Chapter 3 (Hill/Petrucci/McCreary/Perry
Stoichiometry: Chemical Calculations
This chapter deals with quantitative relationships in compounds and between compounds in
chemical reactions. These quantitative relationships are what we call stoichiometry.
So far, the formulas for chemical compounds give us specific stoichiometric information about
how to “make” that compound on the nanoscale. We will extend this idea to the macroscale in
this chapter.
Molecular Masses
Recall: periodic table gives us the average mass (u) of an atom of a specified element
molecular mass: the mass of one molecule in mass units (u); operationally, molecular mass is
the sum of the atomic masses of the atoms in a molecule, based upon the molecular formula for
the compound
Example. What is the molecular mass of water, H2O?
molecular mass = (2 atoms H)(1.00797 u/atom H)
+ (1 atom O)(15.9994 u/atom O)
molecular mass =18.015 u
Formula Masses
formula mass: the mass of one formula “unit” of an ionic compound (u); operationally, formula
mass is the same as the molecular mass for a molecular compound and is based upon the formula
of the ionic compound
Example.
What is the formula mass of water, Na2SO4?
formula mass = (2 atoms Na)(22.9898 u/atom Na)
+ (1 atom S)(32.066 u/atom S)
+ (4 atoms O)(15.9994 u/atom O)
formula mass = 142.043 u
Molecular and Formula Masses: See examples on pp. 76-78 (Hill)
From Nanoscale to Macroscale: The Mole
Need to scale from mass units (u) to grams ... how.? i.e. how many water molecules are needed
to go from 18.015 u to 18.015 g?
1 molecule = 18.015 u
# molecules = 18.015 g
Fact: 1 u = 1.66054 x 10-24 g (see p. 42, Hill)
# H2O molecules = (1 H2O molecule/ 18.015 u)
x (1 u H2O /1.66054 x 10-24 g H2O)
x 18.015 g H2O
# H2O molecules = 1/ 1.66054 x 10-24 = ??
The Mole
# H2O molecules = 6.022 x 1023 H2O molecules !!
This number is what we get, regardless of the molecule/compound used – it represents the
number of molecules required to scale up from u to grams, i.e. from molecular (nanoscale) to
gram quantities (macroscale).
We call the number of molecules (particles) we just calculated Avogadro’s number, after
Amedeo Avogadro.
Definition: 6.022 x 1023 particles = 1 mole (mol)
The Molar Mass
molar mass: the mass of 1 mol of any element or compound, expressed in grams; the molecular
mass of an element or compound, expressed in grams (g)
Units: g substance/mol substance
Example.
What is the molar mass of Fe?
1 mol Fe = 55.847 g Fe
See periodic table for the molar masses of monatomic elements.
Example.
Calculate the molar mass (M) of molecular oxygen, O2 gas. 1 mol O = 15.9994 g O
M = (1 mol O2)(2 mol O/1 mol O2)(15.9994 g O/1 mol O)
M = 31.9988 g O2 /mol O2
Figure 3.1, p. 79 (Hill)
A stoichiometric “picture” of the reaction of C with O2 to make CO2. Nanoscale and macroscale
views!
Moles
Thus, 1 mol Fe = 6.022 x 1023 Fe atoms = 55.847 g Fe
We can use these relationships as conversion factors:
(1) mols atoms
(2) mols grams
(3) atoms grams
Example 3.3, p. 80 (Hill). 1 mol Na = 22.99 g Na
Convert 0.0750 mol Na to grams Na:
g Na = (0.0750 mol Na)(22.99 g Na/1 mol Na) = 1.72 g Na
See also, Exercises 3.3A and 3.3B on p. 81
The Molar Mass: Molecules and Moles of Compounds
See also, figures/schematics in Section 3.3 on p. 81 (Hill)
The Molar Mass: Direct Calculation
molar mass: the mass of 1 mol of compound, expressed in grams; the molecular mass of a
compound, expressed in grams (g)
Units: g compound/mol compound
Example.
What is the molar mass of glucose, C6H12O6?
Solution. g/mol C6H12O6 = (6 mol C)(12.011 g C/mol C)
+ (12 mol H)(1.00797 g H/mol H)
+ (6 mol O)(15.9994 g O/mol O)
g/mol C6H12O6 = 72.066 + 12.09564 + 95.9964 = 180.158 g/mol
Look at Examples 3.4 and 3.5, pp. 82-83 (Hill)
Also, see Exercises 3.4 A, 3.4B, 3.5A and 3.5B
Mass Percent Composition – From Chemical Formulas
mass percent composition: the number of grams of any element in a compound per 100 g of
that compound
Units: %
Example.
Grain alcohol is ethanol, C2H6O – what is the %C, %H and %O in ethanol?
First calculate molar mass:
g C2H6O/mol = 2(12.011) +6(1.00797) + 1 (15.9994)
g C2H6O/mol = 24.022 + 6.0478 + 15.9994 = 46.069 g/mol
Mass Percent Composition from Chemical Formulas
%C = g C/mol C2H6O = 24.022 g C/ 46.069 g/mol) x 100
%C = 52.14352 = 52.143 %
%H = g H/mol C2H6O = 6.0478 g H/ 46.069 g/mol) x 100
%H = 13.127699 = 13.127 %
%O = g O/mol C2H6O = 15.9994 g O/ 46.069 g/mol) x 100
%O = 34.729210 = 34.729 %
See Examples and Exercises on pp. 84-86 (Hill)
Chemical Formulas – From Mass Percent
mass percent composition: the number of grams of any element in a compound per 100 g of
that compound
Units: %
We can determine the empirical (simplest whole-number) formula for a compound from its
mass percent composition data.
1. Assume 100 g of the compound
2. Mass % = g for each element
3. Convert each mass to mols of element
4. Divide by smallest (-er) number of mols
5. Find simplest mol ratio
See Example 3.9, p. 87 (Hill)
1. Assume 100 g of the compound
2. Mass % = g for each element
3. Convert each mass to mols of element
4. Divide by smallest (-er) number of mols
5. Find simplest mol ratio
6. Adjust mol ratios to closest whole-numbers by multiplying by a small integer
7. Use your common sense – don’t round off too casually!!
See and discuss Example 3.10, p.88
See Exercise 3.10A, p. 88 (Hill)
Chemical Formulas – From Mass Percent: Elemental Analysis by Combustion
If we can burn a hydrocarbon sample, we can collect the CO2 and the H2O produced from the C
and the H in the sample and use the data to compute %C and %H in the sample. A simple
combustion “train” is used to collect the carbon dioxide and the water produced.
Example. Calculate the mass % C in a 0.1000 g sample if combustion produces 0.1953 g CO2.
Calculate: CO2 44.010 g/mol and C 12.011 g/mol
?g C = (0.1953 g CO2)(1 mol CO2/44.010 g CO2)(1 mol C/1 mol CO2)(12.011 g C/1 mol C)
?g C = 0.05330 g
%C = (g C/g sample) x 100 = 0.05330 g/0.1000 g) x 100 == 53.30%
In Class Assignment
Work Exercises 3.12A and 3.12B, p. 91 with partner – to be turned in at end of class.
Chemical Reactions – Chemical Equations
chemical equation: shorthand notation for a chemical process with information about (1) both
reactant and and product stoichiometry as well as (2) reaction stoichiometry
General Form
A(state) + B(state) C(state) + D(state)
reactants
products
state = (g) gas; (l) liquid; (s) solid; (aq) water solution (aq = “aqueous”)
means “react(s) to produce”
Example: N2(g) + 3 H2(g) 2 NH3(g)
Nanoscale translation: 1 molecule of nitrogen gas reacts with 3 molecules of hydrogen gas to
produce 2 molecules of ammonia gas
Macroscale translation: 1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce
2 moles of ammonia gas
Example: 1 N2(g) + 3 H2(g) 2 NH3(g)
stoichiometric coefficients representing molecules (nanoscale) or moles (macroscale) are shown
in red
numerical subscripts, shown in blue, represent the stoichiometric quantities of each element
present: atoms of the element in 1 molecule or moles of the element in 1 mole of the compound
For each chemical species in the chemical equation, the product of the coefficient times the
subscript gives the number of atoms (or moles of atoms) of that element.
Balancing Chemical Equations
balanced chemical equation: a chemical equation with equal numbers of atoms of each type on
each side of the reaction arrow in the equation
Example. 4 Fe + 3 O2 2 Fe2O3
Balanced or Unbalanced?
CaCN2 + H2O + CaCO3 + NH3
Balanced !!
3 CaCN2 + H2O + CaCO3 + 2 NH3
Balancing by Inspection – A Four Step Process
1. Balance elements shown in just one compound on each side of the equation first!
2. Balance polyatomic ions as a group
3. Balance free element species (like O2 or Na) last
4. Use fractional coefficients if necessary – then convert to whole-number coefficients!
See and discuss Examples and Exercises, pp. 94-97
Working with Reaction Stoichiometry (pp. 97-103)
Consider the reaction shown on p. 98 in Hill:
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)
(balanced)
This tells us: 1 mol C3H8 = 5 mol O2 = 3 mol CO2 = 4 mol H2O
See Example 3.17
Work through Exercise 3.17A in class
Exercise 3.17A
(a) mols CO2 formed from 0.529 mol C3H8
1 mol C3H8 = 5 mol O2 = 3 mol CO2 = 4 mol H2O
mol CO2 = (0.529 mol C3H8)(1 mol C3H8/ 3 mol CO2)
mol CO2 = 1.59 mol
(b) mols H2O formed from 76.2 mol C3H8
1 mol C3H8 = 5 mol O2 = 3 mol CO2 = 4 mol H2O
mol H2O = (76.2 mol C3H8)(4 mol H2O/1 mol C3H8)
mol H2O = 305 mol H2O
(c) mols CO2 formed from 1.010 mol O2
1 mol C3H8 = 5 mol O2 = 3 mol CO2 = 4 mol H2O
mol CO2 = (1.010 mol O2)(3 mol CO2/5 mol O2)
(answer in back of Hill is wrong!)
mol CO2 = 0.606 mol CO2
Working with Reaction Stoichiometry and Mass (p. 99)
To calculate grams of B from grams of A for a balanced chemical equation,
grams B = (grams A)(1 mol A/g A)(mol B/mol A)(grams B/1 mol B)
See and work through Example 3.18, p. 99
See and work Exercises 3.18A and 3.18B in class
Working with Reaction Stoichiometry: “Limiting Reactants” (pp. 101-103)
When the amounts of two reactants are “given”
Must identify which reactant is present in the smaller stoichiometric amount! This reactant is
called the limiting reactant.
First, a simple example on the nanoscale: N2 + 3 H2 2 NH3
How many NH3 molecules could you make from 18 N2 and 57 H2?
Answer: 36 NH3 molecules – why?
1 mol N2 = 3 mol H2 = 2 mol NH3
How many NH3 molecules could you make from 18 N2 and 57 H2?
(1) # NH3 molecules from N2 = (18 N2)(2 NH3/1 N2) = 36
(2) # NH3 molecules from H2 = (57 H2)(2 NH3/3 H2) = 38
The limiting reactant is the one that produces the fewer number of product molecules (or
mols of product) when we do the calculations this way.
See and work through Example 3.20 (pp. 102-103)
Exercise 3.20A (p.103)
FeS(s) + 2 HCl(aq) FeCl2(aq) + H2S(g)
How many grams of H2S could you make from 10.2 g HCl and 13.2 g FeS?
Strategy: need amounts of both reactants in mols – then calculate mols of product both ways –
finally calculate grams of product from smaller number of mols of product possible
Calculate molar masses for each relevant chemical compound:
FeS 87.913 g/mol
HCl 36.461 g/mol
H2S 34.082 g/mol
Calculate mol H2S from both FeS and HCl:
mol H2S (HCl) = (10.2 g HCl)(1 mol HCl/ 36.461 g HCl)(1 mol H2S/2 mol HCl) = 0.140
mol H2S (FeS) = (13.2 g FeS)(1 mol FeS/ 87.913 g FeS)(1 mol H2S/1 mol FeS) = 0.150
We see that only 0.140 mols H2S can be made from the HCl, compared to 0.150 mols from FeS,
so HCl is the limiting reactant – finish calculating grams of H2S possible (the theoretical yield)
from this (0.140 mols)
grams H2S (HCl) = (0.140 mol H2S)(34.082 g H2S/mol H2S)
grams H2S (FeS) = 4.77 g H2S possible
The excess reagent is FeS – now, how much remains?
FeS
87.913 g/mol; HCl
36.461 g/mol; H2S
34.082 g/mol
Since 0.140 mols of H2S were made, we need to first calculate how much FeS was used ….
mols FeS used = (0.140 mol H2S)(1 mol FeS/1 mol H2S)
mols FeS used = 0.140 mols FeS
grams FeS used = (0.140 mols FeS)(87.913 g FeS/1 mol FeS) = 12.31 = 12.3 grams
Recall: How many grams of H2S could you make from 10.2 g HCl and 13.2 g FeS?
grams FeS remaining = 13.2 g – 12.3 g = 0.9 g “excess”
Yields of Chemical Reactions
The “yield” in a chemical reaction is the number of mols or grams of product theoretically
(calculation) or actually produced in the reaction when we run it in the lab on in the chemical
plant.
theoretical yield: the calculated (maximum possible) yield
actual yield: the laboratory yield
actual yield ≤ theoretical yield (note reasons!!)
impure starting materials; product loss in processing; side-reactions that don’t produce desired
product
% Yield = (actual yield/theoretical yield) x 100
Example. In the reaction in Exercise 3.21A (p. 105), the theoretical yield of the product,
isopentyl acetate (banana flavoring), was found to be 29.77 g. The actual yield obtained in the
lab by a student was 26.8 g. Calculate the % yield for the reaction.
Answer: % Yield = (actual yield/theoretical yield) x 100
% Yield = (26.8 g/29.77 g) x 100 = 90.0% yield
See also Example 3.21 (p. 105) and Exercises 3.21A and 3.21B
Reactions in Solutions and Solution Stoichiometry
To make a solution, we need something to dissolve ( a solute) and a dispersing medium in which
to dissolve it (a solvent).
Solute + Solvent D Solution
Example. A chemist dissolved 300 g of NaCl in water to make a pickle brine. The NaCl
“dissolved,” and the resulting solution contains NaCl as Na+ ions and as Cl- ions. The NaCl is the
solute and the water is the solvent.
Solution Concentration – Molar Concentration
Dilute and Concentrated. Solutions may have a relatively small amount of solute per given
amount of solvent (dilute solutions) or a lot of solute per given amount of solvent (concentrated
solutions).
We could specify how many grams of the solute we put into a certain number of grams or of mL
of solvent.
Another way!
Molarity or Molar Concentration.
molar concentration (M) = mols solute/liter solution
M = mol/L = mol/1000 mL
Example. A solution is prepared by dissolving 1.58 mols of glucose and diluting it to 2.00 L.
What is the molar concentration of glucose in this solution?
Answer:
M = mol glucose/L solution
M = (1.58 mol/2.00 L solution)
M = 0.790 mol/L = 0.790 M
Another Example. Calculate the molarity of a solution in which 2.69 g of C12H22O11 (table
sugar) is dissolved in water and diluted to a final volume of 225 mL of solution?
Answer: Since molarity is mol/L, we’ll need the molar mass of sucrose, C12H22O11 …. 342.3
g/mol!
Then,
M = (2.69 g/225 mL)(1 mL/10-3 L)(1 mol/ 342.3 g)
M = 3.49 x 10-2 mol/L = 3.49 x 10-2 M
You should review: Examples on pp. 108 and 109
Another type of problem: Molarity of solutions for which we know the density and the mass
percent composition of the solute … like Example 3.25, p. 109
Here, we use the density, percent composition and the solute molar mass as conversion factor:
M = (g solute/100 g solution)(g solution/mL solution)(mol solute/g solute)
Exercise 3.25A, p. 109
A bottle of formic acid is 90.0% (w/w) formic acid, HCOOH, and has a density of 1.20 g/mL).
What is the molarity of formic acid in this solution?
Solution. First, translate the information given …
90.0% formic acid: 90.0 g formic acid/100 g solution
1.20 g/mL: 1.20 g solution/1 mL solution
Get molar mass of HCOOH: 46.03 g formic acid/mol
Information we will need.
90.0% formic acid: 90.0 g formic acid/100 g solution
1.20 g/mL: 1.20 g solution/1 mL solution
molar mass of HCOOH: 46.03 g formic acid/mol
Solution. M = mol formic acid/L solution
M = (90.0 g formic acid/100 g solution)(1.20 g solution/1 mL solution)(1 mL/10-3 L)
x (46.03 g formic acid/mol)
M = 23.5 mol formic acid/L = 23.5 M
Solution Concentration – Dilution of Solutions
If we add solvent to an existing solution, what happens?
Same amount of solute spread out over more solvent .. how does this affect the concentration of
the solute … more concentrated or less concentrated? Less concentrated!!
mols solute in original soln. = mols solute in dil. soln.
mols = molarity x volume
…
Why?
Because mol/L x L = mol
or, mol = M x V
Then for dilution of solutions,
M(conc) x V(conc) = M(dilute) x V(dilute) or, M1 x V1 = M2 x V2
You should review: Example 3.26 on pp. 110 and 111
Note: a stock solution is a concentrated version of the solution we take for dilution
Exercise 3.26A. How many mL of 10.15 M stock NaOH solution are needed to make 15.0 L of
0.315 M NaOH?
M1 x V1 = M2 x V2
M1 = 10.15 M V1 = ?
M2 = 0.315 M
V2 = 15.0 L
Substituting: (10.15 M)(V1) = (0.315 M)(15.0 L)
V1 = 0.466 L x (1 mL/10-3 L) = 466 mL
Chemical Reactions in Solutions – Putting It All Together
Consider a balanced equation, where there are species designated as (aq), like in Example 3.27,
p. 112
CaCO3(s) + 2 HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)
The Question: How many g of CaCO3 are consumed by 225 mL of 3.25 M HCl? We will need
the molar mass of CaCO3 …. 106.00 g/mol
Solution. g CaCO3 = (225 mL)(3.25 mol HCl/1000 mL)(1 mol CaCO3/2 mol HCl)
x (106.00 g CaCO3/mol CaCO3)
g CaCO3 = 36.6 g
Chemical Reactions in Solutions – Another Example from the Book
Consider a scenario like Exercise 3.27A, p. 112
NaHCO3(s) + HCl(aq) NaCl(aq) + H2O(l) + CO2(g)
The Question: (a) How many g of CO2 are produced by the reaction between 235 mL of 1.22 M
HCl and 175 mL of 1.55 M NaHCO3 ? Molar mass of CO2 = 44.01 g/mol
First, we have to approach this like a limiting reactant problem, which it is! calculate possible
mol CO2 from each reactant …
mol CO2 from NaHCO3 = (175 mL)(1.55 mol NaHCO3/L)
x (10-3 L/mL)(1 mol CO2/1 mol NaHCO3) = 0.271 mol CO2
mol CO2 from HCl = (235 mL)(1.22 mol HCl/L)(10-3 L/mL)(1 mol CO2/1 mol HCl)
= 0.287 mol CO2
Therefore, we can only make 0.271 mol CO2 !
NaHCO3(s) + HCl(aq) NaCl(aq) + H2O(l) + CO2(g) 44.01 g/mol
Since we can make only 0.271 mol CO2 ,
g CO2 = (0.271 mol CO2)(44.01 g CO2/mol CO2)
g CO2 = 11.9 g CO2
(b) What is the molarity of the NaCl produced? Assume that the total solution volume after
reaction is 175 + 235 = 410 mL = 0.410 L
or,
(1) mol NaCl = (0.175 L)( 1.55 mol NaHCO3/L)(1 mol NaCl/1 mol NaHCO3)
= 0.271 mol NaCl
M = mol NaCl/L = (0.271 mol NaCl/0.410 L) = 0.661 M NaCl
Why did we base our calculation on NaHCO3 instead of HCl? Because NaHCO3 was also the
limiting reagent for NaCl!