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Chapter 3

Nathan G.

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Laboratory Inquiry in Chemistry (Brooks / Cole Laboratory Series for Genera...

Laboratoty Inquiry in Chemistry for CHM 114

Stoichiometry: Calculations with Chemical Formulas and Equations Chapter 3 Chapter 3 Chemical Equations Balancing, Types of Equations Compound Composition Atom mass; Atomic and formula weight % Composition Empirical Formulas Mole Amounts; Molar Mass Mole Calculations Amounts in chemical reactions Chemical Equations Chemical equations help us to describe chemical reactions. BEFORE AFTER Types of Reactions (More in Ch.4) Combustion (of a hydrocarbon with O2) Combination Reactions Decomposition Reactions Single Displacement Reactions (Ch.4) Double Displacement Reactions (Ch.4) Combustion Reactions in Air Hydrocarbons react with oxygen to give carbon dioxide and water: CH3CH2CH3(g) + O2(g) CO2(g) + H2O(g) Balance it! Combustion Reactions in Air Hydrocarbons react with oxygen to give carbon dioxide and water: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) Balance it! Combination Reactions: A + B C Mg(s) + O2(g) ? Combination and Decomposition Reactions Combination and Decomposition Reactions Combination Reactions: A + B C Mg(s) + O2(g) MgO(s) Combination and Decomposition Reactions Combination Reactions: A + B C 2Mg(s) + O2(g) 2MgO(s) Fig 3.5 Combination and Decomposition Reactions Decomposition Reactions: C A + B Sodium azide decomposes to its elements: NaN3(s) ? (N3- is a polyatomic ion) Combination and Decomposition Reactions Decomposition Reactions: C A + B Sodium azide decomposes to its elements: NaN3(s) Na(s) + N2(g) (N3- is a polyatomic ion) Balance it! Combination and Decomposition Reactions Decomposition Reactions: C A + B Sodium azide decomposes to its elements: 2NaN3(s) 2Na(s) + 3N2(g) (N3- is a polyatomic ion) Decomposition Reactions Another example: 2HgO(s) 2Hg(l) + O2(g) Rxn Decomposition Reactions Some compounds decompose to more stable compounds: 2H2O2(l) 2H2O(l) + O2(g) CaCO3(s) CaO(s) + CO2(g) Chemical Composition The quantitative significance of formulas Atomic Weight Molecular Weight / Formula Weight Percent Composition Empirical Formulas What is the Mass of an Atom? When we describe the mass of atoms of an element, we must consider all of the naturally occurring isotopes. Carbon is composed of 98.8925 % C-12 (12.00000 amu) 1.1080 % C-13 (13.00335 amu) Its weighted average is 12.011 amu. Its atomic weight is 12.011 amu. What is the Atomic Weight of Copper? Atomic weight is different than mass number. What is the Average Atomic Weight of Copper? (62.94 amu)(0.6917) = 43.5356 (64.93 amu)(0.3083) = 20.0179 63.55 amu (4 sig figs) Formula Weight The formula weight of a substance is the sum of the atomic weights of the atoms that make up the formula. Mg(OH)2 2(16.00) amu 2(1.008) amu + 24.31 amu 58.33 amu = formula wt. Mg(OH)2 Molecular Weight The formula weight of a molecular compound is also called its molecular weight. H2O 1.008 amu 1.008 amu + 16.00 amu 18.02 amu = molecular wt. of H2O Percent Composition Mass Percent Composition = (#Atoms of Element)(Atomic Weight) x100 Formula Weight of Compound Percent Composition What is the mass %Cl in MgCl2? Formula weight = (2 x 35.45amu) + 24.31 amu 95.21 amu Mass of Cl: 2 x 35.45amu = 70.90amu Mass %Cl = Percent Composition What is the mass %Mg in MgCl2? 74.47% Cl 25.53% Mg Group Work How would you go about determining the mass of carbon in 5.0 g CCl4? Group Work How would you go about determining the mass of carbon in 5.0 g CCl4? %C = (12.011amu/153.811 amu)x100 = 7.8089% C g Carbon = (0.078089)(5.0g) = 0.39 g Group Work Which of the following has the greatest %O? Explain. MgO CaO Na2O K2O % Composition What is the percent carbon in C2H4? What is the percent carbon in C3H6? % Composition What is the percent carbon in C2H4? % C = 85.63% What is the percent carbon in C3H6? % C = 85.63% Empirical Formula Different compounds with the same percent composition have the same empirical formula. The empirical formula shows the ratio of atoms of each element as the smallest whole-number ratio. What is the empirical formula for Benzene (C6H6)? Group Work Which of the following are empirical formulas? CH4 C2H2 N2O5 C2H6 Al2O3 Group Work Which of the following are empirical formulas? CH4 C2H2 N2O5 C2H6 Al2O3 Mole Scale We usually measure substances on scales billions of times larger than the amu scale. It is more useful to measure masses in grams. 1 amu = 1.66054 x 10-24 g 1 gram = 6.02214 x 1023 amu Mole Scale The mass of one C-12 atom is exactly 12 amu. The mass of 6.02214x1023 C-12 atoms is 12.0000 grams. The mass of exactly 1 mole of C-12 atoms is exactly 12 grams. Avogadro’s Number 6.02214199 x 1023 Avogadro’s number: 6.02214x1023 particles per mole (of anything) Avogadro’s Number 6.02214199 x 1023 1 mole = 6.022 x 1023 C-12 atoms = 6.022 x 1023 C atoms = 6.022 x 1023 H2O molecule = 6.022 x 1023 NaCl formula units Avogadro’s Number Use Avogadro’s number as a conversion factor to convert between moles and number of molecules or formula units: 1 mole = 6.022x1023 molecules or formula units Moles 1 mole of H2O Contains 6.022x1023 H2O molecules Contains 2x(6.022x1023) H atoms Contains 1x(6.022x1023) O atoms Avogadro’s Number Group Work 1. How many carbon atoms are there in 1.0 mole of CH4? 2. How many hydrogen atoms are there in 1.0 mole of CH4? 3. How many hydrogen atoms are there in 0.10 mole of CH4? Molar Mass 1 mole of C-12 has a mass of 12.00 g of C has a mass of 12.01 g of H has a mass of 1.008 g of H2O has a mass of 18.02 g of Ca(NO3)2 has a mass of 164.098 g Molar mass = mass of 1 mole of that substance Molar Mass If you write the units on molar mass as X g/mol, then molar mass can be used as a conversion factor for converting between moles of a substance and grams of a substance. e.g. The molar mass of H2O is 18.02 g/mol Moles to Grams What is the mass (in grams) of 0.500 mole of oxygen gas? Molar mass of oxygen gas = ? O2 = 32.00 g/mol Grams to Moles How many moles are contained in 36.0 grams of water? Molar mass = 18.02g/mol Using Moles Moles provide bridge from macroscopic scale to atomic scale Group Work Convert 0.037 mol CCl4 to grams. (MMCCl4 = 153.81 g/mol) 2. Convert 10.50 grams of N2 gas to number of N2 molecules. Group Work Convert 0.037 mol CCl4 to grams. (MMCCl4 = 153.81 g/mol) 5.7 g CCl4 Convert 10.50 grams of N2 gas to number of N2 molecules. 0.3747 mol N2 2.257x1023 N2 molecules Group Work How many hydrogen atoms are in 18.02 grams of water? Determining the Empirical Formula Can we determine the formula for a compound from the % composition? Empirical formula % Composition (mole ratio) (mass ratio) Determining the Empirical Formula for a Compound Given that the % composition of a compound is 30.45% N and 69.56% O, what is the formula for the compound? 1. Assume 100 g of compound 2. Convert g of each element to moles. 3. Divide or multiply each by the same number to get whole-number subscripts. Determining the Empirical Formula for a Compound Given that the % composition of a compound is 30.45% N and 69.56% O, what is the formula for the compound? 1. Assume 100 g of compound 30.45 g N 69.56 g O Determining the Empirical Formula for a Compound Given that the % composition of a compound is 30.45% N and 69.56% O, what is the formula for the compound? 2. Convert g of each element to moles. (30.45 g N)(1 mol/14.01g N) = 2.173 mol N (69.56 g O)(1 mol/16.00g O) = 4.348 mol O Determining the Empirical Formula for a Compound Given that the % composition of a compound is 30.45% N and 69.56% O, what is the formula for the compound? 3. Divide or multiply each by the same number to get whole-number subscripts. (2.173 mol N)/(2.173) = 1 mol N (4.348 mol O )/(2.173) = 2 mol O Determining the Empirical Formula for a Compound Given that the % composition of a compound is 30.45% N and 69.56% O, what is the formula for the compound? 3. Divide or multiply each by the same number to get whole-number subscripts. (2.173 mol N)/(2.173) = 1 mol N (4.348 mol O )/(2.173) = 2 mol O The empirical formula is NO2 Determining the Molecular Formula for a Compound Now that we know that the empirical formula is NO2, can we determine the molecular formula? Only if we know the molecular weight. Determining the Molecular Formula for a Compound If the molecular weight of the compound is given to us as 46.01 amu, then the molecular formula is NO2. If the molecular weight of the compound is given to us as 92.02 amu, then the molecular formula is N2O4. Models of NO2 and N2O4 Same % Composition Same Empirical Formula Different molecular weights Quantitative Information from Balanced Equations Stoichiometry Stoichiometry - quantitative relationships between reactants and products. Mole-Mole Mass-Mass Limiting reagents (reactants) Percent Yield Use Balanced Equations N2(g) + 3H2(g) 2NH3(g) 1 molecule 3 molecules 2 molecules Use Balanced Equations N2(g) + 3H2(g) 2NH3(g) 1 molecule 3 molecules 2 molecules 2 molecules Use Balanced Equations N2(g) + 3H2(g) 2NH3(g) 1 molecule 3 molecules 2 molecules 2 molecules 6 molecules 4 molecules 500 molecules Use Balanced Equations N2(g) + 3H2(g) 2NH3(g) 1 molecule 3 molecules 2 molecules 2 molecules 6 molecules 4 molecules 500 molecules 1500 molecules 1000 molecules Use Balanced Equations N2(g) + 3H2(g) 2NH3(g) 1 molecule 3 molecules 2 molecules 2 molecules 6 molecules 4 molecules 500 molecules 1500 molecules 1000 molecules 6.02x1023 (3)x 6.02x1023 (2)x 6.02x1023 Use Balanced Equations N2(g) + 3H2(g) 2NH3(g) 1 molecule 3 molecules 2 molecules 2 molecules 6 molecules 4 molecules 500 molecules 1500 molecules 1000 molecules 6.02x1023 (3)x 6.02x1023 (2)x 6.02x1023 1 mole 3 moles 2 moles Use Balanced Equations N2(g) + 3H2(g) 2NH3(g) 1 molecule 3 molecules 2 molecules 2 molecules 6 molecules 4 molecules 500 molecules 1500 molecules 1000 molecules 6.02x1023 (3)x 6.02x1023 (2)x 6.02x1023 1 mole 3 moles 2 moles 0.4 mol Use Balanced Equations N2(g) + 3H2(g) 2NH3(g) 1 molecule 3 molecules 2 molecules 2 molecules 6 molecules 4 molecules 500 molecules 1500 molecules 1000 molecules 6.02x1023 (3)x 6.02x1023 (2)x 6.02x1023 1 mole 3 moles 2 moles 0.4 mol 1.2 mol 0.8 mol Mole Ratios N2(g) + 3H2(g) 2NH3(g) What is the relationship between NH3 and H2? What is the mole ratio? If 1.2 mol H2 reacts how much NH3 will form? Mole Ratios N2(g) + 3H2(g) 2NH3(g) What is the mole ratio for determining the moles of N2 that will react with 1.2 mol of H2? Mass Relationships N2(g) + 3H2(g) 2NH3(g) What mass (grams) of H2 will react with 1.00 gram of N2? Mass Relationships N2(g) + 3H2(g) 2NH3(g) What mass (grams) of H2 will react with 1.00 gram of N2? Group Work 2H2(g) + O2(g) 2H2O(l) What mass (grams) of O2 will react with 5.0g of H2? What mass of H2O will be produced? Group Work Complete combustion of octane occurs by the following balanced equation. 2C8H18(l) + 25O2(g) 16CO2(g) + 18H2O(g) 1. How many moles O2 will react with 3.0 mol of octane? 2. How many grams of O2 will react with 3.0 mol of octane? Group Work Complete combustion of octane occurs by the following balanced equation. 2C8H18(l) + 25O2(g) 16CO2(g) + 18H2O(g) 1. How many moles O2 will react with 3.0 mol of octane? 37.5 mol O2 2. How many grams of O2 will react with 3.0 mol of octane? 1200 grams O2 Group Work You set a pure piece of aluminum out on the counter. After a few days you determine that it has gained a mass of 0.030 grams. You attribute this to a chemical reaction with the oxygen in the air. 1. Write a balanced equation for the combination reaction of Al and O2. 2. Calculate the mass of aluminum metal that reacted. Limiting Reactants Suppose that you react hydrogen and oxygen to form water. 2H2(g) + O2(g) 2H2O(g) The mole ratio of H2 to O2 is 2/1. 2H2(g) + O2(g) 2H2O(g) How many moles of H2O will form when the following are mixed? 1. 4 mol H2 and 2 mol O2 2. 2 mol H2 and 2 mol O2 3. 10 mol H2 and 4 mol O2 2H2(g) + O2(g) 2H2O(g) How many moles of H2O will form when the following are mixed? 1. 4 mol H2 and 2 mol O2 4 mol H2O 2. 2 mol H2 and 2 mol O2 2 mol H2O 3. 10 mol H2 and 4 mol O2 8 mol H2O Which reactant limits the amount of H2O formed? 2H2(g) + O2(g) 2H2O(g) What mass of H2O can be produced when 5.0 grams of H2 are combined with 10.0 grams of O2? 2H2(g) + O2(g) 2H2O(g) What mass of H2O can be produced when 5.00 grams of H2 are combined with 10.0 grams of O2? Moles H2 = 5.00 g ÷ 2.016 g/mol = 2.48 mol H2 Moles O2 = 10.0 g ÷ 32.00 g/mol = 0.313 mol O2 2H2(g) + O2(g) 2H2O(g) What mass of H2O can be produced when 5.00 grams of H2 are combined with 10.0 grams of O2? Moles H2 = 5.00 g ÷ 2.016 g/mol = 2.48 mol H2 Moles O2 = 10.0 g ÷ 32.00 g/mol = 0.313 mol O2 O2 is the limiting reactant because there is not enough O2 to react with the given amount of H2. 2H2(g) + O2(g) 2H2O(g) What mass of H2O can be produced when 5.00 grams of H2 are combined with 10.0 grams of O2? Moles H2 = 5.00 g ÷ 2.016 g/mol = 2.48 mol H2 Moles O2 = 10.0 g ÷ 32.00 g/mol = 0.313 mol O2 (0.313 mol O2)(2 mol H2O/1 mol O2)(18.02 g/mol) = 11.3 g H2O Limiting Reactants on a Molecular Scale What is the limiting reactant in this reaction? Percent Yield Theoretical yield is what is calculated from the limiting reactant. Actual yield is what is actually obtained when the reaction is run. Group Work In the following reaction, 2.0 mol of H2 were combined with 1.0 mole of I2. H2(g) + I2(s) 2HI(g) What is the theoretical yield? (*Find limiting reactant first) 2. If the percent yield for this reaction was determined to be 65%, what is the actual yield (in moles)? 3. How many grams of HI were actually produced? Group Work In the following reaction, 2.0 mol of H2 were combined with 1.0 mole of I2. H2(g) + I2(s) 2HI(g) 1. What is the theoretical yield? 2.0 mol HI 2. If the percent yield for this reaction was determined to be 65%, what is the actual yield (in moles)? 1.3 mol HI 3. How many grams of HI were actually produced? 165 g (160 g)