg LiOH LiOH (s) + CO 2 (g) Li 2 CO 3 (s) + H 2 O (l) Balanced = 2LiOH (s) + CO 2 (g) Li 2 CO 3 (s) + H 2 O (l) LiOH molar mass = 23.95 g/mol 41.8 mol LiOH = 1.00 kg LiOH/ 23.95 g LiOH 1mol CO 2 / 2mol LiOH 41.8 mol LiOH x (1mol CO 2 / 2mol LiOH) = 20.9 mol CO 2 20.9 mol CO 2 x (44.0 g CO 2 /1mol CO 2 ) = 920. g CO 2 absorbed

11. 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

12. Limiting Reagent NH 3(g) + CuO (s) N 2(g) + Cu (s) + H 2 O (g) 18.1 g NH 90.4 g CuO Balanced = 2NH 3(g) + 3CuO (s) N 2(g) + 3Cu (s) + 3H 2 O (g) Moles = 18.1 g NH 3 x (1 mol NH 3 / 17.03 g NH 3 ) = 1.06 mol NH 3 90.4 g CuO x (1 mol CuO/ 79.55 g CuO) = 1.14 mol CuO Mole Ratio = 3 mol CuO/ 2 mol NH 3 Moles CuO required to react with 1.06 mol NH 3 =1.06 mol NH 3 x (3 mol/ 2 mol) = 1.59 mol CuO 1.59 mol CuO>1.14 mol CuO CuO is limiting reagent

13. 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 N 2 /10.6 g N 2 x 100% = 62.5% yield

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