stoichiometry by kate mckee. stoichiometry main ideas: ●atomic mass ●mole ●stoichiometric...
TRANSCRIPT
![Page 1: Stoichiometry by Kate McKee. Stoichiometry Main Ideas: ●Atomic Mass ●Mole ●Stoichiometric Problems o Percent Composition o Determining Formula o Amounts](https://reader035.vdocument.in/reader035/viewer/2022081420/56649cda5503460f949a3965/html5/thumbnails/1.jpg)
Stoichiometry
by Kate McKee
![Page 2: Stoichiometry by Kate McKee. Stoichiometry Main Ideas: ●Atomic Mass ●Mole ●Stoichiometric Problems o Percent Composition o Determining Formula o Amounts](https://reader035.vdocument.in/reader035/viewer/2022081420/56649cda5503460f949a3965/html5/thumbnails/2.jpg)
Stoichiometry
Main Ideas: ● Atomic Mass● Mole● Stoichiometric Problems
o Percent Compositiono Determining Formulao Amounts of Reactants and Productso Limiting Reagento Percent Yield
● Balancing Equations
![Page 3: Stoichiometry by Kate McKee. Stoichiometry Main Ideas: ●Atomic Mass ●Mole ●Stoichiometric Problems o Percent Composition o Determining Formula o Amounts](https://reader035.vdocument.in/reader035/viewer/2022081420/56649cda5503460f949a3965/html5/thumbnails/3.jpg)
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
![Page 4: Stoichiometry by Kate McKee. Stoichiometry Main Ideas: ●Atomic Mass ●Mole ●Stoichiometric Problems o Percent Composition o Determining Formula o Amounts](https://reader035.vdocument.in/reader035/viewer/2022081420/56649cda5503460f949a3965/html5/thumbnails/4.jpg)
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
![Page 5: Stoichiometry by Kate McKee. Stoichiometry Main Ideas: ●Atomic Mass ●Mole ●Stoichiometric Problems o Percent Composition o Determining Formula o Amounts](https://reader035.vdocument.in/reader035/viewer/2022081420/56649cda5503460f949a3965/html5/thumbnails/5.jpg)
Mole - Avogadro's Number
6.022 × 1023
● 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 go Molar mass of carbon
![Page 6: Stoichiometry by Kate McKee. Stoichiometry Main Ideas: ●Atomic Mass ●Mole ●Stoichiometric Problems o Percent Composition o Determining Formula o Amounts](https://reader035.vdocument.in/reader035/viewer/2022081420/56649cda5503460f949a3965/html5/thumbnails/6.jpg)
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
![Page 7: Stoichiometry by Kate McKee. Stoichiometry Main Ideas: ●Atomic Mass ●Mole ●Stoichiometric Problems o Percent Composition o Determining Formula o Amounts](https://reader035.vdocument.in/reader035/viewer/2022081420/56649cda5503460f949a3965/html5/thumbnails/7.jpg)
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 compoundPercent Composition of Water (H2O)
Mass of H = 2 mol H x 1.0079 g/mol = 2.0158 gMass 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%
![Page 8: Stoichiometry by Kate McKee. Stoichiometry Main Ideas: ●Atomic Mass ●Mole ●Stoichiometric Problems o Percent Composition o Determining Formula o Amounts](https://reader035.vdocument.in/reader035/viewer/2022081420/56649cda5503460f949a3965/html5/thumbnails/8.jpg)
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 empiricalMolar Mass known = 98.96 g/mol2.021 mol Cl / 2.021 mol Cl = 1 Molecular formula = 98.96 g/49.48 g = 22.021 mol C / 2.021 mol C = 1 Cl2C2H4 4.04 mol H / 2.021 mol H = 2Empirical = ClCH2 = 49.48 g/mol
![Page 9: Stoichiometry by Kate McKee. Stoichiometry Main Ideas: ●Atomic Mass ●Mole ●Stoichiometric Problems o Percent Composition o Determining Formula o Amounts](https://reader035.vdocument.in/reader035/viewer/2022081420/56649cda5503460f949a3965/html5/thumbnails/9.jpg)
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 productso write unbalanced equationo balance with trial and error ( same # each type of atom appears on
both reactant and product sides)
C2H3Br + 3O2 CO + 2H2O + HBr
2C2H3Br + 3O2 4CO + 2H2O + 2HBr
![Page 10: Stoichiometry by Kate McKee. Stoichiometry Main Ideas: ●Atomic Mass ●Mole ●Stoichiometric Problems o Percent Composition o Determining Formula o Amounts](https://reader035.vdocument.in/reader035/viewer/2022081420/56649cda5503460f949a3965/html5/thumbnails/10.jpg)
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 LiOHLiOH(s)+ CO2 (g) Li2CO3 (s) + H2O(l)
Balanced = 2LiOH(s)+ CO2 (g) Li2CO3 (s) + H2O(l)
LiOH molar mass = 23.95 g/mol41.8 mol LiOH = 1.00 kg LiOH/ 23.95 g LiOH
1mol CO2/ 2mol LiOH41.8 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
![Page 11: Stoichiometry by Kate McKee. Stoichiometry Main Ideas: ●Atomic Mass ●Mole ●Stoichiometric Problems o Percent Composition o Determining Formula o Amounts](https://reader035.vdocument.in/reader035/viewer/2022081420/56649cda5503460f949a3965/html5/thumbnails/11.jpg)
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 equationo determine moleso determine mole ratioo determine moles required to react with each othero compare amount required with amount present to determine limiting
reagento check with mole ratio
![Page 12: Stoichiometry by Kate McKee. Stoichiometry Main Ideas: ●Atomic Mass ●Mole ●Stoichiometric Problems o Percent Composition o Determining Formula o Amounts](https://reader035.vdocument.in/reader035/viewer/2022081420/56649cda5503460f949a3965/html5/thumbnails/12.jpg)
Limiting Reagent
NH3(g)+ CuO(s) N2(g)+ Cu(s)+ H2O(g) 18.1 g NH 90.4 g CuO
Balanced = 2NH3(g)+ 3CuO(s) 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 CuOMole 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 CuOCuO is limiting reagent
![Page 13: Stoichiometry by Kate McKee. Stoichiometry Main Ideas: ●Atomic Mass ●Mole ●Stoichiometric Problems o Percent Composition o Determining Formula o Amounts](https://reader035.vdocument.in/reader035/viewer/2022081420/56649cda5503460f949a3965/html5/thumbnails/13.jpg)
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
![Page 14: Stoichiometry by Kate McKee. Stoichiometry Main Ideas: ●Atomic Mass ●Mole ●Stoichiometric Problems o Percent Composition o Determining Formula o Amounts](https://reader035.vdocument.in/reader035/viewer/2022081420/56649cda5503460f949a3965/html5/thumbnails/14.jpg)
You’re Welcome
● You’re Welcome