Stoichiometry: Chemical Calculations NaCl, salt Buckyball, C 60 Ethanol, C 2 H 6 O.
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Transcript of Stoichiometry: Chemical Calculations NaCl, salt Buckyball, C 60 Ethanol, C 2 H 6 O.
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Stoichiometry: Chemical Calculations
Stoichiometry: Chemical Calculations
NaCl, saltNaCl, saltNaCl, saltNaCl, salt
Buckyball, CBuckyball, C6060
Ethanol, CEthanol, C22HH66OO
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Just to Review…Please convert 12.4 fluid ounces to cm3
1 fl. oz. = 0.0295735 L
1 cm3 = 1 mL
Convert 4.5 centuries to seconds.
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The Carbon-12 ScaleThe Carbon-12 Scale• The atomic mass of an element indicates
how heavy, on average, an atom of an element is when compared to an atom of another element– Unit is the atomic mass unit (amu)
• The standard scaled based on the carbon-12 isotope
• Mass of one 12C atom = 12 amu (exactly)• Note that 12C and C-12 mean the same thing
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Figure 3.1 – Mass Spectrometer
• A mass spectrometer is used to determine atomic masses
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Figure 3.1 – Mass Spectrometer
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Isotopic Abundance
• Most elements exist in nature as a mixture of two or more isotopes.
• To determine the mass of an element, we must know the mass of each isotope and the atom percent of the isotopes (isotopic abundance)
• The mass spectrometer can determine the isotopic abundance
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Avogadro’s NumberAvogadro’s Number• A sample of any element with a mass equal
to its atomic mass contains the same number of atoms, NA, regardless of the identity of the element.
– NA = 6.022 X 1023 Avogadro’s #
• It represents the number of atoms of an element in a sample whose mass in grams is numerically equal to the atomic mass of the element.
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Amadeo Avogadro & Avogadro’s Number
6.02214199 x 1023
(6.022 x 1023 to 4 s.f.)
Avogadro’s NumberAvogadro’s Number
There is Avogadro’s number of particles in a mole of any substance.There is Avogadro’s number of particles in a mole of any substance.
Amadeo AvogadroAmadeo Avogadro1776-18561776-1856
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Examples:
• 6.022 x 1023 H atoms in 1.008 g atomic mass H = 1.008 amu
• 6.022 x 1023 S atoms in 32.07 g atomic mass S = 32.07 amu
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Mole: A Perspective on SizeThe Green Pea Analogy:
A Dramatic Reading
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Examples of Mole Quantities
EOS
1 mole of stars in the universe = 6.022 x 1023 stars1 mole of pennies = 6.022 x 1023 pennies
(beats the lottery!)
1 mole of glucose molecules = 6.022 x 1023 molecules1 mole of helium atoms = 6.022 x 1023 atoms1 mole of potassium ions (K+) = 6.022 x 1023 ions
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Example:water - H2O
Molecular MassMolecular mass: the sum of the atomic masses of all atoms in a molecular formula.
- (units are amu or u)
2 Hydrogen atoms2(1.01 amu)
1 Oxygen atom
+ 16.00 amu
= 18.02 amu
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Sample Problem: Glucose
Glucose - C6H12O6
= 6(12.01 u) + 12(1.01 u) + 6(16.00 u)= 180.18 u
EOS
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Formula Mass
Formula mass is the sum of the masses of the atoms or ions present in a formula unit – the unit for an ionic formula
Na+ Cl-
Na+
Na+Cl-
Cl-
Cl-Na+
Crystal ofsodium chloride
One Na+ and one Cl– make a formula unit for sodium chloride
The mass of one formula unit is:= 22.99 amu + 35.45 amu= 58.44 amu
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Sample Problem
Example 3.1 Determine the formula or molecular mass for each of the following:
CaI2 (NH4)2S
Al(NO3)3 C6H12O6
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Mole--DefinitionChemistry is a quantitative
science—we need a “counting unit” aka the
1 mole is the amount of 1 mole is the amount of substance that contains as substance that contains as many particles (atoms, many particles (atoms, molecules) as there are in molecules) as there are in 12.0 g of 12.0 g of 1212C.C.
1 mole is the amount of 1 mole is the amount of substance that contains as substance that contains as many particles (atoms, many particles (atoms, molecules) as there are in molecules) as there are in 12.0 g of 12.0 g of 1212C.C.
MOLE!MOLE!
518 g of Pb, 2.50 mol
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One-mole Amounts
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AnalogiesWe can group items and count by grouping:
12 eggs = 1 dozen eggs
We can also group items and
count by weighing:
Grass seed and nails —Purchased by the POUND, not by the item.
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Molar Mass
The molar mass, MM, in grams/mole, is numerically equal to the sum of the masses (in amu) of the atoms in the formula
Molar mass is the mass of one mole of a particular substance.
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Molar Masses of Some Substances
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Equivalencies
1 mole of any substance contains Avogadro’s number of particles and the mass on the periodic table expressed in grams.
1 mol of C = 12.01 g of C= 6.022 x 1023 atoms of C
1 mol of O2 = 32.00 g of O2= 6.022 x 1023 molecules
1 mol of NaCl = 58.44 g of NaCl= 6.022 x 1023 formula units of NaCl
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The Mole and Reactions
Example: consider the formation of carbon dioxide
EOS
Answer: one doesn’t!
At the molecular level ...
Problem: how does one mass out a single carbon atom? Note that the mass in grams is ~2.00 x 10–23 g!
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The Solution ...
EOS
For carbon, mass out:
2.0 × 10–23 g atom–1 × 6.0 × 1023 atoms mol–1
= 12 g C
Use a measurable amount – molar quantities
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MOLECULAR WEIGHT VS. MOLAR MASS
MOLECULAR WEIGHT VS. MOLAR MASS
Molecular weight = sum of the atomic
weights of all atoms in the molecule (in
amu)
Molar mass = molecular weight in grams
(in g/mol)
We will use molar mass in all problems in
this chapter!!!!
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Equivalencies
same number different entities macroscopic amounts EOS
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Must Memorize Elements THAT EXIST AS DIATOMIC MOLECULES Must Memorize Elements THAT
EXIST AS DIATOMIC MOLECULES
Remember Mr. BrINClHOFRemember Mr. BrINClHOFOROR
HON17 !!!HON17 !!!
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Practice problems1. Calcium carbonate, CaCO3, is the
principal mineral found in marble and limestone. How many moles are in 188.0 g of CaCO3?
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Practice problems
2. What is the mass, in grams, of 0.329 mol of spearmint oil, C10H14O?
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Practice problems3. Find the mass of a single lead atom.
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Practice problems
4. How many individual lead atoms are in a 1.000 g sample of this metal?
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Practice problems5. a. Calculate the number of moles of aluminum in
a solid cube that measures 3.40 cm on a side. (d=2.70 g/cm3).
b. How many atoms of aluminum are in the same sample?
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Practice problems6. a. How many molecules of oxygen, O2,
are in 0.00100 grams of this gas?
b. How many atoms?
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Molar Mass
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Definition: Describes the proportion of elements in a compound using a percent
Equal to the mass of each element present in 100 g of compound!
Percent Composition by Mass
Mass of elementPercent composition of an element = 100
Mass of compound
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Sample ProblemExample: Sodium carbonate is a compound used in the manufacture of soap and glass. Determine the percent composition by mass of each element in this compound.
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Sample ProblemExample : Determine the percent by
mass of water in Al2(SO4)3∙18H2O.
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Sample ProblemExample: Magnetite, Fe2O3, is one of the principal iron containing ores. How much elemental iron can be obtained from a metric ton (103 kg) of this ore, assuming 100 % recovery? (hint: first find % iron in Fe2O3)
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Determining FormulasDetermining Formulas
In chemical analysis we determine the % by weight of each element in a given amount of pure compound and derive the EMPIRICAL or SIMPLEST formula.
PROBLEM: A compound of B and H is 81.10% B. What is its empirical formula?
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Empirical Formula Calculations
Percent to massMass to molesDivide by Small
Multiply ‘til whole
% g moles empirical formulaDivide by smallest
mole value
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Percent to mass: (always assume 100 g sample!)
A compound of B and H is 81.10% B. What is its empirical formula?A compound of B and H is 81.10% B. What is its empirical formula?
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Mass to moles:
81.10 g B • 1 mol
10.81 g = 7.502 mol B81.10 g B •
1 mol10.81 g
= 7.502 mol B
18.90 g H • 1 mol
1.008 g = 18.75 mol H18.90 g H •
1 mol1.008 g
= 18.75 mol H
A compound of B and H is 81.10% B. What is its empirical formula?A compound of B and H is 81.10% B. What is its empirical formula?
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Now, recognize that atoms combine in the ratio of small whole numbers.
Find the ratio of moles of elements in the compound by “dividing by small”
A compound of B and H is 81.10% B. What is its empirical formula?A compound of B and H is 81.10% B. What is its empirical formula?
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But we need a whole number ratio!
Must multiply each ratio by the smallest integer available to obtain whole numbers (multiply ‘til whole)
A compound of B and H is 81.10% B. What is its empirical formula?A compound of B and H is 81.10% B. What is its empirical formula?
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Multiply ‘til Whole Hints
Common Possible Endings:
.33 x 3
.25 x 4
.67 x 3
.50 x 2
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Sample ProblemExample: Bicarbonate of soda is used in products like Alka-Seltzer and generally relieves an upset stomach. Determine the empirical formula of this compound based on the following percent composition: 27.36% Na, 1.200%H, 14.30% C, 57.14% O.
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Sample ProblemExample: A 25.00 gram sample of an orange compound contains 6.64 g of potassium, 8.84 g of chromium, and 9.52 g of oxygen. Find its empirical formula.
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A compound of B and H is 81.10% B. Its empirical formula is B2H5. What is its molecular formula?
Is the molecular formula B2H5, B4H10,
B6H15, B8H20, etc.?
BB22HH66 is one example of this class of compounds. is one example of this class of compounds.
B2H6
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We need to do an EXPERIMENT to find the MOLAR MASS.
Here experiment gives 53.3 g/molCompare with the mass of B2H5 = 26.66 g/unit
Find the ratio of these masses.53.3 g/mol
26.66 g/unit of B2H5 =
2 units of B2H51 mol
53.3 g/mol26.66 g/unit of B2H5
= 2 units of B2H5
1 mol
Multiply all of the subscripts by the ratio and obtain the molecular formula:
Molecular formula = B4H10
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Sample ProblemExample : A certain compound has the empirical formula C2H4O. Its molar mass is about 90 g/mole. What is the molecular formula?
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Sample ProblemExample: A hydrate of magnesium iodide has the formula MgI2 ∙ X H2O. To determine the value of X, a student heats a sample of the hydrate until all the water is gone. A 1.628 g sample of hydrate is heated to constant mass of 1.072 g. What is the value of X?
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Writing and Balancing Chemical Equations
• All chemical reactions have two parts:
• ReactantsReactants - the substances you start with (on left side of arrow)
• Products Products - the substances you end up with (on right side of arrow)
• The reactants turn into the products.
• Reactants Products
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In a chemical reaction…
• The way atoms are joined is changed.• Atoms aren’t created or destroyed; they just
combine together in new ways.• Can be described using sentences, symbols or
word equations:
Example:
Copper reacts with chlorine to form copper (II) chloride.
Copper + chlorine copper (II) chloride
Cu + Cl2 CuCl2
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Symbols used in Equations• The arrow separates the reactants from the
products; means “reacts” or “yields”
• The plus sign = “and”
• Subscripts are used to describe the number of atoms in a FORMULA.
• Coefficients are used to describe the number of molecules or formula units in the REACTION. They are the only things changed when balancing a reaction.
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States of Matter• Solid--(s) after the formula
• Precipitate -- a solid formed in a reaction
• Gas--(g) after the formula
• Liquid—(l) after the formula
• Aqueous-- (aq) after the formula - dissolved in water.
used after a product indicates a gas (same as (g))
used after a product indicates a solid or precipitate (same as (s))
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Other Symbols used in Equations
• indicates a reversible reaction (More later)
• show that heat is supplied to the reaction
• is used to indicate a catalyst used or supplied, in this case, platinum.
heat ,
Pt
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Chemical Equations
EOS
Example: consider the formation of water
H2(g) + O2(g) H2O(g)
Law of Conservation of Mass must be obeyed …
therefore, equations must be balanced.
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Balancing Equations
Chemical “bookkeeping” of atoms involved in the reaction:
H2(g) + O2(g) H2O(g)
H – 2 O – 2Reactants
H – 2 O – 1Products
COEFFICIENTS must be added so reactant atoms EQUAL product atoms!
Note the imbalance in oxygen atoms
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Hints & Tips for Balancing Equations
• Take one element at a time, working from left to right except for H and O. Save H for next to last and O for last.
• IF EVERYTHING BALANCES EXCEPT FOR O, and there is no way to balance O with a whole number, double all the coefficients and try again. (Because O is a diatomic element)
• (Shortcut) polyatomic ions that appear on both sides of the equation should be balanced as independent units
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Balancing Equations
EOS
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Balancing Equations Practice
Balance the following chemical equations using the appropriate coefficients:
____ Al(s) + _____ Br2 (l) _____ Al2Br6 (s)
____ Na3PO4 + ____ Fe2O3 ____ Na2O + ____ FePO4
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Types of EquationsSynthesis or Combination
Equation in Symbols: A + B ABSample Equation:
2Cu (s) + O2 (g) 2 CuO (s)
Predicting Products:
Elements Compounds OR
Compounds
More Complex Compounds
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Types of EquationsDecomposition
Equation in Symbols: AB A + BSample Equation:
2 CuO (s) 2Cu (s) + O2 (g)
Predicting Products:Compounds Elements OR More Complex Compounds
Compounds
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Types of EquationsSingle Replacement
Equation in Symbols: A + BC AB + C• Metal replacing metal• Nonmetal replacing nonmetal
Sample Equation:
Mg (s) + CuCl2 (aq) Cu (s) + MgCl2 (aq)
Predicting Products:Cations replace cations (can also have anions replacing anions)
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Types of EquationsDouble Replacement – 2 ionic compounds
Equation in Symbols: AX + BY BX + AY
Sample Equation: 2AgNO3(aq) + CuCl2 (aq) Cu (NO3)2 (aq) + 2AgCl (s)
Predicting Products:Cations switch places; solid formed
(must be driving force)
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Types of EquationsCombustion
Equation in Symbols: CxHy + O2 CO2 + H2O
Sample Equation:
CH4(g) + O2 (g) CO2 (g) + H2O (l)
Predicting Products:
Hydrocarbons react to form CO2 and
H2O
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Stoichiometric Equivalence and Reaction Stoichiometry
CS2 + 3O2 CO2 + 2 SO2
Interpretation in terms of moles:
1 mole of CS2 + 3 moles of O2 form:
1 mole of CO2 + 2 moles of SO2
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Stoichiometric Equivalence and Reaction Stoichiometry
CS2 + 3O2 CO2 + 2 SO2
Conversion factors extracted from balanced equation:
1 mole of CS2 3 moles of O2
etc.
3 moles of O2 1 mole of CO2
These ratios are called MOLE RATIOS!
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Stoichiometric EquivalentsCoefficients from a balanced chemical equation show molar equivalents of reactants and products
==> form conversion factors
In the formation of water:2 mol H2 = 1 mol O2
2 mol H2 = 2 mol H2O
1 mol O2 = 1 mol H2O
2H2 + O2 2H2O
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Sample Problem
Example 3.12 Using the equation below, determine:
CS2 + 3O2 CO2 + 2SO2
• the number of moles of oxygen required to react with 1.38 mol of carbon disulfide
• the number of moles of SO2 produced from 1.38 moles of carbon disulfide.
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Sample ProblemExample 3.13
For 2NH3 + H2SO4 (NH4)2SO4 determine:
• the mass of product possible when 1.43 mol of NH3 are reacted with an excess of sulfuric acid.
• the mass of NH3 required to react completely with 35.00 g of sulfuric acid.
• the mass of sulfuric acid required to form 1000 grams of product.
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Welcome to Mole Island
1 mole = 22.4 L @ STP
1 mol = molar mass
1 mol = 6.02 x 1023 particles
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Stoichiometry Island Diagram
Mass
Particles
Volume Mole Mole
Mass
Known UnknownSubstance A Substance B
Stoichiometry Island Diagram
Volume
Particles
M
V
P
Mass Mountain
Liter Lagoon
Particle Place
Mole Island
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Stoichiometry Island Diagram
Mass
Particles
Volume Mole Mole
Mass
Volume
Particles
Known Unknown
Substance A Substance B
1 mole = molar mass (g)Use coefficientsfrom balanced
chemical equation1 mole = 22.4 L @ STP
1 mole =
6.022 x
1023 partic
les
(atoms o
r molecu
les)
1 mole = 22.4 L @ STP
1 mole = 6.022 x 10 23 particles
(atoms or molecules)
1 mole =
molar m
ass (g
)
(gases) (gases)
Volume(liquids)
USE
DENSITY
Volume (liquids)
US
E
DE
NSI
TY
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Sample Problem
Example : How many milliliters of liquid water can be produced by the combustion of 775 mL of octane with oxygen? Assume that the volumes of the octane and the water are measured at 20oC where the densities are 0.7025 g/mL for octane and 0.9982 g/mL for water.
C8H18(l) + O2(g) CO2(g) + H2O(l)
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Reaction Yields
Theoretical yieldTheoretical yield – predicted amount of product formed from the limiting reagent, based only on the stoichiometry of the reaction
If all worked perfectly ...
2H2(g) + O2(g) 2 H2O(g)Example: 1 mol H2 will produce 1 mol of water
Actual yield – amount of product produced
In practice, actual < theoretical: errors, poor technique, etc. ...
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Limiting Reactants (Reagents)
EOS
Definition: Chemical reactant that is completely consumed in a reaction and therefore limits the quantity of product formed.
**Depends on stoichiometry of reaction
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Limiting Reagents (cont.)
EOS
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Limiting Reagents (cont.)How many meals can be made from 105 sandwiches, 202 cookies, and 107 oranges?
1105 105
1
1202 101
2
1107 107
1
mealsandwiches x meals
sandwich
mealcookies x meals
cookies
mealoranges x meals
orange
Cookies limit the total number of whole meals with excess sandwiches and oranges
Excess Reactant = Reactant left over when limiting reactant is used up
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Steps for Determining Limiting Reactant
1. Write a balanced equation.2. Convert to moles!3. Take first reactant, calculate theoretical
yield of any product.4. Repeat #3 for second reactant.5. Compare results. Whichever reactant
gives the LEAST amount of the product is the limiting reactant and determines the theoretical yield. The other is in excess.
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Sample Problem
CS2(l) + 3 O2(g) CO2(g) + 2 SO2(g)
Determine the theoretical yield of product if one starts with 1.20 mol of CS2 and 3.83 mol O2
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Sample ProblemExample: For the following reaction, determine
the theoretical yield of product if one starts with 105 g of CS2 and 145 g of O2
CS2(l) + 3O2(g) CO2(g) + 2SO2(g)
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Percent Yield
EOS
100actua
theorpercent y
eticie d
all x
l
Reaction yields are expressed as a ratio in the form of a percentage:
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ExampleFor the following reaction, determine the theoretical yield if one starts with 1.20 g of antimony and 2.40 g of iodine.
2 Sb(s) + 3 I2(s) 2 SbI3(s)
Determine theoretical yield first:
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ExampleFor the following reaction, determine the theoretical yield if one starts with 1.20 g of antimony and 2.40 g of iodine.
2 Sb(s) + 3 I2(s) 2 SbI3(s)Determine theoretical yield first:
If 3.00 g of product are actually formed, what is the percent yield?
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Acknowledgements
• Thomson/Brooks Cole (Textbook Publishers)
• Mark P. Heitz, State University of New York at Brockport, Prentice Hall Publishers (EOS Slides)