Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations
Matter Matter is anything that has mass and takes up space 2
Composition of Matter Atom number of protons = atomic number (Z) number of neutrons = mass number atomic number = A - Z 3
Periodic Table A systematic organization of the elements Elements are arranged in order of atomic number 4
The quantity of materials consumed and produced in chemical reactions STOICHIOMETRY 5
CHEMICAL EQUATIONS 6
Chemical Equations Chemical equations are concise representations of chemical reactions 7
What Is in a Chemical Equation? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) Reactants appear on the left side of the equation 8
What Is in a Chemical Equation? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) Products appear on the right side of the equation 9
What Is in a Chemical Equation? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) The states of the substance is written in parentheses to the right of each compound (g) = gas; (l) = liquid; (s) = solid; (aq) = in aqueous solution 10
What Is in a Chemical Equation? CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) Coefficients balance the equation to follow the law of conservation of mass 11
Balancing Equations Hydrogen and oxygen can make water OR hydrogen peroxide: H 2 (g) + O 2 (g) H 2 O(l) H 2 (g) + O 2 (g) H 2 O 2 (l) 2 H 2 (g) + O 2 (g) 2 H 2 O(l) 12
Balancing Equations: Stoichiometry The study of the mass relationships in chemistry Based on the Law of Conservation of Mass (Antoine Lavoisier, 1789) Total mass of substances present at the beginning of a chemical reaction is equal to the mass at the end Mass = X Mass = X 13
EXAMPLE Write a balanced equation for the reaction between solid silicon dioxide and solid carbon to produce solid silicon carbide and carbon monoxide gas. WRITING BALANCED CHEMICAL EQUATIONS 1.Write the unbalanced equation by writing chemical formulas for each of the reactants and products. (If the unbalanced equation is provided in the problem, skip this step and go to Step 2.) 2.If an element occurs in only one compound on both sides of the equation, balance it first. If there is more than one such element, balance metals before nonmetals. SOLUTION SiO 2 (s) + C(s) SiC(s) + CO(g) BEGIN WITH Si SiO 2 (s) + C(s) SiC(s) + CO(g) 1 Si atom 1 Si atom Si is already balanced. BALANCE O NEXT SiO 2 (s) + C(s) SiC(s) + CO(g) 2 O atoms 1 O atom To balance O, put a 2 before CO(g). SiO 2 (s) + C(s) SiC(s) + 2 CO(g) 2 O atoms 2 O atoms 14
EXAMPLE Continued 3.If an element occurs as a free element on either side of the chemical equation, balance it last. Always balance free elements by adjusting the coefficient on the free element. 4.If the balanced equation contains coefficient fractions, change these into whole numbers by multiplying the entire equation by the appropriate factor. 5.Check to make certain the equation is balanced by summing the total number of each type of atom on both sides of the equation. BALANCE C SiO 2 (s) + C(s) SiC(s) + 2 CO(g) 1 C atom 1 C + 2 C = 3 C atoms To balance C, put a 3 before C(s). SiO 2 (s) + 3 C(s) SiC(s) + 2 CO(g) 3 C atoms 1 C + 2 C = 3 C atoms This step is not necessary in this example. Proceed to Step 5. SiO 2 (s) + 3 C(s) SiC(s) + 2 CO(g) The equation is balanced. 15
EXAMPLE Continued SKILLBUILDER Write a balanced equation for the reaction between solid chromium(iii) oxide and solid carbon to produce solid chromium and carbon dioxide gas. Answer: 16
Chemical Equations Chemical equations must Be consistent with experimental facts Contain only reactants and products involved in reaction Must follow law of conservation of mass Equations must be balanced Number of type of atoms in reactants must equal products Use the smallest whole numbers possible 17
TYPES OF CHEMICAL REACTION 18
Three Types of Reactions 1. Combination reactions 19
Combination Reaction Two or more substances react to form one product A + B C 2 Mg(s) + O 2 (g) 2 MgO(s) 20
Three Types of Reactions 1. Combination reactions 2. Decomposition reactions 21
Decomposition Reaction One substance breaks down into two or more substances C A + B 2 NaN 3 (s) 2 Na(s) + 3 N 2 (g) 22
Three Types of Reactions 1. Combination reactions 2. Decomposition reactions 3. Combustion reactions 23
Combustion Reactions Combustion reactions generally rapid reactions that produce a flame involve oxygen in the air as a reactant Example: CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) 24
QUANTIFYING SUBSTANCES 25
Formula weight Formula Weight Sum of the atoms mass in a chemical formula Calculated for compounds and pure substances Quantitative significance of a formula Calculate the formula weight of calcium chloride, CaCl 2 Ca: 1(40.08 amu) + Cl: 2(35.453 amu) 110.99 amu 26
Molecular Weight Molecular Weight Sum of the atomic mass of the atoms in a molecule Calculate the molecular weight of ethane, C 2 H 6 C: 2(12.011 amu) + H: 6(1.00794 amu) 30.070 amu 27
Mass Percent Mass percent of an element Number of grams of the element present in 100 grams of the compound Percent composition by mass of water is 88.81% oxygen and 11.19% hydrogen 88.81 g oxygen and 11.19 g hydrogen in 100 g H 2 O 28
Mass Percent Mass percent of an element Number of grams of the element present in 100 grams of the compound Formula: number of atoms mass of element % element 100 mass of compound sample 29
EXAMPLE MASS PERCENT COMPOSITION Calculate the mass percent of Cl in freon-114 (C 2 Cl 4 F 2 ). SORT You are given the molecular formula of freon-114 and asked to find the mass percent of Cl. STRATEGIZE You can use the information in the chemical formula to substitute into the mass percent equation and obtain the mass percent Cl. GIVEN: C 2 Cl 4 F 2 FIND: Mass % Cl SOLUTION MAP RELATIONSHIPS USED Mass percent of element X = 30
EXAMPLE Continued MASS PERCENT COMPOSITION SOLVE Calculate the molar mass of freon-114 and substitute the values into the equation to find mass percent Cl. SOLUTION 4 Molar mass Cl= 4(35.45 g) = 141.8 g Molar mass C 2 Cl 4 F 2 = 2(12.01) + 4(35.45) + 2(19.00) = 24.02 + 141.8 + 38.00 = 69.58% 31
EXAMPLE USING MASS PERCENT COMPOSITION AS A CONVERSION FACTOR The FDA recommends that adults consume less than 2.4 g of sodium per day. How many grams of sodium chloride can you consume and still be within the FDA guidelines? Sodium chloride is 39% sodium by mass. SORT You are given the mass of sodium and the mass percent of sodium in sodium chloride. When mass percent is given, write it as a fraction. Percent means per hundred, so 39% sodium indicates that there are 39 g Na per 100 g NaCl. You are asked to find the mass of sodium chloride that contains the given mass of sodium. GIVEN: FIND: g NaCl SOLUTION MAP STRATEGIZE Draw a solution map that starts with the mass of sodium and uses the mass percent as a conversion factor to get to the mass of sodium chloride. RELATIONSHIPS USED 39 g Na : 100 g NaCl 32
EXAMPLE Continued SOLVE Follow the solution map to solve the problem, beginning with grams Na and ending with grams of NaCl. The amount of salt you can consume and still be within the FDA guideline is 6.2 g NaCl. USING MASS PERCENT COMPOSITION AS A CONVERSION FACTOR SOLUTION. Twelve and a half packets of salt contain 6.2 g NaCl. 33
Mole Difficult to count individual molecules due to its small size Chemist use a mole to work with smaller number 6.02 10 23 atoms or molecules Chemist counting unit 34
Avogadro s Number Moles provide a bridge from the molecular scale to the real-world scale Conversion factor 1 mole = 6.022 X 10 23 objects Objects Particles Molecules Formula units Atoms 35
Avogadro s Number: Mole Relationships One mole of atoms, ions, or molecules equals Avogadro s number of those particles Number of atoms or ions in molecules or formula units contain Avogadro s number times the number of each element in the compound 36
Avogadro s Number: Significant Figures Limiting factor for significant figure determination should NEVER be Avogadro s number Most accurate value of Avogadro s number to date is 6.0221415 x 10 23 6.022 x 10 23 is sufficient for most calculations 37
Molar Mass Molar mass is the mass of 1 mol of a substance (i.e., g/mol) Atomic mass for an element on the periodic table (g/mol) Formula weight is equal to the molar mass (in g/mol) 38
Calculate Molar Mass What is the mass in grams of 1.000 mol of glucose, C 6 H 12 O 6? Given : Chemical formula and 1 mole of glucose Find: mass of glucose formula mass: 6 C atoms = 6 (12.0 amu) = 72.0 amu 12 H atoms = 12( 1.0 amu) = 12.0 amu 6 O atoms = 6(16.0 amu) = 96.0 amu 72.0 amu + 12.0 amu + 96.0 amu = 180.0 amu Because glucose formula mass is 180.0 amu, one mole of this substance has a mass of 180.0 g. So the molar mass is 180.0 g/mol 39
Molar Mass Molar mass (g/mol) is numerically equal to atomic mass (amu) Units do not have same value 6.022 x 10 23 amu = 1.000 g 40
Conversions Using Moles Moles provide a bridge from the molecular scale to the real-world scale 41
Convert Grams to Moles Calculate the number of moles of glucose (C 6 H 12 O 6 ) in 5.380 g of C 6 H 12 O 6 Know: 5.380 g of C 6 H 12 O 6 and chemical formula Find: moles of glucose molar mass: 72.0 g/mol + 12.0 g/mol + 96.0 g/mol = 180.0 g/mol Conversion factor: 1 mol C6H12O6 180.0 g C H 6 12 O 6 Equation: 1 mol C H O 5.380 g C H mol C H O 6 12 6 6 12O6 0. 02989 180.0 g C6H12O6 6 12 6 42
Calculate Number of Molecules How many glucose molecules are in 5.23 g of C 6 H 12 O 6? Know: 5.23 g of C 6 H 12 O 6 and chemical formula Find: # of glucose molecules Calculate molar mass: 72.0 g/mol + 12.0 g/mol + 96.0 g/mol = 180.0 g/mol Conversion factor: 23 Equation: 1 mol C H O 1 mol C6H 180.0 g C H 6 6.022 10 12 O6 O 12 molecules 23 6 12 6 6 12 6 22 5.23 g C6H12O6 1.75 10 molecules C6H12O6 180.0 g C6H12O6 1 mol C6H12O6 6 C H 6.02 10 molecules C6H 1 mol C H O O 6 12 6 12 O 6 43
44 Calculate Number of Atoms How many oxygen atoms are in 5.23 g of C 6 H 12 O 6? Know: 1.75 x 10 22 molecules C 6 H 12 O 6 (from previous slide) and chemical formula Find: # O atoms Conversion factor: Equation: 6 O atoms 1 molecule C H 6 12 O 6 22 6 atoms O 1.75 10 molecules C6H12O6 1.05 10 1 molecule C H O 6 12 6 23 atoms O
Convert Moles to Atoms Calculate the number of H atoms in 0.350 mol of C 6 H 12 O 6 Know: 0.350 mol of C 6 H 12 O 6 Find: # H atoms Conversion factor: Equation: 23 6.02 10 molecules C6H 1 mol C H O 6 12 6 and chemical formula 12 O 6 12 H atoms 1 molecule C H 23 6.02 10 molecules C6H12O6 12 H atoms 0.350 mol C6H12O6 2.53 10 1 mol C H O 1 molecule C H O 6 12 6 6 12 6 6 12 24 O 6 H atoms 45
EMPIRICAL FORMULAS FROM ANALYSES 46
Determining Empirical Formulas Empirical formula from experimental data 1. Direct analysis Percent composition Mass of each element present 47
Determining Empirical Formulas Steps in determining the empirical formula from the percent composition 48
Determining Empirical Formulas an Example The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA. Step 1: Assuming 100.00 g of para-aminobenzoic acid, C: 61.31 g H: 5.14 g N: 10.21 g O: 23.33 g 49
Determining Empirical Formulas an Example Step 2: Calculate moles of each element 1 mol 12.01 g C: 61.31 g = 5.105 mol C 1 mol 1.01 g H: 5.14 g = 5.09 mol H 1 mol 14.01 g N: 10.21 g = 0.7288 mol N 1 mol 16.00 g O: 23.33 g = 1.458 mol O 50
Determining Empirical Formulas an Example Step 3: Calculate the mole ratio by dividing by the smallest number of moles: 5.105 mol C C: 5.105 mol = 7.005 7 0.7288 mol 5.09 mol H 0.7288 mol N 1.458 mol O H: 5.09 mol 0.7288 mol = 6.984 7 N: 0.7288 mol 0.7288 mol = 1.000 O: 1.458 mol 0.7288 mol = 2.001 2 51
Determining Empirical Formulas an Example Mole ratio calculated are the subscripts for the empirical formula: C 7 H 7 NO 2 52
Determining Empirical Formulas Empirical formula from experimental data 1. Direct analysis Percent composition Mass of each element present 2. Indirect analysis Combustion analysis 53
Combustion Analysis Compounds containing C, H, and O are routinely analyzed through combustion in a chamber like the one shown in figure above C is determined from the mass of CO 2 produced H is determined from the mass of H 2 O produced O is determined by the difference after C and H have been determined 54
55 Determine Empirical Formula by Combustion Analysis Isopropyl alcohol, a substance sold as rubbing alcohol, is composed of C, H, and O. Combustion of 0.255 g of isopropyl alcohol produces 0.561 g of CO 2 and 0.306 g of H 2 O. Determine the empirical formula of isopropyl alcohol. Know: What atoms reactant contains; Amount of CO 2 (0.561g) and H 2 O (0.306g) produced upon combustion of 0.255g of alcohol 1 st : Calculate g carbon present in CO 2 : Know: Molar mass of CO 2 is 44.0 g/mol and 0.561 g CO 2 present Conversion factor: Equation: 1 mol C 1 mol CO 2 12 g C 1 mol C 1 mol CO2 1 mol C 12 g C 0.561 g CO2 0. 153 g C 44.0 g CO 1 mol CO 1 mol C 2 2 1 mol CO 44.0 g CO 2 2
Determine Empirical Formula by Combustion Analysis 2 nd : Calculate g H present in H 2 O: Know: 18.0 g/mol H 2 O, 0.306 g H 2 0 and chemical forumula Conversion factor: 2 mol H 1 mol H O 2 1.01 g H 1 mol H 1 mol H2O 18.0 g H O 2 Equation: 3 rd : Calculate g O: 1 mol H2O 2 mol H 1.01 g H 0.306 g H2O 0. 0343 g H 18.0 g H O 1 mol H O 1 mol H Know: 0.255 g alcohol and mass of C and H present Equation: mass of O = mass of sample (mass of C + mass of H) = 0.255 g (0.153 g + 0.0343 g) = 0.068 g O 2 2 56
57 Determine Empirical Formula by Combustion Analysis 4 th : Calculate number of moles in C, H and O: Know: 0.153 g C, 0.0343 g H, and 0.068 g O in alcohol Conversion Factor: Equations: 1 mol C 12.0 g C 1 mol H 1.01 g H 1 mol C 0.153 g C 0. 0128 mol C 12.0 g C 1 mol O 16.0 g O 1 mol H 0.0343 g H 0. 0340 mol 1.01 g H H 1 mol O 0.068 g O 0. 0043 mol O 16.0 g O
58 Determine Empirical Formula by Combustion Analysis 5 th : Compare relative number of moles to find empirical formula Know: 0.0128 mol C, 0.0340 mol H, and 0.0043 mol O in alcohol Equations: 0.0128 0.0340 0.0043 C : 3.0 H : 7. 9 O : 1 0.0043 0.0043 0.0043 Empirical Formula: C 3 H 8 O
Determining Molecular Formulas Molecular formula from experimental data 1. Determine empirical formula 2. Determine molecular formula mass 3. Determine molecular formula Molecular formula = (empirical formula) x x is a whole number x = molecular formula mass / empirical formula mass Number of atoms in a molecular formula is a multiple of the number of atoms in an empirical formula 59
Determining a Molecular Formula an Example The empirical formula of a compound was found to be CH. It has a molar mass of 78 g/mol. What is its molecular formula? Solution: Mass of empirical formula = 12 g + 1 g = 13 g Whole-number multiple = 78 g / 13 g = 6 The molecular formula is C 6 H 6 60
QUANTITATIVE RELATIONSHIPS 61
Quantitative Relationships: Chemical Formulas Chemical formula has two interpretations 1. Microscopic-level Indicates number of atoms of each element present in one molecule or formula unit of substance Subscripts indicate number of atoms present 2. Macroscopic-level Indicates number of moles of atoms of each element present in one mole of a substance Subscripts indicate numbers of moles of atoms present Microscopic Macroscopic CC 2 2 H 6 6 CC 2 H 2 6 6. 22 atoms of of carbon 22 moles of of carbon 66 atoms atoms of of hydrogen hydrogen 66 moles moles of of hydrogen hydrogen 62
Quantitative Relationships: Chemical Formulas Example: How many moles of carbon are in 2.65 moles of C 10 H 20 O? Know: 2.65 moles of C 10 H 20 O and chemical formula Find: moles of carbon Conversion factor: moles carbon Equation: 10 1 mole C 10 10 moles C 2.65 moles C10H20O 26. 5 moles C 1 mole C H O H 20 O 10 20 63
Quantitative Relationships: Chemical Equations Coefficients in the balanced equation show Relative numbers of molecules of reactants and products Relative numbers of moles of reactants and products Can be converted to mass Quantitative relationship among reactants and products is called stoichiometry 64
Stoichiometric Calculations Stoichiometric calculations compare two different materials Use the MOLE RATIO from the balanced equation 65
An Example of a Stoichiometric Calculation How many grams of water can be produced from 1.00 g of glucose? C 6 H 12 O 6 (s) + 6 O 2 (g) 6 CO 2 (g) + 6 H 2 O(l) The first step is to convert grams of substance A to moles substance A There is 1.00 g of glucose to start Use molar mass for conversion factor 66
An Example of a Stoichiometric Calculation How many grams of water can be produced from 1.00 g of glucose? C 6 H 12 O 6 (s) + 6 O 2 (g) 6 CO 2 (g) + 6 H 2 O(l) Second step is to convert moles of one substance in the equation to moles of another substance Use MOLE RATIO from the balanced equation 67
An Example of a Stoichiometric Calculation How many grams of water can be produced from 1.00 g of glucose? C 6 H 12 O 6 (s) + 6 O 2 (g) 6 CO 2 (g) + 6 H 2 O(l) Third step is to convert moles to grams of substance B Use molar mass for conversion factor 68
LIMITING REACTANTS AND YIELDS 69
Limiting Reactants Limiting reactant is the reactant present in the smallest stoichiometric amount It is the reactant that will run out of first H 2 is the limiting reactant O 2 would be the excess reagent 70
Limiting Reactants Use limiting reactant in all stoichiometry calculations to determine amount of product(s) or amount of reactant(s) used in a reaction 71
Yields Theoretical yield is the maximum amount of product that can be made It is the amount of product possible as calculated through stoichiometry Actual yield (experimental yield) is the amount actually produced and measured 72
Percent Yield Percent yield Compares the actual yield to the theoretical yield Percent yield = 100 actual yield theoretical yield 73
EXAMPLE LIMITING REACTANT AND THEORETICAL YIELD FROM INITIAL MOLES OF REACTANTS Consider this reaction: 2 Al(s) + 3 Cl 2 (g) 2 AlCl 3 (s) If you begin with 0.552 mol of aluminum and 0.887 mol of chlorine, what is the limiting reactant and theoretical yield of AlCl 3 in moles? SORT You are given the number of moles of aluminum and chlorine and asked to find the limiting reactant and theoretical yield of aluminum chloride. GIVEN: 0.552 mol Al 0.887 mol Cl 2 FIND: limiting reactant theoretical yield of AlCl 3 STRATEGIZE Draw a solution map that shows how to get from moles of each reactant to moles of AlCl 3. The reactant that makes the least amount of AlCl 3 is the limiting reactant. The conversion factors are the stoichiometric relationships (from the balanced equation). SOLUTION MAP 74
EXAMPLE LIMITING REACTANT AND THEORETICAL YIELD FROM INITIAL MOLES OF REACTANTS Continued SOLVE Follow the solution map to solve the problem. SOLUTION Because the 0.552 mol of Al makes the least amount of AlCl 3, Al is the limiting reactant. The theoretical yield is 0.552 mol of AlCl 3. 75
EXAMPLE LIMITING REACTANT AND THEORETICAL YIELD FROM INITIAL MOLES OF REACTANTS Continued SKILLBUILDER Limiting Reactant and Theoretical Yield from Initial Moles of Reactants Consider the reaction: 2 Na(s) + F 2 (g) 2 NaF(s) If you begin with 4.8 mol of sodium and 2.6 mol of fluorine, what is the limiting reactant and theoretical yield of NaF in moles? Answer: 76
77
EXAMPLE THE MOLE CONCEPT CONVERTING BETWEEN GRAMS AND MOLES Calculate the number of moles of sulfur in 57.8 g of sulfur. SORT Begin by sorting the information in the problem. You are given the mass of sulfur and asked to find the number of moles. STRATEGIZE Draw a solution map showing the conversion from g S to mol S. The conversion factor is the molar mass of sulfur. SOLVE Follow the solution map to solve the problem. Begin with 57.8 g S and use the conversion factor to determine mol S. GIVEN: FIND: SOLUTION MAP RELATIONSHIPS USED SOLUTION 78
EXAMPLE CONVERTING BETWEEN MOLES AND NUMBER OF ATOMS A silver ring contains 1.1 10 22 silver atoms. How many moles of silver are in the ring? SORT You are given the number of silver atoms and asked to find the number of moles. STRATEGIZE Draw a solution map, beginning with silver atoms and ending at moles. The conversion factor is Avogadro s number. GIVEN: FIND: SOLUTION MAP RELATIONSHIPS USED 79
EXAMPLE Continued CONVERTING BETWEEN MOLES AND NUMBER OF ATOMS SOLVE Follow the solution map to solve the problem. Beginning with 1.1 10 22 Ag atoms, use the conversion factor to determine the moles of Ag. SOLUTION 80
EXAMPLE THE MOLE CONCEPT CONVERTING BETWEEN GRAMS AND NUMBER OF ATOMS How many aluminum atoms are in an aluminum can with a mass of 16.2 g? SORT You are given the mass of aluminum and asked to find the number of aluminum atoms. STRATEGIZE The solution map has two steps. In the first step, convert from g Al to mol Al. In the second step, convert from mol Al to the number of Al atoms. The required conversion factors are the molar mass of aluminum and the number of atoms in a mole. GIVEN: FIND: SOLUTION MAP RELATIONSHIPS USED SOLVE Follow the solution map to solve the problem, beginning with 16.2 g Al and multiplying by the appropriate conversion factors to arrive at Al atoms. SOLUTION 81
EXAMPLE THE MOLE CONCEPT CONVERTING BETWEEN GRAMS AND MOLES FOR COMPOUNDS Calculate the mass (in grams) of 1.75 mol of water. SORT You are given moles of water and asked to find the mass. GIVEN: FIND: STRATEGIZE Draw a solution map showing the conversion from mol H 2 O to g H 2 O. The conversion factor is the molar mass of water, which you can determine by summing the atomic masses of all the atoms in the chemical formula. SOLUTION MAP RELATIONSHIPS USED SOLVE Follow the solution map to solve the problem. Begin with 1.75 mol of water and use the molar mass to convert to grams of water. SOLUTION 82
EXAMPLE THE MOLE CONCEPT CONVERTING BETWEEN MASS OF A COMPOUND AND NUMBER OF MOLECULES What is the mass of 4.78 10 24 NO 2 molecules? SORT You are given the number of NO 2 molecules and asked to find the mass. STRATEGIZE The solution map has two steps. In the first step, convert from molecules of NO 2 to moles of NO 2. In the second step, convert from moles of NO 2 to mass of NO 2. The required conversion factors are the molar mass of NO 2 and the number of molecules in a mole. GIVEN: FIND: SOLUTION MAP RELATIONSHIPS USED 83
EXAMPLE Continued THE MOLE CONCEPT CONVERTING BETWEEN MASS OF A COMPOUND AND NUMBER OF MOLECULES SOLVE Using the solution map as a guide, begin with molecules of NO 2 and multiply by the appropriate conversion factors to arrive at g NO 2. SOLUTION 84
EXAMPLE CHEMICAL FORMULAS AS CONVERSION FACTORS CONVERTING BETWEEN GRAMS OF A COMPOUND AND GRAMS OF A CONSTITUENT ELEMENT Carvone (C 10 H 14 O) is the main component of spearmint oil. It has a pleasant aroma and mint flavor. Carvone is added to chewing gum, liqueurs, soaps, and perfumes. Calculate the mass of carbon in 55.4 g of carvone. SORT You are given the mass of carvone and asked to find the mass of one of its constituent elements. STRATEGIZE Base the solution map on: Grams Mole Mole Grams. You need three conversion factors. The first is the molar mass of carvone. The second conversion factor is the relationship between moles of carbon and moles of carvone from the molecular formula. GIVEN: FIND: SOLUTION MAP RELATIONSHIPS USED 85
EXAMPLE Continued CHEMICAL FORMULAS AS CONVERSION FACTORS CONVERTING BETWEEN GRAMS OF A COMPOUND AND GRAMS OF A CONSTITUENT ELEMENT The third conversion factor is the molar mass of carbon. SOLVE Follow the solution map to solve the problem, beginning with g C 10 H 14 O and multiplying by the appropriate conversion factors to arrive at g C. SOLUTION 86
For the reaction X Y, X is referred to as the a. yield. b. reactant. c. product. d. coefficient. 87
Hydrocarbons burn to form a. H 2 O and CO 2. b. charcoal. c. methane. d. O 2 and H 2 O. 88
2 NaN 3 2 Na + 3 N 2 This is an example of a reaction. a. decomposition b. combination c. combustion d. replacement 89
The percentage yield of a reaction is 100% (Z), where Z is a. theoretical yield/actual yield. b. calculated yield/actual yield. c. calculated yield/theoretical yield. d. actual yield/theoretical yield. 90
How many atoms of Mg, O, and H are represented by the notation 3 Mg(OH) 2? a. 1 Mg, 2 O, and 2 H b. 2 Mg, 2 O, and 2 H c. 6 Mg, 6 O, and 6 H d. 3 Mg, 6 O, and 6 H 91
Which has more mass, a mole of water (H 2 O) or a mole of glucose (C 6 H 12 O 6 )? a. Mole of glucose b. Mole of water 92
Which contains more molecules, a mole of water or a mole of glucose? a. Mole of water b. Mole of glucose c. Requires Avogadro s number to answer question d. They both contain the same number of molecules 93
Could the empirical formula determined from chemical analysis be used to tell the difference between acetylene, C 2 H 2, and benzene, C 6 H 6? a. Yes, because the empirical formula is specific to the molecule. b. Yes, because the empirical formula tells the number of atoms in the molecule. c. No, because the empirical formula cannot be determined for a liquid. d. No, because the empirical formulas for C 2 H 2 and C 6 H 6 will be the same. 94
In the example, 1.00 g of C 4 H 10 reacts with 3.59 g of O 2 to form 3.03 g of CO 2. Using only addition and subtraction, calculate the amount of H 2 O produced. a. 1.56 g H 2 O b. 3.12 g H 2 O c. 5.00 g H 2 O d. 7.62 g H 2 O 95
The formula mass of any substance is also known as a. Avogadro s number. b. atomic weight. c. density. d. molar mass. 96
Ribose has a molecular weight of 150 grams per mole and the empirical formula CH 2 O. The molecular formula of ribose is a. C 4 H 8 O 4. b. C 5 H 10 O 5. c. C 6 H 14 O 4. d. C 6 H 12 O 6. 97