Download 3.1 Balancing Chemical Equations

Chapter 3: Formulas, Equations and Moles
Chapter Goals
Be able to:
write and balance chemical equations
calculate molar mass
convert between grams, moles, and formula units
determine limiting reagents and theoretical yields of reactions in
grams and moles
know to prepare solutions of known molarity by dissolving a solid
in a solvent, and by diluting a more concentrated solution
convert between molarity, volume, moles, and grams for a
solution
understand titrations
determine the percent composition and empirical formula of a
compound
use combustion analysis data to obtain the empirical formula of a
compound containing carbon, hydrogen, and one other element.
determine the molecular formula of a compound from empirical
formula and molar mass.
Balancing Procedure: (4 Steps)
3.1 Balancing Chemical Equations
Balanced Equation
the numbers and types of atoms on both sides of the
arrow (equations) are the same
- law of mass conservation
reactants
limestone
Calcium carbonate
products
quicklime + gas
calcium oxide + carbon dioxide
CaCO3(s)
CaO(s) + CO2(g)
Formula unit
one unit; whether atom, ion or molecule corresponding to
a given formula
Now balance oxygen
1) Write the unbalanced equation using the correct formula for
each reactant and product
Example:
C5H12 + O2
so, need an
CO2 + H2O
2) Find suitable coefficients
numbers placed before formulas to indicate how many
formula units of each substance are required to balance
the equation
Example (cont.): Let s try beginning with carbon
C5H12 + O2
CO2 + H2O
C5H12 + O2
CO2 +
Dr. C. Bottaro Chem 1010 Fall 2005
H2O
C5H12 +
in front of O2 on LHS.
O2
CO2 +
H2O
3.) If necessary, reduce the coefficients to their smallest
whole-number values through division by a common
divisor
Example (cont.): the ratio
is the smallest ratio.
4.) Check you answer by making sure that the numbers
and kinds of atoms are the same on both sides of the
equation (mass balance)
LS
RS
C
C
H
H
O
O
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3.2 Chemical Symbols on a Different Level
Examples:
Balance the following equations:
C6H12O6
C2H6O +
Fe + O2
Fe2O3
NH3 + Cl2
N2H4 +
KClO3 + C12H22O11
Chemical symbols represent both a microscopic and
macroscopic level.
CO2
Microscopic level - chemical formulas represent the behavior of (a
few) atoms and molecules.
Macroscopic level - chemical formulas and chemical equation
represent the large-scale (gram scale) behavior of atoms and
molecules that give rise to observable properties.
Still in the same ratio but instead of 2 atoms forming one
molecule it might be 2 million atoms forming 1 million molecules
NH4Cl
KCl +
CO2 +
H2 O
3.3 Avogadro s Number and the Mole
balanced equation
demonstrates the ratio of atoms or molecules reactants required
to form certain products
Needed to deal with macroscopic behavior of chemical requires
that the number (or coefficient) ratio from balanced equation be
converted to a mass ratio.
Molecular mass (MM, molecular weight, MW, formula mass)
considered average mass of a substance because it doesn t
separate the mass of the individual isotopes from each other
The mass of a molecule is just the sum of the masses of the
atoms (taken from the periodic table) making up the molecule.
e.g.
Formula mass sum of atomic masses of all atoms in a formula
unit of any compound, molecule or ion
- doesn t necessarily refer to a molecule
Dr. C. Bottaro Chem 1010 Fall 2005
Macroscopic behaviour is what we observe when we
observe physical properties like melting point, phase (i.e.
solid, liquid or gas)
Proportions in molecules, ions and reactions are the same
on the macro- and microscopic (atomic) level
Equal numbers of different molecules will always have a mass
ratio equal to their molecular or formula mass ratio.
Consider this reaction
C2H4 (g) + HCl (g)
C2H5Cl (g)
Mass ratio
molecular masses
Whole number ratio 1
:
1
(coefficients from balanced equation 1 C2H4 reacts with 1 HCl)
This lead to the invention of the mole concept.
Mole (mol) - SI base unit for measuring an amount of a substance
based on the amount a particular substance that would result in the
mass in grams will be equal to the atomic, molecular or formula
weight in amu.
Molar mass the mass of one mole of any substance and is
equal to the formula mass in grams
e.g. Cl atomic mass = 35.483 amu molar mass = 35.483 g/mol
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Avogadro s Number (NA)
Practice Problems
One mole of any substance contains 6.02 x 1023 units of that
substance.
D-alanine, an amino acid used by organisms to make proteins,
is represented below. (red = O; gray = C; blue = N; white = H)
Avogadro s Number: NA,
represents the
of an
element in a sample whose mass in grams is numerically equal to
the atomic mass of the element
is the numerical value assigned to the unit, 1 mole.
NA = 6.022 x 1023
also applies to molecules and ions.
e.g.
1 mol HCl =
molecules with molar mass =
1 mol NaCl =
molecules with molar mass =
Write the formula for D-alanine.
Calculate its molar mass.
g
g
How many molecules of D-alanine is there in 0.5 mol?
3.4 Stoichiometry: Chemical Arithmetic
the coefficients in a balanced equation indicate the numbers
of moles of substances in a reaction
essential to be able to convert from moles to mass and visa
versa
General Relationship:
m=M x n
where: m = mass in grams
MM = Molar mass (g/mol or g mol-1)
n = amount in moles
Mass to Moles and Moles to Mass Interconversion
Moles to Mass
3.5 Yields of Chemical Reactions
actual (or experimental) yield
the quantity of material you
theoretical yield
maximum amount of product that
from a chemical reaction
obtain in the lab
be obtained
Note: the ACTUAL YIELD should always be less than the
THEORETICAL YIELD
percent yield =
x
100%
Mass to Moles
the efficiency of a chemical reaction as a ratio of actual
yield to theoretical yield
# of mol x grams = # of grams
1 mol
# of grams x 1 mol = moles
grams
Dr. C. Bottaro Chem 1010 Fall 2005
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Conversions between Moles and Grams
Practice Problem:
If 10.0 grams of CH4 and 20.0 g of
Cl2 are used in the reaction
CH4 + 4Cl2
CCl4 + 4HCl
1) Which one is the limiting
reagent?
2) Which is in excess?
3) What is the theoretical yield of
carbon tetrachloride?
a) in moles
b) in grams
4) How much HCl in grams would
be produced if the percent yield
for this reaction was 65%?
Limiting Reagents
3.6 Reactions with Limiting Amounts
of Reactants
limiting reagent (reactant)
excess reagent (or reactant)
supplied in excess to ensure complete reaction (consumption)
of limiting reagent
Steps for determining limiting reagent
1) calculate the amount of product that would be formed if the
first reactant was completely consumed
2) repeat this calculation for the second reactant
i.e. calculate how much product would be formed if all that
reactant were consumed
3) choose the smaller of the two amounts calculated in steps
(1) and (2).
- This is the theoretical yield of the product;
3.7 Concentrations of Reactants in
Solution: Molarity
Molarity (or concentration of a solution)
Describes the
dissolved in a
of solution
Solute - usually a solid or liquid dissolved in a solvent
expressed as moles of solute per liter of solution
molarity (M) =
Practice Problem: Lithium oxide can be used to absorbed water
as shown in this reaction: Li2O (s) + H2O (g)
2 LiOH (s)
If you have 15.0 kg of Li2O and need to absorb 15.0 kg of H2O:
(a) Do you have enough Li2O (which is the limiting reagent?)
(b) How many grams of the excess reagent are consumed?
(c) How many grams of lithium hydroxide are formed?
Dr. C. Bottaro Chem 1010 Fall 2005
(L)
Note: the symbol [ ] usually used to represent molarity of a
species in solution
e.g. [HCl] = 0.100 M 0.100 mol/L (mol L-1, mol HCl/ L)
Importance:
i) allows calculation of no. of moles of solute in a given
volume of solution
ii) allows calculation of volume of solution containing a given
# of moles of solute
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3.8 Diluting Concentrated Solutions
Chemicals are sometimes bought and stored as
concentrated solutions that must be diluted before use:
concentrated solution + solvent
diluted solution
when diluting solutions the number of moles of solute
remains the same; only the volume is changed by adding
more solvent
Since
and,
initial moles of solute (ni) = final moles of solute (nf)
Then,
Mi Vi = Mf Vf
Rearrangement of the equation allows for the calculation of
the molarity of the diluted (Mf) solution:
3.9 Solution Stoichiometry
in considering mass relations in
solution, the number of moles is
calculated by multiplying volume by
molarity. n=V x M
=VM
you can also determine the volume of
one solution required to react with a
known volume of the other.
There are many iterations of this
problem:
you may be given all the
concentrations (M) and will need to
determine the relationship between
volumes from the stoichiometry.
or you may be given all the volumes and only the concentration
of one solution and are required to determine the unknown
concentration again all based on stoichiometry.
Dr. C. Bottaro Chem 1010 Fall 2005
Examples
1. How many moles of solute are present in 250 mL of 0.45
M Na2CO3?
2.
How many grams of solute would you use to prepare
750.0 mL of 1.75 M KMnO4?
3.
What volume of 18.0 M H2SO4 is required to prepare
500.0 mL of 0.150 M aqueous H2SO4?
4.
What is the final concentration if 75.0 mL of 3.50 M
glucose is diluted to a volume of 300.0 mL?
3.10 Titration
a procedure for determining the
of a
solution by allowing a
with a
of another substance
whose concentration is
.
by finding the volume of the standard solution that reacts with
the measured volume of the first solution, the concentration of
the first solution can be calculated
Note: the reaction must go to completion and have a
yield of 100%
Acid-Base Titration
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3.11 Percent Composition and
Empirical Formulas
Percent Composition from a Formula
percent composition
- another way of expressing molecular composition
- is the mass of each element in a compound relative to
the total mass of the compound (i.e. expressed in
terms of the mass percent of each element)
Generally:
% element =
X 100 %
empirical formula
formula in which the atom ratio is the simplest wholenumber ratio possible
e.g.
Compound
Hydrogen peroxide
Benzene
Ethylene
Propane
Molecular
Formula
H2O2
C6H6
C2H4
C3H8
Empirical
Formula
HO
CH
CH2
C3H8
** important characteristics of formulas - the subscripts in a
formula represent not only the atom ratio in which the
different elements are combined, but also the mole ratio
molecular formula
of the simplest formula (i.e.
actual no. of atoms in molecule)
may be obtained only if the molar mass of a substance is
known from experiment
Calculating ratio of molecular mass to empirical
formula mass gives the required multiple to convert
from an empirical formula to molecular formula:
Multiple =
Molecular Mass
Empirical Formula Mass
e.g.
-
Empirical Formula: CH3
Dr. C. Bottaro Chem 1010 Fall 2005
Data for determining simplest formula can be expressed as
either :
i) masses of constituent elements in a weighed sample of the
compound
ii) mass percents of the elements in the compound
iii) masses of products obtained by reaction of a weighed
sample of the compound
Note: if mass percents of elements in a compound are given,
then assume a 100-g sample and calculate the mass of
each element in that sample
Practice Problem:
Nicotine was analyzed to be 74.0% carbon and 8.65%
hydrogen and 17.35% N by mass. Its molar mass is 162
g/mol.
What is the empirical formula and molecular formula for the
compound?
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3.12 Determining Empirical Formulas:
Elemental Analysis
Combustion Analysis
method used for the determination of empirical formulas,
particularly useful for compounds containing carbon and hydrogen
compound burned in the presence of O2 to yield its
products
3.13 Determining Molecular
Masses: Mass Spectrometry
mass spectrometer
used to determine both atomic and molecular masses
produces ions (usually of molecules) whose mass can be
detected
CO2 and H2O if the compound is a hydrocarbon
products are separated and weighed
For the combustion of methane:
from masses of CO2 and H2O, the # of moles of carbon
and hydrogen in the products can be calculated and thus
the C:H mole ratio can be determined
provides only the empirical formula need more
information to get molecular formula e.g. molecular mass
mass spectrum:
graph of ion mass versus intensity the number
masses of various ions versus the relative number of
those ions produced in the instrument
contains ions of numerous different masses with the
heaviest ion generally due to the ionized molecule
itself
measuring the mass of this ion, the molecular mass
of the molecule can be determined
Dr. C. Bottaro Chem 1010 Fall 2005
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