4 Calculations and the Chemical Equation GENERAL CHEMISTRY General, Organic and Biochemistry 7th Edition Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

4.1 The Mole Concept and Atoms Atoms are exceedingly small Unit of measurement for mass of an atom is atomic mass unit (amu) carbon-12 assigned the mass of exactly 12 amu 1 amu = 1.66  10-24 g Periodic table gives atomic weights in amu

4.1 The Mole Concept and Atoms Mass of Atoms What is the atomic weight of one atom of fluorine? Answer: 19.00 amu What would be the mass of this one atom in grams? Chemists usually work with much larger quantities It is more convenient to work with grams than amu when using larger quantities 4.1 The Mole Concept and Atoms

The Mole and Avogadro’s Number A practical unit for defining a collection of atoms is the mole 1 mole of atoms = 6.022  1023 atoms This is called Avogadro’s number This has provided the basis for the concept of the mole 4.1 The Mole Concept and Atoms

4.1 The Mole Concept and Atoms To make this connection we must define the mole as a counting unit The mole is abbreviated mol A mole is simply a unit that defines an amount of something Dozen defines 12 Gross defines 144 2 4.1 The Mole Concept and Atoms

4.1 The Mole Concept and Atoms Atomic Mass The atomic mass of one atom of an element corresponds to: The average mass of a single atom in amu The mass of a mole of atoms in grams 1 atom of F is 19.00 amu 19.00 amu/atom F 1 mole of F is 19.00 g 19.00 g/mole F 4.1 The Mole Concept and Atoms =19.00 g/mol F

Calculating Atoms, Moles, and Mass We use the following conversion factors: Density converts grams  milliliters Atomic mass unit converts amu grams Avogadro’s number converts moles  number of atoms Molar mass converts grams  moles 2 4.1 The Mole Concept and Atoms

Strategy for Calculations Map out a pattern for the required conversion Given a number of grams and asked for number of atoms Two conversions are required Convert grams to moles 1 mol S/32.06 g S OR 32.06 g S/1 mol S Convert moles to atoms mol S  (6.022  1023 atoms S) / 1 mol S 4.1 The Mole Concept and Atoms

Practice Calculations Calculate the number of atoms in 1.7 moles of boron. Find the mass in grams of 2.5 mol Na (sodium). Calculate the number of atoms in 5.0 g aluminum. Calculate the mass of 5,000,000 atoms of Au (gold) 4.1 The Mole Concept and Atoms

Interconversion Between Moles, Particles, and Grams 4.1 The Mole Concept and Atoms

4.2 The Chemical Formula, Formula Weight, and Molar Mass 3 Chemical formula - a combination of symbols of the various elements that make up the compound Formula unit - the smallest collection of atoms that provide two important pieces of information The identity of the atoms The relative number of each type of atom

4.2 The Chemical Formula, Formula Weight and Molar Mass Consider the following formulas: H2 – 2 atoms of hydrogen are chemically bonded forming diatomic hydrogen, subscript 2 H2O – 2 atoms of hydrogen and 1 atom of oxygen, lack of subscript means one atom NaCl – 1 atom each of sodium and chlorine Ca(OH)2 – 1 atom of calcium and 2 atoms each of oxygen and hydrogen, subscript outside parentheses applies to all atoms inside 4.2 The Chemical Formula, Formula Weight and Molar Mass

4.2 The Chemical Formula, Formula Weight and Molar Mass Consider the following formulas: (NH4)3SO4 – 3 ammonium ions and 1 sulfate ion Ammonium ion contains 1 nitrogen and 4 hydrogen Sulfate ion contains 1 sulfur and 4 oxygen Compound contains 3 N, 12 H, 1 S, and 4 O CuSO4.5H2O This is an example of a hydrate - compounds containing one or more water molecules as an integral part of their structure 5 units of water with 1 CuSO4 4.2 The Chemical Formula, Formula Weight and Molar Mass

Comparison of Hydrated and Anhydrous Copper Sulfate 4.2 The Chemical Formula, Formula Weight and Molar Mass Hydrated copper sulfate Anhydrous copper sulfate Marked color difference illustrates the fact that these are different compounds

Formula Weight and Molar Mass Formula weight - the sum of the atomic weights of all atoms in the compound as represented by its correct formula expressed in amu What is the formula weight of H2O? 16.00 amu + 2(1.008 amu) = 18.02 amu Molar mass – mass of a mole of compound in grams / mole Numerically equal to the formula weight in amu What is the molar mass of H2O? 18.02 g/mol H2O 4 4.2 The Chemical Formula, Formula Weight and Molar Mass

Formula Unit 4.2 The Chemical Formula, Formula Weight and Molar Mass Formula unit – smallest collection of atoms from which the formula of a compound can be established 4.2 The Chemical Formula, Formula Weight and Molar Mass What is the molar mass of (NH4)3PO4? 3(N amu) + 12(H amu) + P amu + 4(O amu)= 3(14.01) + 12(1.008) + 30.97 + 4(16.00)= 149.10 g/mol (NH4)3PO4

4.2 The Chemical Formula, Formula Weight and Molar Mass Molar mass - The mass in grams of 1 mole of atoms What is the molar mass of carbon? 12.01 g/mol C This means counting out a mole of Carbon atoms (i.e., 6.022  1023) they would have a mass of 12.01 g One mole of any element contains the same number of atoms, 6.022  1023, Avogadro’s number 4.2 The Chemical Formula, Formula Weight and Molar Mass

4.5 Calculations Using the Chemical Equation Calculating quantities of reactants and products in a chemical reaction has many applications Need a balanced chemical equation for the reaction of interest The coefficients represent the number of moles of each substance in the equation 9

4.5 Calculations Using the Chemical Equation General Principles Chemical formulas of all reactants and products must be known Equation must be balanced to obey the law of conservation of mass Calculations of an unbalanced equation are meaningless Calculations are performed in terms of moles Coefficients in the balanced equation represent the relative number of moles of products and reactants 4.5 Calculations Using the Chemical Equation

Using the Chemical Equation Examine the reaction: 2H2 + O2  2H2O Coefficients tell us? 2 mol H2 reacts with 1 mol O2 to produce 2 mol H2O What if 4 moles of H2 reacts with 2 moles of O2? It yields 4 moles of H2O 4.5 Calculations Using the Chemical Equation

Using the Chemical Equation 2H2 + O2  2H2O The coefficients of the balanced equation are used to convert between moles of substances How many moles of O2 are needed to react with 4.26 moles of H2? Use the factor-label method to perform this calculation 4.5 Calculations Using the Chemical Equation

Use of Conversion Factors 2H2 + O2  2H2O 1 2 2.13 mol O2 4.5 Calculations Using the Chemical Equation Digits in the conversion factor come from the balanced equation

Conversion Between Moles and Grams 9 Requires only the formula weight Convert 1.00 mol O2 to grams Plan the path Find the molar mass of oxygen 32.0 g O2 = 1 mol O2 Set up the equation Cancel units 1.00 mol O2  32.0 g O2 1 mol O2 Solve equation 1.00  32.0 g O2 = 32.0 g O2 moles of Oxygen grams of Oxygen 4.5 Calculations Using the Chemical Equation

Conversion of Mole Reactants to Mole Products Use a balanced equation C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) 1 mol C3H8 results in: 5 mol O2 consumed 1 mol C3H8 /5 mol O2 3 mol CO2 formed 1 mol C3H8 /3 mol CO2 4 mol H2O formed 1 mol C3H8 /4 mol H2O This can be rewritten as conversion factors 4.5 Calculations Using the Chemical Equation

Calculating Reacting Quantities Calculate grams O2 reacting with 1.00 mol C3H8 Use 2 conversion factors Moles C3H8 to moles O2 Moles of O2 to grams O2 Set up the equation and cancel units 1.00 mol C3H8  5 mol O2  32.0 g O2 = 1 mol C3H8 1 mol O2 1.00  5  32.0 g O2 = 1.60 x 102 g O2 4.5 Calculations Using the Chemical Equation moles C3H8 moles Oxygen grams Oxygen

Calculating Grams of Product from Moles of Reactant Calculate grams CO2 from combustion of 1.00 mol C3H8 Use 2 conversion factors Moles C3H8 to moles CO2 Moles of CO2 to grams CO2 Set up the equation and cancel units 1.00 mol C3H8  3 mol CO2  44.0 g CO2 = 1 mol C3H8 1 mol CO2 1.00  3  44.0 g CO2 = 1.32  102 g CO2 4.5 Calculations Using the Chemical Equation moles C3H8 moles CO2 grams CO2

Relating Masses of Reactants and Products Calculate grams C3H8 required to produce 36.0 grams of H2O Use 3 conversion factors Grams H2O to moles H2O Moles H2O to moles C3H8 Moles of C3H8 to grams C3H8 Set up the equation and cancel units 36.0 g H2O  1 mol H2O  1 mol C3H8  44.0 g C3H8 18.0 g H2O 4 mol H2O 1 mol C3H8 36.0  [1/18.0]  [1/4]  44.0 g C3H8 = 22.0 g C3H8 4.5 Calculations Using the Chemical Equation grams H2O moles H2O moles C3H8 grams C3H8

Calculating a Quantity of Reactant Ca(OH)2 neutralizes HCl Calculate grams HCl neutralized by 0.500 mol Ca(OH)2 Write chemical equation and balance Ca(OH)2(s) + 2HCl(aq) CaCl2(s) + 2H2O(l) Plan the path Set up the equation and cancel units 0.500 mol Ca(OH)2  2 mol HCl  36.5 g HCl 1 mol Ca(OH)2 1 mol HCl Solve equation 0.500  [2/1]  36.5 g HCl = 36.5 g HCl 4.5 Calculations Using the Chemical Equation moles Ca(OH)2 moles HCl grams HCl

A Visual Example of the Law of Conservation of Mass 4.5 Calculations Using the Chemical Equation

General Problem-solving Strategy 4.5 Calculations Using the Chemical Equation

4.5 Calculations Using the Chemical Equation Sample Calculation Na + Cl2  NaCl Balance the equation Calculate the moles Cl2 reacting with 5.00 mol Na Calculate the grams NaCl produced when 5.00 mol Na reacts with an excess of Cl2 Calculate the grams Na reacting with 5.00 g Cl2 2Na + Cl2  2NaCl 4.5 Calculations Using the Chemical Equation

Theoretical and Percent Yield Theoretical yield - the maximum amount of product that can be produced Pencil and paper yield Actual yield - the amount produced when the reaction is performed Laboratory yield Percent yield: 10 4.5 Calculations Using the Chemical Equation = 125 g CO2 actual  100% = 97.4% 132 g CO2 theoretical

4.5 Calculations Using the Chemical Equation Sample Calculation If the theoretical yield of iron was 30.0 g and actual yield was 25.0 g, calculate the percent yield: 2 Al(s) + Fe2O3(s)  Al2O3(aq) + 2Fe(aq) [25.0 g / 30.0 g]  100% = 83.3% Calculate the % yield if 26.8 grams iron was collected in the same reaction 4.5 Calculations Using the Chemical Equation