Download Chapter #3 : Stoichiometry

Chapter #3 : Stoichiometry
MOLE
¥ The Mole is based upon the definition:
¥ The amount of substance that contains as
many elementary parts (atoms, molecules,
or other?) as there are atoms in exactly
¥ 12 grams of carbon -12.
¥ 1 Mole = 6.022045 x 1023 particles
Mole - Mass Relationships of Elements
Element Atom/Molecule Mass
1 atom of H = 1.008 amu
Mole Mass Number of Atoms
1 mole of H = 1.008 g = 6.022 x 1023 atoms
1 atom of Fe = 55.85 amu 1 mole of Fe = 55.85 g = 6.022 x 1023 atoms
1 atom of S = 32.07 amu
1 mole of S = 32.07 g = 6.022 x 1023 atoms
1 atom of O = 16.00 amu
1 mole of O = 16.00 g = 6.022 x 1023 atoms
1 molecule of O2 = 32.00 amu
1 mole of O2 = 32.00 g = 6.022 x 1023 molecule
1 molecule of S8 = 2059.52 amu
1 mole of S8 = 2059.52 g = 6.022 x 1023 molecules
Molecular Mass - Molar Mass ( M )
The Molecular mass of a compound expressed in amu is
numerically the same as the mass of one mole of the
compound expressed in grams.
For water: H2O
Molecular mass = (2 x atomic mass of H ) + atomic mass of O
= 2 ( 1.008 amu) + 16.00 amu = 18.02 amu
Mass of one molecules of water = 18.02 amu
Molar mass = ( 2 x atomic mass of H ) + atomic mass of O
= 2 ( 1.008 g ) + 16.00 g = 18.02 g
18.02 g H2O = 6.022 x 1023 molecules of water = 1 mole H2O
Calculating the Number of Moles and Atoms in a
Given Mass of Element
Problem: Tungsten (W) is the element used as the filament in lig ht
bulbs, and has the highest melting point of any element
3680oC. How many moles of tungsten, and atoms of the
element are contained in a 35.0 mg sample of the metal?
Plan: Convert mass into moles by dividing the mass by the atomic
weig ht of the metal, then calculate the number of atoms by
multiplying by Avog adro s number!
Solution: Converting from mass of W to moles:
1 mol W
183.9 g W
Moles of W = 35.0 mg W x
NO. of W atoms = 1.90 x 10 - 4 mol W x
=
=
6.022 x 10 23 atoms
1 mole of W
mol
mol
=
atoms of Tungsten
Calculating the Moles and Number of
Formula Units in a given Mass of Cpd.
Problem: Trisodium Phosphate is a component of some detergents.
How many moles and formula units are in a 38.6 g sample?
Plan: We need to determine the formula, and the molecular mass from
the atomic masses of each element multiplied by the coefficients.
Solution: The formula is Na 3PO 4. Calculating the molar mass:
M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen =
= 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g /mol
= 68.97 g/mol + 30.97 g/mol + 64.00 g /mol = 163.94 g/mol
Converting mass to moles:
Moles Na3PO 4 = 38.6 g Na3PO 4 x (1 mol Na3PO 4)
163.94 g Na3PO 4
=
mol Na3PO4
Formula units = 0.23545 mol Na 3PO 4 x 6.022 x 1023 formula units
1 mol Na3PO 4
=
formula units
Flow Chart of Mass Percentage Calculation
Moles of X in one
mole of Compound
M (g / mol) of X
Mass (g) of X in one
mole of compound
Divide by mass (g) of one mole
of compound
Mass fraction of X
Multiply by 100
Mass % of X
Calculating Mass Percentage and Masses of
Elements in a Sample of a Compound - I
Problem: Sucrose (C12H 22O11) is common table sug ar.
( a) What is the mass percent of each element in sucrose?
( b) How many grams of carbon are in 24.35 g of sucrose?
(a) Determining the mass percent of each element:
mass of C = 12 x 12.01 g C/mol =
144.12 g C/mol
mass of H = 22 x 1.008 g H/mol =
22.176 g H/mol
mass of O = 11 x 16.00 g O/mol =
176.00 g O/mol
342.296 g/mol
Finding the mass fraction of C in Sucrose & % C :
Total mass of C
144.12 g C
Mass Fraction of C =
=
mass of 1 mole of sucrose
342.30 g Cpd
=
%
To find mass % of C = 0.421046 x 100% =
Calculating Mass Percents and Masses of
Elements in a Sample of Compound - II
(a) continued
Mass % of H =
mol H x M of H x 100% = 22 x 1.008 g H x 100%
mass of 1 mol sucrose
342.30 g
=
%H
Mass % of O =
mol O x M of O x 100% = 11 x 16.00 g O x 100%
mass of 1 mol sucrose
342.30 g
=
%O
(b) Determining the mass of carbon:
Mass (g) of C = mass of sucrose X( mass fraction of C in sucrose)
Mass (g) of C = 24.35 g sucrose X 0.421046 g C
1 g sucrose
=
gC
Mol wt and % composition of NH4NO3
¥
¥
¥
2 mol N x 14.01 g/mol = 28.02 g N
4 mol H x 1.008 g /mol = 4.032 g H
3 mol O x 15.999 g /mol = 48.00 g O
g/mol
28.02g N2
x 100%
80.05g
=
%N
%H =
4.032g H2 x 100% =
80.05g
%H
%O =
48.00g O2
x 100% =
80.05g
%N =
%O
99.997%
Calculate the Percent Composition
of Sulfuric Acid H2SO4
Molar Mass of Sulfuric acid =
2(1.008g ) + 1(32.07g) + 4(16.00g) = 98.09 g/mol
%H =
2(1.008g H2)
x 100% =
98.09g
%H
%S =
1(32.07g S)
x 100% =
98.09g
% S
%O =
4(16.00g O)
x 100% =
98.09 g
% O
100.00%
Check =
Empirical and Molecular Formulas
Empirical Formula - The simplest formula for a compound
that agrees with the elemental analysis! The
smallest set of whole numbers of atoms.
Molecular Formula - The formula of the compound as it
exists, it may be a multiple of the Empirical
formula.
Steps to Determine Empirical Formulas
Mass (g) of Element
M (g/mol )
Moles of Element
use no. of moles as subscripts
Preliminary Formula
change to integer subscripts
Empirical Formula
Some Examples of Compounds with the same
Elemental Ratio s
Empirical Formula
Molecular Formula
CH2(unsaturated Hydrocarbons)
C2H4 , C3H6 , C4H8
OH or HO
H2O2
S
S8
P
P4
Cl
Cl2
CH2O (carbohydrates)
C6H12O6
Determining Empirical Formulas from
Masses of Elements - I
Problem: The elemental analy sis of a sample compound g ave the
following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is the
empirical formula and name of the compound?
Plan: First we have to convert mass of the elements to moles of the
elements using the molar masses. Then we construct a preliminary
formula and name of the compound.
Solution: Finding the moles of the elements:
1 mol Na
=
22.99 g Na
1
mol
Cr
Moles of Cr = 6.420 g Cr x
=
52.00 g Cr
1 mol O
Moles of O = 7.902 g O x
=
16.00 g O
Moles of Na = 5.678 g Na x
mol Na
mol Cr
mol O
Determining Empirical Formulas from
Masses of Elements - II
Constructing the preliminary formula:
Na 0.2469 Cr 0.1235 O0.4939
Converting to integer subscripts (dividing all by smallest subscript):
Na 1.99 Cr 1.00 O4.02
Rounding off to whole numbers:
Na 2CrO 4
Sodium Chromate
Determining the Molecular Formula from
Elemental Composition and Molar Mass - I
Problem: The sugar burned for energy in cells of the body is Glucose
(M = 180.16 g/mol), elemental analy sis shows that it contains
40.00 mass % C, 6.719 mass % H, and 53.27 mass % O.
(a) Determine the empirical formula of g lucose.
(b) Determine the Molecular formula.
Plan: We are only given mass %, and no weig ht of the compound so w e
will assume 100g of the compound, and % becomes grams, and
we can do as done previously with masses of the elements.
Solution:
Mass Carbon = 40.00% x 100g/100% = 40.00 g C
Mass Hy drogen = 6.719% x 100g/100% = 6.719g H
Mass Oxyg en = 53.27% x 100g /100% = 53.27 g O
99.989 g Cpd
Determining the Molecular Formula from
Elemental Composition and Molar Mass - II
Converting from G rams of Elements to moles:
Moles of C = Mass of C x 1 mole C = 3.3306 moles C
12.01 g C
1 mol H
Moles of H = Mass of H x
= 6.6657 moles H
1.008 g H
Moles of O = Mass of O x 1 mol O = 3.3294 moles O
16.00 g O
Constructing the preliminary formula C 3.33 H 6.67 O 3.33
Converting to integer subscripts, divide all subscripts by the smallest:
C 3.33/3.33 H 6.667 / 3.33 O3.33 / 3.33 = CH2O
Determining the Molecular Formula from
Elemental Composition and Molar Mass - III
(b) Determining the Molecular Formula:
The formula weig ht of the empirical formula is:
1 x C + 2 x H + 1 x O = 1 x 12.01 + 2 x 1.008 + 1 x 16.00 = 30.03
Whole-number multiple =
=
M of Glucose
empirical formula mass
180.16
= 6.00 = 6
30.03
Therefore the Molecular Formula is:
C 1 x 6H 2 x 6O 1 x 6 =
C6H12O6
=
Adrenaline is a very Important
Compound in the Body - I
¥ Analy sis gives :
¥
C = 56.8 %
¥
H = 6.50 %
¥
O = 28.4 %
¥
N = 8.28 %
¥ Calculate the
Empirical Formula !
Adrenaline - II
¥
¥
¥
¥
¥
¥
¥
¥
¥
¥
Assume 100g!
C = 56.8 g C/(12.01 g C/ mol C) = 4.73 mol C
H = 6.50 g H/( 1.008 g H / mol H ) = 6.45 mol H
O = 28.4 g O/(16.00 g O/ mol O) = 1.78 mol O
N = 8.28 g N/(14.01 g N/ mol N) = 0.591 mol N
Divide by 0.591 =
C = 8.00 mol C = 8.0 mol C
or
H = 10.9 mol H = 11.0 mol H
O = 3.01 mol O = 3.0 mol O C8H11O3N
N = 1.00 mol N = 1.0 mol N
Ascorbic acid ( Vitamin C ) - I
contains C , H , and O
¥ Upon combustion in excess oxyg en, a 6.49 mg
sample yielded 9.74 mg CO 2 and 2.64 mg H2O
¥ Calculate it s Empirical formula!
¥ C: 9.74 x10 -3g CO 2 x(12.01 g C/44.01 g CO2)
=
gC
¥ H: 2.64 x10-3g H2O x (2.016 g H2/18.02 gH 2O)
=
gH
¥ Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg
=
mg O
Vitamin C combustion - II
¥ C = 2.65 x 10-3 g C / ( 12.01 g C / mol C ) =
=
mol C
¥ H = 0.295 x 10 -3 g H / ( 1.008 g H / mol H ) =
=
mol H
¥ O = 3.54 x 10-3 g O / ( 16.00 g O / mol O ) =
=
mol O
¥ Divide each by
¥ C = 1.00
¥ H = 1.32
¥ O = 1.00
Multiply each by 3
= 3.00 = 3.0
= 3.96 = 4.0
= 3.00 = 3.0
C3H4O3
Determining a Chemical Formula from
Combustion Analysis - I
Problem: Ery throse (M = 120 g/mol) is an important chemical
compound as a starting material in chemical synthesis, and
contains Carbon Hydrogen, and Oxyg en. Combustion
analysis of a 700.0 mg sample yielded 1.027 g CO2 and
0.4194 g H2O.
Plan: We find the masses of Hy drogen and Carbon using the mass
fractions of H in H2O, and C in CO 2. The mass of Carbon and
Hy drogen are subtracted from the sample mass to get the mass
of Oxyg en. We then calculate moles, and construct the empirical
formula, and from the given molar mass we can calculate the
molecular formula.
Determining a Chemical Formula from
Combustion Analysis - II
Calculating the mass fractions of the elements:
Mass fraction of C in CO 2 =
mol C x M of C
=
mass of 1 mol CO2
= 1 mol C x 12.01 g C/ 1 mol C =
44.01 g CO 2
0.2729 g C / 1 g CO2
mol H x M of H
=
mass of 1 mol H2O
2 mol H x 1.008 g H / 1 mol H
=
= 0.1119 g H / 1 g H2O
18.02 g H2O
Mass fraction of H in H 2O =
Calculating masses of C and H
Mass of Element = mass of compound x mass fraction of element
Determining a Chemical Formula from
Combustion Analysis - III
Mass (g) of C = 1.027 g CO2 x 0.2729 g C = 0.2803 g C
1 g CO2
0.1119 g H
= 0.04693 g H
Mass (g) of H = 0.4194 g H2O x
1 g H2O
Calculating the mass of O:
Mass (g) of O = Sample mass -( mass of C + mass of H )
= 0.700 g - 0.2803 g C - 0.04693 g H = 0.37277 g O
Calculating moles of each element:
C = 0.2803 g C / 12.01 g C/ mol C = 0.02334 mol C
H = 0.04693 g H / 1.008 g H / mol H = 0.04656 mol H
O = 0.37277 g O / 16.00 g O / mol O = 0.02330 mol O
C0.02334H0.04656O0.02330 = CH2O formula weig ht = 30 g / formula
120 g /mol / 30 g / formula = 4 formula units / cpd = C4H8O4
Atoms
Molecular
Formula
Avogadro s
Number
Molecules
6.022 x 1023
Moles
Moles
Chemical Equations
Qualitative Information:
Reactants
Products
States of Matter: (s) solid
(l) liquid
(g) gaseous
(aq) aqueous
2 H2 (g) + O2 (g)
2 H2O (g)
Balanced Equations
¥ mass balance (atom balance)- same number of each
element
(1) start with simplest element
(2) progress to other elements
(3) make all whole numbers
(4) re-check atom balance
1 CH4 (g) + O2 (g)
1 CO2 (g) + H2O (g)
1 CH4 (g) + O2 (g)
1 CO2 (g) + 2 H2O (g)
1 CH4 (g) + 2 O2 (g)
1 CO2 (g) + 2 H2O (g)
¥charge balance (no spectator ions)
Ca 2+ (aq) + 2 OH- (aq) + Na +
Ca(OH)2 (s) + Na +
Information Contained in a Balanced Equation
View ed in
terms of:
Reactants
Products
2 C2H6 (g) + 7 O 2 (g) = 4 CO2 (g) + 6 H2O(g) + Energy
Molecules
2 molecules of C 2H6 + 7 molecules of O 2 =
4 molecules of CO2 + 6 molecules of H2O
Amount (mol)
Mass (amu)
Mass (g)
2 mol C2H6 + 7 mol O2 = 4 mol CO 2 + 6 mol H2O
60.14 amu C2H6 + 224.00 amu O 2 =
176.04 amu CO 2 + 108.10 amu H2O
60.14 g C2H 6 + 224.00 g O2 = 176.04 g CO 2 + 108.10 g H2O
Total Mass (g)
284.14g = 284.14g
Balancing Chemical Equations - I
Problem: The hydrocarbon hexane is a component of G asoline that
burns in an automobile engine to produce carbon dioxide and
water as well as energy . Write the balanced chemical
equation for the combustion of hexane (C6H 14).
Plan: Write the skelton equation from the words into chemical
compounds with blanks before each compound. begin the
balance with the most complex compound first, and save oxyg en
untill last!
Solution:
O2 (g)
CO 2 (g) +
H2O(g) + Energ y
C6H 14 (l) +
Begin with one Hexane molecule which say s that we will get 6 CO2 s!
1 C6H 14 (l) +
O2 (g)
6 CO 2 (g) +
H2O(g) + Energ y
Balancing Chemical Equations - II
The H atoms in the hexane will end up as H2O, and we have 14
H atoms, and since rach water molecule has tw o H atoms, we will g et
a total of 7 water molecules.
1 C6H 14 (l) +
O2 (g)
6 CO 2 (g) + 7 H2O(g) + Energ y
Since oxyg en atoms only come as diatomic molecules
(two O atoms, O 2),we must have even numbers of oxyg en atoms on the
product side. We do not since w e have 7 water molecules! Therefore
multiply the hexane by 2, giving a total of 12 CO2 molecules, and
14 H2O molecules.
2 C6H 14 (l) +
O2 (g)
12 CO 2 (g) + 14 H2O(g) + Energ y
This now gives 12 O 2 from the carbon dioxide, and 14 O atoms from the
water, which will be another 7 O 2 molecules for a total of 19 O 2 !
2 C6H14 (l) + 19 O2 (g)
12 CO2 (g) +14 H2O(g) + Energy
Chemical Equation Calc - I
Atoms (Molecules)
Avogadro s
Number
6.02 x 1023
Molecules
Reactants
Products
Chemical Equation Calc - II
Mass
Atoms (Molecules)
Avogadro s
Number
Reactants
6.02 x 1023
Molecules
Moles
Molecular
g/mol
Weight
Products
Sample Problem: Calculating Reactants and
Products in a Chemical Reaction - I
Problem: Given the following chemical reaction between Aluminum
Sulfide and water, if we are given 65.80 g of A l2S3: a) How many moles
of water are required for the reaction? b) What mass of H2S & Al(OH)3
would be formed?
Al 2S 3 (s) + 6 H2O(l)
2 Al(OH)3 (s) + 3 H2S(g)
Plan: Calculate moles of Aluminum Sulfide using it s molar mass, then
from the equation, calculate the moles of Water, and then the moles of
Hydrog en Sulfide, and finally the mass of Hy drogen Sulfide using
it s molecular weight.
Solution:
a) molar mass of Aluminum Sulfide = 150.17 g / mol
moles Al2S3 =
65.80 g Al2S3
=
150.17 g Al2S3/ mol Al2S 3
moles Al2S 3
Calculating Reactants and Products in a
Chemical Reaction - II
a) cont.
0.4382 moles Al2S 3 x
moles H2O
6 moles H 2O =
1 mole Al2S3
b) 0.4382 moles Al2S3 x 3 moles H 2S = 1.314 moles H2S
1 mole Al2S3
molar mass of H 2S = 34.09 g / mol
mass H 2S = 1.314 moles H 2S x 34.09 g H2S =
g H2S
1 mole H 2S
0.4382 moles Al2S3 x 2 moles Al(OH)3 = 0.4764 moles Al(OH)3
1 mole Al2S3
molar mass of Al(OH)3 = 78.00 g / mol
mass Al(OH)3 = 0.4764 moles Al(OH )3 x 78.00 g Al(OH)3 =
1 mole Al(OH)3
=
g Al(OH)3
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - I
Problem: Calcium Phosphate could be prepared in the following
reaction sequence:
4 P4 (s) + 10 KClO3 (s)
P4O10 (s) + 6 H2O (l)
2 H3PO4 (aq) + 3 Ca(OH)2 (aq)
4 P4O10 (s) + 10 KCl (s)
4 H3PO4 (aq)
6 H2O(aq) + Ca 3(PO4)2 (s)
Given: 15.5 g P4 and sufficient KClO3 , H 2O and Ca(OH )2. What mass
of Calcium Phosphate could be formed?
Plan: (1) Calculate moles of P 4.
(2) Use molar ratios to get moles of Ca3(PO 4)2.
(3) Convert the moles of product back into mass by using the
molar mass of Calcium Phosphate.
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - II
Solution:
1 mole P 4
moles of Phosphorous = 15.50 g P4 x 123.88 g P4= 0.1251 mol P4
For Reaction #1 [ 4 P4 (s) + 10 KClO4 (s)
4 P4O10 (s) + 10 KCl (s) ]
For Reaction #2 [ 1 P4O10 (s) + 6 H 2O (l)
4 H 3PO 4 (aq) ]
For Reaction #3 [ 2 H3PO 4 + 3 Ca(OH)2
1 Ca 3(PO 4)2 + 6 H 2O]
0.1251 moles P4 x 4 moles P 4O 10 x 4 moles H 3PO 4 x 1 mole Ca 3(PO 4)2
4 moles P4
1 mole P 4O10
2 moles H3PO 4
= 0.2502 moles Ca3(PO4)2
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - III
Molar mass of Ca3(PO 4)2 = 310.18 g mole
moles Ca3(PO4)2 x
mass of product =
=
310.18 g Ca3(PO 4)2
1 mole Ca3(PO 4)2 =
g Ca 3(PO4)2
Limiting Reactant Problems
aA + bB + cC
dD + eE + f F
Steps to solve
1) Identify it as a limiting Reactant problem - Information on the:
mass, number of moles, number of molecules, volume and
molarity of a solution is g iven for more than one reactant!
2) Calculate moles of each reactant!
3) Divide the moles of each reactant by the coefficient (a,b,c etc....)!
4) Which ever is smallest, that reactant is the limiting reactant!
5) Use the limiting reactant to calculate the moles of product
desired then convert to the units needed (moles, mass, volume,
number of atoms etc....)!
Limiting Reactant Problem:
Sample Problem 3.11 ( p 110-111)
Problem: A fuel mixture used in the early days of rocketry is composed
of two liquids, hydrazine (N2H 4) and dinitrog en tetraoxide (N2O4). They
ignite on contact ( hypergolic!) to form nitrogen g as and water vapor.
How many grams of nitrog en gas form when exactly 1.00 x 102 g N2H4
and 2.00 x 10 2 g N2O4 are mixed?
Plan: First write the balanced equation. Since amounts of both reactants
are given, it is a limiting reactant problem. Calculate the moles of each
reactant, and then divide by the equation coefficient to find which is
limiting and use that one to calculate the moles of nitrogen gas, then
calculate mass using the molecular weight of nitrogen g as.
Solution:
2 N 2H4 (l) + N 2O4 (l)
3 N 2 (g) + 4 H2O (g) + Energy
Sample Problem 3.11 cont.
molar mass N2H4 = ( 2 x 14.01 + 4 x 1.008 ) = 32.05 g/mol
molar mass N2O4 = ( 2 x 14.01 + 4 x 16.00 ) = 92.02 g/mol
Moles N2H4 =
1.00 x 102 g
32.05 g /mol
Moles N2O4 =
2.00 x 102 g
= 2.17 moles N 2O 4
92.02 g /mol
= 3.12 moles N 2H 4
dividing by coefficients 3.12 mol / 2 = 1.56 mol N2H4
2.17 mol / 1 = 2.17 mol N2O4
3 mol N2
Nitrogen yielded = 3.12 mol N2H4 =
2 mol N2H4
Limiting !
= 4.68 moles N2
Mass of Nitrogen = 4.68 moles N2 x 28.02 g N2 / mol =
g N2
Acid - Metal Limiting Reactant - I
¥ 2Al(s) + 6HCl(g)
2AlCl3(s) + 3H 2(g)
¥ Given 30.0g Al and 20.0g HCl, how many moles of
Aluminum Chloride will be formed?
¥ 30.0g Al / 26.98g Al/mol Al = 1.11 mol Al
¥
1.11 mol Al / 2 = 0.555
¥ 20.0g HCl / 36.5g HCl/mol HCl = 0.548 mol HCl
¥
O.548 mol HCl / 6 = 0.0913
¥ HCl is smaller therefore the Limiting reactant!
Acid - Metal Limiting Reactant - II
¥ since 6 moles of HCl yield 2 moles of AlCl3
¥ 0.548 moles of HCl will yield:
¥ 0.548 mol HCl / 6 mol HCl x 2 moles of
mol of AlCl3
¥
AlCl3 =
Ostwald Process Limiting Reactant Problem
¥ What mass of NO could be formed by the reaction 30.0g of
Ammonia g as and 40.0g of Oxy gen gas?
4NO (g) + 6 H 2O(g)
¥ 4NH 3 (g) + 5 O2 (g)
¥ 30.0g NH 3 / 17.0g NH 3/mol NH 3 = 1.76 mol NH 3
1.76 mol NH 3 / 4 = 0.44 mol NH 3
¥ 40.0g O 2 / 32.0g O2 /mol O2 = 1.25 mol O2
1.25 mol O 2 / 5 = 0.25 mol O 2
¥ Therefore Oxyg en is the Limiting Reagent!
¥ 1.25 mol O 2 x 4 mol NO = 1.00 mol NO
5 mol O 2
¥ mass NO = 1.00 mol NO x 30.0 g NO
1 mol NO
=
g NO
Chemical Reactions in Practice: Theoretical,
Actual, and Percent Yields
Theoretical yield: The amount of product indicated by the
stoichiometrically equivalent molar ratio in the balanced equation.
Side Reactions: These form smaller amounts of different products that
take away from the theoretical yield of the main product.
Actual yield: The actual amount of product that is obtained.
Percent yield: (%Yield)
% Yield =
A ctual Yield
x 100
Theoretical Yield
Percent Yield Problem:
Problem: Given the chemical reaction between Iron and water to form
the iron oxide, Fe3O 4 and H ydrog en g as g iven below. If 4.55g of Iron is
reacted with sufficent water to react all of the Iron to form rust, w hat is
the percent yield if only 6.02g of the oxide are formed?
Plan: Calculate the theoretical yield and use it to calculate the percent
yield, using the actual y ield.
Solution:
3 Fe + 4 H O
Fe O + 4 H
(s)
2
(l)
3
4 (s)
2 (g)
4.55 g Fe
= 0.081468 mol = 0.0815 mol
55.85 g Fe
1 mol Fe3O4
mol Fe
0.0815 mol Fe x
= 0.0272 mol Fe3O4
3 mol Fe
231.55 g Fe3O4
0.0272 mol Fe3O4 x
= 6.30 g Fe3O4
1 mol Fe3O4
Actual
Yield
Percent Yield =
x 100% = 6.02 g Fe3O4 x 100% =
Theoretical Yield
6.30 g Fe3O4
%
Percent Yield / Limiting Reactant Problem - I
Problem: Ammonia is produced by the Haber Process using Nitrog en
and Hy drogen Gas. If 85.90g of Nitrogen are reacted with
21.66 g Hy drogen and the reaction yielded 98.67 g of
ammonia what was the percent yield of the reaction.
2 NH3 (g)
N2 (g) + 3 H2 (g)
Plan: Since we are given the masses of both reactants, this is a limiting
reactant problem. First determine which is the limiting reagent
then calculate the theoretical yield, and then the percent yield.
Solution: Moles of Nitrogen and Hydrogen:
Divide by coefficient
85.90 g N2
moles N2 =
= 3.066 mol N2 to g et limiting:
28.02 g N2
3.066 g N2
= 3.066
1 mole N2
1
21.66 g H2
moles H 2 =
= 10.74 mol H2
2.016 g H2
10.74 g H2
= 3.582
1 mole H 2
3
Percent Yield/Limiting Reactant Problem - II
Solution Cont.
N2 (g) + 3 H2 (g)
2 NH3 (g)
We have 3.066 moles of Nitrogen, and it is limiting, therefore the
theoretical yield of ammonia is:
2 mol NH 3
= 6.132 mol NH3
3.066 mol N2 x
1 mol N2
(Theoretical Yield)
17.03 g NH 3
= 104.427 g NH3
1 mol NH 3
(Theoretical Yield)
A ctual Yield
Percent Yield =
x 100%
Theoretical Yield
6.132 mol NH 3 x
Percent Yield =
98.67 g NH3
104.427 g NH3
x 100% =
94.49 %
Molarity (Concentration of Solutions)= M
M=
Moles of Solute =
Liters of Solution
Moles
L
solute = material dissolved into the solvent
In air , Nitrog en is the solvent and oxygen, carbon dioxide, etc.
are the solutes.
In sea water , Water is the solvent, and salt, mag nesium chloride, etc.
are the solutes.
In brass , Copper is the solvent (90%), and Zinc is the solute(10%)
Preparing a Solution - I
¥ Prepare a solution of Sodium Phosphate by
dissolving 3.95g of Sodium Phosphate into water
and diluting it to 300.0 ml or 0.300 l !
¥ What is the Molarity of the salt and each of the
ions?
¥ Na3PO4 (s) + H2O(solvent) = 3 Na+(aq) + PO4-3(aq)
Preparing a Solution - II
¥ Mol wt of Na3PO4 = 163.94 g / mol
¥ 3.95 g / 163.94 g/mol = 0.0241 mol Na3PO4
¥ dissolve and dilute to 300.0 ml
¥ M = 0.0241 mol Na3PO4 / 0.300 l = 0.0803 M
Na3PO4
¥ for PO4-3 ions = 0.0803 M
¥ for Na+ ions = 3 x 0.0803 M = 0.241 M
Make a Solution of Potassium Permanganate
Potassium Permanganate is KMnO4 and has a molecular mass
of 158.04 g / mole
Problem: Prepare a solution by dissolving 1.58 grams of KMnO 4
into sufficient water to make 250.00 ml of solution.
1.58 g KMnO4 x
Molarity =
1 mole K MnO 4
= 0.0100 moles KMnO 4
158.04 g KMnO 4
0.0100 moles KMnO4
= 0.0400 M
0.250 liters
Molarity of K+ ion = [K+] ion = [MnO 4-] ion = 0.0400 M
Dilution of Solutions
¥ Take 25.00 ml of the 0.0400 M KMnO4
¥ Dilute the 25.00 ml to 1.000 l - What is the
resulting Molarity of the diluted solution?
¥ # moles = Vol x M
¥ 0.0250 l x 0.0400 M = 0.00100 Moles
¥ 0.00100 Mol / 1.00 l = 0.00100 M
Chemical Equation Calc - III
Mass
Atoms (Molecules)
Avogadro s
Number
6.02 x 1023
Reactants
Molarity
moles / liter
Solutions
Molecules
Moles
Molecular
g/mol
Weight
Products
Calculating Mass of Solute from a Given
Volume of Solution
Volume (L) of Solution
Molarity M = (mol solute / Liters of solution) = M/L
Moles of Solute
Molar Mass (M) = ( mass / mole) = g/mol
Mass (g) of Solute
Calculating Amounts of Reactants and
Products for a Reaction in Solution
Al(OH)3 (s) + 3 HCl (aq)
Mass (g) of Al(OH)3
M (g/mol)
Moles of Al(OH)3
molar ratio
Moles of HCl
3 H2O(l) + AlCl3 (aq)
Given 10.0 g Al(OH)3, what volume of
1.50 M HCl is required to neutralize the
base?
10.0 g Al(OH)3
78.00 g/mol
= 0.128 mol Al(OH)3
0.128 mol Al(OH )3 x 3 moles HCl =
moles Al(OH)3
0.385 Moles HCl
Vol H Cl = 1.00 L HCl
x 0.385 Moles HCl
M ( mol/L)
1.50 Moles HCl
Volume (L) of HCl
Vol HCl = 0.256 L = 256 ml
Solving Limiting Reactant Problems in
Solution - Precipitation Problem - I
Problem: Lead has been used as a glaze for pottery for years, and can be
a problem if not fired properly in an oven, and is leachable from the
pottery. Vinegar is used in leaching tests, followed by Lead precipitated
as a sulfide. If 257.8 ml of a 0.0468 M solution of Lead nitrate is added
to 156.00 ml of a 0.095 M solution of Sodium sulfide, what mass of
solid Lead Sulfide will be formed?
Plan: It is a limiting-reactant problem because the amounts of two
reactants are given. After writing the balanced equation, determine the
limiting reactant, then calculate the moles of product. Convert moles of
product to mass of the product using the molar mass.
Solution: Writing the balanced equation:
Pb(NO3)2 (aq) + Na 2S (aq)
2 NaNO3 (aq) + PbS (s)
Volume (L) of
Pb(NO3)2
solution
Volume (L)
of Na2S
solution
Multiply by
M (mol/L)
Multiply by
M (mol/L)
Amount (mol)
of Pb(NO3)2
Amount (mol)
of Na2S
Molar Ratio
Molar Ratio
Amount (mol)
of PbS
Amount (mol)
of PbS
Choose the lower number
of PbS and multiply by
M (g/mol)
Mass (g) of PbS
Volume (L) of
Pb(NO3)2
solution
Volume (L)
of Na2S
solution
Multiply by
M (mol/L)
Devide by
equation
coeficient
Multiply by
M (mol/L)
Amount (mol)
of Pb(NO3)2
Smallest
Molar Ratio
Amount (mol) Devide by
of Na2S
equation
coeficient
Amount (mol)
of PbS
Mass (g) of PbS
Solving Limiting Reactant Problems in
Solution - Precipitation Problem - II
Moles Pb(NO 3)2 = V x M = 0.2578 L x (0.0468 Mol/L) =
= 0.012065 Mol Pb +2
Moles Na2S = V x M = 0.156 L x (0.095 Mol/L) = 0.01482 mol S -2
Therefore Lead Nitrate is the Limiting Reactant!
Calculation of product yield:
Moles PbS = 0.012065 Mol Pb+2x
1 mol PbS
= 0.012065 Mol Pb +2
1 mol Pb(NO3)2
0.012065 Mol Pb+2 = 0.012065 Mol PbS
0.012065 Mol PbS x 239.3 g PbS
1 Mol PbS
=
g PbS