chapter 3 - chemical calculations
TRANSCRIPT
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Stoichiometry
Atomic Mass
The Mole concept
Molar Mass Percent Composition of Compounds
Determination of Formula of Compounds
Writing and Balancing Chemical Equations Interpreting balance equations, and
Reaction Stoichiometry and Calculations
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Atomic Masses
Absolute massesof atoms cannot be obtained
too small to measure the mass directly;
Relative atomic massesare used insteadmasses
relative to a chosen standard or reference. Carbon-12 is used as atomic mass referenceit is
assigned an atomic mass of 12 uexactly;
Other atoms are assigned masses relative to that of
carbon-12;
Relative atomic massesare determined using mass
spectrometer;
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A Schematic Diagram of Mass
Spectrophotometer
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Isotope Mass of CO2
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Mass Spectrum of Chlorine
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Atomic Mass Spectrum of Mercury
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Calculation of Relative Atomic Masses
Example-1:
An atomic mass spectrum gives atomic mass ratio of
oxygen atom to carbon-12 as 1.3329:1. If the atomic
mass of carbon-12 is exactly 12 u, what is the atomic
mass of oxygen?
Atomic mass of oxygen = 1.3329 x 12 u
= 15.995 u
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Calculation of Average Atomic Masses
Example-2:
Chlorine is composed of two stable naturally occurring
isotopes: chlorine-35 (75.76%; 34.9689 u) and
chlorine-37 (24.24%; 36.9659 u). What is the average
atomic mass of copper?
Atomic mass of chlorine
= (0.7576 x 34.9689 u) + (0.2424 x 36.9659 u)
= 35.45 u(as given in the periodic table)
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Calculation of Average Atomic Masses
Example-3:
Copper is composed of two naturally occurring
isotopes: copper-63 (69.09%; 62.93 u) andcopper-65 (30.91%; 64.93 u). What is the
average atomic mass of copper?
Atomic mass of copper= (0.6909 x 62.93 u) + (0.3091 x 64.93 u)
= 63.55 u(as given in the periodic table)
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Exercise #1: Relative Atomic Mass
A mass spectrometer computed the atomic mass
ratio of fluorine to carbon-12 as 1.5832-to-1. If the
atomic mass of carbon-12 is 12 u(exactly), whatis the atomic mass of fluorine in u?
(Answer: 18.998 u)
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Exercise #2: Average Atomic Mass
Natural boron is composed of two isotopes:
19.78% boron-10 (atomic mass = 10.0129 amu)
and 80.22% boron-11 (atomic mass = 11.0093amu). What is the average atomic mass of
naturally occurring boron?
(Answer: 10.81 u)
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Molar Quantity
The Mole:
A quantity that contains the Avogadros
number of items;
Avogadros number = 6.022 x 1023
12.01 g of carbon contains the Avogadros
number of carbon atoms.1 mole of carbon = 12.01 g
1 carbon atom = 12.01 u(or amu)
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Gram-Atomic Mass
Mass of 1 carbon-12 atom = 12 u(exactly);
Mass of 1 mole of carbon-12 = 12 g;
Mass of 1 oxygen atom = 16.00 u
Mass of 1 mole of oxygen = 16.00 g
Gram-atomic mass= mass (in grams) of 1 mole ofan elementthat is, the mass (in grams) thatcontains the Avogadros number of atoms of that
element.
gram-atomic massisthe molar massof an elementin grams.
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Atomic Mass & Gram-Atomic Mass
Examples:
Element Atomic mass Gram-atomic mass
Carbon 12.01 u 12.01 g/molOxygen 16.00 u 16.00 g/mol
Aluminum 26.98 u 26.98 g/mol
Silicon 28.09 u 28.09 g/molGold 197.0 u 197.0 g/mol
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Molecular Mass and Molar Mass
Molecular mass= the mass of a molecule inu;
Molar mass = the mass of one mole of an element
or a compound, expressed in grams.
Examples:Molecular Mass Molar Mass
N2 28.02 u 28.02 g/mol
H2O 18.02 u 18.02 g/molC8H18 114.22 u 114.22 g/mol
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Calculating Molar Mass
Calculating the molar mass of sucrose, C12H22O11:
(12 x 12.01 g) + (22 x 1.008 g) + (11 x 16.00 g)
= 342.3 g/mole
Molar mass of ammonium hydrogen phosphate,
(NH4)2HPO4:
(2 x 14.01 g) + (9 x 1.008 g) + (1 x 30.97 g) + (4 x 16.00 g)
= 132.06 g/mole
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Percent Composition of a Compound
Composition of aluminum sulfate, Al2(SO4)3:
Molar mass of Al2(SO4)3=
(2 x 26.98 g) + (3 x 32.06 g) + (12 x 16.00 g) = 342.14 g/mole
Mass percent of Al = (53.96 g/342.14 g) x 100% = 15.77%
Mass percent of S = (96.18 g/342.14 g) x 100% = 28.11%
Mass percent of O = (192.0 g/342.14 g) x 100% = 56.12%
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Formula of Compounds
Empirical Formula
A chemical formula that represents a simple whole
number ratio of the number of moles of elements in thecompound. Examples: MgO, Cu2S, CH2O, etc.
Molecular Formula
A formula that shows the actual number of atoms of
each type in a molecule.
Examples: C4H10, C6H6, C6H12O6.
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Empirical Formula-1
Empirical formula from composition:
Example: A compound containing carbon, hydrogen, and oxygen hasthe following composition (by mass percent): 68.12% C, 13.73%H, and 18.15% O, Determine its empirical formula.
Solution:
Use mass percent to calculate mole and mole ratio of C:H:O
Mole of C = 68.12 g x (1 mol C/12.01 g) = 5.672 mol C
Mole of H = 13.73 g x (1 mol H/1.008 g) = 13.62 mol H
Mole of O = 18.15 g x (1 mol O/16.00 g) = 1.134 mol O
Divide all moles by mole of O (smallest value) to get simple ratio:5.672 mol C/1.134 mol O = 5; 13.62 mol H/1.134 mol O = 12, and
1.134 mol O/1.134 mol O = 1;
Mole ratio: 5C:12H:1O Empirical formula = C5H12O
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Empirical Formula-2
Empirical formula from mass of elements in a sample of compound
Example: When 1.96 g of phosphorus is burned, 4.49 g of a
phosphorus oxide is obtained. Calculate the empirical formula of
the phosphorus oxide.
Solution:
Calculate moles of P and O in sample and obtain a simple mole ratio;
Mole of P = 1.96 g P x (1 mol/30.97 g) = 0.0633 mol P;
Mole of O = (4.49 g1.96 g) x (1 mol/16.00 g) = 0.158 mol O;
Divide by mole of P (smaller value) to get a simple mole ratio:0.0633 mol P/0.0633 = 1 mol P; 0.158 mol O/0.0633 = 2.5 mol O
Mole ratio: 1 mol P to2.5 mol O, OR 2 mol P to5 mol O
Empirical formula = P2O5
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Empirical Formula-3
Empirical formula from data of combustion reaction:
Example: A compound is composed of carbon, hydrogen, and oxygen. When 2.32g of this compound is burned in excess of oxygen, it produces 5.28 g of CO2gasand 2.16 g of water. Calculate the composition (in mass percent) of thecompound and determine its empirical formula.
Solution:
Find mass of C, H, and O in the sample and then calculate their masspercent:
Mass of C = 5.28 g CO2x (12.01 g C/44.01 g CO2) = 1.44 g
Mass % of C = (1.44 g C/2.32 g sample) x 100% = 62.1%
Mass of H = 2.16 g H2O x (2 x 1.008 g/18.02 g H2O) = 0.24 g
Mass % of H = (0.242 g H/2.32 g sample) x 100% = 10.4%Mass of O = 2.32 g sample1.44 g C0.24 g H = 0.64 g
Mass % of O = 10062.1% C10.4% H = 27.5%
Derive empirical formula from these mass percent composition
(next slide)
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Empirical Formula-3
Empirical formula from data of combustion (continued):
Calculate mole and simple mole ratio from calculated mass of each
element:
Mole of C = 1.44 g C x (1 mol/12.01 g) = 0.12 molMole of H = 0.242 g x (1 mol/1.008 g) = 0.24 mol
Mole of O = 0.64 g x (1 mol/16.00 g) = 0.04 mol
Divide all moles by mole of O (smallest mole) to obtain a simple ratio:
0.12 mol C/0.04 = 3 mol C; 0.24 mol H/0.04 = 6 mol H;
0.04 mol O/0.04 = 1 mol O
Simple molar ratio: 3 mol C : 6 mol H : 1 mol O
Empirical formula: C3H6O
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Molecular Formula
Molecular formula is derived from empirical formula and molecularmass, which is obtained independently
Empirical formula = CxHyOz;molecular formula = (CxHyOz)n,
where n = (molecular mass/empirical formula mass)
Example:
A compound has an empirical formula C3H6O and its molecularformula is 116.2 u. What is the molecular formula?
Solution:
Empirical formula mass = (2 x 12.01 u) + (6 x 1.008 u) + 16.00 u
= 58.1 u
Molecular formula = (C3H6O)n; where n = (116.2 u/58.1 u) = 2
Incorrect molecular formula = (C3H6O)2;
Correct molecular formula = C6H12O2
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Exercise #3: Determination of Formulas
A 2.00-gram sample of phosphorus is completelyreacted with oxygen gas, which yields 4.58 g of
product that is composed of only phosphorus and
oxygen. In separate analyses, the compound is
found to have molar mass of about 284 g/mol.
(a) Determine the empirical and molecular
formulas of the compound. (b) Write an equation
for the reaction of phosphorus with oxygen gas.
(Answer: P2O5; P4O10; 4P + 5 O2P4O10)
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Chemical Equation #1
Description of reaction:
Iron reacts with oxygen gas and forms solid
iron(III) oxide:
Identity: reactants = iron (Fe) and oxygen gas
(O2); product = iron(III) oxide
Chemical equation: Fe(s)+ O2(g)
Fe2O3(s) Balanced equation:
4Fe(s)+ 3 O2(g)2Fe2O3(s)
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Chemical Equation #2
Description of reaction:
Phosphorus reacts with oxygen gas to form
solid tetraphosphorus decoxide.
Equation: P(s)+ O2(g)P4O10(s)
Balanced eqn.: 4P(s)+ 5 O2(g)P4O10(s)
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Chemical Equation #3
Description of reaction:
Propane gas (C3H8) is burned in air (excess of oxygen)
to form carbon dioxide gas and water vapor;
Identity: reactants = C3H8(g)and O2(g);products = CO2(g)and H2O(g);
Equation: C3H8(g)+ O2(g)CO2(g)+ H2O(g);
Balanced equation:
C3H8(g)+ 5 O2(g)3CO2(g)+ 4H2O(g)
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Chemical Equation 4
Description of reaction:Ammonia gas (NH3) reacts with oxygen gas to
form nitrogen monoxide gas and water vapor;
Equation: NH3(g)+ O2(g)NO(g)+ H2O(g);
Balancing the equation:
2NH3(g)+5/2O2(g)2NO(g)+ 3H2O(g);
Multiply throughout by 2 to get rid of the fraction:
4NH3(g)+ 5 O2(g)4NO(g)+ 6H2O(g);
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Balancing Chemical Equations
Rules for balancing equations:
1. Use smallest integer coefficients in front of each reactants andproducts as necessary; coefficient 1 need not be indicated;
2. The formula of the substances in the equation MUST NOT be
changed.
Helpful steps in balancing equations:
1. Begin with the compound that contains the most atoms or types ofatoms.
2. Balance elements that appear only once on each side of the arrow.
3. Next balance elements that appear more than once on either side.
4. Balance free elements last.
5. Finally, check that smallest whole number coefficients are used.
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Stoichiometry
Stoichiometry = the quantitative relationships betweenone reactant to another, or between a reactant andproducts in a chemical reaction.
Interpreting balanced equations:
Example: C3H8(g)+ 5 O2(g) 3CO2(g)+ 4H2O(g);
The equation implies that:
1 C3H8molecule reacts with 5 O2molecules to produce
3 CO2molecules and 4 H2O molecules; OR1 mole of C3H8reacts with 5 moles of O2to produce 3
moles of CO2and 4 moles of H2O.
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Stoichiometric Calculations
Mole-to-mole relationship: Example: In the following reaction, if 6.0 moles of octane, C8H18, is
completely combusted in excess of oxygen gas, how many moles ofCO2and H2O, respectively, will be formed? How many moles of O2does it consumed?
Reaction: 2C8H18(l) + 25 O2(g) 16CO2(g) + 18H2O(g)
Calculations:
Mole CO2formed = 6.0 mol C8H18x (16 mol CO2/2mol C8H18) = 48 moles
Mole H2O formed = 6.0 mol C
8H
18x (18 mol H
2O/2mol C
8H
18) = 54 moles
Mole O2consumed = 6.0 mol C8H18x (25 mol O2/2mol C8H18) = 75 moles
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Stoichiometric Calculations
Mass-to-mole-to-mole-to-mass relationship: Example-1: In the following reaction, if 690 g of octane, C8H18, is
completely combusted in excess of oxygen gas, how many grams of
CO2are formed?
Reaction: 2C8H18(l)+ 25 O2(g) 16CO2(g)+ 18H2O(g)
Calculation-1:
Moles C8H18reacted = 690 g C8H18x (1 mol/114.2 g) = 6.0 moles
Moles CO2formed = 6.0 mol C8H18x (16 mol CO2/2 mol C8H18)
= 48 moles CO2
Mass of CO2formed = 48 mol CO2x (44.01 g/mol) = 2.1 x 103g
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Stoichiometric Calculations
Mass-to-mole-to-mole-to-mass relationship: Example-2: In the following reaction, if 690 g of octane, C8H18, is
completely combusted in excess of oxygen gas, how many grams of
H2O are formed?
Reaction: 2C8H18(l)+ 25 O2(g) 16CO2(g)+ 18H2O(g)
Calculation-2:
Moles C8H18reacted = 690 g C8H18x (1 mol/114.2 g) = 6.0 moles
Moles H2O formed = 6.0 mol C8H18x (18 mol H2O/2 mol C8H18)
= 54 moles CO2
Mass of H2O formed = 54 mol H2O x (18.02 g/mol) = 970 g
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Stoichiometric Calculations
Mass-to-mole-to-mole-to-mass relationship: Example-3: In the following reaction, if 690 g of octane, C8H18, is
completely combusted in excess of oxygen gas, how many grams of
oxygen gas are consumed?
Reaction: 2C8H18(l)+ 25 O2(g) 16CO2(g)+ 18H2O(g)
Calculation-3:
Moles C8H18reacted = 690 g C8H18x (1 mol/114.2 g) = 6.0 moles
Moles O2consumed = 6.0 mol C8H18x (25 mol O2/2 mol C8H18)
= 75 moles O2
Mass of H2O formed = 75 mol O2x (32.00 g/mol) = 2.4 x 103g g
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Stoichiometry Involving Limiting Reactant
Limiting reactant
one that got completely consumed in a
chemical reaction before the other
reactants.
Product yields depend on the amount of
limiting reactant
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A Reaction Stoichiometry
Example:
In the reaction: 2Cu(s)+ S(s) Cu2S(s),
2 moles of copper are required to react completely with1 mole of sulfur, which will produce 1 mole of
copper(I) sulfide.
If a reaction is carried out using 1 mole of copper and 1
mole of sulfur, then copper will be the limiting reactant
and sulfur is in excess. Only 0.5 mole of copper(I)
sulfide is obtained.
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Exercise #4: Stoichiometry Calculations
Ammonia is produced from the reaction with
hydrogen according to the following equation:
N2(g) + 3H2(g) 2NH3(g)
If 25 N2molecules are reacted with 60 H2
molecules in a sealed container, which molecules
will be completely consumed? How many NH3
molecules are formed?
(Answer: H2; 40 NH3molecules)
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Exercise #5: Limiting Reactants and
Reaction Yields
Ammonia is produced in the following reaction:
N2(g) + 3H2(g) 2NH3(g)
(a) If 118 g of nitrogen gas is reacted with 31.5 g of
hydrogen gas, which reactant will be completely consumed
at the end of the reaction? (b) How many grams of the
excess reactant will remain (unreacted)? (c) How many
grams ammonia will be produced when the limiting
reactant is completely reacted and the yield is 100%?
(Answer: (a) N2; (b) 6.0 g; (c) 143.4 g of NH3)
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Theoretical, Actual and Percent Yields
Theoretical yield:
yield of product calculated based on the stoichiometry
of balanced equation and amount of limiting reactant
(assuming the reaction goes to completion and the
limiting reactant is completely consumed).
Actual Yield:
Yield of product actually obtained from experiment
Percent Yield = (Actual yield/Theoretical yield) x 100%
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Exercise #6: Limiting Reactant & Yields
In an ammonia production, the reactor is charged with N2
and H2gases at flow rates of 805 g and 195 g per minute,
respectively, at 227oC, and the reaction is as follows:
N2(g) + 3H2(g) 3 NH3(g)(a) What is the rate (in g/min) that ammonia is produced if
the yield is 100%? (b) If the reaction produces 915 g of
NH3per minute, calculate the percentage yield of the
reaction.
(Answer: (a) 978.5 g/min; (b) Yield = 93.5%)