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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 analysis shows that it contains
40.00 mass % C, 6.719 mass % H, and 53.27 mass % O.
(a) Determine the empirical formula of glucose.
(b) Determine the Molecular formula.
Plan: We are only given mass %, and no weight of the compound so we
will assume 100g of the compound, and % becomes grams, and
we can do as done previously with masses of the elements.
Solution:
Mass Carbon =
Mass Hydrogen =
Mass Oxygen =
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 analysis shows that it contains
40.00 mass % C, 6.719 mass % H, and 53.27 mass % O.
(a) Determine the empirical formula of glucose.
(b) Determine the Molecular formula.
Plan: We are only given mass %, and no weight of the compound so we
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 Hydrogen = 6.719% x 100g/100% = 6.719g H
Mass Oxygen = 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 Grams of Elements to moles:
Moles of C =
Moles of H =
Moles of O =
Constructing the preliminary formula:
Converting to integer subscripts, ÷ all subscripts by the smallest:
Determining the Molecular Formula from
Elemental Composition and Molar Mass - II
Converting from Grams 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, ÷ all subscripts by the smallest:
C 3.33/3.33 H 6.667 / 3.33 O3.33 / 3.33 = C1H2O1 = CH2O
Determining the Molecular Formula from
Elemental Composition and Molar Mass - III
(b) Determining the Molecular Formula:
The formula weight of the empirical formula is:
Whole-number multiple =
M compound
empirical formula mass
Therefore the Molecular Formula is:
=
Determining the Molecular Formula from
Elemental Composition and Molar Mass - III
(b) Determining the Molecular Formula:
The formula weight 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 g/mol
Whole-number multiple =
=
M of the compound
empirical formula mass
=
180.16 g/mol
emp.mol
= 6.00 = 6
30.03 g/emp. mol
mol
Therefore the Molecular Formula is:
C1x6H2x6O1x6 =
C6H12O6
Adrenaline is a very Important
Compound in the Body - I
• Analysis gives :
•
C = 56.8 %
•
H = 6.50 %
•
O = 28.4 %
•
N = 8.28 %
• Calculate the
Empirical Formula !
Adrenaline - II
• Assume 100g!
• C=
• H=
• O=
• N=
Divide by smallest
•
•
•
•
C=
H=
O=
N=
=>
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 smallest (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
Fig. 3.5
m
CnHm + (n+ )O2(g)
2
m
n CO2(g) +
H2O(g)
2
Ascorbic acid ( Vitamin C ) - I contains
LINUS PAULING
only C , H , and O
• Upon combustion in excess oxygen, a 6.49 mg
sample yielded 9.74 mg CO2 and 2.64 mg H2O.
• Calculate its Empirical formula!
• Mass C:
• Mass H:
• Mass Oxygen =
Ascorbic acid ( Vitamin C ) - I
contains only C , H , and O
• Upon combustion in excess oxygen, a 6.49 mg
sample yielded 9.74 mg CO2 and 2.64 mg H2O.
• Calculate its Empirical formula!
• Mass C: 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2)
= 2.65 x 10-3 g C
• Mass H: 2.64 x10-3g H2O x (2.016 g H/18.02 gH2O)
= 2.95 x 10-4 g H
• Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg
= 3.54 mg O = 3.54x10-3 g O
Vitamin C combustion - II
• Moles C =
• Moles H =
• Moles O =
• Divide each by smallest:
• Moles C =
• Moles H =
• Moles O =
Vitamin C combustion - II
• C = 2.65 x 10-3 g C / ( 12.01 g C / mol C ) =
= 2.21 x 10-4 mol C
• H = 0.295 x 10-3 g H / ( 1.008 g H / mol H ) =
= 2.92 x 10-4 mol H
• O = 3.54 x 10-3 g O / ( 16.00 g O / mol O ) =
= 2.21 x 10-4 mol O
•
•
•
•
Divide each by smallest (2.21 x 10-4 ):
C = 1.00
Multiply each by 3: C = 3.00 = 3.0
H = 1.32
(to get ~integers)
H = 3.96 = 4.0
O = 1.00
O = 3.00 = 3.0
C3H4O3
Determining a Chemical Formula from
Combustion Analysis - I
Problem: Erthrose (M = 120 g/mol) is an important chemical
compound used often as a starting material in chemical
synthesis, and contains Carbon, Hydrogen, and Oxygen.
Combustion analysis of a 700.0 mg sample yielded:
1.027 g CO2 and 0.4194 g H2O.
Plan: We find the masses of Hydrogen and Carbon using the mass
fractions of H in H2O, and C in CO2. The mass of Carbon and
Hydrogen are subtracted from the sample mass to get the mass
of Oxygen. 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 CO2 =
Mass fraction of H in H2O =
Calculating masses of C and H:
Mass of Element = mass of compound x mass fraction of element
Determining a Chemical Formula from
Combustion Analysis - II
Calculating the mass fractions of the elements:
mol C x M of C
Mass fraction of C in CO2 =
=
mass of 1 mol CO2
= 1 mol C x 12.01 g C/ 1 mol C = 0.2729 g C / 1 g CO2
44.01 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 H2O =
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 =
Mass (g) of H =
Calculating the mass of O:
Calculating moles of each element:
C=
H=
O=
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
Mass (g) of H = 0.4194 g H2O x
= 0.04693 g H
1 g H2 O
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 weight = 30 g / formula
120 g /mol / 30 g / formula = 4 formula units / cpd = C4H8O4
Chemical Equations
Qualitative Information:
Reactants
Products
States of Matter: (s) solid
(l) liquid
(g) gaseous
(aq) aqueous
2 H2 (g) + O2 (g)
But also Quantitative Information!
2 H2O (g)
Methane/oxygen reaction
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
Table 3.2 (P 66) Information Conveyed by the
Balanced Equation for the Combustion of Methane
Reactants
Products
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
1 molecule CH4 +
2 molecules of O2
1 molecule CO2 +
2 molecules H2O
1 mol CH4 molecules +
2 mol O2 molecules
1 mol CO2 molecules +
2 mol H2O molecules
6.022 x 1023 CH4 molecules +
2 x (6.022 x 1023) O2 molecules
16g CH4 + 2 (32g) O2
80g reactants
6.022 x 1023 CO2 molecules +
2 x (6.022 x 1023) H2O molecules
44g CO2 + 2 (18g) H2O
80g products
Flare in a natural
gas field
STOP
this
waste!
Eric McFarland is the President and
C.E.O. for GRT, Inc., a Professor of
Chemical Engineering at the University
of California Santa Barbara. He holds a
Ph.D. from the Massachusetts Institute
of Technology, as well as an M.D. from
Harvard Medical School. From 19961998, he was a founding Director of
Source: Stock
BostonTechnologies
Symyx
How to Balance Equations
• Mass Balance (or Atom Balance)- same number of
each element on each side of the equation:
(1) start with simplest element (or largest molecule)
(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)
• Make charges balance. (Remove “spectator” ions.)
Ca2+ (aq) + 2 OH- (aq) + Na+
Ca(OH)2 (s) + Na+
DEMO: Methane Bubbles
Information Contained in a Balanced Equation
Viewed in
terms of:
Reactants
Products
2 C2H6 (g) + 7 O2 (g) = 4 CO2 (g) + 6 H2O(g) + Energy
Molecules
Amount (mol)
Mass (amu)
Mass (g)
Total Mass (g)
Information Contained in a Balanced Equation
Viewed in
terms of:
Reactants
Products
2 C2H6 (g) + 7 O2 (g) = 4 CO2 (g) + 6 H2O(g) + Energy
Molecules 2 molecules of C2H6 + 7 molecules of O2 =
4 molecules of CO2 + 6 molecules of H2O
Amount (mol) 2 mol C2H6 + 7 mol O2 = 4 mol CO2 + 6 mol H2O
60.14 amu C2H6 + 224.00 amu O2 =
176.04 amu CO2 + 108.10 amu H2O
Mass (amu)
Mass (g)
60.14 g C2H6 + 224.00 g O2 = 176.04 g CO2 + 108.10 g H2O
Total Mass (g)
284.14g = 284.14g
Balancing Chemical Equations - I
Problem: The hydrocarbon hexane is a component of Gasoline 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 (C6H14).
Plan: Write the skeleton equation, converting the words into chemical
compounds, with blanks before each compound. Begin the
element balance, putting 1 on the most complex compound first,
and save oxygen until last!
Solution:
Balance any elements that this forces:
Balancing Chemical Equations - I
Problem: The hydrocarbon hexane is a component of Gasoline 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 (C6H14).
Plan: Write the skeleton equation, converting the words into chemical
compounds, with blanks before each compound. Begin the
element balance, putting 1 on the most complex compound first,
and save oxygen until last!
Solution:
C6H14 (l) +
O2 (g)
CO2 (g) +
H2O(g) + Energy
Begin with one C6H14 molecule which says that we will get 6 CO2’s!
1 C6H14 (l) +
O2 (g)
6 CO2 (g) +
H2O(g) + Energy
Balancing Chemical Equations - II
Next balance H atoms:
1 C6H14 (l) +
O2 (g)
6 CO2 (g) +
H2O(g) + Energy
Balance O atoms last:
C6H14 (l) +
C6H14 (l) +
O2 (g)
O2 (g)
CO2 (g) +
H2O(g) + Energy
CO2 (g) +
H2O(g) + Energy
Balancing Chemical Equations - II
The H atoms in the hexane will end up as H2O, and we have 14
H atoms, and since each water molecule has two H atoms, we will get
a total of 7 water molecules.
1 C6H14 (l) +
O2 (g)
6 CO2 (g) + 7 H2O(g) + Energy
Since oxygen atoms only come as diatomic molecules
(two O atoms, O2),we must have even numbers of oxygen atoms on the
product side. We do not since we have 7 water molecules! Therefore
multiply the hexane by 2, giving a total of 12 CO2 molecules, and
14 H2O molecules.
2 C6H14 (l) +
O2 (g)
12 CO2 (g) + 14 H2O(g) + Energy
This now gives 12 O2 from the carbon dioxide, and 14 O atoms from the
water, which will be another 7 O2 molecules for a total of 12+7 =19 O2 !
2 C6H14 (l) + 19 O2 (g)
12 CO2 (g) +14 H2O(g) + Energy
Molecular model: Balanced equation
C2H6O(aq) + 3 O2(g)
2 CO2 (g) + 3 H2O(g) + Energy
(NH4)2Cr2O7(s) → _ Cr2O3(s) + _ N2(g) + _ H2O(g)
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 Al2S3: a) How many moles
of water are required for the reaction? b) What mass of H2S & Al(OH)3
would be formed?
2 Al(OH)3 (s) + 3 H2S(g)
Al2S3 (s) + 6 H2O(l)
Plan: Calculate moles of Aluminum Sulfide using its molar mass, then
from the equation, calculate the moles of Water, and then the moles of
Hydrogen Sulfide, and finally the mass of Hydrogen Sulfide using
it’s molecular weight.
Solution:
a) molar mass of Aluminum Sulfide =
moles Al2S3 =
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 Al2S3: a) How many moles
of water are required for the reaction? b) What mass of H2S & Al(OH)3
would be formed?
2 Al(OH)3 (s) + 3 H2S(g)
Al2S3 (s) + 6 H2O(l)
Plan: Calculate moles of Aluminum Sulfide using its molar mass, then
from the equation, calculate the moles of Water, and then the moles of
Hydrogen Sulfide, and finally the mass of Hydrogen Sulfide using
it’s molecular weight.
Solution:
a) molar mass of Aluminum Sulfide = 150.17 g / mol
moles Al2S3 =
65.80 g Al2S3
= 0.4382 moles Al2S3
150.17 g Al2S3/ mol Al2S3
Calculating Reactants and Products in a
Chemical Reaction - II
a) cont.
H2O: 0.4382 moles Al2S3 x ___ moles H2O
1 mole Al2S3
b)H2S: 0.4382 moles Al2S3 x ___ moles H2S
1 mole Al2S3
molar mass of H2S =
mass H2S =
Al(OH)3: 0.4382 moles Al2S3 x
molar mass of Al(OH)3 =
mass Al(OH)3 =
Calculating Reactants and Products in a
Chemical Reaction - II
a) cont.
0.4382 moles Al2S3 x
6 moles H2O = 2.629 moles H2O
1 mole Al2S3
b) 0.4382 moles Al2S3 x 3 moles H2S = 1.314 moles H2S
1 mole Al2S3
molar mass of H2S = 34.09 g / mol
mass H2S = 1.314 moles H2S x 34.09 g H2S = 44.81 g H2S
1 mole H2S
0.4382 moles Al2S3 x 2 moles Al(OH)3 = 0.8764 moles Al(OH)3
1 mole Al2S3
molar mass of Al(OH)3 = 78.00 g / mol
mass Al(OH)3 = 0.8764 moles Al(OH)3 x 78.00 g Al(OH)3 =
1 mole Al(OH)3
= 68.36 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) + Ca3(PO4)2 (s)
Given: 15.5 g P4 and sufficient KClO3 , H2O and Ca(OH)2. What mass
of Calcium Phosphate could be formed?
Plan: (1) Calculate moles of P4.
(2) Use molar ratios to get moles of Ca3(PO4)2.
(3) Convert the moles of product back into mass by using the
molar mass of Calcium Phosphate.
DEMO: Match reaction- Somorjai
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - II
Solution:
moles of P4 =
For Reaction #1 [ 4 P4 (s) + 10 KClO3 (s)
For Reaction #2 [ 1 P4O10 (s) + 6 H2O (l)
For Reaction #3 [ 2 H3PO4 + 3 Ca(OH)2
moles Ca3(PO4)2 =
moles P4 x
4 P4O10 (s) + 10 KCl (s) ]
4 H3PO4 (aq) ]
1 Ca3(PO4)2 + 6 H2O]
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - II
Solution:
1 mole P4
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)
For Reaction #2 [ 1 P4O10 (s) + 6 H2O (l)
For Reaction #3 [ 2 H3PO4 + 3 Ca(OH)2
4 P4O10 (s) + 10 KCl (s) ]
4 H3PO4 (aq) ]
1 Ca3(PO4)2 + 6 H2O]
0.1251 moles P4 x 4 moles P4O10 x 4 moles H3PO4 x 1 mole Ca3(PO4)2
4 moles P4
1 mole P4O10
2 moles H3PO4
= 0.2502 moles Ca3(PO4)2
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - III
Molar mass of Ca3(PO4)2 = 310.18 g mole
Mass of Ca3(PO4)2 product =
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - III
Molar mass of Ca3(PO4)2 = 310.18 g mole
310.18 g Ca3(PO4)2
mass of product = 0.2502 moles Ca3(PO4)2 x
=
1 mole Ca3(PO4)2
= 77.61 g Ca3(PO4)2
Balanced
reaction!
Defines
stoichiometric
ratios!
Unbalanced (i.e., non-stoichiometric)
mixture!
Limited by syrup!
Molecular model: N2 molecules require
3H2 molecules for the reaction
N2 (g) + 3 H2 (g)
2 NH3 (g)
Figure 3.9: Hydrogen and Nitrogen reacting to
form Ammonia, the Haber process
Limiting Reactant Problem:
A Sample Problem
Problem: A fuel mixture used in the early days of rocketry is composed
of two liquids, hydrazine (N2H4) and dinitrogen tetraoxide (N2O4). They
ignite on contact ( hypergolic!) to form nitrogen gas and water vapor.
How many grams of nitrogen gas form when exactly 1.00 x 102 g N2H4
and 2.00 x 102 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 gas.
Solution:
2 N2H4 (l) + N2O4 (l)
3 N2 (g) + 4 H2O (g) + Energy
Sample Problem 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
2g
1.00
x
10
Moles N2H4 =
32.05 g/mol
=
2.00 x 102 g
Moles N2O4 =
=
92.02 g/mol
Divide by coefficients to see how many moles per mole as written, to
decide which is limiting reagent.
N2 yield in moles =
Mass of N2 =
Sample Problem 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
2g
1.00
x
10
Moles N2H4 =
32.05 g/mol
= 3.12 moles N2H4
2.00 x 102 g
Moles N2O4 =
= 2.17 moles N2O4
92.02 g/mol
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 = 131 g N2
Acid - Metal Limiting Reactant - I
• 2Al(s) + 6HCl(g)
2AlCl3(s) + 3H2(g)
Consider the reaction above. If we react 30.0 g Al and
20.0 g HCl, how many moles of aluminum chloride
will be formed?
• 30.0 g Al
• 20.0g HCl
• Limiting reactant = one w/ fewer “equivalents” =
Acid - Metal Limiting Reactant - I
• 2Al(s) + 6HCl(g)
2AlCl3(s) + 3H2(g)
Consider the reaction above. If we react 30.0 g Al and
20.0 g HCl, how many moles of aluminum chloride
will be formed?
• 30.0 g Al / (26.98 g Al / mol Al) = 1.11 mol Al
1.11 mol Al / 2 = 0.555 equivalents Al
• 20.0g HCl / (36.5gHCl / mol HCl) = 0.548 mol HCl
O.548 mol HCl / 6 = 0.0913 equivalents HCl
• HCl is smaller therefore the Limiting reactant!
Acid - Metal Limiting Reactant - II
• Since 6 moles of HCl yield 2 moles of AlCl3
________ moles of HCl will yield:
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 x (2 moles of AlCl3 / 6 mol HCl)
= 0.183 mol of AlCl3
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 given for more than one reactant!
2) Calculate moles of each reactant!
3) Divide the moles of each reactant by stoic. coefficient (a,b,c etc...)!
4) Whichever 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....)!
Ostwald Process Limiting Reactant Problem
• What mass of NO could be formed by the reaction 30.0g of
ammonia gas and 40.0g of oxygen gas w/ the rxn below?
4NH3 (g) + 5 O2 (g)
4NO(g) + 6 H2O(g)
• 30.0g NH3
• 40.0g O2
• _______ fewer, therefore ______ is the Limiting Reagent!
• Moles NO formed =
• Mass NO =
Ostwald Process Limiting Reactant Problem
• What mass of NO could be formed by the reaction 30.0g of
ammonia gas and 40.0g of oxygen gas w/ the rxn below?
4NH3 (g) + 5 O2 (g)
4NO(g) + 6 H2O(g)
• 30.0g NH3 / (17.0g NH3/mol NH3) = 1.76 mol NH3
1.76 mol NH3 / 4 = 0.44 eq. NH3
• 40.0g O2 / (32.0g O2 /mol O2) = 1.25 mol O2
1.25 mol O2 / 5 = 0.25 eq O2
• Oxygen fewer, therefore oxygen is the Limiting Reagent!
• 1.25 mol O2 x 4 mol NO = 1.00 mol NO
5 mol O2
• mass NO = 1.00 mol NO x 30.0 g NO
1 mol NO
= 30.0 g NO
Flowchart : Solving a stoichiometry problem
involving masses of reactants
This mole ratio is called the
ratio of “stoichiometric coefficients”
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 =
or
Actual Yield (mass or moles)
x 100%
Theoretical Yield (mass or moles)
Actual yield = Theoretical yield x (% Yield / 100%)
Percent Yield Problem:
Problem: Given the chemical reaction between Iron and water to form
the iron oxide, Fe3O4 and hydrogen gas given below. If 4.55 g of iron is
reacted with sufficent water to react all of the iron to form rust, what is
the percent yield if only 6.02 g of the oxide are formed?
Plan: Calculate the theoretical yield and use it to calculate the percent
yield, using the actual yield.
Solution:
3 Fe + 4 H O
Fe O + 4 H
(s)
2
(l)
3
Moles Fe =
Theoretical moles Fe3O4 =
Theoretical mass Fe3O4 =
Percent Yield = Actual Yield x 100% =
Theoretical Yield
4 (s)
2 (g)
Percent Yield Problem:
Problem: Given the chemical reaction between Iron and water to form
the iron oxide, Fe3O4 and hydrogen gas given below. If 4.55 g of iron is
reacted with sufficent water to react all of the iron to form rust, what is
the percent yield if only 6.02 g of the oxide are formed?
Plan: Calculate the theoretical yield and use it to calculate the percent
yield, using the actual yield.
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 Fe = 0.0815 mol Fe
55.85 g Fe
mol Fe
0.0815 mol Fe x 1 mol Fe3O4 = 0.0272 mol Fe3O4
3 mol Fe
0.0272 mol Fe3O4 x 231.55 g Fe3O4 = 6.30 g Fe3O4
1 mol Fe3O4
Percent Yield = Actual Yield x 100% = 6.02 g Fe3O4 x 100% =
6.30 g Fe3O4
Theoretical Yield
95.6 %
Percent Yield / Limiting Reactant Problem - I
Problem: Ammonia is produced by the Haber process using nitrogen
and hydrogen Gas. If 85.90g of nitrogen are reacted with
21.66 g hydrogen and the reaction yielded 98.67 g of
ammonia what was the percent yield of the reaction.
N2 (g) + 3 H2 (g)
2 NH3 (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 N2 =
moles H2 =
eqs N2 =
eqs H2 =
Percent Yield / Limiting Reactant Problem - I
Problem: Ammonia is produced by the Haber process using nitrogen
and hydrogen Gas. If 85.90g of nitrogen are reacted with
21.66 g hydrogen and the reaction yielded 98.67 g of
ammonia what was the percent yield of the reaction.
N2 (g) + 3 H2 (g)
2 NH3 (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
moles N2 = 85.90 g N2 = 3.066 mol N2 to get eqs. of each:
28.02 g N2
3.066 g N2 = 3.066eq
1 mole N2
1
moles H2 = 21.66 g H2 = 10.74 mol H2
2.016 g H2
10.74 g H2 = 3.582eq
1 mole H2
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:
mol NH3 = 3.066 mol N2 x
mass NH3 =
Percent Yield =
Percent Yield =
Actual Yield x 100%
Theoretical Yield
98.67 g NH3
g NH3
x 100% =
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 NH3 = 6.132 mol NH
3.066 mol N2 x
3
1 mol N2
(Theoretical Yield)
17.03 g NH3
= 104.427 g NH3
1 mol NH3
(Theoretical Yield)
Actual Yield x 100%
Percent Yield =
Theoretical Yield
6.132 mol NH3 x
Percent Yield =
98.67 g NH3
104.427 g NH3
x 100% = 94.49 %