Stoichiometry notes
The Mole Concept
A. Mole
1. definition: a word that means a number
2. like a dozen means 12
3. mole = 6.02 x 1023 [this number is known as Avogadro’s #]
4. useful for discussion of very small items (atoms, molecules...)
B. Relationships
1. moles to atoms of elements
a. 1 mole of any element = 6.02 x 1023 atoms of that element
b. 2.0 mol Na x 6.02 x 1023 atoms/ 1 mol Na = 1.2 x 1024 atoms Na
c. 4.5x1023 atom He x 1 mol He/6.02 x 1023 atoms He = .75 mol He
2. moles to grams of element
a. 1 mole of any element = atomic mass of element in grams (from PT)
b. 1 mole Na = 23.0 g
c. 2.5 mol C x 12.0 g C/ 1 mol C = 30. g C
d. 120 g Ca x 1 mol Ca/ 40.1 g Ca = 3.0 mol Ca
3. moles to molecules
a. 1 mole of any compound = 6.02 x 1023 molecules of that compound
b. 1.0 mol of NaOH = 6.02 x 1023 MOLECULES of NaOH ---(actually
should be formula units since NaOH is ionically bonded and not a true
molecule)
c. 2.0 mol KCl x 6.02 x 1023 molecules (formula units) = 12 x 1023 form unit
d. 6.0 x 1024 molecules water x 1 mol water/ 6.02 x 1023 molecules= 10. mol
water
4. mole to grams of molecules
a. 1 mole of any molecule or compound = molecular (formula) mass of that
compound (from PT)
b. 1 mol Water = 18.0 g water
c. 2.0 mol water x 18.0 g water/ 1 mol water = 36 g water
d. 65g water x 1 mol water/ 18.0 g water = 3.6 mol water
5. moles to atoms of a molecule
a. relationship between atoms within a molecule is given by the subscript
b. 1 molecule C6H5O = 6 atoms C, 5 atoms H, 1 atom O
c. 1 mol C6H5O = 6 mol C, 5 mol H, 1 mol O
d. # atoms of each element in 3.2 mol H2O?
3.2 mol H2O x (6.02 X 1023 molecules H2O/ 1 mol H2O) x (2 atoms H/1
molecule H2O) = 3.9 x 1024 atoms H {and similarly 2.0 x 1024 atoms O}
e. alternately the problem may be solved this way
3.2 mol H2O x (2 mol H/1 mol H2O) x (6.02 x 1023 atoms H/ 1 mol H) =
3.9 x 1024 atoms H
6. mole to particles
a. ionic substances may be identified by their particles
b. NaCl → Na+ + Cl- - for each NaCl compound (formula unit) 2 particles will
exist (1 Na and 1 Cl)
c.1 mol K2SO4 = (2K+ and 1 SO4-) 3 mole particles
d. 2.0 mol K2SO4 = ? ions and ? ions potassium
2.0 mol K2SO4 x (3 mol ions/1 mol K2SO4 ) x (6.02 x 1023 ions/1 mol ions) =
3.6 x 1024 ions total
2.0 mol K2SO4 x (2 mol K+ ions/ 1 mol K2SO4) X (6.02 x 1023 ions/ 1 mol ions)
= 2.4 x 1024 K+ ions
7. magnitude issues (in general)
a. when asked for # of VERY SMALL items like atoms, molecules, ions...
the answer will have VERY LARGE EXPONENT
b. when asked for # moles , magnitude will be relatively small - from
fraction to 2 digit numbers
c. when asked for grams, magnitude is moderate - tens to hundreds to
thousands
Formulas
A. Definition of terms
1. molecular mass: the mass in amu of one molecule of a covalent molecule 2. atomic mass: the mass in amu of one atom of an element - an average of
the isotopes of the element
3. formula mass-the mass in amu of one formula unit of an ionic compound
4. MOLAR MASS - mass of 1 mole of a substance in grams - can be used
collectively for any and all of the three
above definitions:
find the molar mass of Fe, H2O, FeCl3
find the atomic mass of Fe, Li, Ca....
find the molecular mass of H2O, C6H6, CO2
find the formula mass of NaCl, KBr, BaS
5. relative molecular mass (Mr)= mass of a molecule relative to hydrogen=
unit less diatomic fluorine molecules have an Ar = 38 -- note that these are
for single molecules or below single atoms -- not moles of atoms or molecules
6. relative atomic mass (Ar) = mass of an atom relative to hydrogen= unit
less Fluorine atoms or 19 x heavier than H atoms and have an Ar = 19
** relative masses are unit less - found in the same way as other masses
7. hydrated compounds -- have water molecules attached to compound with
typical formulas of CuCl2·6H20.
B. Writing formulas and naming compounds
1. molecules
a. acids
1. binary
a. name: hydro____ic acid
b. example HBr = hydrobromic A
c. formula - assume H is +1/ charge on other element
2. ternary
a. name ___ous acid or _______ic acid
b. if polyatomic ion from which acid is derived ends in
“ate” - “ic acid”
“ ite” - “ous acid”
c. example - H2SO4 -from sulfate - sulfuric A
H2SO3 - from sulfite - sulfurous Acid
b. non acid
1. naming - use prefixes mono, di, tri, tetra, penta, hexa
2. exceptions - do not use mono prefix on first element
3. examples: CO = carbon dioxide (not monocarbon monoxide)
4. N2O4 = dinitrogen tetroxide
2. ionic compounds
a. binary
1. name = element_ element”ide”
2. NaCl = sodium chloride
3. formula - look for charges and use zero-sum rule
4,. Na +1 Cl
-1
= NaCl, Ba
+2
F -1 = BaF2
b. ternary
1. know your polyatomic ions
acetate CH3COO -1 (C2H3O2-1)
hydroxide OH -1
CN -1 cyanide
HSO4 -1 hydrogen sulfate (bisulfate)
SO4 -2 sulfate
SO3 -1 sulfite
NO3 -1 nitrate
NO2 -1 nitrite
NH4 +1 ammonium
PO4 -3 phosphate
HPO4 -2 hydrogen phosphate
H2PO4 -1 dihydrogen phosphate
CO3 -2 carbonate
HCO3 -1 hydrogen carbonate (bicarbonate)
MnO4 -1 permanganate3
Cr2O7 -2 dichromate
CrO4 -2 chromate
C2O4 -2 oxalate
2. name = element___polyatomic (no change in ending)
3. example CaC2O4 = calcium oxalate
4. formulas ---ions and zero-sum rule ****watch placement of ( )
( ) must be used with > 1 polyatomic ion but MUST not be used
with only 1 polyatomic
5. example Na+1 HPO4
-2
= Na2HPO4
6. transition metals must have the charge of the metal ion written in
the name since > 1 oxidation state (charge) is possible
7. example FeO = iron (II) oxide, Fe2O3 = iron(III)oxide
find the charge on ion by reverse zero-sum rule --remember compounds do not have an overall charge
C. Calculations
1. definition of terms
a. empirical formulas: simplest whole number ratio of atoms of each
element in a compound * CH for benzene,
b. molecular formula: the ACTUAL number of atoms of each
element in a compound P4O10
c. % composition: the relative amounts of elements (by mass) in a
compound
2. determination of terms
a. empirical formulas
1. from experiment:
2.475 g of copper oxide is found to have 2.199 g Cu it has 2.475 - 2.199g =
0.277 g oxygen
O = 0.277g x 1 mol O/ 16 g O = 0.01731 mol O
Cu = 2.199 g Cu x 1 mol Cu/ 63.55 g Cu = 0.03460 mol Cu
So Cu.03460O divide each subscript by the smaller (.01731) = Cu1.999 O1 so
Cu2O
2. if the math works out to be a fractional subscript, multiply by the
smallest integers until the numbers are whole ie: Fe1O1.3 (by 2) Fe2O2.6 (NO)
or (by 3) Fe3O3.9 = Fe3O4
b. molecular formulas
empirical formula P2O5 and molar mass is 280. g
the formula mass of P2O5 is 140. g and molar mass is 280.g so 280.g/140.g =
2; so each subscript is doubled and molecular formula is P4O10
c. % composition:
CO2 : C = 12.0g x1 + O = 16.0g x 2 = 44.0g total
%C = 12.0g C/ 44.0 g x 100% =27.3 % C and 72.7 % O
Equations
A.Parts of a chemical equation
1. reactants
2. products
3. ---> or ⇄
4. phase subscripts (s), (l), (g), (aq), arrow up , arrow down
5. distinguish between coefficient (before formula) and subscript
B. Balancing equations
1. # of each element on each side of equation must equal
2. obeys law of conservation of matter ( must account for all matter in the
equation)
3. hints
a. polyatomic ions may be counted as a whole or as individual elements
b. keep H and O until last when balancing
c. NO FRACTIONAL COEFFICIENTS
4. amounts of substances in balanced equations = stoichiometry
C. Net Ionic equations
1. Ionic equations may be written as IONS when soluble (refer to solubility
rules), not compounds
2. any ion found in the reactant and product of a reaction is a spectator and
need not be included in a net ionic equation
3. net ionic equation gives the most important parts of the reaction and does
not include parts that do not change
4. examples
molecular equation: Ag + NaCl ----> AgCl + Na
ionic equation: Ag + Na+ + Cl- ----> AgCl + Na
net ionic equation: Ag + Cl- ---> AgCl(s)
D. solubility rules (in water)
1. nitrate salts are soluble
2. alkali metal salts and ammonium salts are soluble
3. chlorides, bromides and iodide soluble EXCEPT Ag+1, Pb+2, Hg2+2
4. sulfates soluble EXCEPT Ba+2, Pb+2, Hg+2, Ca+2
5. hydroxides are insoluble EXCEPT Na+1 and K+1
6. most sulfide, carbonate, chromate and phosphate are slightly soluble
****slightly soluble substances will ppt (solid forming) in reactions
Mass and Gaseous Volume Relationships in Rxns
A. Experimental and theoretical yields
1. balanced equations give the mole to mole ratio of reactants and products
2. dimensional analysis may be used to predict quantities of reactants and
products
3.see moles notes above
4. theoretical yield problem
How much water can be produced when 3.0 g of hydrogen and an unlimited
amount of oxygen is available?
the rxn: 2 H2 + O2 ---> 2 H2O
3.0 g H2 x (1 mol H2/ 2.02g H2) x (2 mol H+O/2 mol H2 )x (18.0 g H2O /1 mol
H2O) = 27 g
B. Limiting Reagents
1. When one reactant is in excess (not completely consumed) another
reagent is considered limiting. This means that the limiting reagent
determines the yield of the reaction (not the excess reagent)
2. simple example -- making sandwiches if you have 16 slices of bread and 16
slices of cheese (1 sandwich = 2 slices bread + 1 slice cheese)
The bread is the LIMITING REAGENT, since only 8 sandwiches can
be made and 8 slices of cheese are in excess.
in equation form 2 bread + 1 cheese ----> 1 sandwich
so... 16 bread x 1 sandwich/ 2 bread = 8 sandwiches
. ... 16 cheese x 1 sandwich / 1 cheese = 16 sandwiches (impossible
with bread given)
3. a real example
consider : KI + H2SO4 <--> I2 + K2SO4
What mass of potassium sulfate is produced when 147 g sulfuric acid, 332 g
potassium iodide are combined?
a. first balance the equation:
2 KI + H2SO4 ---> I2 + K2SO4
b. next find # of moles of each reactant
1.50 moles sulfuric acid, and 2.00 moles KI
c. next find # moles of one reactant needed to completely use the other
reactant
2.00 mole KI x 1 mole H2SO4 /2 mole KI = 1.00 mol of H2SO4 needed
d. Compare the moles of that reactant that you HAVE to the moles that you
calculated that you NEED. If you have MORE reactant than you NEED, that
reactant is the Excess and the other reactant is the Limiting Reactant – OR -If you
have LESS reactant than you NEED, then that reactant is the limiting reagent and
the other reactant is the excess.
For H2SO4: Have 1.50 moles, Need 1.00 moles – Have more than need
therefore H2SO4 is the excess and KI is Limiting Reagent.
e. Use LR to solve question
2.00 mol KI x (1 mol K2SO4/2 mol KI ) x (174.3g K2SO4/1 mol K2SO4)=
174 g K2SO4
f. to find the mass of excess reagent left over, subtract the amount that is
needed of the LR from the original amount (HAVE) of excess reagent.
1.50 mol H2SO4 – 1.00 mol H2SO4 = 0.50 mol H2SO4 in excess
4. The yield of product calculated from the limiting reagent = the
theoretical yield.
C. Gaseous volume relationships
1. Avogadro’s law = a given volume of any gas has the same number of
particles.
2. When conditions of pressure and temperature are constant, the balanced
equation relates the ratio of VOLUMES of GASES (as well as moles of
substances).
3. N2(g) + 3 H2(g) ---> 2 NH3(g)
1 mole nitrogen 3 moles hydrogen --> 2 moles ammonia
1 volume nitrogen + 3 volumes hydrogen ---> 2 volumes ammonia (at
constant T, P)
10 ml nitrogen + 30 ml hydrogen ---> 20 ml ammonia
4. under standard conditions STP - 1 atm (101.325 kPa) and 273 K the volume
of 1 mole gas=22. 4 liters.
5. using the ammonia equation above
How many moles of ammonia can be produced from 3.4 L hydrogen at STP?
3.4 L H2 x (1 mol H2/ 22.4 L H2) X (2 mole NH3/ 3 mole H2) = .10 mol NH3
Solutions
A. Definitions of terms
1. solute - part of the solution that is being dissolved - dissolvee
2. solvent - part of the solution that is doing the dissolving - dissolver
3. solution- a mixture of substances wherein one part dissolves ( is solvated)
in another.
4. concentration - a description of the relative amounts of solute and solvent
5. concentrated = much solute in solvent (vague concept)
6. dilute - little solute in solvent (vague concept)
B. specifying concentrations
1. units = MOLARITY
2. MOLARITY = moles solute / liter of solution
3. brackets represent concentration [ ]
4. note that volume of solution is given NOT volume of solvent
5. volumetric flasks are used for accurate molar concentrations
C calculations
1. finding [ ]
a. given 2.0 moles of solute in 3.0 liters of solution
2.0 moles/ 3.0 liters of solution = .67 mol/L
b. given 60. grams of NaOH in 2.0 liters solution
60 g NaOH x (1 molNaOH/ 40 g NaOH) / 2.0 L = .75 mol/L
NaOH
2. finding amount of solute in 450 ml a 3.0 mol/L solution of NaOH ,
(? mass NaOH)
3.0 moles NaOH/1 L soln x .450 L soln = 1.4 moles NaOH x 40. g /1
mol = 56 g NaOH
3 finding volume of solution given 3.6 moles of NaOH what volume of 2.0
mol/L solution
3.5 moles NaOH x 1 liter/ 2.0 moles = 1.8 liters of solution
4. dilution problems: you want 450 ml of 2.4 mol/L NaOH and are given 6.2
mol/L NaOH
Find the # of moles needed, then the volume of the concentrated
substance needed to have that number of moles, then dilute to
volume
.450 l x 2.4 moles NaOH/ 1L= 1.08 moles NaOH x 1 l/ 6.2 moles = .17 L
so 170 ml of concentrated NaOH is diluted with water to 450 ml.
D. Stoichiometry problems with solutions
H2SO4 (aq) + 2NaOH(aq) ===> 2 HOH + Na2SO4
1. given a 55 ml of 3.4 M H2SO4, what volume of 5.2 M NaOH is needed to
react?
3.4 moles H2SO4 / 1 liter x .055L = .187 mol H2SO4
.187 mol H2SO4 x 2 moles NaOH/ 1 mol H2SO4 = .374 mol NaOH
.374 mol NaOH x 1 L soln/ 5.2 mol NaOH = .072 L NaOH
More calculations
A. % yield, % error
% yield = -exp/ theo X 100 %
% error = Theo - exp/ theo x 100%
% yield + % error = 100%
Example: When 18 moles of CO2 are produced in the above example
then % yield = 18/ 20 x 100% = 90% yield
% error = 20 -18 / 20 x 100% = 10 % error
B. Molecular mass problems
1. Ethylene glycol , antifreeze, is composed of 38.7%C, 9.7% H and 51.6% O by
mass. Its molar mass = 62.1 g/mol:
what is emp form
what is molec form
38.7g C x 1 mol/12 g =3.225 mol C
9.7 g H x 1 mol/ 1 g = 9.7 mol H
51.6g O x 1 mol/16 g = 3.225 mol O
C3.225/ 3/.225 H9.7/ 3.225 O3.225/ 3.225 = C1H2.8O1 = CH3O
emp mass = 12.0 g + (3x1.01 g) + 16.0 g = 31.0 g
mm = 62.1 g / 31 g = 2 so multiply all subscripts by 2
molec form = C2H6O2
2. Caproic acid (dirty sock odor) is composed of C, H and O. Combustion of a 0.225 g
sample of this compound produces 0.512 g CO2 and 0.209 g HOH.
a. What is the emp form of the acid?
b. acid has mm of 116 g/mol, what is molecular formula?
.512 g CO2 x 1 mol CO2/44.0g CO2 x 1mol C/1 mol CO2 x 12.0g C/1 mol C
=.140 g C
.209 g HOH x (1 mol HOH/18.0 g HOH) x (2 mol H/1 mol HOH ) x (1.01g H/1
mol H) = .0232 g H
mass of O is the leftover mass so .225 g - .140 g -.0232 g = .062 g O
then find moles of each element, put as subscript and find emp form
.140 g C x 1mol C/12.0 g C = .0117 mol C
.0232 g H x 1 mol H/ 1.01 g H = .0230 mol H
.062 g O x 1 mol O/ 16.0 g O = .0039 mol O
C.01163/.00389 H .02322/.00389 0 .00389/ .00389 = C 2.9 H 5.9 O 1
so emp form = C3H6O
and 116 g / 58 g = 2
molec form = C6H1202
3. another Stoichiometry example
C6H12O6(s) + 6 O2(g) ----> 6 CO2(g) + 6 H2O(l)
? grams of water produced by combustion of 1.0 g glucose?
1.0 g C6H12O6 x 1 mol/180.0 g x 6 mol HOH/1 mol glucose X 18 g HOH/1 mol
HOH = 0.60 g water
***always change from one substance to another by using MOLES
MOLES -- Grams of a single substance