Page 1 of 3
(continued)
Thus the required conversion factor from Al to I2 is
3 mol I2
2 mol Al
Now we can determine the moles of I2 required by 1.30 moles of Al:
mol I2
1.30 mol Al 3 1.95 mol I2
2 mol Al
We know the moles of I2 required to react with the 1.30 mol of Al (35.0 g).
The next step is to convert 1.95 mol of I2 to grams so we will know how
much to weigh out. We do so by using the molar mass of I2. The atomic
mass of iodine is 126.9 g (for 1 mol of I atoms), so the molar mass of I2 is
2 126.9 g/mol 253.8 g/mol mass of 1 mol of I2
Now we convert the 1.95 mol of I2 to grams of I2.
253.8 g I2
1.95 mol I2 495 g I2
mol I2
We have solved the problem. We need to weigh out 495 g of iodine (contains I2 molecules) to react exactly with the 35.0 g of aluminum.
This problem can be summarized as follows:
2Al(s)
n
2AlI3(s)
495 g
I2
35.0 g
Al
26.98
g/mol
1.30 mol
Al
3I2(s)
253.8
g/mol
3 mol I2
2 mol Al
1.95 mol
I2
Self-Check Exercise 9.4
Calculate the mass of AlI3(s) formed by the reaction of 35.0 g Al(s) with
495 g I2(s).
Mass Calculations Using Scientific Notation
Objective: To carry out mass calculations that involve scientific notation.
S
o far in this chapter, we have spent considerable time “thinking through”
the procedures for calculating the masses of reactants and products in
chemical reactions. We can summarize these procedures in the following steps.
262
Chapter 9 Chemical Quantities
Page 2 of 3
Steps for Calculating the Masses of Reactants
and Products in Chemical Reactions
STEP 1 Balance the equation for the reaction.
STEP 2 Convert the masses of reactants or products to moles.
STEP 3 Use the balanced equation to set up the appropriate mole
ratio(s).
STEP 4 Use the mole ratio(s) to calculate the number of moles of the
desired reactant or product.
STEP 5 Convert from moles back to mass.
The process of using a chemical equation to calculate the relative masses
of reactants and products involved in a reaction is called stoichiometry (pronounced stoý.ke.oḿ.ĕtry). Chemists say that the balanced equation for a
chemical reaction describes the stoichiometry of the reaction.
We will now consider a few more examples that involve chemical stoichiometry. Because real-world examples often involve very large or very small
masses of chemicals that are most conveniently expressed by using scientific
notation, we will deal with such a case in Example 9.6.
Example 9.6
Stoichiometric Calculations: Using Scientific Notation
CHEMISTRY
For a review of writing formulas of ionic compounds,
see Chapter 4.
Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment. The products are solid lithium
carbonate and liquid water. What mass of gaseous carbon dioxide can
1.00 103 g of lithium hydroxide absorb?
Solution
Step 1 Using the description of the reaction, we can write the unbalanced
equation
LiOH(s) CO2(g) n Li2CO3(s) H2O(l)
The balanced equation is
2LiOH(s) CO2(g) n Li2CO3(s) H2O(l)
Check this for yourself.
Step 2 We convert the given mass of LiOH to moles, using the molar mass
of LiOH, which is 6.941 g 16.00 g 1.008 g 23.95 g.
1 mol LiOH
1.00 103 g LiOH 41.8 mol LiOH
23.95 g LiOH
Step 3 The appropriate mole ratio is
1 mol CO2
2 mol LiOH
(continued)
9.4 Mass Calculations Using Scientific Notation
263
Page 3 of 3
(continued)
PROBLEM SOLVING
Carrying extra significant
figures and rounding off only
at the end gives an answer of
919 g CO2.
Step 4 Using this mole ratio, we calculate the moles of CO2 needed to react with the given mass of LiOH.
1 mol CO2
41.8 mol LiOH 20.9 mol CO2
2 mol LiOH
Step 5 We calculate the mass of CO2 by using its molar mass (44.01 g).
44.01 g CO2
20.9 mol CO2 920. g CO2 9.20 102 g CO2
1 mol CO2
Thus 1.00 103 g of LiOH(s) can absorb 920. g of CO2 (g).
We can summarize this problem as follows:
2LiOH(s)
CO2(g)
n
1.00 103 g
LiOH
Grams of
CO2
Use molar mass of
LiOH
Use molar mass
of CO2
Moles of
LiOH
Use mole
ratio between
CO2 and LiOH
Li2CO3(s) H2O(l)
Moles of
CO2
Astronaut Sidney M. Gutierrez
changes the lithium hydroxide
The conversion string is
cannisters on space shuttle
Columbia. The lithium hydroxide
1 mol CO2
1 mol LiOH
44.01 g CO2
1.00 103 g LiOH is used to purge carbon dioxide
23.95 g LiOH
2 mol LiOH
1 mol CO2
from the air in the shuttle’s
cabin.
9.20 102 g CO2
Self-Check Exercise 9.5
Hydrofluoric acid, an aqueous solution containing dissolved hydrogen fluoride, is used to etch glass by reacting with the silica, SiO2, in the glass to
produce gaseous silicon tetrafluoride and liquid water. The unTOP TEN
balanced equation is
Chemicals Used in the United
States (annually)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
264
Chemical
Amount (kg)
Sulfur acid (H2SO4)
Nitrogen (N2)
Oxygen (O2)
Ethylene (C2H4)
Lime (CaO)
Ammonia (NH3)
Phosphoric acid (H3PO4)
Sodium hydroxide (NaOH)
Propylene (C3H7)
Chlorine (Cl2)
4.33 1010
3.09 1010
2.43 1010
2.13 1010
1.87 1010
1.61 1010
1.19 1010
1.19 1010
1.17 1010
1.14 1010
Chapter 9 Chemical Quantities
HF(aq) SiO2(s) n SiF4(g) H2O(l)
a. Calculate the mass of hydrogen fluoride needed to react with
5.68 g of silica. Hint: Think carefully about this problem.
What is the balanced equation for the reaction? What is
given? What do you need to calculate? Sketch a map of the
problem before you do the calculations.
b. Calculate the mass of water produced in the reaction described in part a.