Department of Chemistry CHEM1010 General Chemistry *********************************************** Instructor: Dr. Hong Zhang Foster Hall, Room 221 Tel:

CHEM1010/General Chemistry _________________________________________ Chapter 6. (L21)-Chemical Accounting Today’s Outline..Review of chemical sentences and balancing chemical reactions..Volume relationships in chemical reactions and equations..One of the most important numbers in chemistry: Avogadro’s number..Counting molecules, the unit of mole..Formula masses..Molar mass..Molar volume

Chapter 6. (L21)-Chemical Accounting Building Your Chemical Vocabulary Ca 2+ : calcium cation Mg 2+ : magnesium cation Both ions are the major contributors to the hardness of water Bottle water: Some cations are added to bring good, agreeable, inviting taste.

Chapter 6. (L21)-Chemical Accounting Volume relationships in chemical reactions and equations Gas reactions, new chemical facts: hydrogen gas + oxygen gas  water steam 2 volumes 1 volume 2 volumes Volume ratio: 2:1:2 2H 2 + O 2 = 2H 2 O Reaction coefficient ratio: 2:1:2 Interesting agreement

Chapter 6. (L21)-Chemical Accounting Volume relationships in chemical reactions and equations Gas reactions, new chemical facts: hydrogen gas + nitrogen gas  water steam 3 volumes 1 volume 2 volumes Volume ratio: 3:1:2 3H 2 + N 2 = 2NH 3 Reaction coefficient ratio: 3:1:2 Interesting agreement

Chapter 6. (L21)-Chemical Accounting Volume relationships in chemical reactions and equations Gas reactions: The law of combining volumes When all measurements are made at the same temperature and pressure, the volumes of gaseous reactants and products are in a small whole-number ratio. The law was formulized by Joseph Louis Gay- Lussac, a French scientist ( ) Key: There must be some relationship between the numbers of molecules and the volumes of gaseous reactants and products.

Chapter 6. (L21)-Chemical Accounting Volume relationships in chemical reactions and equations: Avogadro’s hypothesis How to explain the Law of Combining Volumes? Avogadro’s hypothesis: Equal volumes of all gases, when measured at the same temperature and pressure, contain the same number of molecules

Chapter 6. (L21)-Chemical Accounting Volume relationships in chemical reactions and equations: Avogadro’s hypothesis Avogadro’s hypothesis means that if we weigh equal volumes of several gases, say two gases, then the ratio of their masses should be the same mass ratio of the molecules themselves Suppose 1 volume of H 2 has 10 molecules of H 2, then, by Avogadro’s hypothesis, 1 volume of O 2 should also have 10 molecules of O 2, then Mass of 1 volume H 2 = 10 molecules×Mass-H 2 /molecule Mass of 1 volume O 2 = 10 molecules×Mass-O 2 /molecule Mass of 1 vol. H 2 /Mass of 1 vol. O 2 = Mass-H 2 /Mass-O 2

Chapter 6. (L21)-Chemical Accounting One of the most important numbers in chemistry: Avogadro’s number Avogadro’s number: The number of atoms in a 12-g sample of carbon-12 is called Avogadro’s number Avogadro’s number has been experimentally determined to be ×10 23 But, 6.02×10 23 is sufficiently enough for our purpose in general chemistry

Chapter 6. (L21)-Chemical Accounting Counting molecules, the unit of mole In chemistry, because 1 molecule or 1 atom or 1 ion has too small masses, it is inconvenient to work with just a few molecules, on the other hand, the common quantities of chemicals contain so many molecules or atoms or ions, it is inconvenient to count molecules or atoms or ions directly. (100 dollar bill, $50, 20, 10, 5, 1 bills, cents) So, in chemistry, we count molecules or atoms or ions using the unit of mole (like chemical dozen) The unit of mole is the number of molecules or atoms or ions, not the total mass of the molecules or atoms or ions (number, not mass) Mole is a unit close to the scope in our life.

Chapter 6. (L21)-Chemical Accounting Counting molecules, the unit of mole Definition: 1 mole is an amount of substance that contains the same number of elementary units (molecules, or atoms, or ions) as there are atoms in exactly 12 g of carbon-12. By the definition of Avogadro’s number, we know that 1 mole is the amount of substance that contains the same number of elementary units (molecules, or atoms, or ions) exactly as the Avogadro’s number, that is, 6.02×10 23.

Chapter 6. (L21)-Chemical Accounting Formula masses Definition: Formula mass of a molecule or ion is the sum of the masses of each of the atoms represented by the formula Example: Formula mass of O 2 = 16.0u×2 = 32.0u Formula mass of SO 2 = 32.1u u×2 = 64.1u Formula mass of CO 2 = 12.0u u×2 = 44.0u

Chapter 6. (L21)-Chemical Accounting Molar mass Definition: The molar mass of a substance is the mass of 1 mole of that substance in the unit of gram. It is numerically equivalent to the atomic mass of the atom or the formula mass of the molecule of concern. Examples: mass of 1 mole Na = 23.0 g Na mass of 1 mole CO 2 = 44.0 g CO 2 mass of 1 mole O 2 = 32.0 g O 2 mass of 1 mole CO 3 2- = 60.0 g CO 3 2-

Chapter 6. (L21)-Chemical Accounting Molar mass Conversion from moles to mass Example: How many grams of CO 2 are in mole CO 2 ? mass of 1 mole CO 2 = 44.0 g CO 2 ? g CO 2 = mole×44.0 g CO 2 /1 mole CO 2 = 22.0 g CO 2 Tip: The conversion factor lies at the molar mass of concern.

Chapter 6. (L21)-Chemical Accounting Molar mass Conversion from mass to moles Example: How many moles of CO 2 are in 11.0 g CO 2 ? mass of 1 mole CO 2 = 44.0 g CO 2 ? mole CO 2 = 11.0 g×1 mole CO 2 /44.0 g CO 2 = 0.25 mole CO 2 Tip: The conversion factor lies at the molar mass of concern.

Chapter 6. (L21)-Chemical Accounting Molar volume Definition: The volume occupied by 1 mole of gas is the molar volume under certain conditions Standard condition: 1 atmosphere (atm) (sea level air pressure) 0 ºC (water freezing point) Important number: 1 mole of gas has the volume of 22.4 L under the standard conditions. No matter what the molecule is. Althoughmass of 1 mole CO 2 = 44.0 g CO 2 mass of 1 mole O 2 = 32.0 g O 2 under the standard conditions, 1 mole of CO 2 has a volume of 22.4 L 1 mole of O 2 has a volume of 22.4 L

Chapter 6. (L21)-Chemical Accounting Molar volume Density of gas under the standard conditions: Under the standard conditions, mass of 1 mole CO 2 = 44.0 g CO 2 mass of 1 mole O 2 = 32.0 g O 2 1 mole of CO 2 has a volume of 22.4 L 1 mole of O 2 has a volume of 22.4 L Examples: Density of CO 2 at 1 atm 0 ºC = 44.0 g/22.4 L = 1.96 g/L Density of O 2 at 1 atm 0 ºC = 32.0 g/22.4 L = 1.43 g/L

Chapter 6. (L21)-Chemical Accounting Molar volume Conversion from density of gas under the standard conditions to molar mass for the gas: Examples: Given Density of CO 2 at 1 atm 0 ºC = 1.96 g/L, what is the molar mass of CO 2 ? molar mass of CO 2 = (1.96 g/L)(22.4 L/mole) = 43.9 = ~ 44 g/mole

Chapter 6. (L21)-Chemical Accounting Quiz Time For gaseous reactions, the volume ratios for the reactants and products are equal to (a) the mass ratios of the reactants and products; (b) the ratios of valence electron pairs for the reactants and products; (c) the coefficients ratios for the reactants and products as shown in the stoichiometry for the reaction; (d) the ratios of the molecular sizes for the reactants and products.

Chapter 6. (L21)-Chemical Accounting Quiz Time Avogadro’s number is the number of (a) the number of electrons of the molecules; (b) the number of protons of the molecules; (c) the number of atoms in a 12-g sample of carbon-12 atom; (d) the number of valence electrons of the molecules.

Chapter 6. (L21)-Chemical Accounting Quiz Time Avogadro’s number is the number for (a) the volume of atoms or molecules; (b) the mass of atoms or molecules; (c) the number of atoms or molecules under a certain condition; (d) the size of atoms or molecules.

Chapter 6. (L21)-Chemical Accounting Quiz Time Avogadro’s number is obtained by (a) artificial assignment; (b) theoretical calculation; (c) exact measurements through specially designed experiments; (d) an educated guess.

Chapter 6. (L21)-Chemical Accounting Quiz Time Avogadro’s number is numerically equal to (a) 6.20×10 23 ; (b) 6.02×10 32 ; (c) 2.06×10 23 ; (d) 6.02×10 23.

Chapter 6. (L21)-Chemical Accounting Quiz Time The unit of mole concerns (a) the volume of atoms or molecules; (b) the mass of atoms or molecules; (c) the number of atoms or molecules under a certain condition; (d) the size of atoms or molecules.

Chapter 6. (L21)-Chemical Accounting Quiz Time The unit of mole is defined as (a) the volume of atoms or molecules; (b) the mass of atoms or molecules; (c) 1 mole being an amount of substance that contains the same number of elementary units (molecules, or atoms, or ions) as there are atoms in exactly 12 g of carbon-12; (d) the size of atoms or molecules.

Chapter 6. (L21)-Chemical Accounting Quiz Time 1 mole of molecular substances contains (a) various numbers of molecules depending on temperature and pressure; (b) arbitrary numbers of molecules; (c) a number of molecules depending on the position the molecule has in the periodic table; (d) always the same number of molecules, that is, 6.02×10 23 molecules.

Chapter 6. (L21)-Chemical Accounting Quiz Time 1 mole of molecular substances has (a) the same mass no matter what molecule is of concern; (b) arbitrary mass; (c) the mass numerically equal to its molecular formula mass; (d) the mass of 6.02×10 23.

Chapter 6. (L21)-Chemical Accounting Quiz Time M olar mass of a substance is (a) the mass of the protons of that substance; (b) the mass of the electrons of that substance; (c) the mass of 1 mole of that substance of concern, which is numerically equal to the atomic mass or formula mass; (d) the mass of the neutrons of that substance.

Chapter 6. (L21)-Chemical Accounting Quiz Time The formula mass of N 2 is (a) 14.0u; (b) 1.4.0u; (c) 28.0u; (d) 2.8u.

Chapter 6. (L21)-Chemical Accounting Quiz Time The formula mass of CO is (a) 12.0u; (b) 16.0u; (c) 28.0u; (d) 24.0u.

Chapter 6. (L21)-Chemical Accounting Quiz Time The formula mass of NaHCO 3 is (a) 24.0u; (b) 23.0u; (c) 60.0u; (d) 84.0u.

Chapter 6. (L21)-Chemical Accounting Quiz Time 1 mole of NaHCO 3 is (a) 24.0 g NaHCO 3 ; (b) 23.0 g NaHCO 3 ; (c) 60.0 g NaHCO 3 ; (d) 84.0 g NaHCO 3.

Chapter 6. (L21)-Chemical Accounting Quiz Time 1 mole of SO 2 is (a) 6.40 g SO 2 ; (b) 32.0 g SO 2 ; (c) 64.0 g SO 2 ; (d) 16.0 g SO 2.

Chapter 6. (L21)-Chemical Accounting Quiz Time 0.25 mole of SO 2 is (a) 25.0 g SO 2 ; (b) 32.0 g SO 2 ; (c) 64.0 g SO 2 ; (d) 16.0 g SO 2.

Chapter 6. (L21)-Chemical Accounting Quiz Time 96 g SO 2 is (a) 0.5 mole SO 2 ; (b) 1.0 mole SO 2 ; (c) 2.0 moles SO 2 ; (d) 1.5 moles SO 2.

Chapter 6. (L21)-Chemical Accounting Quiz Time 96 g S is (a) 0.5 mole S; (b) 1.0 mole S; (c) 2.0 moles S; (d) 3.0 moles S.

Chapter 6. (L21)-Chemical Accounting Quiz Time 0.25 mole of S is (a) 12.0 g S; (b) 32.0 g S; (c) 96 g S; (d) 24.0 g S.

Chapter 6. (L21)-Chemical Accounting Quiz Time The standard condition in chemistry is commonly referred to (a) 0 atm and 0 ºC; (b) 100 atm and 100 ºC; (c) 1 atm and 1 ºC; (d) 1 atm and 0 ºC.

Chapter 6. (L21)-Chemical Accounting Quiz Time At the standard condition, 1 mole of gas has (a) various volumes depending on the molecule of the gas of concern; (b) no volume at all; (c) 22.4 liter or L; (d) 44.2 L.

Chapter 6. (L21)-Chemical Accounting Quiz Time At the standard condition, 1 mole of CO gas has a density of (a) 28.0g/L; (b) 0.0 g/L; (c) 22.4 g/L; (d) 1.25 g/L.