The Mole & Stoichiometry
Dr. Ron Rusay
Fall 2007
© Copyright 2007 R.J. Rusay
Stoichiometry - Mole - Mass
Relationships in Chemical Reactions
Stoichiometry:
•Writing and Balancing Chemical Equations
•Calculating the amounts of Reactant and Product
Chemical Reactions
Atoms, Mass & Balance: eg. Zn + S -->
QuickTime™ and a
Cinepak decompressor
are needed to see this picture.
Chemical Equation
•
Representation of a chemical reaction:
_ C2H5OH + _ O2  _ CO2 + _ H2O
reactants
products
• C=2; H =5+1=6; O=2+1
•
C=1; H=2; O=2+1
1 C2H5OH + 3 O2  2 CO2 + 3 H2O
Chemical Equation
• C2H5OH + 3 O2  2 CO2 + 3 H2O
The equation is balanced and the reaction can be
completely stated as:
1
mole of ethanol reacts with 3 moles of
oxygen
 to produce 2 moles of carbon dioxide and 3
moles of water.
The Chemical Equation:
Mole & Masses
• C2H5OH + 3 O2  2 CO2 + 3 H2O
 46g
(1 mole) of ethanol reacts with 3 moles
of oxygen to produce 2 moles of carbon
dioxide and 3 moles of water.
 How
many grams of carbon dioxide and
water are respectively produced?
The Chemical Equation:
Moles & Masses
• C2H5OH + 3 O2  2 CO2 + 3 H2O
 46g
(1 mole) of ethanol requires 3 moles of
oxygen to produce 2 moles of carbon
dioxide and 3 moles of water.
 How
many grams of oxygen are needed to
react with all of the 46g (1 mole) of ethanol ?
 How
many grams of oxygen are needed to
react with 15.3g of ethanol in a 12oz. beer ?
Chemical Stoichiometry
 Stoichiometry
is the study of chemicals
and their quantities that are consumed
and produced in chemical reactions.
 It
quantitatively relates the behavior of
atoms and molecules to observable
chemical change and measurable mass
effects.
QUESTION
The fuel in small portable lighters is butane (C4H10). Suppose after
using such a lighter for a few minutes (perhaps to encourage your
favorite concert performer to play one more encore) you had used 1.0
gram of fuel. How many moles of butane would this be?
1.
2.
3.
4.
58 moles
0.077 moles
1.7  10–24 moles
0.017 moles
ANSWER
Choice 4 shows the answer for a proper grams to mole conversion.
You need to know the molar mass of butane: 4  12g/mol = 48 for
carbon + 10  1g/mol = 10.0 for hydrogen. Total = 48.0 + 10.0 =
58.0 g/mol. Next; 1.0 gram of butane  1 mol/58.0 g = 0.017 mol
Section 3.3: Molar Mass
QUESTION
Agriculturally, the following equation is important because it is used
to make millions of tons of urea. When the equation is balanced,
how many hydrogen atoms will be present on both sides of the
equation? Also, how many moles of NH3 would be needed to react
completely with 0.5 moles of CO2?
NH3 + CO2  CO(NH2)2 + H2O
1.
2.
3.
4.
Three; two
Three; one
Six; one
Six ; two
ANSWER
Choice 3 answers both stoichiometry questions correctly. By
placing a coefficient of 2 in front of NH3 six hydrogen atoms would
be represented on the left side of the equation. This equals the six
on the right side. Next, the balanced equation shows a 2:1
relationship between NH3 and CO2. So, if only 0.50 mole of CO2
were present then one mole of NH3 would be needed to maintain
the 2:1 reacting ratio.
0.50 mole CO2  2 mol NH3/1mol CO2 = 1.0 mol NH3
Section 3.7: Balancing Chemical Equations
Section 3.8: Stoichiometric Calculations: Amounts of Reactants
and Products
Combustion Analysis
CnHm + ( n + m ) O2 (g) ---> n CO(g) + m H2O(g)
2
2
Combustion Analysis Calculation
Ascorbic Acid ( Vitamin C )
• Combustion of a 6.49 mg sample in excess oxygen,
yielded 9.74 mg CO2 and 2.64 mg H2O
• Calculate it’s Empirical formula!
• C: 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2)
= 2.65 x 10-3 g C
• H: 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O)
= 2.92 x 10-4 g H
• Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg
= 3.54 mg O
Vitamin C: Calculation
(continued)
• 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 2.21 x 10-4
• C = 1.00 Multiply each by 3 = 3.00 = 3.0
• H = 1.32
= 3.96 = 4.0
• O = 1.00
= 3.00 = 3.0
C3H4O3
Combustion Problem
Problem: Erythrose (MM = 120.0 g/mol) is an important chemical
compound in chemical synthesis.
It contains Carbon, Hydrogen and Oxygen. Combustion
analysis of a 700.0 mg sample yielded 1.027 g CO2 and
0.4194 g H2O.
Calculate the molecular formula of erythrose and send your
answer to Dr. R.by e-mail before the next class. It’s all or nothing
… send just the formula. I trust your methodology.
Erythrose Solution
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
Erythrose Solution
0.2729 g C
Mass (g) of C = 1.027 g CO2 x 1 g CO2
= 0.2803 g C
0.1119 g H
Mass (g) of H = 0.4194 g H2O x 1 g H2O = 0.04693 g H
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 / cmpd = C4H8O4