3. Mass Relationships in Chemical Reactions
atomic mass: mass of an atom in atomic mass units (amu)
atomic mass unit: by definition, the mass of a
12
6 C-atom
is 12 u
(the textbook uses amu instead of u in violation of international convention)
mass of a
12
6 C-atom
example:
16
O
= 1.9926 × 10
−23
g =⇒ 1 u = 1.6605 × 10−24 g
15.9949 u
periodic table reports the atomic mass of carbon as 12.011 u;
why??
naturally occurring carbon contains some
13
C
atomic mass = average of the isotopic masses, weighted according to the naturally occurring abundances of the isotopes
abundances in %, on a number basis
atomic mass =
% abundance isotope 1
100
+
× m isotope 1
% abundance isotope 2
100
× m isotope 2
+···
GChem I
3.1
40
example:
Ar: 39.9624 u, 99.600%;
38
Ar: 37.96272 u, 0.063%
36
Ar: 35.96755 u, 0.337%;
m Ar = 0.99600 × 39.9624 u + 0.00337 × 35.96755 u + 0.00063 × 37.96272 u
= 39.8025504 u + 0.12121064 u + 0.023916514 u
= 39.803 u + 0.121 u + 0.024 u
= 39.948 u
bulk sample of an element contains an astronomically large
number of atoms
chemical counting unit
a mole, symbol mol, is the amount of substance that contains
the same number of elementary entities as there are carbon-12
atoms in exactly 12 g of carbon-12
N A = 6.022136736 × 1023 mol−1 Avogadro’s constant
N A = 6.022 × 1023 mol−1 = 6.022 × 1023
GChem I
1
mol
3.2
molar mass M of an element = mass in grams of 1 mol of
atoms of that element
M = ( amu)
g
mol
examples:
sodium: atomic mass = 22.9898 u
MNa = 22.9898
g
mol
copper: atomic mass = 63.546 u
MCu = 63.546
g
mol
[inclass problems]
GChem I
3.3
formula mass: mass of a formula unit in amu
molecular mass: mass of a molecule in amu
obtained by adding the atomic masses of the atoms in a formula
unit or a molecule
mole of a compound = N A molecules or N A formula units
molar mass of a compound = mass in grams of one mole of molecules or formula units
mole of elements: H2 , O2 , N2 , F2 , Cl2 , Br2 , I2 , P4 , S8
difference: atomic mass versus molecular mass
example: What is the molecular mass of N2 O5 ?
M N2 O5 = 2 × 14.0067 u + 5 × 15.9994 u = 108.0104 u
What is the molar mass of N2 O5 ?
MN2O5 = 108.0104 g mol−1
What is the mass of 0.65 moles of N2 O5 ?
m = n M = 0.65 mol × 108.0104 g mol−1 = 70. g
GChem I
3.4
composition of compounds
percent composition (by mass): mass % of elements (atoms) in
a compound
(i) given the formula of the compound =⇒ theoretical % composition
% -X =
n X · MX
Mcompound
· 100%
(ii) % composition =⇒ empirical formula
+ molecular or molar mass =⇒ molecular formula
take 100 g of compound
nX =
% -X
MX
divide the numbers of mole n X by the smallest of them ⇒ ratio
of the number atoms in the formula unit ⇒ empirical formula
“molecular formula =
molar mass
×
formula mass
empirical formula”
[inclass problems]
GChem I
3.5
reactants −−→ products
chemical reaction
evidence:
color change
evolution of gas
formation of a solid (precipitate) within a clear solution
evolution or absorption of heat
formula expression
H2 + O2 −→ H2 O
chemical equation
balanced
2 H2 + O2 −→ 2 H2 O
conservation of atoms =⇒ balance
2 H2 + 1 O2 −→ 2 H2 O
stoichiometric coefficients
balancing often by trial and error; only the stoichiometric coefficients can be changed to balance the formula expression
tips:
GChem I
3.6
(1) Balance elements which occur only in one reactant and one
product.
(2) If polyatomic ions remain intact, balance the number of polyatomic ions of a given type.
example:
Al + CuSO4 −→ Al2 (SO4 )3 + Cu
Al + 3 CuSO4 −→ Al2 (SO4 )3 + Cu
2 Al + 3 CuSO4 −→ Al2 (SO4 )3 + 3 Cu
(3) Balance free elements last.
states of matter
2 C6 H14 O4 (l) + 15 O2 (g) −→ 12 CO2 (g) + 14 H2 O(l)
chemical reaction in solution; common solvent: water = aqueous solution; denoted by (aq)
2 Al(s) + 3 CuSO4 (aq) −→ Al2 (SO4 )3 (aq) + 3 Cu(s)
reaction conditions
∆
2 Ag2 O(s) −→ 4 Ag(s) + O2 (g)
GChem I
3.7
decomposition reaction; ∆: high temperature; heat the reactant
mixture
Rh
2 CO(g) + 2 NO(g) −→ 2 CO2 (g) + N2 (g)
catalytic reaction
stoichiometry
2 H2 + O2 −→ 2 H2 O
2 mol H2 + 1 mol O2 −→ 2 mol H2 O
stoichiometric factor: conversion factor
here:
2 mol H2 O
,
2 mol H2
2 mol H2 O
,
1 mol O2
2 mol H2
1 mol O2
and the inverses
If all reactants are completely and simultaneously consumed,
they are present in stoichiometric proportions.
a A + b B −→ p P + · · ·
initial mass of reactants A and B is given: m A , m B
stoichiometric calculations: m −→ n
GChem I
3.8
nA =
mA
MA
,
nB =
mB
MB
if n A /n B = a/b , then A and B are present in stoichiometric
proportions
Commonly: One reactant is converted completely into products,
while some quantity of all other reactants remains unreacted.
The reactant that is completely consumed is called the limiting
reactant or limiting reagent.
The limiting reactant determines the amount of product
that can be formed.
The other reactants are said to be in excess.
Determine the limiting reagent:
a A + b B + c C + · · · −→ p P + · · ·
m A −→ n A ;
m B −→ n B ;
m C −→ n C ;
...
Determine the amount of product (moles) that each reactant can
make.
n P(A) =
GChem I
p
nA;
a
n P(B) =
p
nB;
b
n P(C) =
p
nC;
c
...
3.9
Choose the smallest value.
This is the maximum amount that can be made, since the corresponding reactant is completely consumed = limiting reagent.
(B)
(A)
(C)
So if n P is less than n P and is less than n P , then B is the
(B)
limiting reagent and at most n P moles of the product P can be
made
that amount is the theoretical yield
calculated from the balanced chemical equation
actual yield: amount of product actually produced (experiment)
the actual yield is generally lower than the theoretical yield:
competing reactions, side reactions, back reaction, etc.
percent yield:
percent yield =
actual yield
theoretical yield
× 100%
if percent yield ≈ 100%, the reaction is complete or the reaction
is quantitative
GChem I
3.10