Transcript Stichiometry

Stoichiometry
Molar relationships in chemical
reactions
3.1 Chemical Equations:
A Review
• Law of conservation of mass
• Relationship between reactant and
products produces a balanced
chemical equation
• Reactants  Products
A+B C+D
• Balancing chemical eqns
Balancing Chemical Equations
• Must have correct chemical formulas
• Change coefficients as needed to balance
equation
• Do not change subscripts of chemical
formulas
• Indicate physical states by writing in
parentheses (s), (l), (g), (aq)
• Begin with an element that appears in only
one reactant and product
• If possible, do not begin with O or H
Balancing Chemical Equations
1) ____ N2 + ____ H2  ____ NH3
2) ____ KClO3  ____ KCl + ____ O2
3) ____ NaCl + ____ F2  ____ NaF + ____ Cl2
4) ____ H2 + ____ O2  ____ H2O
5) ____ Pb(OH)2 + ____ HCl  ____ H2O + ____ PbCl2
6) ____ AlBr3 + ____ K2SO4  ____ KBr + ____ Al2(SO4)3
7) ____ CH4 + ____ O2  ____ CO2 + ____ H2O
8) ____ C3H8 + ____ O2  ____ CO2 + ____ H2O
9) ____ C8H18 + ____ O2  ____ CO2 + ____ H2O
10) ____ FeCl3 + ____ NaOH  ____ Fe(OH)3 + ____NaCl
Balancing Chemical Equations
11) ____ P + ____O2  ____P2O5
12) ____ Na + ____ H2O  ____ NaOH + ____H2
13) ____ Ag2O  ____ Ag + ____O2
14) ____ S8 + ____O2  ____ SO3
15) ____ CO2 + ____ H2O  ____ C6H12O6 + ____O2
16) ____ K + ____ MgBr  ____ KBr + ____ Mg
17) ____ HCl + ____ CaCO3  ____ CaCl2 + ____H2O + ____ CO2
18) ____ HNO3 + ____ NaHCO3  ____ NaNO3 + ____ H2O + ____ CO2
19) ____ H2O + ____ O2  ____ H2O2
20) ____ NaBr + ____ CaF2  ____ NaF + ____ CaBr2
3.2 Patterns of Chemical Reactivity
• Because groups of elements have similar
chemical behavior, the periodic table can
be used to predict reactions involving
elements
• Example:
Na (s) + H2O (l)  NaOH (aq) + H2 (g)
• Alkalai metals + water  metal hydroxides + H2
Types of Chemical Reactions
• Synthesis (Combination) reactions
• Two or more substances combine to form
a new substance
• Often involve elements combining to form
a compound
• A + B  C
4Na (s) + O2 (g)  2Na2O (s)
sodium oxide
Types of Chemical Reactions
• Decomposition reactions
• A single substance breaks down into two or
more smaller substances
• A  B + C
H2CO3 (aq)  CO2 (g) + H2O (l)
carbonic acid
Types of Chemical Reactions
•
•
•
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Combustion reactions
Hydrocarbons + O2  CO2 + H2O
Cpds containing C, H, O  CO2 + H2O
Involve…
– O2 as reactant
– Release of heat & light
– Often H2O is a product
3.3 Formula Weights
• Atomic mass unit (amu)
• A unit invented to describe extremely
small masses
• Defined as 1/12 the mass of an atom of
12C
• 1 amu = 1.66054 x 10-24 g
Average Atomic Masses
• Atomic masses given on the PT are
weighted averages of all the isotopes of
that element
• Equals the sum of the products of the
(mass) x (frequency) of the isotopes
• mavg = Σ (m1f1) + (m2f2) + (m3f3) + …..
Average atomic mass
• mavg = Σ (m1f1) + (m2f2) + (m3f3) + …..
• Example
• Chlorine occurs as two isotopes:
• Isotope amu
freq
• 35Cl
34.964 75.53%
• 37Cl
36.966 24.47%
• Atomic mass of chlorine =
(m x f)
26.41
+9.05
35.46
Formula & Molecular Weights
• Formula mass refers to ionic cpds
• Molecular mass refers to covalent cpds
• Is the sum of atomic masses of the
molecule
Calculate mol mass of Ca (NO3)2
Percent Composition of Formulas
• Percentage by mass contributed by each
element in the formula
mass of elementin thecpd
% mass 
molarmass of thecpd
Determine% oxygenin nitricacid, HNO3

16g 3
%O
 0.76
63g
% O  76%
Percent Composition
• Determine the % composition of all
elements in sucrose, C12H22O11
• %C = (12 x 12) / 342 = 42.0%
• %H = (1 x 22) / 342 = 6.4%
• %O = (16 x 11) / 342 = 51.6%
3.4 The Mole
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•
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•
Is a counting unit
Avagadro’s number = 6.022 x 1023
Interconvert between moles & particles
Molar mass = the mass of one mole of a
substance
• Molar mass of elements = atomic mass in
grams
Converting between particles,
moles, and grams
• Particles and grams cannot be
interconverted directly.
• The mole is the bridge between particles
and grams.
molar mass
Avagadro ' s




particles
moles
grams




Avagadro ' s
molar mass
3.5 Empirical Formula Analysis
• To determine empirical formula from
percent composition data
• Empirical formula gives simplest ratio of
atoms in a compound
• Therefore, always convert to moles when
calculating empirical formulas
• Sample exercise 3.13
Combustion Analysis
• Determines empirical formula based upon
analysis of products of combustion
• Example: 18.4 g of a compound of CHO
produced 41.7 g CO2 and 19.65 g H2O
• CxHyOz + O2  CO2 + H2O
18.4 g
41.7g
19.65 g
• All carbon in CO2 came from the CHO
• All hydrogen in H2O came from CHO
Combustion Analysis
1. Determine amt of C in CHO by determining
amt of C in the CO2 produced.
%C in CO2 x g CO2 = g C in CHO
.2729 x 41.17 = 11.23 g C = 0.935 mol C
Combustion Analysis
2. Determine amt of H in CHO by determining amt
of H in the H2O produced.
%H in H2O x g H2O = g H in CHO
.1119 x 19.65 = 2.20 g H = 2.18 mol H
Combustion Analysis
• Determine amt of O in CHO by determining amt
of H in the H2O produced
• By law of conservation of mass,
grams of CHO = gC + gH + gO
• So grams O = grams CHO – gC – gH
• 18.40 – 11.23 – 2.20 = 4.97 gO = .311 mol O
• Mole ratio
C 0.935 H 2.18 O 0.311
• Empirical formula C3H7O
3.6 Quantitative Information from
Balanced Equations
• Balanced chemical equations are like specific
recipes
• Balanced equations tell us ….
– the relative amounts of reactants (ingredients)
– the relative amounts of products
– other relevant conditions needed (e.g. heat)
• The unit of measure in a balanced equation is
the mole
– Indicated by coefficients
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Stoichiometric Calculations
The coefficients in the balanced equation give
the ratio of moles of reactants and products.
3.6 Quantitative Information from
Balanced Equations
• Mole ratio used to calculate actual molar
amounts in a balanced eqn
• Sample Problem 3.16
• Determine mass of water produced from
combustion of 1.00 g glucose
• C6H12O6 + __ O2  __ CO2 +__ H2O
• C6H12O6 + 6 O2  6 CO2 + 6 H2O
Solving Stoichiometry Problems
Sample problem
• Determine mass of water produced from
combustion of 1.00 g glucose
• C6H12O6 + 6 O2  6 CO2 + 6 H2O
• Determine moles glucose
• Determine moles water
• Determine grams water
3.7 Limiting Reactants
• Stoichiometrically least abundant reactant
• Completely consumed in the reaction
• Determines theoretical yield of the
reaction
• When all reactants are present in
stoichiometric amounts, all are limiting
reagents
Limiting Reactants
• The limiting reactant is the reactant present in
the smallest stoichiometric amount.
– In other words, it’s the reactant you’ll run out of first (in
this case, the H2).
Limiting Reactants
In the example below, the O2 would be the
excess reagent.
Limiting Reagents
• How many moles of ammonia can be
produced when 3.0 mol nitrogen and 6.0
mol of hydrogen are permitted to react?
• Write & balance the equation
• Determine the LR to determine yield
– Reactant  least yield is the LR
• What is the excess reagent? How much is
left over?
Theoretical & Actual Yields
• Theoretical Yield
• Quantity of product produced when 100% of
limiting reagent reacts
• Determined from balanced equation
• Actual Yield
• Quantity of product actually produced
• No reaction is 100% efficient, therefore….
• Theoretical Yield does not equal Actual Yield
Sample Problem 3.20
• 2 C6H12 + 5 O2  2 H2C6H8O4 + 2 H2O
• Given 25.0 g cyclohexane and excess
oxygen, determine theoretical yield of
adipic acid
Percent Yield
• Expresses the efficiency of a reaction
• Is ratio of actual yield to theoretical yield
• Percent Yield = Actual Y/ Theroretical Y
%Y = AY / TY
Sample Problem 3.20 (cont)
• 2 C6H12 + 5 O2  2 H2C6H8O4 + 2 H2O
• Theoretical yield = 43.5 g adipic acid (from
first part)
• If actual yield of adipic acid was 33.5 g,
determine percent yield of reaction
• %Y = Actual yield / Theoretical yield
• %Y = 33.5g/43.5g = .77 = 77%