Stoichiometry by Kate McKee. Stoichiometry Main Ideas: ●Atomic Mass ●Mole ●Stoichiometric Problems o Percent Composition o Determining Formula o Amounts.

Stoichiometry

by Kate McKee

Stoichiometry

Main Ideas: ● Atomic Mass● Mole● Stoichiometric Problems

o Percent Compositiono Determining Formulao Amounts of Reactants and Productso Limiting Reagento 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 × 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

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 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%

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

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

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

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

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

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