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
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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
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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