# Chemical Accounting: Mass and Volume Relationships

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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

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

C(s) + O2(g)  CO2(g) (s) solid (l) liquid (g) gas (aq) aqueous or in water Chapter 6

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

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

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

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

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

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

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

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

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

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

Chapter 6

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

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

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

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

Chapter 6

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

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

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

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

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

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

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

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

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

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

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

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

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

Chapter 6

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

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

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

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

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

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

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

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

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

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

End of Chapter 6 Chapter 6

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

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

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

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

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

Chapter 6

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

Chapter 6

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

Chapter 6

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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