Transcript Chapter 3 Stoichiometry

Chapter 3
Stoichiometry
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Stoichiometry: The study of quantities of
materials consumed and produced in
chemical reactions
Atomic Masses: Are determined by
comparing with 12C (carbon-12 scale). By
definition, carbon-12 is assigned a mass of
exactly 12 atomic mass units (amu) and the
masses of all other atoms are given relative
to this standard
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Measurement of Atomic Masses: Mass
spectrometer is used to measure the atomic
masses. Charged particles are bent in a
magnetic field and the deflection depends on
the masses which causes the ions to separate.
A comparison of the positions where the
ions hit the detector plate gives very accurate
values of their relative masses
Schematic Diagram of a Mass Spectrometer
Atomic Masses
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Isotopes: Elements occur in nature as
mixtures of isotopes.
Carbon = 98.89 % 12C & 1.11 % 13C
Carbon atomic mass = (0.9889)(12 amu) +
(0.0111)(13.0034) = 12.01 amu
Average atomic mass = [(relative abundance
of each isotope) x (the mass of each isotope)]
Lithium consists of:
7.50 % 6Li which has an accurate weight of
6.015121 amu and 92.50 % 7Li which has an
accurate weight of 7.016003 amu. Av. At.
Mass?
Av. At. Mass = (0.0750) x (6.015121
amu/atom) + (0.9250) x (7.016003 amu/atom)
= (0.4511) amu + (6.4898) amu
= 6.941 amu
Neon Gas
Mass Spectrum of Natural Copper
The Mole
Mole: The amount of a substance that contains
as many entities as there are in exactly 12 g of
carbon-12.
1 mole of anything = 6.022 x 1023 units of that
thing (Avogadro’s number)
The mole is defined as 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 carbon = 12.01g of carbon =
6.022X1023 atoms C
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How many Li atoms are there in 3.0 g of lithium?
3.0 g Lithium x (1 mol Lithium)/(6.941g) x (6.022 x
1023 atom)/(1mol Lithium) = 2.6 x 1023 atoms of Li.
Cobalt is a metal that is added to steel to improve its
resistance to corrosion. Calculate both the number of
moles in a sample of cobalt containing 5.00 x 1020
atoms and the mass of the sample.
5.00 x 1020 atoms Co x (1 mol Co)/ 6.022 x
1023atoms Co)
= 8.30 x 10-4 mol Co
8.30 x 10-4 mol Co x (58.93 g Co)/(1 mol Co)
= 4.89 x 10-2 g Co
Example
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How much does 2.0 x 1024 atoms of Ne weigh?
2.0 x 1024 atoms Ne x (1 mol Ne)/(6.022X1023
atoms Ne) x (20.18 g Ne)/(1 mol Ne) = 67g
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8.46 g B = ? Moles
8.46 g B x (1 mol B)/(10.81 g B) = 0.783 moles
of B
Molar Mass
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Molar Mass: A substance’s molar mass is the mass in
grams of one mole of the substance
Molar mass of a substance is numerically equal to its
formula mass
Molar mass of CO2 = 12.01 g + 2 x 16.00 g = 44.01 g
Molar mass of Al2(SO4)3 —
2 Al = 2 x 26.98 = 53.96; 3 S = 3 x 32.06 = 96.18
12 O = 12 x 16.00 = 192.0; Total = 342.14 g
1 mole of Al2(SO4)3 = 342.1 g
How many atoms of Al are in 52 g of Al2(SO4)3?
52 g x (1 mol)/(342.1 g) x (6.022 x 1023 molecules)/(1
mol) x (2 Al atom)/(1 molecule) = 1.8 x 1023 atoms
Percent Composition of Compounds
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If the formula of the compound is known, the %
composition can be calculated. Elemental % = (mass
of element in compound)/mass of compound) x 100
Compute the mass percent of each element in
Penicillin F which has the formula C14H20N2SO4.
C: 14 mol x 12.01 g/mol = 168.1 g
H: 20 mol x 1.008 g/mol = 20.16 g
N: 2 mol x 14.01 g/mol = 28.02 g
S: 1 mol x 32.07 g/mol = 32.07 g
O: 4 mol x 16.00 g/mol = 64.00 g
Example
Mass of 1 mol C14H20N2SO4 = 312.4 g
% C = (168.1 g)/(312.4 g) x 100 % = 53.81 %
% H = (20.16 g)/(312.4 g) x 100 % = 6.453 %
% N = (28.02 g)/(312.4 g) x 100 % = 8.969 %
% S = (32.07 g)/(312.4 g) x 100 % = 10.27 %
% O = (64.00 g)/(312.4 g) x 100% = 20.49 %
Determining the Formula of a compound
If we know the % composition we can find
the formula. Molecular formula = (empirical
formula)n where n=integer
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Empirical formula = CH = simplest formula
Molecular formula = (CH)6 = C6H6
For some compounds (like H2O, NaCl, NH3)
the empirical and the molecular formulas are
the same, for others it is different (empirical
formula = CH4, the molecular formula = C2H8)
Schematic Diagram of the Combustion Device Used
to Analyze Substances for Carbon and Hydrogen
How do you find an empirical (simplest)
formula?
1. Convert % to grams of each element
2. Convert % grams to moles of atoms
3. Divide by the lowest number of moles
4. If needed adjust for fractions
Example: A white powder determined to have 43.64
% P and 56.36 % O (Simplest way is to consider 100
g sample)
Step 1: 43.64 % P
43.64 g P
56.36 % O
56.36 g O
Step 2: 43.64 g P x (1 mol P)/(30.97 g) = 1.409 mol
56.36 g O x (1 mol O)/(16.00 g) = 3.523 mol
Step 3: P1.409/1.409 = 1; O3.523/1.409 = 2.5
Step 4: P1O2.5 x 2 = P2O5  empirical formula
What is the true molecular formula?
 Obtain the empirical formula
 Compute the mass of the empirical formula
 Calculate the ratio (integer) of the molar
mass/empirical formula mass
 Multiply the empirical formula subscript by
the integer gives the molecular formula
For the above compound, molecular weight =
290 g/mole, What is the molecular formula?
Empirical weight—
2 P = 2 x 30.97 = 61.94 g
5 O = 5 x 16.00 = 80.00 g
Total = 141.94 g  141.9 g
N = 290/142  2.04  must be integer  2
Molecular formula = (P2O5 )2 = P4O10
(tetraphosphorus decaoxide)
Chemical Equations
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A chemical change involves a reorganization of the atoms in
one or more substances.
Chemical equation is a representation of a chemical reaction
with the reactants on the left side of an arrow and the
products on the right side
C2H5OH + 3O2  2CO2 + 3H2O
Reactants
Products
Atoms have been reorganized. Bonds have been broken and
new ones have been formed
In a chemical reaction, atoms are neither created nor
destroyed
All atoms present in the reactants must be accounted for
among the products (i.e. # atoms need to be same on both
sides of the arrow)
The above equation is balanced.
The Meaning of a Chemical Equation
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CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
The nature of the reactants and products (physical states)
The relative numbers of reactants and products (# moles)
The relative numbers of the reactants and products in a
reaction are indicated by the coefficients in the balanced
equation
1 mole of methane reacts with 2 moles of oxygen to
produce 1 mole of carbon dioxide and 2 moles of water
1 molecule CH4 + 2 molecule O2 produce 1 molecule CO2
+ 2 molecule H2O
16 g CH4 + 2(32 g) O2 produce 44 g CO2 + 2(18 g) H2O
80 g reactants produce 80 g products
Balancing Chemical Equations
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The total number of atoms of any element must be
the same on both sides of the chemical equation
The identities of the reactants and products must not
be changed
The formulas of the compounds must never be
changed in balancing a chemical equation
The subscripts in a formula cannot be changed, nor
can atom be added or subtracted from a formula
Need to balance the equation by inspection, i.e. by
trial and error
Writing and Balancing the Equation for a
Chemical Reaction
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Determine the reactants, the products and the
physical states
Write the unbalanced equation that summarizes the
reaction
Balance the equation by inspection
(NH4)2Cr2O7(s)  Cr2O3(s) + N2(g) + H2O(g)
2N, (4x2)H, 2Cr, 70  2Cr, 30, 2N, 2H, 10
(NH4)2Cr2O7(s)  Cr2O3(s) + N2(g) + 4 H2O(g)
The equation is balanced
Stoichiometric Calculations: Amounts of
Reactants and Products
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Balance the equation for the reaction
Convert mass to moles
Set up appropriate mole ratios from
balanced equation
Use mole ratios to calculate moles of
desired substituent
Convert moles to grams, if necessary
Example
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If 2.85 g of NH3 are burned, how much H2O will be
formed?
4NH3 + 5O2  4NO + 6H2O
2.85 g NH3 x (1 mol NH3)/(17.04 g) = 0.167 mol
NH3
0.167 mol NH3 x (6 mol H2O)/(4 mol NH3) = 0.251
mol H2O
0.251 mol H2O x (18.02 g H2O)/(1 mol H2O) = 4.52
g H2O
Combined:
2.85 g NH3 x (1 mol)/(17.04 g) x (6 H2O)/(4 NH3) x
(18.02 g)/(1 mol H2O) = 4.52 g H2O
Calculations Involving a Limiting Reactant
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The limiting reactant is the reactant that is consumed
first, limiting the amounts of products formed.
Steps for Solving Stoichiometry Problems:
Write and balance the equation
Convert the known masses of substances to moles
Determine which reactant is limiting
Using the amount of the limiting reactant and the
appropriate mole ratios, compute the number of
moles of the desired product
Convert the moles to grams, using the molar mass
Three Different Stoichiometric
Mixtures of Methane and Water
Mixture of CH4 and H2O Molecules
Methane and Water Reacting
Hydrogen and Nitrogen React to Form Ammonia
According to the Equation N2 + 3H22NH3
Example
18.1g of gaseous NH3 is reacted at high temperature with 90.4g
of solid CuO to produce nitrogen gas, solid copper and water
vapor. Find the limiting reactant. How many grams of N2 will be
formed?
2 NH3 (g) + 3CuO (s)  N2 (g) +3Cu (s) + 3H2O (g)
18.1g NH3 X (1 mol NH3 )/(17.03g NH3) = 1.06 mol NH3
90.4g CuO X (1 mol CuO)/(79.55g CuO) = 1.14 mol CuO
1.06 mol NH3 X (3 mol CuO)/(2 mol NH3) = 1.59 mol CuO
1.14 mol CuO is actually present, the amount of CuO is limiting,
CuO will run out before NH3 does. Since CuO is limiting reactant,
we need to use CuO to calculate the amount of N2 formed.
1.14 mol CuO X (1mol N2)/(3mol CuO) X
(28.0g N2)/(1mol N2) = 10.6g N2
..Continued
Percent yield = (Actual yield)/(Theoretical yield) X 100 %
Actual yield: Amount of product actually formed
Theoretical yield: The amount of product predicted from
the reactants
If in the above example, 6.63 grams of nitrogen actually
produced instead of the predicted 10.6 grams what is the
percent yield?
Percent yield = (Actual yield)/(Theoretical yield) X 100 %
= (6.63g)/(10.6g) X 100 % = 62.5 %
Example
68.5kg CO (g) is reacted with 8.60kg H2 (g) to produce
methanol. Calculate the theoretical yield of
methanol. If 3.57 X 104g CH3OH is actually
produced, what is the percent yield of methanol?
2 H2 (g) + CO (g)  CH3OH (l)
68.5kg CO X (1000g CO)/(1kg CO) X (1mol
CO)/(28.02g CO)
= 2.44 X 103 mol CO
8.60kg H2 X (1000g H2)/(1kg H2) X (1mol
H2)/(2.016g H2)
= 4.27 X 103 mol H2
2.44 X 103 mol CO X (2mol H2)/(1mol CO)
= 4.88 X 103 mol H required
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..Continued
Mol H2 present is less than mol H2 required, so H2 is
limiting reagent so we need to use H2 for theoretical
yield.
4.27 X 103 mol H2 X (1mol CH3OH)/(2mol H2) X
(32.04g CH3OH)/(1mol CH3OH) = 6.84 X 104g
CH3OH
% yield = (3.57 X 104g CH3OH)/(6.84 X 104g
CH3OH) X 100 %
= 52.2 %
Summary
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Stoichiometry: Quantities of materials consumed or
produced
Atomic masses: Based on exactly 12 amu to a 12C atom
Mole: A mole is a unit of measure
Avogadro’s number: 6.022 X 1023
Molar mass: Mass in grams of one mole
Formula of a compound: Empirical formula & Molecular
formula
Chemical equation: representation of a chemical reaction
Meaning of a chemical equation: physical states & #moles
Balancing Chemical equation
Stoichiometric calculation: mass to mole conversion
Limiting reactant: which limits the amount of product
formed