Transcript Chapter 3

STOICHIOMETRY
The Mole
Atomic and Molecular Masses
Chemical Formula
Stoichiometry
ATOMIC MASS
• Atomic mass standard is based on the atom
Carbon-12 or 12C (6 p+, 6 n0, 6 e-).
• One C-12 atom weighs exactly 12 amu
• 1 amu = atomic mass unit = 1.661E-27 kg
ATOMIC WEIGHT
• Atomic weight (mass) of an element is
defined as a weighted average over all
naturally occurring isotopes of the element;
this is the number on bottom on each
element box on the Periodic Table.
Calculate the Atomic Weight of
Boron
• Boron has two naturally occurring isotopes.
10B has a mass of M = 10.0129 amu and
1
abundance of A1 = 19.78%. 11B has a mass
of M2 = 11.0093 amu and abundance of A2
= 80.22%.
• At. Wt = Σ (Mi x Ai)/100 =
[10.0129 x 19.78 + 11.0093 x 80.22]/100 =
10.81 amu (see PT)
MOLE
• The number of C-12 atoms in exactly 12
grams of pure C-12
• 6.022E+23 items
• Avogadro’s Number, N
• A mole of an element has a mass equal to its
average atomic weight (mass).
• 1 mol of naturally occurring Boron has a
mass of 10.81 g.
•MOLECULE and MOLAR
MASS
• Molecule = arrangement of atoms
chemically bonded together.
• Molar Mass = Sum of atomic masses of
constituent atoms in one molecule (amu) or
one mole of molecules (gram).
• Use atomic and molar masses to the 1/100
place.
CHEMICAL FORMULA
• Qualitative description of the constituent
elements in a molecule or ion.
– C12H22O11 contains C, H and O
– SO42- contains S and O
• Quantitative description of the relative
numbers (subscripts) of atoms of each
element.
– Can be used to determine % composition or
mass %.
TYPES OF CHEMICAL
FORMULAS
• Chemical - shows type and number of
atoms (shorthand notation)
• Structural - shows chemical bonds (Fig
2.16)
• Ball and Stick - shows spatial arrangement,
3D (Fig 3.7 and 2.18)
• Space filling - shows space atoms fill, 3D
(Fig 2.17, also p 95)
Figure 3.7 The Two Forms of
Dichloroethane
Computer-Generated Molecule of
Caffeine
CONVERSIONS
• Grams to Moles
Divide by Molar Mass*
* = Atomic or Molar Mass
• Moles to Grams
x by Molar Mass
• Grams to amu
Divide by N
• amu to Grams
x by N
• Moles to #Units** x by N
• #Units to Moles
Divide by N
** = Atoms or Molecules
DETERMINATION OF A
CHEMICAL FORMULA
• A chemical formula can be determined from
the
– Mass of each element in the formula
– % Mass of each element in the formula (%
Composition)
– Number of moles of each element in the
formula
– Elemental analysis by combustion
CHEMICAL FORMULAS
• EMPIRICAL - includes all atoms in
molecule in correct smallest integer ratios
• MOLECULAR - includes all atoms in
molecule in actual numbers and correct
ratios; can be determined from the empirical
formula and molar mass.
CHEMICAL REACTION
• A chemical reaction involves
rearrangements of atoms; breaking initial
chemical bonds (in the reactants) and
making new chemical bonds (in the
products).
• R1 + R2  P1 + P2 + P3
• Methane burns in oxygen to form carbon
dioxide and water
CHEMICAL EQUATION
• Shorthand symbolic notation for a chemical
reaction
– CH4(g) + O2 (g)  H2O(ℓ) + CO2(g) Note that this
reaction is NOT BALANCED
• Qualitative aspect
– identity of reactants [R] and products [P]; use study of
nomenclature to write equations
– Identify the state of matter for each [R] and [P]
– identify reaction type
CHEMICAL EQUATION (2)
• Quantitative aspect
– how much reactant is consumed and how much
product is formed
– coefficients must be consistent with the Law of
Conservation of Mass; atoms are neither
created nor destroyed in a chemical reaction.
– i.e. chemical equation must be balanced
• CH4(g) + 2O2 (g)  2H2O(ℓ) + CO2(g)
Note that this reaction is BALANCED
STOICHIOMETRY
• Quantitative relationships in a chemical
reaction based on a BALANCED chemical
equation.
• Relationships between R(eactant)1 and R2
or R1 and P(roduct)2 or P1 and P2
C(s) + 2H2(g)  CH4(g)
Formation of methane
• One atom of solid carbon reacts with two
molecules of gaseous hydrogen to produce one
molecule of gaseous methane.
• One mole of solid carbon reacts with two moles
of hydrogen gas to produce one mole of
methane gas.
• 12.0 g of C reacts with 4.0 g of H2 to produce 16.0
g of CH4. Note conservation of mass: 12+4 = 16
STOICHIOMETRIC
COEFFICIENTS
• We will use mole interpretation for
stoichiometric coefficients (SC), the
coefficients in front of Rs and Ps. I.e., SCs
represent # of moles of each R and P
• Provide quantitative (i.e. mole)
relationships between R and P.
• Can be used to determine amount of mass
of each R and P (using mol to g conversion)
MOLE RATIOS
• A mole ratio is a ratio of Stoichiometric
Coefficients from a balanced chemical eqn.
• These ratios are conversion factors from
amt of R1 to amt of R2, amt of P2 to amt of
R1, etc
C(s) + 2H2(g)  CH4(g)
1 mol 2 mol 
1 mol
12.0 g 4.0 g 
16.0 g
• How many g of carbon are needed to react with
10.0 g of hydrogen? How much CH4is formed
• g-H2  mol-H2  mol-C  g-C
• [10.0g H2 /2.0g H2/mol]x[1 molC/2 mol H2]
x[12.0g C/mol] = 30.0 g C
• [10.0g H2 /2.0g H2/mol]*[1 molCH4/2 mol H2] *
[16.0g CH4/mol] = 40.0 g CH4
• Is mass conserved?
Calculating Mass of Reactants and
Products
REACTION YIELD
• In the previous example, say that only 32.0
g of CH4 were produced due to side
reactions and waste.
• We define the percent or reaction yield as
[actual yield/theoretical yield]x100
• This gives % yield = [32.0/40.0] x 100 =
80.0%
LIMITING REACTANT
• Find the actual moles of each reactant. Use the
balanced chem eqn to determine how many mol of
R2 is required to react completely with R1. Do
you have enough R2? If not, R2 = limiting
reactant = LR and R1 = reactant in excess = XS.
• Always use the LR to solve the stoichiometric
problem to find the amount of product formed.
• Calculate the amount of XS left over.
• Calculate the grams of methane formed when 18.5
g carbon and 2.9 g hydrogen react.
Solving a
Stoichiom.
Problem
Involving
Masses of
Reactants
and
Products