#2 Exam Range 16 59 out of 60 Mean score 46.4/60 = 77.3% Curve = 6% > 60 % in A/B range May apply another curve at the end of the semester Example: someone with 48/60 (80%) gets an adjusted score of 86% (high B) Web postings exam keys, curves file, results histograms, your grades up to now Homework chapter 6 See WebCT, Due Tuesday November 13 Before starting Chapter 6, lets look at videos relevant to Chapter 5. After all, Chapter 5 material will covered on the final exam
Chapter 6
Chemistry for Changing Times
Chemical Accounting
Chemical Sentences: Equations
Chemical equations represent the sentences in the language of chemistry. They are the means of communicating a chemical change using the symbols and formulas to represent the elements and compounds involved in a chemical reaction.
Chemical Sentences: Equations
Reactants are the species present before the reaction. Products are the species present after the reaction. Reactants Products
The arrow () means yield(s) or react(s) to produce. See example on next slide
Chemical Sentences: Equations
The following are used to denote the states of matter of a species in an equation: (s) (l) (g) (aq) = = = = solid liquid gas aqueous solution
C(s) + O2(g) CO2(g) {@ coal burning power plants}
Chemical Sentences: Equations
Coefficients are numbers used to balance a chemical equation. Never change the subscripts.
Hydrogen fuel Hydrogen economy
Balance equations by trial and error AND/OR use this strategy: 1) If an element occurs in just one substance on each side of the equation, try balancing that element first 2) Balance any reactants or products that exist as free elements last ALWAYS check your result! Lets Balance: Fe2O3 + C CO2 + Fe C3H8 + O2
Conservation of atoms
Iron ore plus coal (steel industry) Burning propane
CO2 + H2O
Volume Relationships in Chemical Equations
Law of Combined Volumes: When all measurements are made at the same temperature and pressure, the volumes of gaseous reactants and products are in a small wholenumber ratio.
Joseph Louis Gay-Lussac 1809
Reaction above is the Haber Process
Volume Relationships in Chemical Equations Avogadros Hypothesis (1911): Volumes of all gases, when measured at the same temperature and pressure, contain the same number of molecules.
EXAMPLE 6.3
Volume Relationships of Gases
What volume of oxygen is required to burn 0.556 L of propane 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 each volume of C3H8(g) requires 5 volumes of O2(g). Thus, we use 5 L O2(g)/1 L C3H8(g) as the ratio to find the volume of oxygen required. ? L O2(g) = O.556 L C3 H8(g) x Exercise 6.3A Using the equation in Example 6.3, calculate the volume of CO2(g) produced when 0.492 L of propane is burned if the two gases are compared at the same temperature and pressure. Exercise 6.3B If 10.0 L each of propane and oxygen are combined at the same temperature and pressure, which gas will be over left after reaction? What volume of that gas will remain? 5 L O2(g) 1 L C3 H8(g) = 2.78 L O2(g)
Avogadros Number
Avogadros number is defined as the number of atoms in a 12-g sample of carbon-12 and is:
6.02 x
23 10
See article #21 for biography of Avogadro
The Mole A mole (mol) is defined as the amount of a substance that 23 particles. contains 6.02 x 10
October 23 each year is Mole Day! link: http://www.moleday.org/ (article #22)
The Mole
Formula mass is the average mass of a formula unit relative to that of a carbon-12 atom.
It is simply the sum of the atomic masses for all atoms in a formula. If the formula represents a molecule, often the term molecular mass is used
Example: H2O A mole of a substance is its atomic or molecular mass in grams
EXAMPLE 6.4
Calculating Molecular Masses
Calculate (a) the molecular mass of nitrogen dioxide (NO2) an amber colored gas that is a constituent of smog, and (b) the formula mass of ammonium sulfate [(NH4)2SO4] a fertilizer commonly used by home gardeners.
Solution a. We start with the molecular formula: NO2. Then, to determine the molecular mass, we need only to add the atomic mass of nitrogen to twice the atomic mass of oxygen. 1 x atomic mass of N = 1 x 14.0 u = 14.0 u 2 x atomic mass of O = 2 x 16.0 u = 32.0 u Formula mass of NO2 = 46.0 u Using a calculator, we need only write down the final answer, 46.0 u. That is, we have no need to record the numbers 14.0 and 32.0. We must make certain that all the atoms in the formula unit are accounted for, which means paying particular attention to all the subscripts and parentheses in the formula. The (NH4)2 means that both the N and the H4 must be multiplied by 2that is, the formula indicates a total of two N atoms and eight H atoms. Combining the atomic masses, we have 2 x atomic mass of N = 2 x 14.0 u 8 x atomic mass of H = 8 x 1.01 u 1 x atomic mass of S = 1 x 32.0 u 4 x atomic mass of O = 4 x 16.0 u Formula mass of (NH4)2SO4 = 28.0 u = 8.08 u = 32.0 u = 64.0 u = 132.1 u
b.
EXAMPLE 6.5
Mole-to-Mass Conversions
How many grams of N2 are in 0.400 moles N2? Solution The molecular mass of N2 is 2 x 14.0 u = 28.0 u. The molar mass of N2 is therefore 28.0 g/mol. Using the molar mass as a conversion factor (red), we have ? g N2 = 0.400 mol N2 x Exercise 6.5 Calculate the mass, in grams, of (a) 0.0728 mol silicon, (b) 55.5 mol H2O, and (c) 0.0728 mol Ca(H2PO4)2. 28.0 g N2 1 mol N2 = 11.2 g N2
EXAMPLE 6.6
Mass-to-Mole Conversions
Calculate the number of moles of Na in a 62.5-g sample of sodium metal. Solution The molar mass of Na is 23.0 g/mol. To convert from a mass in grams to an amount in moles, we must use the inverse of the molar mass as a conversion factor (1 mol Na/23.0 g Na) to get the proper cancellation of units. When we start with grams, we must have grams in the denominator of our conversion factor (red). ? mol Na = 62.5 g Na x Exercise 6.6 Calculate the amount, in moles, of (a) 3.71 g Fe, (b) 165 g butatne, C4H10, and (c) 0.100 mol Mg(NO3)2. 1 mol Na 23.0 g Na = 2.72 mol Na