12.2 USING MOLES
MASS-MASS RELATIONSHIPS
Stoichiometry
is the study of quantitative relationships in chemical reactions. A
basic idea used in solving stoichiometric problems is the mole concept. If you
are given the mass of one substance and know the balanced equation, you can
calculate the reactants needed or the products produced because the equation
shows relative number of moles of reactants and products.
Moles and Particles
A
general procedure for mass-mass problems uses the following steps.
1. Write a balanced equation.
Identify the known and calculate its molar mass.
2. Convert from mass of given
material to moles.
(1MOLE
OF KNOWN / MOLAR MASS OF KNOWN)
3. Determine the mole ratio from
the coefficients of the balanced equation and convert from moles of given
material to moles of required material.
(COEFFICIENT {MOLES} OF UNKNOWN /
COEFFICIENT{MOLES} OF KNOWN)
4. Express the moles of required
material in grams.
(MOLAR
MASS OF UNKNOWN / 1 MOLE OF UNKNOWN)
The setup for a mass-mass calculation follows the
format given below.
(start with grams given)-> (grams to moles)-> (use mole
ratio)-> (moles to grams)-> (end with grams required)
Solving Mass-Mass Problems
EXAMPLE
Calculate
the mass of HCl needed to react with 10.0 g Zn.
Solving Process:
Step
1. Begin with the balanced equation.
Zn(s) + 2HCl(aq)ŕ ZnCl2(aq)
+ H2(g)
Step
2. Convert grams of zinc to moles.
Step
3. Determine the mole ratio that exists between Zn and HCl and convert
from
moles Zn to moles HCl.
1 mole Zn reacts with 2 moles HCl
Step
4. Convert moles of HCl to grams of HCl.
grams HCl =
= 11.2 g HCl
Note
that the conversion ratios are chosen and arranged so all the units divide out
except the desired unit, in this case, grams of HCl. Since all the ratios are
equal to 1,multiplying by one of them, or by all of them, changes only the
units of the answer.
EXAMPLE
Calculate
the mass of O2 produced if 2.50 g KClO3 are completely
decomposed by heating.
Solving Process:
Step
1. Write the balanced equation.
2KClO3(s)ŕ
2KCl(s) + 3O2(g)
Step
2. Convert mass of KClO3 to moles.
Step
3. Determine the mole ratio that exists between KClO3 and O2.
2 moles KClO3 yields 3 moles O2
Step
4. Convert moles of O2 to grams.
grams O2
AVOGADRO’S
PRINCIPLE AND MOLAR VOLUME
What is
the relationship between the mass of a gas and its volume? Avogadro’s
principle states that equal volumes of all
gases, measured under the same conditions of pressure and temperature, contain
the same number of particles or moles. One mole of any gas has a mass
equal to its molecular mass.
For
example:
1 mole N2
= 28.0 g N2 =
6.02 X 1023 molecules of N2
1 mole
CO2 = 44.0 g CO2 =
6.02 X 1023 molecules of CO2
The
volume of one mole of a gas at STP is the molar
volume of the gas. One mole of a
gas at STP occupies 22.4 liters (L). This volume is the same for all
gases at STP. The mass and volume of any gas are related as follows.
1 mole of any gas = molecular
mass = 22.4 L
STP:
STANDARD TEMPERATURE AND PRESSURE FOR A GAS
TEMPERATURE = 0.00 oC =
273 K
PRESSURE = 1 ATMOSPHERE = 101 kPa
Mole Relationships
EXAMPLE
How many
grams of carbon dioxide, CO2, will occupy a volume of 500.0 mL at
STP?
Solving
Process:
The
conversion equalities are
1 mol CO2 =
22.4 L CO2 (STP)
1 mol CO2 =
44.0 g CO2
= 0.982 g
MASS-GAS
VOLUME PROBLEMS
Many
chemical reactions involve gases. It is often necessary to know the volume of
gas involved with a known mass of material in a reaction. Problems of this type
are similar to mass-mass problems, however one additional piece of information
is needed. In mass-volume problems, mass is changed to moles of the desired
substance and then converted to volume using the relationship:
1 mole of any gas = 22.4 L
of that gas at STP
The
reverse calculation may also be done. Volume is changed to moles and
moles
are changed to mass.
EXAMPLE
Calculate
the volume of oxygen produced at STP by the decomposition of 10.0 g of
potassium chlorate, KClO3.
Solving
Process:
Write
the balanced equation.
2KClO3(s)ŕ
2KCl(s) + 3O2(g)
Start
with the known mass of KClO3 given in the problem and convert to
volume of oxygen at STP.
EXAMPLE
A
student performs an experiment involving the reaction of magnesium metal
with
hydrochloric acid to form hydrogen gas. From the given data, calculate the mass
of magnesium.
1. volume
of hydrogen gas formed 42.0
mL
2. temperature
of hydrogen 20.0°C
3. pressure
99.3
kPa
4. vapor
pressure of water 2.3
kPa
5. pressure
of dry hydrogen (99.3
- 2.3) 97.0 kPa
VOLUME-VOLUME
PROBLEMS
It is
possible to calculate the volume of a gas in a reaction when the volume of
another gas in the reaction is known. Two methods can be used in solving these
volume-volume problems. The first method is the same as the mass-mass or
mass-volume method.
Use the
following steps.
Step
1. Convert the given volume to moles.
Step
2. From the balanced equation, convert the moles of given substance to
moles of required substance.
Step
3. Convert the moles of required substance back to its volume.
In each
case, the temperature and pressure must be taken into consideration.
Two
methods of solving for gas volume are given. The second method usually involves
only an inspection and simple mental calculation. It can be used easily when
the temperature and pressure remain constant.
EXAMPLE
If 6.00
L of oxygen are available to burn carbon disulfide, CS2, how many L
of carbon dioxide are produced? The products of the combustion of carbon
disulfide are carbon dioxide and sulfur dioxide.
Solving
Process:
Balance
the equation for this reaction.
CS2(l)
+ 3O2(g) ŕ
CO2(g) + 2SO2(g)
Convert
6.00 L O2 to L of CO2.
Therefore,
6.00 L O2 will produce 2.00 L CO2. Note that the changes
to and from moles divide out.
Alternate
Method: Temperature and Pressure Constant
The
balanced equation indicates the relative number of moles of reactant and
product.
The
coefficients also indicate the relative volumes of the gases at constant
temperature and pressure. The relationship is a result of the principle stated
in Avogadro’s hypothesis: equal volumes of gases at the same
temperature and pressure contain the same number of particles. If the
gases are measured at the same temperature and pressure then 3 volumes O2:1
volume CO2 or 3 L O2:1 L CO2. The volume of CO2
will be one-third the volume of O2. Since the O2 volume
is 6.00 L, the CO2 volume is 2.00 L. Conversion ratios could be used
as follows.
LIMITING
REACTANTS
Many
reactions continue until one of the reactants is consumed. The reactant
that is
used up first is called the limiting reactant. The
other reactant is said to be in excess. When discussing limiting reactants, we
will deal only with nonreversible reactions. It is possible to determine
whether a material is in excess or is deficient in a reaction by experiment or
by calculation.
Limiting
reactant problems are most easily solved by comparing the moles of
the
reactants present using the following steps.
Step
1. Write a balanced equation.
Step
2. Change both given quantities to moles.
Step
3. From the balanced equation, determine the moles of required substance
that each given quantity will produce.
Step
4. Complete the problem using the quantity that yields the lesser amount
of product. This reactant is the limiting reactant.
EXAMPLE
If 40.0
g of H3PO4 react with 60.0 g of MgCO3,
calculate the volume of CO2 produced at STP.
Solving
Process:
Step
1. Write the balanced equation.
2H3PO4(aq) +
3MgCO3(s) ŕ
Mg3(PO4)2(s) +
3CO2(g) + 3H2O(l)
Step
2. Change grams of reactant to moles of reactant.
Step 3. From the balanced equation
determine the moles of CO2 that will be
produced by each reactant.
The
limiting reactant produces the lesser amount of product, so in this
case, H3PO4
is the limiting reactant.
Step
4. Use the limiting reactant to complete the problem.
Therefore,
40.0 g of H3PO4 will produce 13.7 L of CO2
measured at standard temperature and pressure.
The same
approach for finding limiting reactants can also be used in mass-mass or
volume-volume problems.
NONSTANDARD
CONDITIONS
Gas
volume changes dramatically when pressure or temperature change.
The
molar volume is 22.4 L only at STP. If the experimental conditions are
different from STP in a problem, it is still necessary to calculate the gas
volume at STP.
The
secret to success in these problems is to remember that the central step
(moles
of given to moles of unknown) must take place at STP. Thus, if you are given a
volume of gas at other than STP, you must convert to STP before performing the
moles to moles step in the solving process. On the other hand, if you are
requested to find the volume of a gas at conditions other than STP, you must
convert the volume after the moles to moles step.
EXAMPLE
How many
grams of ammonium sulfate must react with excess sodium hydroxide to produce
408 mL of ammonia measured at 27°C and 98.0 kPa?
Solving
Process:
Write
the balanced equation.
(NH4)2SO4(s)
+ 2NaOH(aq) ŕ Na2SO4(aq)
+ 2NH3(g) + 2H2O(l)
Convert
408 mL NH3 at 27°C and 98.0 kPa to STP and then convert to g of
(NH4)2SO4.
Since the temperature decreases, the volume decreases and the absolute
temperature ratio is
EXAMPLE
What volume of hydrogen collected
over water at 27°C and 97.5 kPa is produced by the reaction of 3.00 g of Zn
with an excess of sulfuric acid? The vapor pressure of water at 27°C is 3.6
kPa.
Solving Process:
Step 1. Write the balanced equation.
Zn(s) + H2SO4(aq)
ŕ ZnSO4(aq) + H2(g)
Step 2. Convert from grams of Zn to liters
of dry H2 at 27°C and 97.5 kPa. Then convert to liters of H2 at STP
by using the absolute temperature and pressure ratios.
Step 3. The liters of H2 at STP
must be converted to the conditions given in the problem. As the temperature is
increased, the volume will increase, so
the absolute temperature ratio is
Step 4. As pressure is decreased volume will
increase, so the pressure ratio is
This ratio must be corrected for the
vapor pressure of water, which is
3.6 kPa at this temperature. The
corrected ratio is
IDEAL GAS EQUATION
The ideal gas equation combines
the four physical variables (pressure, volume, temperature, and number of
particles) for gases into one equation. Remember that an ideal gas is composed
of point masses that do not take up space, and these masses are not attracted
to each other at all. All real gases deviate somewhat from the gas laws since
the molecules of real gases are not point masses (they take up space) and they
attract one another.
The ideal gas equation is PV =
nRT, where P is the pressure in kilopascals. V
is the volume in cubic decimeters
and T is the temperature in kelvin. The n represents the number
of moles of a gas. With these units, the value of the constant R is 8.31
L . kPa/mol . K. There are other values of R depending
upon the units used to derive R.
We can use the ideal gas equation to
determine the molecular mass (M) of
a gas. The number of moles (n)
of any species is equal to its mass (m) divided by the molecular mass (M).
Thus, the ideal gas equation can also be written as follows.
How many moles of gas will a 1250-mL
flask hold at 35.0°C and a pressure of 95.4 kPa?
Solving Process:
The ideal gas equation can be solved
for the number of moles, n, of a substance.
Before we can substitute the known
values into the ideal gas equation, 35.0°C must be converted to 308.2 K. We get
the following expression.
The solution is 0.0466 mol. Note
that all other units in the problem divide out.
EXAMPLE
A flask has a volume of 258 mL. A
gas with mass 1.475 g is introduced into the flask at a temperature of 302.0 K
and a pressure of 9.86 X 104
Solving Process:
The number of moles, n, of a
substance is equal to mass, m, divided by the molecular mass, M. Therefore,
the ideal gas equation may be written
Remember that the units of volume,
pressure, temperature, and quantity of gas must be consistent with the value of
R.
MASS PERCENTS
The mass percent of elements
in a compound gives the relative amount of
each element present. The percent of
an element in a compound is determined by the following equation.
To calculate mass percent:
(a) calculate the total mass for each
element,
(b) calculate the formula mass for the
entire compound,
(c) divide the total mass of each
element by the formula mass of the compound, and
(d) multiply by 100%.
EXAMPLE
Find the mass percent of nitrogen in
ammonium nitrate, NH4NO3, an important source of nitrogen
in fertilizers.
Solving Process:
Calculate the formula mass; then
find the percentage.
MOLECULAR AND EMPIRICAL FORMULAS
The empirical formula of a
compound is the smallest whole number ratio
of the number of atoms of each
element in the substance. The molecular formula gives the actual number
of atoms in the molecule. For instance, CH2 is the empirical formula
for the series of molecular compounds C2H4, C3H6,
C4H8, and so on.
There is a definite relationship
between the empirical and the molecular formula.
Note that the molecular formula is
always a whole number multiple of the
empirical formula. As can be seen in
Table 12-2 the empirical formula and the molecular formula are not
always the same.
EMPIRICAL FORMULAS
Earlier, we used the formula of a
compound to determine its mass percents.
Now we reverse the procedure and
determine the empirical formula from the
mass percents. The elements in
compounds combine in simple whole number ratios of atoms. To determine an
empirical formula, masses of elements are converted to moles and then a ratio
of moles is determined.
EXAMPLE
Determine the empirical formula for
sodium sulfite. Sodium sulfite contains
36.5% sodium, 25.4% sulfur, and
38.1% oxygen.
Solving Process:
A percentage indicates a part of one
hundred. Therefore, the percentage composition data indicates that there are
36.5 g Na, 25.4 g S, and 38.1 g O in 100 g of compound.
Step 1. Find the number of moles.
Step
2. Determine the ratio of moles.
EXAMPLE
What
is the empirical formula of a compound that contains 53.73% Fe and
46.27%
S?
Solving
Process:
There
are 53.73 g Fe and 46.27 g S in 100 g of compound.
Step
1. Find the number of moles.
In
the previous example problem, the relative numbers of atoms were small
whole
numbers and we could write the formula directly from them. The ratio 1 to 1.5
must be expressed in terms of whole numbers, since a fractional part of an atom
does not exist. By multiplying both numbers in the ratio by two, we obtain two
atoms Fe and three atoms S. The empirical formula is Fe2S3.
MOLECULAR
FORMULAS
The
molecular formula indicates not only the ratio of the atoms of the elements in
a compound but also the actual number of atoms of each element in one molecule
of the compound.
The
molecular formula calculation is the same as the empirical formula calculation,
except that the molecular mass is used in an additional step. Remember, the
molecular formula is always a whole number multiple of the empirical formula.
EXAMPLE
An
organic compound is found to contain 92.25% carbon and 7.75% hydrogen.
If
the molecular mass is 78 u, what is the molecular formula?
Solving
Process:
Determine
the empirical formula.
