1. Chemical Accounting: Mass and Volume Relationships
Chapter 6
1 April 2017
Chemistry for Changing Times 10th edition
Hill/Kolb
Chapter 6
Chemical Accounting: Mass and Volume Relationships
Daniel Fraser
University of Toledo, Toledo OH
©2004 Prentice Hall
CHE 105

2. Chemical Equations C + O2  CO2 Substances on left Substances on right
Reactants
Starting materials
Substances on right
Products
+ “and”
 “reacts to produce”
Chapter 6

3. Reading a Chemical Reaction
C(s) + O2(g)  CO2(g)
(s) solid
(l) liquid
(g) gas
(aq) aqueous or in water
Chapter 6

4. Balancing Chemical Equations
Must have the same number of each element on both sides of the equation
Chapter 6

5. Suggestions for Balancing Chemical Equations
If element occurs once on each side, balance first
Balance any reactants or products that exist as free elements last
CANNOT change subscripts
CANNOT add/delete products or reactants
Chapter 6

6. Example 6.1  
Balance the following equation, which represents the chemical reaction
involved when an airbag deploys:
NaN3
N2 + Na
Solution
We can see initially that the sodium atoms are balanced, but the nitrogen atoms
are not. For this sort of problem, we will use the concept of the least common
multiple. There are three nitrogen atoms on the left (reactants side) and two on
the right (products side). The least common multiple of 2 and 3 is 6. Therefore,
we need three N2 and two NaN3:
We now have two sodium atoms on the left. We can get two on the right by
placing the coefficient 2 in front of Na:
2 NaN3
2 Na + 3 N2
(balanced)
Checking, we count two Na atoms and six N atoms on each side. The
equation is balanced.
Chapter 6

7. Example 6.1 (cont.) Exercise 6.1A
The reaction between hydrogen and nitrogen to give ammonia, called the
Haber process, is typically the first step in the industrial production of fertilizer.
Balance the following reaction for the Haber process:
H2 + N2
NH3
Exercise 6.1B
Iron ores such as Fe2O3 are smelted, by reaction with carbon, to produce metallic
iron and carbon dioxide. Balance the following reaction:
Fe2O3 + C
CO2 + Fe
Chapter 6

8. Example 6.2
When a fuel such as methane is burned in sufficient air, the products are carbon
dioxide and water. Balance the following equation for this combustion:
CH4 + O2
CO2+ H2O
(not balanced)
In this equation, oxygen appears in two different products and by itself in O2;
we leave the oxygen for last and balance the other two elements first. Carbon is
already balanced, with one atom on each side of the equation. For hydrogen,
the least common multiple of 2 and 4 is 4, and so we place the coefficient 2 in
front of H2O to balance hydrogen. Now we have four hydrogen atoms on each side:
CH4 + O2
CO2 + 2 H2O
(not balanced)
Solution
Now for the oxygen. There are four oxygen atoms on the right. If we place a 2
in front of O2 on the left, the oxygen atoms balance.
CH4 + 2 O2
CO2 + 2 H2O
(balanced)
The equation is balanced. Generally, when you have a chemical species that
only contains one type of element, it is easier to balance that element last.
Chapter 6

9. Example 6.2 (cont.) Exercise 6.2
Butane is another common fuel. Balance the following equation describing
the combustion of butane:
C4H10 + O2
CO2 + H2O
Does it takes more oxygen (per molecule) to burn butane than it does to burn
methane? How much more CO2 is produced?
Chapter 6

10. Avogadro’s Number
Number of carbon-12 atoms in a 12-g sample of carbon-12
6.022 x 1023
Chapter 6

11. Mole Similar to a dozen EXCEPT 1 mol has Avogadro’s number of items
Abbreviated mol
Easier to weigh out moles instead of molecules
1 mol H2O has how many moles of O? H?
Chapter 6

12. Formula Masses
Average mass of a formula unit relative to that of a carbon-12 atom
C formula mass =
O formula mass =
Chapter 6

13. Molar Mass Mass of 1 mol of a substance
Different for every compound
Numerically equivalent to formula mass
Units of gram/mole
Can be used to convert between moles and mass
Chapter 6

14. Chapter 6

15. Molar Volume Volume occupied by 1 mol of gas
Standard temperature and pressure (STP)
1 atm pressure and 0°C
1 mole of gas has volume of 22.4 L
Chapter 6

16. Mole to Mass Relationships
Balanced chemical reaction gives molar ratios
Chapter 6

17. Stoichiometry
Relationship between reactants and products in a chemical reaction
Converts between moles of different substances in a reaction
Chapter 6

18. Mass Relationships in Chemical Reactions
Cannot weigh out mol of compounds, only masses
Must have balanced chemical equation
Chapter 6

19. Chapter 6

20. Example 6.4 Solution Exercise 6.4
Calculate the molecular mass of sulfur dioxide (SO2), an irritating gas formed when
sulfur is burned.
Solution
We think about the problem in the following way. Add the atomic mass of
sulfur to twice the atomic mass of oxygen.
1 x the atomic mass of S = 1 x 32.1u = 32.1u
2 x the atomic mass of O = 2 x 16.0 u = 32.0 u
Formula mass of SO = 64.1 u
Exercise 6.4
Calculate the formula mass of (a) C6H4Cl2 and (b) H3PO4.
Chapter 6

21. Example 6.5 Solution Exercise 6.5
Calculate the formula mass of sodium azide, NaN3, used in automobile airbags.
Solution
To determine a formula mass, we add the atomic masses of the constituent
elements:
1 x atomic mass of Na = 1 x 23.0 u = 23.0 u
3 x atomic mass of N = 3 x 14.0 u = 42.0 u
Formula mass of NaN = 65.0 u
Exercise 6.5
Calculate the formula mass of (a) NaHCO3 (the main ingredient in baking
soda) and (b) (NH4)2SO4 (a fertilizer commonly used by home gardeners).
Remember, everything within the parentheses must be multiplied by 2.
Chapter 6

22. Example 6.6 Solution Exercise 6.6
How many grams of N2 are in mol N2?
Solution
The molar mass of N2 is 28.0 g/mol. Therefore:
= 11.2 g N2
28.0 g N2
1mol N2
? g N2 = mol N2 x
Calculate the mass, in grams, of (a) 55.5 mol H2O and (b) 55 mol uranium.
Exercise 6.6
Chapter 6

23. Example 6.7 Solution Exercise 6.7 65 g NaN3
Calculate the number of moles of NaN3 in a 10.0-g sample of the solid.
Solution
In this case we need the molar mass of NaN3. In Example 6.5 we calculated 65.0 u
as the formula mass of NaN3. The molar mass is then 65.0 g/mol.
To convert from a mass in grams to an amount in moles, we must use the inverse of the
molar mass (1 mol NaN3/65.0 g NaN3) to get the proper cancellation of units. When we
start with grams, we must have grams in the denominator of our conversion factor:
? mol NaN3 = 10.0 g NaN3 x
= mol NaN3
65 g NaN3
1 mol NaN3
Calculate the amount, in moles, of (a) 3.71 g Fe and (b) 165 g butane, C4H10.
Exercise 6.7
Chapter 6

24. Example 6.10 Solution Exercise 6.10
Nitrogen monoxide (nitric oxide), one of the air pollutants discharged by
internal combustion engines, combines with oxygen to form nitrogen dioxide,
a reddish-brown gas that irritates the respiratory system and eyes. The
equation for this reaction is
2 NO + O2
2 NO2
State the molecular, molar, and mass relationships indicated by the equation.
Solution
Molecular: Two molecules of NO react with one molecule of O2 to form
two molecules of NO2.
Molar: 2 mol of NO react with 1 mol of O2 to form 2 mol of NO2.
Mass: 60.0 g of NO react with 32.0 g of O2 to form 92.0 g of NO2.
Hydrogen sulfide, a gas that smells like rotten eggs, burns in air to produce
sulfur dioxide and water according to the equation
Exercise 6.10
2 H2S + 3 O2
2 SO2 + 2 H2O
State the molecular, molar, and mass relationships indicated by this equation.
Chapter 6

25. Example 6.11 Solution Exercise 6.11
When mol of propane is burned in a plentiful supply of oxygen, how
many moles of oxygen is consumed?
C3H8 + 5 O2
3 CO2 + 4 H2O
Solution
The equation tells us that 5 mol O2 is required to burn 1 mol C3H8 . We can write
1 mol C3H mol O2
where we use the symbol to mean “is chemically equivalent to.” From this
relationship we can construct conversion factors to relate moles of oxygen
to moles of propane. The possible conversion factors are
1 mol C3H8
5 mol O2
and
Multiply the given quantity (0.105 mol C3H8) by the factor on the right to get
an answer with the desired units (moles of oxygen):
5 mol O2
1 mol C3H8
? mol O2 = mol C3H8 x
= mol O2
For the combustion of propane in Example 6.11: (a) How many moles of
carbon dioxide is formed when mol of C3H8 is burned? (b) How many moles
of water is produced when 76.2 mol of C3H8 is burned? (c) How many moles of
carbon dioxide is produced when mol of O2 is consumed?
Exercise 6.11
Chapter 6

26. Example 6.12
Calculate the mass of oxygen needed to react with 10.0 g of carbon in the
reaction that forms carbon dioxide.
Solution
The balanced equation is
C + O2
CO2
The molar masses are 2 x g/mol for O2 and 12.0 g/mol for C.
We convert the mass of the given substance, carbon, to an amount in moles:
? mol C = 10.0 g C x
1 mol C
12.0 g C
= mol C
We use coefficients from the balanced equation to establish the
stoichiometric factor that relates the amount of oxygen to that
of carbon:
1 mol C
0.833 mol C x
1 mol O2
= mol O2
Chapter 6

27. Example 6.12 (cont.) Exercise 6.12A Exercise 6.12B
We convert from moles of oxygen to grams of oxygen:
0.833 mol O2 x
32.0 g O2
1 mol O2
= 26.7 g O2
We can also combine the five steps into a single setup. Note that the units in the
denominators of the conversion factors are chosen so that each cancels the unit
in the numerator of the preceding term:
(The slightly different answers are due to rounding in the intermediate steps.)
Exercise 6.12A
Calculate the mass of oxygen (O2) needed to react with g of nitrogen (N2) in
the reaction that forms nitrogen dioxide.
Exercise 6.12B
Calculate the mass of carbon dioxide formed by burning 775 g of each of
(a) methane (CH4) and (b) butane (C4H10 ). Assume excess oxygen is available.
Chapter 6

28. Example 6.13
The decomposition of sodium azide produces nitrogen gas that is used to
inflate automobile airbags. What mass of nitrogen, in grams, can be made from
60.0 g of sodium azide?
Solution
We start by writing the balanced chemical equation, which we did in
Example 6.1:
2 NaN3
2 Na + 3 N2
The molar mass of NaN3 is 65.0 g/mol and the molar mass of N2 is
2 x 14.0 = 28.0 g/mol.
We convert the mass of the given substance, sodium azide, to an amount
in moles:
60.0 g NaN3 x
1 mol NaN3
65.0 g NaN3
= mol NaN3
We use coefficients from the balanced equation to establish the
stoichiometric factor that relates the amount of nitrogen gas to that of
sodium azide:
0.923 mol NaN3 x
3 mol N2
2 mol NaN3
= 138 mol N2
Chapter 6

29. Example 6.13 (cont.) Exercise 6.13A Exercise 6.13B
We convert from moles of nitrogen gas to grams of nitrogen gas:
1.38 mol N2 x
28.0 g N2
1 mol N2
= 38.6 g N2
As is usually the case, all of the steps outlined above can be combined into a
single setup:
60.0 g NaN3 x
1 mol NaN3
65.0 g NaN3 x
3 mol N2
2 mol NaN3
28.0 g N2
1 mol N2
= g N2
Notice that the units of the numerator in one stoichiometric factor are the
units in the denominator of the next stoichiometric factor. In this way, the
correct cancellation of units occurs and the units of the final numerator are
the units of your answer.
Exercise 6.13A
Ammonia reacts with phosphoric acid (H3PO4) to form ammonium
phosphate [(NH4)3PO4]. How many grams of ammonia are needed to react
completely with 74.8 g of phosphoric acid?
Exercise 6.13B
a The decomposition of potassium chlorate (KClO3) produces potassium
chloride (KCl) and O2 gas. How many grams of oxygen can be made
from 2.47 g of potassium chlorate?
Chapter 6

30. Solutions Homogeneous mixture of two or more substances
Solute – what is being dissolved
Solvent – what is doing the dissolving
Aqueous solutions – water is solvent
Chapter 6

31. Solubility Soluble: an appreciable quantity dissolves
Insoluble: very little, if any, quantity dissolves
Dilute solution: little solute in a lot of water
Concentrated solution: lots of solute in the solvent
Chapter 6

32. Measurement of Solubility
Molarity (M): amount of solute, in moles, per liter of solution
Chapter 6

33. Chapter 6

34. Percent Concentration
If both solute and solvent are liquids, use percent by volume
Chapter 6

35. Percent by mass is commonly used for commercial solutions
35.0% HCl means 35.0 g HCl for every g of solution
Chapter 6

36. Note the differences between mass percent, percent by volume, and molarity
10% HCl solution is considerably different than 10 M HCl
Require different amounts of HCl
Chapter 6

37. Example 6.18 Solution Exercise 6.18A Exercise 6.18B
Calculate the molarity of a solution made by dissolving 3.50 mol of NaCl in enough
water to produce 2.00 L of solution.
Solution
Molarity (M) =
moles of solute
liters of solution
3.50 mol NaCl
2.00 L solution
= 1.75 M NaCl =
We read 1.75 M NaCl as 1.75 molar NaCl.
Exercise 6.18A
Calculate the molarity of a solution that has mol of NH3 in 5.75 L of solution.
Exercise 6.18B
Calculate the molarity of a solution made by dissolving mol of H3PO4 in enough water
to produce 775 mL of solution.
Chapter 6

38. Example 6.19 Solution Exercise 6.19
What is the molarity of a solution in which 333 g of potassium hydrogen carbonate is
dissolved in enough water to make 10.0 L of solution?
Solution
First, prepare a setup to convert from mass of KHCO3 to moles of KHCO3:
333 g KHCO3 x
1 mol KHCO3
100.1 g KHCO3
= 3.33 g KHCO3
Now use this value as the numerator in the defining equation for molarity. The solution
volume, 10.0 L, is the denominator:
molarity x
3.33 mol KHCO3
10.0 L solution
= .333 M KHCO3
Exercise 6.19
Calculate the molarity of each of the following solutions:
a mol of H2SO4 in 2.00 L of solution
b mol of KI in 2.39 L of solution
c mol of HF in 752 mL of solution. (HF is used for etching glass.)
Chapter 6

39. Example 6.20 Solution Exercise 6.20
How many grams of NaCl is required to prepare L of typical over-the-counter saline
solution (about 0.15 M NaCl)?
Solution
First we calculate the moles of NaCl:
15 mol NaCl
0.500 L solution x
1 L solution
= mol NaCl
Then we use the molar mass to calculate the grams of NaCl:
58.4 g NaCl
0.075 mol NaCl x
1 mol NaCl
= 4.4 g NaCl
Exercise 6.20
How many grams of potassium hydroxide is required to prepare each of the following solutions?
a L of 6.00 M KOH
b mL of 1.00 M KOH
Chapter 6

40. Example 6.21 Solution Exercise 6.21A Exercise 6.21B
Concentrated hydrochloric acid has a concentration of 12.0 M HCl. How many milliliters
of this solution would one need to get mol of HCl?
Solution
liters of HCl solution =
moles of solute
molarity
0.425 mol HCl
12.0 M HCl
12.0 mol HCl/L
= L = =
We would need L (35.4 mL) of the solution to have mol. Remember that
molarity is moles per liter of solution, not per liter of solvent.
Exercise 6.21A
How many milliliters of 15.0 M aqueous ammonia (NH3 ) solution do you need to get
0.445 mol of NH3 ?
Exercise 6.21B
How many grams of HNO3 are in 500 mL of rain that has a concentration
of 2.0 x 10–5 M HNO3?
Chapter 6

41. Example 6.22 Solution Exercise 6.22
Two-stroke engines use a mixture of 120 mL of oil dissolved in enough gasoline to make
4.0 L of fuel. What is the percent by volume of oil in this mixture?
Solution
Percent by volume =
120 mL oil
4000 mL solution
x 100% = 3.0%
Exercise 6.22
What is the percent by volume of a solution made by dissolving 235 mL of ethanol in
enough water to make exactly 500 mL of solution (approximately the concentration of
ethanol in distilled spirits)?
Chapter 6

42. Example 6.23 Solution Exercise 6.23
Describe how to make 775 mL of vinegar (about a 5.0% by volume solution of acetic acid
in water).
Solution
Volume of solute =
percent by volume x volume of solution
100%
Let’s begin by rearranging the equation for percent by volume to solve for volume of solute:
Substituting, we have
5.0% x 775 mL
100%
= 39 mL =
Take mL of acetic acid and add enough water to make 775 mL of solution.
Exercise 6.23
Ethanol used for medicinal purposes is generally of a grade referred to as USP
(an abbreviation of United States Pharmacopoeia, the official publication of
standards for pharmaceutical products). USP ethanol is 95% CH3CH2OH by
volume. Describe how to make 100 mL USP-grade ethanol.
Chapter 6

43. Example 6.24 Solution Exercise 6.24
What is the percent by mass of a solution of 25.0 g of NaCl dissolved in 475 g (475 mL)
of water?
Solution
Percent by mass =
mass of NaCl
mass of solution
x 100%
25.0 g NaCl
500 g solution =
x 100% = 5.00% NaCl
Exercise 6.24
Hydrogen peroxide from the local drugstore is a 3% by weight solution of H2O2 in water.
What is the percent by mass of a solution of 9.40 g of H2O2 dissolved in 335 g (335 mL)
of water?
Chapter 6

44. End of Chapter 6
Chapter 6

45. Avogadro’s hypothesis – equal volumes of gases at constant pressure and temperature have the same number of molecules
Chapter 6

46. Volume Relationships
Law of combining volumes – when all measurements are made at same temperature and pressure, volumes of gaseous reactants and products are in small whole-number ratio
Chapter 6

47. Gas Laws – Kinetic Molecular Theory
All matter is composed of tiny discrete particles called molecules
Molecules in a gas are in rapid constant motion and move in straight lines
Molecules of a gas are tiny compared with distances between gas molecules
There is little attraction between molecules of a gas
Chapter 6

48. Molecules collide with each other, with energy being conserved in the collision
Temperature (T) is a measure of the average kinetic energy of the gas molecules
Chapter 6

49. Boyle’s Law
At constant T, volume (V) of a gas varies inversely with its pressure (P)
V  1/P
Or PV = a
Chapter 6

50. Chapter 6

51. Charles' Law
At constant P, the volume of a fixed amount of gas is directly proportional to its absolute T
V  T
Or V/T = a
Chapter 6

52. Chapter 6

53. Ideal Gas Law PV = nRT P = pressure V = volume n = number of moles
R = gas constant = L atm/mol K
T = absolute temperature
Chapter 6

54. Chapter 6

55. Example 6.3 Solution Exercise 6.3
What volume of oxygen is required to burn L of propane C3H8 if both gases
are measured at the same temperature and pressure?
C3H8(g) + 5 O2(g)
3 CO2(g) + 4 H2O(g)
Solution
The coefficients in the equation indicate that five volumes of O2(g) is required for every
volume of C3H8(g). Thus, we use 5 L O2(g)/1 L C3H8(g) as the ratio to find the
volume of oxygen required.
= 2.78 L O2(g)
5 L O2(g)
1LC3H8(g)
? L O2(g) = L C3H8(g) x
Exercise 6.3
Using the equation in Example 6.3, calculate the volume of CO2(g) produced
when L of propane is burned if the two gases are compared at the same
temperature and pressure.
Chapter 6

56. Example 6.8 Solution Exercise 6.8A Exercise 6.8B
Calculate the density of (a) nitrogen gas and (b) oxygen gas, both at STP.
Solution
a At STP, 1 mol (28.0 g) of N2 occupies 22.4 L:
28.0 g/ mol
22.4 L/ mol
= 1.25 g/L
b Similarly for oxygen gas:
32.0 g/ mol
22.4 L/ mol
= 1.43 g/L
Calculate the density of He at STP.
Exercise 6.8A
Estimate the density of air at STP (assume 78% N2 and 22% O2 ) and compare
this value to the value of He you calculated in part (a).
Exercise 6.8B
Chapter 6

57. Example 6.9 Solution Exercise 6.9A Exercise 6.9B
The density of diethyl ether vapor at STP is 3.30 g/L. Calculate the molar mass
of diethyl ether.
Solution
To solve, we simply multiply the density by the molar volume. The units
of liters cancel and the grams per mole, the units of molar mass, remain.
30.3 g
22.4 L
1 L
1 mol
= 73.9 g/mol x
The density of an unknown gas at STP is 2.30 g/L. Calculate its molar mass.
Exercise 6.9A
An unknown gaseous compound contains only hydrogen and carbon and
its density is 1.34 g/mol at STP. What is the formula for the compound?
Exercise 6.9B
Chapter 6

58. Example 6.13 (cont.)
b. Phosphorus reacts with oxygen to form tetraphosphorus decoxide.
The equation is
P4 + O2
P4O10
(not balanced)
How many grams of tetraphosphorus decoxide can be made from 3.50 g of
phosphorus?
Chapter 6

59. Example 6.14 Solution Exercise 6.14
A gas is enclosed in a cylinder fitted with a piston. The volume of the gas is
2.00 L at atm. The piston is moved to increase the gas pressure to
5.15 atm. Which of the following is a reasonable value for the volume of
the gas at the greater pressure?
0.20 L
0.40 L
1.00 L
16.0 L
Solution
The pressure increase is almost 10-fold. The volume should drop to about
one-tenth of the initial value. We estimate a volume of 0.20 L. (The
calculated value is L.)
Exercise 6.14
A gas is enclosed in a 10.2-L tank at 1208 mmHg. Which of the following
is a reasonable value for the pressure when the gas is transferred to a
30.0-L tank?
25 lb/in.2
300 mmHg
400 mmHg
3600 mmHg
Chapter 6

60. Example 6.15
A cylinder of oxygen has a volume of 2.25 L. The pressure of the gas is 1470
pounds per square inch (psi) at 20°C. What volume will the oxygen occupy at
standard atmospheric pressure (14.7 psi) assuming no temperature change?
Solution
It is most helpful to first separate the initial from the final condition:
Then use the equation V1P1 = V2P2 and solve for the desired volume or pressure. In this
case, we solve for V2 :
V1P1
V2 = P2
2.25 L x 1470 psi
14.7 psi
= 225 L
Because the final pressure in Example 6.15 is less than the initial pressure,
we expect the final volume to be larger than the original volume, and it is.
Chapter 6

61. Example 6.15 (cont.) Exercise 6.15A Exercise 6.15B
A sample of air occupies 73.3 mL at 98.7 atm and 0°C. What volume will the
air occupy at 4.02 atm and 0°C?
Exercise 6.15B
A sample of helium occupies 535 mL at 988 mmHg and 25°C. If the sample
is transferred to a 1.05-L flask at 25°C, what will be the gas pressure in the flask?
Chapter 6

62. Example 6.16  
A balloon indoors, where the temperature is 27°C, has a volume of 2.00 L. What
would its volume be (a) outdoors, where the temperature is –23°C and (b) in a hot car
parked in the sun where the temperature is 47°C? (Assume no change in pressure
in either case.)
Solution
First, convert all temperatures to the Kelvin scale:
T(K) = T (°C) + 273
The initial temperature is ( ) = 300 K, and the final temperatures are
(a) (– ) = 250 K and (b) ( ) = 320 K.
a We start by separating the initial from the final condition:
Solving the equation V1 T1 V2 T2 =
Chapter 6

63. Example 6.16 (cont.) Exercise 6.16A for V2 ,we have V1T2 V2 = T1
2.00 L x 250K
300 K
As we expected, the volume decreased because the temperature decreased.
b We have the same initial conditions as in (a), but different final conditions:
In this case, since the temperature increases, the volume must also increase:
V1T2 T1
V2 =
2.00 L x 320K
300 K
= 2.13 L
Exercise 6.16A
a A sample of oxygen gas occupies a volume of 2.10 L at 25°C. What volume will
this sample occupy at 150°C? (Assume no change in pressure.)
Chapter 6

64. Example 6.16 (cont.) Exercise 6.16B
b A sample of hydrogen occupies 692 L at 602°C. If the pressure is held constant,
what volume will the gas occupy after being cooled to 23°C?
Exercise 6.16B
At what Celsius temperature will the initial volume of oxygen in Exercise 6.16A occupy
0.750 L? (Assume no change in pressure.)
Chapter 6

65. Example 6.17 Solution Exercise 6.17A Exercise 6.17B
Use the ideal gas law to calculate (a) the volume occupied by 1.00 mol of nitrogen gas at
244 K and 1.00 atm pressure, and (b) the pressure exerted by mol of oxygen in a
15.0-L container at 303 K.
Solution
a We start by solving the ideal gas equation for V:
V =
nRT P
1.00 mol
1.00 atm
L • atm
mol • K x
x 244 K = 20.0 L
b Here we solve the ideal gas equation for P:
P =
nRT V
P =
0.500 mol
15.0 L
L • atm
mol • K
x 303 K = atm x
Exercise 6.17A
Determine (a) the pressure exerted by mol of oxygen in an 18.0-L container at
40°C, and (b) the volume occupied by mol of nitrogen gas at 25°C and atm.
Exercise 6.17B
Determine the volume of nitrogen gas produced from the decomposition of 130 g
sodium azide (about the amount in a typical automobile air bag) at 25°C and 1 atm.
Chapter 6

66. Chapter 6 Chapter 6 1 April 2017 06-01 Title:
Balanced Chemical Reaction Equation
Caption:
The reaction between hydrogen and oxygen is exactly balanced when four hydrogen molecules and two oxygen molecules form two water molecules.
Notes:
Note that the number of H atoms and O atoms are the same on both sides of the seesaw.
Chapter 6
CHE 105

67. Chapter 6 Chapter 6 1 April 2017 06-02 Title:
Balancing a Chemical Reaction Equation
Caption:
Incorrect and correct forms of a balanced chemical reaction equation.
Notes:
The integrity of the molecular formulas of the products must be maintained.
Chapter 6
CHE 105

68. Chapter 6 Chapter 6 1 April 2017 06-03 Title: Law of Combining Volumes
Caption:
Illustration of Gay-Lussac's law of combining volumes for gaseous reactions.
Notes:
The number of flasks for each reactant and product is the same as its stoichiometric coefficient in the balanced chemical reaction equation.
Chapter 6
CHE 105

69. Chapter 6 Chapter 6 1 April 2017 06-04 Title: Avogadro's Law Caption:
Two gases of equal volume contain the same number of molecules.
Notes:
This law is independent of the gas's molecular formula.
Chapter 6
CHE 105

70. Chapter 6 Chapter 6 1 April 2017 06-04-01UN Title: Avogadro's Number
Caption:
Illustration of the size of Avogadro's number.
Notes:
Avogadro's number is hard to grasp.
Chapter 6
CHE 105

71. Chapter 6 Chapter 6 1 April 2017 06-07 Title: Mole–Mass Relationships
Caption:
The mass of a sample of matter that corresponds to one mole of that substance depends on its chemical identity.
Notes:
Molar mass varies with each pure substance.
Chapter 6
CHE 105

72. Chapter 6 Chapter 6 1 April 2017 06-08 Title: Stoichiometry Caption:
A balanced chemical reaction equation allows us to determine mass relationships among reactants and products.
Notes:
stoichiometric relationships
Chapter 6
CHE 105

73. Chapter 6 Chapter 6 1 April 2017 06-10 Title: Boyle's Law Caption:
Kinetic molecular theory helps to explain Boyle's law.
Notes:
Note the inverse relationship between pressure and volume.
Chapter 6
CHE 105

74. Chapter 6 Chapter 6 1 April 2017 06-11 Title: Graph of Boyle's Law
Caption:
A graphic representation of Boyle's law showing the inverse relationship between pressure and volume.
Notes:
If one compresses a balloon filled with air, the pressure inside increases and the balloon will burst.
Chapter 6
CHE 105

75. Chapter 6 Chapter 6 1 April 2017 06-12 Title: Graph of Charles's Law
Caption:
Graphic representation of Charles's law showing the direct relationship between volume and temperature.
Notes:
This is the principle behind hot-air ballooning.
Chapter 6
CHE 105