Ch 3 Stoichiometry.pdf

Chapter 3: Stoichiometry of Formulas and Equations 2.8 Molecular Masses from Chemical Formulas 3.1 The Mole, Molar Mass, Amount 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing Chemical Equations 3.4 Stoichiometry: Calculating Quantities of Reactant and Product ©McGraw-Hi...

Chapter 3: Stoichiometry of Formulas and Equations 2.8 Molecular Masses from Chemical Formulas 3.1 The Mole, Molar Mass, Amount 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing Chemical Equations 3.4 Stoichiometry: Calculating Quantities of Reactant and Product ©McGraw-Hill Education. 2.8 Molecular Masses from Chemical Formulas Molecular Mass (MM, M) = sum of atomic masses (amu) For the H2O molecule: M = (2 x atomic mass of H) + (1 x atomic mass of O) = (2 x 1.008 amu) + (1 x 16.00 amu) = 18.02 amu – In this course, we read masses off the periodic table to 4 significant figures. For ionic compounds we refer to a formula mass since ionic compounds do not consist of molecules. ©McGraw-Hill Education. Sample Problem 2.16: Problem and Plan Calculating the Molecular Mass of a Compound PROBLEM: Using the periodic table, calculate the molecular (or formula) mass of: a) tetraphosphorous trisulfide b) ammonium nitrate PLAN: Write the formula and then multiply the number of atoms by the respective atomic masses. Add the masses for each compound. ©McGraw-Hill Education. The Modern Periodic Table Figure 2.10 ©McGraw-Hill Education. Sample Problem 2.16: Solution SOLUTION: a) P4S3: M = (4 x atomic mass of P) + (3 x atomic mass of S) = (4 x 30.97 amu) + (3 x 32.07 amu) = 220.09 amu b) NH4NO3: Formula mass = (2 x atomic mass of N) + (4 x atomic mass of H) + (3 x atomic mass of O) = (2 x 14.01 amu) + (4 x 1.008 amu) + (3 x 16.00 amu) = 80.05 amu ©McGraw-Hill Education. Using Molecular Depictions to determine Formula, Name, and Mass for a compound Sample Problem 2.17: Problem and Plan PROBLEM: Each scene represents a binary compound. Determine its formula, name, and molecular (or formula) mass. PLAN: Each compound contains only two elements. Find the simplest whole number ratio of atoms in each compound and use this formula to determine the name and the molecular (or formula) mass. ©McGraw-Hill Education. Sample Problem 2.17: Solution SOLUTION: a) There is 1 brown Na+ for every green F-, so the formula is NaF, an ionic compound, which is named sodium fluoride. Formula mass = (1 x atomic mass of Na) + (1 x atomic mass of F) = 22.99 amu + 19.00 amu = 41.99 amu b) There are 3 green F for every blue N, so the formula is NF3, a covalent compound, which is named nitrogen trifluoride. Molecular mass = (1 x atomic mass of N) + (3 x atomic mass of F) = 14.01 amu + (3 x 19.00) = 71.01 amu ©McGraw-Hill Education. The Mole (Amount of Substance) The mole (a physical property whose unit is 1 mol) is the amount of a substance that contains the same number of entities as there are atoms in exactly 12.00 g of carbon-12. – The term “entities” refers to atoms, ions, molecules, formula units, or electrons – in fact, any type of particle. – amount refers to counting entities by weighing the mass of sample One mole contains 6.022x1023 entities (to 4 significant figures). – This number is called the Avogadro’s number and is abbreviated as NA. – The conversion factor is 1 𝑚𝑜𝑙 = 6.022 × 1023 – In rational form, the Avogadro’s Number as a conversion factor is 6.022 × 1023 1 23 𝑁𝐴 = = 6.022 × 10 = 6.022 × 1023 𝑚𝑜𝑙 −1 1 𝑚𝑜𝑙 𝑚𝑜𝑙 ©McGraw-Hill Education. Why is the Mole Useful? The Mole concept allows us to count extremely small molecules, atoms, ion pairs or any particle by measuring the mass of the sample. Real life example. You want to buy 1,000 steel balls of 0.25 inch in diameter sold in bulk from a hardware store. At the store, the clerk will weigh steel balls of total mass 𝑚 = 1,000 × 𝑚1 𝑏𝑎𝑙𝑙. Counting to 1,000 is time consuming and tedious. The number of moles of a substance (formula) tells us both the number of objects and the mass of substance: 1 𝑚𝑜𝑙 12𝐶 = 6.022 × 1023 12𝐶 = 12.00 𝑔 12𝐶 Exercise. How many 12𝐶 atoms are there in 1.25 g of 12𝐶? 6.022 × 1023 𝑁 = 1.25𝑔 × = 6.28 × 1022 (3 𝑠𝑖𝑔. 𝑓𝑖𝑔𝑠) 12.00𝑔 ©McGraw-Hill Education. Why is the Mole Useful? – cont’d The Mole concept allows us to talk and compare number of molecules in different samples without using very large (“astronomic”) numbers. There are 2 conversion factors in the definition of mole: 1 𝑚𝑜𝑙 12𝐶 = 6.022 × 1023 12𝐶 = 12.00 𝑔 12𝐶 Exercise. How many 12𝐶 mol are there in 1.25 g of 12𝐶? 1 𝑚𝑜𝑙 𝑁 = 1.25𝑔 × = 0.104 𝑚𝑜𝑙 (3 𝑠𝑖𝑔. 𝑓𝑖𝑔𝑠) 12.00𝑔 It is more convenient to say that there are 0.104 mol of 12𝐶 in 1.25 g of pure sample than 6.28 × 1022 atoms of this isotope in the same sample. ©McGraw-Hill Education. One Mole of Some Familiar Substances Fig 3.1. One mol of: Cu (63.55 g), liquid water (18.02 g), NaCl (58.44 g), sucrose (or sugar, 342.3 g), Al (26.98 g). ©McGraw-Hill Education. © McGraw-Hill Education/Charles Winters/Timeframe Photography, Inc. Molar Mass The molar mass (M) of a substance is the mass per mole of its entities (atoms, molecules, or formula units for ionic compounds). The value of the Molar Mass is identical to the Molecular Mass. The difference between the two properties is the units: – Molecular Mass refers to the mass of 1 molecule in amu; – Molar Mass refers to the mass of 1 mol of molecules in g/mol. For monatomic elements, the molar mass is the same as the atomic mass in grams per mole. The atomic mass is simply read from the Periodic Table: – The atomic mass of Ne = 20.18 amu – The molar mass of Ne = 20.18 g/mol. ©McGraw-Hill Education. Molar Mass, Cont’d For molecular elements and for compounds, the formula is needed to determine the molar mass. The molar mass of O2 = 2 x M of O = 2 x 16.00 g/mol = 32.00 g/mol The molar mass of SO2 = 1 x M of S + 2 x M of O = [32.06 + 2(16.00)] g/mol = 64.06 g/mol K2S: M = 2 x M(K) + 1 x M(S) = 2 x 39.10 g/mol + 32.06 g/mol = 110.26 g/mol ©McGraw-Hill Education. Information Contained in the Chemical Formula of Glucose C6H12O6 Carbon (C) Hydrogen (H) Oxygen (O) Atoms/molecule of compound 6 atoms 12 atoms 6 atoms Moles of atoms/mole of compound 6 mol of C atoms 12 mol of H atoms 6 mol of O atoms Atoms/mole of compound 6(6.022x1023) atoms 12(6.022x1023) atoms 6(6.022x1023) atoms Mass/molecule of compound 6(12.01 amu) = 72.06 amu 12(1.008 amu) = 12.10 amu 6(16.00 amu) = 96.00 amu Mass/mole of compound 72.06 g 12.10 g 96.00 g Molar Mass of C6H12O6, M 72.06 g/mol C + 12.10 g/mol H + 96.00 g/mol O = = 180.16 g/mol C6H12O6 Table 3.1 ©McGraw-Hill Education. Interconverting Moles, Mass, and Number of Chemical Entities # of grams 𝑔 Mass = 𝑚 𝑔 = 𝑛(# of moles) 𝑚𝑜𝑙 × 1 𝑚𝑜𝑙 1 𝑚𝑜𝑙 𝑛 # of moles = 𝑚(mass) 𝑔 × #grams 𝑔 6.022 × 1023 entities 𝑁 # of entities = 𝑛 # of moles 𝑚𝑜𝑙 × 1 𝑚𝑜𝑙 1 𝑚𝑜𝑙 𝑛 (# of moles) = 𝑁(# of entities) × 6.022 × 1023 entities ©McGraw-Hill Education. Mass-mole-number Relationships for Elements Fig 3.2 ©McGraw-Hill Education. Converting Between Mass and Amount of an Element – Sample Problem 3.1 Problem, Plan and Solution PROBLEM: Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342 mol of Ag? PLAN: To convert mol of Ag to mass of Ag in g we need the molar mass of Ag. From the Periodic Table: M(Ag) = 107.9 g/mol SOLUTION: 107.9 𝑔 Ag 0.0342 𝑚𝑜𝑙 Ag × = 𝟑. 𝟔𝟗 𝐠 𝐀𝐠 1 𝑚𝑜𝑙 Ag ©McGraw-Hill Education. Converting Between Number of Entities and Amount of an Element Sample Problem 3.2 - Problem and Plan PROBLEM: Gallium (Ga) is a key element in solar panels, calculators and other light-sensitive electronic devices. How many Ga atoms are in 2.85 x 10-3 mol of gallium? PLAN: To convert mol of Ga to number of Ga atoms we need to use Avogadro’s number. ©McGraw-Hill Education. Sample Problem 3.2 - Solution SOLUTION: 23 Ga atoms 6.022 × 10 2.85 × 10−3 𝑚𝑜𝑙 Ga atoms × 1 𝑚𝑜𝑙 Ga atoms 𝟐𝟏 = 𝟏. 𝟕𝟐 × 𝟏𝟎 𝐆𝐚 𝐚𝐭𝐨𝐦𝐬 ©McGraw-Hill Education. Converting Between Number of Entities and Mass of an Element Sample Problem 3.3 – Problem and Plan PROBLEM: Iron (Fe) is the main component of steel and is therefore the most important metal in society; it is also essential in the body. How many Fe atoms are in 95.8 g of Fe? PLAN: The number of atoms cannot be calculated directly from the mass. We must first determine the number of moles of Fe atoms in the sample and then use Avogadro’s number. From the Periodic Table: M(Fe) = 55.85 g/mol ©McGraw-Hill Education. Sample Problem 3.3 - Solution SOLUTION: 1 𝑚𝑜𝑙 Fe 𝑛 𝑚𝑜𝑙 𝐹𝑒 = 95.8 𝑔 Fe × = 1.72 𝑚𝑜𝑙 Fe 55.85 𝑔 Fe 6.022 × 1023 atoms Fe 𝑁 𝑎𝑡𝑜𝑚𝑠 𝐹𝑒 = 1.72 𝑚𝑜𝑙 Fe × = 𝟏. 𝟎𝟒 × 𝟏𝟎𝟐𝟒 𝐚𝐭𝐨𝐦𝐬 𝐅𝐞 1 𝑚𝑜𝑙 Fe The 2 steps can be combined into one with 2 conversion factors (there is no need to calculate the intermediate value of # of Fe moles): 1 𝑚𝑜𝑙 Fe 6.022 × 1023 atoms Fe 𝑁 𝐹𝑒 𝑎𝑡𝑜𝑚𝑠 = 95.8 𝑔 Fe × × 55.85 𝑔 Fe 1 𝑚𝑜𝑙 Fe = 1.04 × 1024 𝐹𝑒 𝑎𝑡𝑜𝑚𝑠 ©McGraw-Hill Education. Amount-Mass-Number Relationship Fig 3.3 ©McGraw-Hill Education. Sample Problem 3.4A – Problem and Plan Converting Between Number of Entities and Mass of Compound I PROBLEM: Nitrogen dioxide is a component of urban smog that forms from the gases in car exhausts. How many molecules are in 8.92 g of nitrogen dioxide? PLAN: Write the formula for the compound and calculate its molar mass. Use the given mass to calculate first the number of moles and then the number of molecules. ©McGraw-Hill Education. Sample Problem 3.4A - Solution SOLUTION: NO2 is the formula for nitrogen dioxide. ℳ = (1 × ℳ of N) + 2 × ℳ of O = g 46.01 mol g 14.01 mol 1 mol NO2 8.92 g NO2 × = 0.194 mol NO2 46.01 g NO2 6.022 × 1023 molecules NO2 0.194 mol NO2 × 1 mol NO2 = 𝟏. 𝟏𝟕 × 𝟏𝟎𝟐𝟑 𝐦𝐨𝐥𝐞𝐜𝐮𝐥𝐞𝐬 𝐍𝐎𝟐 ©McGraw-Hill Education. +2 16.00g mol = Sample Problem 3.4B – Problem and Plan Converting Between Number of Entities and Mass of Compound II PROBLEM: Ammonium carbonate, (NH4)2CO3, a white solid that decomposes with warming, is used as baking powder. a) How many formula units are in 41.6 g of ammonium carbonate? b) How many O atoms are in this sample? c) How many H atoms are in this sample ? PLAN: Write the formula for the compound and calculate its molar mass. Use the given mass to calculate first the number of moles and then the number of formula units. The number of O atoms can be determined using the formula and the number of formula units. ©McGraw-Hill Education. Sample Problem 3.4B - Solution SOLUTION: (NH4)2CO3 is the formula for ammonium carbonate. ℳ = 2 × ℳ of N + 8 × ℳ of H + 1 × ℳ of C + (3 × ℳ of O) = 2 × 14.01 g/mol + 8 × 1.008 g/mol + 1 × 12.01 g/mol + (3 × 16.00 g/mol) = 𝟗𝟔. 𝟎𝟗 𝐠/𝐦𝐨𝐥 ©McGraw-Hill Education. Sample Problem 3.5 – Solution, Cont’d (a) 41.6 g (NH4 )2 CO3 × 1 mol (NH4 )2 CO3 96.09 g (NH4 )2 CO3 = 0.433 mol(NH4 )2 CO3 6.022 × 1023 formula units (NH4 )2 CO3 0.433 mol (NH4 )2 CO3 × 1 mol (NH4 )2 CO3 = 𝟐. 𝟔𝟏 × 𝟏𝟎𝟐𝟑 𝐟𝐨𝐫𝐦𝐮𝐥𝐚 𝐮𝐧𝐢𝐭𝐬 𝐨𝐟 (𝐍𝐇𝟒 )𝟐 𝐂𝐎𝟑 (b) 2.61 × 1023 formula units (NH4 )2 CO3 × 𝟕. 𝟖𝟑 × 𝟏𝟎𝟐𝟑 𝑶 𝒂𝒕𝒐𝒎𝒔 3 𝑂 𝑎𝑡𝑜𝑚𝑠 1 formula unit (NH4 )2 CO3 (c) 2.61 × 1023 formula units (NH4 )2 CO3 × 8 𝐻 𝑎𝑡𝑜𝑚𝑠 = 𝟐. 𝟎𝟗 × 𝟏𝟎𝟐𝟒 𝑯 𝒂𝒕𝒐𝒎𝒔 1 formula unit (NH4 )2 CO3 ©McGraw-Hill Education. = Mass Percent from the Chemical Formula Each element in a chemical formula contributes a fraction of a compound’s mass. This fraction, multiplied by 100, gives the element’s mass percent: 1) Mass % of element X = atoms of X in formula x atomic mass of X (amu) × 100% molecular (or formula) mass of compound (amu) OR 2) Mass % of element X = moles of X in formula x molar mass of X (g/mol) × 100% Molar Mass (mass (g) of 1 mol of compound) ©McGraw-Hill Education. Sample Problem 3.5 – Problem and Plan Calculating the Mass Percent of Each Element in a Compound from the Formula PROBLEM: The effectiveness of fertilizers depends upon their nitrogen content. Ammonium nitrate is a common fertilizer. What is the mass percent of each element in ammonium nitrate? PLAN: We know the relative amounts (mol) of the elements from the formula and we have to find the mass % of each element. We multiply the amount of each by its molar mass to find its mass. Dividing each mass by the mass of 1 mol of ammonium nitrate gives the mass fraction of each element, and multiplying by 100 gives each mass %. The calculation steps for any element (X) are shown in the road map. ©McGraw-Hill Education. Sample Problem 3.6 – Plan, Cont’d ©McGraw-Hill Education. Sample Problem 3.6 - Solution SOLUTION: The formula is NH4NO3 (see Table 2.5). In 1 mol of NH4NO3 there are 2 mol of N, 4 mol of H, and 3 mol of O. Converting amount (mol) of N to mass (g): We have 2 mol of N in 1 mol of NH4NO3; so 14.01 𝑔 N Mass 𝑔 of N = 2 𝑚𝑜𝑙 N × = 28.02 𝑔 N 1 𝑚𝑜𝑙 N Calculating the mass of 1 mol of NH4NO3: M = (2 x M of N) + (4 x M of H) + (3 x M of O) = (2 x 14.01 g/mol N) + (4 x 1.008 g/mol H) + (3 x 16.00 g/mol O) = 80.05 g/mol NH4NO3 ©McGraw-Hill Education. Sample Problem 3.6 – Solution, Cont’d Finding the mass fraction of N in NH4NO3: total mass of N 28.02 g N Mass fraction of N = = Mass of 1 mol NH4NO3 80.05 g NH4NO3 = 0.3500 Changing to mass %: Mass % of N = mass fraction of N x 100 = 0.3500 x 100 = 35.00 mass % N Combining the steps for each of the other elements in NH4NO3: mol H x M of H Mass % of H = × 100 Mass of 1 mol NH4NO3 1.008 g H 4 mol H x 1 mol H × 100 = 5.037 mass % H = 80.05 g NH4NO3 ©McGraw-Hill Education. Sample Problem 3.6 - Check SOLUTION (continued): Mol O x M of O Mass % of O = × 100 Mass of 1 mol NH4NO3 16.00 g O 3 mol O x 1 mol O × 100 = 80.05 g NH4NO3 = 59.96 mass % O Check: The answers make sense. The mass % of O is greater than that of N because there are more moles of N in the compound and its molar mass is greater. The mass % of H is small because the molar mass of H is small. The sum of the mass percents is 100.00%. Comment: From here on, you should be able to determine the molar mass of a compound, so that calculations will no longer be shown. ©McGraw-Hill Education. Mass Fraction and the Mass of an Element An element always constitutes the same fraction of the mass of a given compound. Mass fraction can also be used to calculate the mass of a particular element in any mass of a compound. Mass of any element in sample = mass of element in 1 mol of compound mass of compound x mass of 1 mol of compound ©McGraw-Hill Education. Sample Problem 3.6 – Problem and Plan Calculating the Mass of an Element in a Compound PROBLEM: Use the information from Sample Problem 3.6 to determine the mass (g) of nitrogen in 650. g of ammonium nitrate. PLAN: The mass fraction of nitrogen in ammonium nitrate gives us the relative mass of nitrogen in 1 mole of ammonium nitrate. We can use this information to find the mass of nitrogen in any sample of ammonium nitrate. ©McGraw-Hill Education. Sample Problem 3.7 - Solution SOLUTION: Finding the mass of N in a given mass of ammonium nitrate, NH4NO3: Mass (g) of N = mass (g) of NH4NO3 x = 650. g NH4NO3 2 mol x M of N (g/mol) mass (g) of 1 mol of NH4NO3 28.02 g N x = 228 g N 80.05 g NH4NO3 CHECK: Rounding shows that the answer is “in the right ballpark”: N is about 1/3 of the mass of NH4NO3 and 1/3 of 700 g is 233 g. ©McGraw-Hill Education. 3.2 Empirical and Molecular Formulas We can determine the empirical formula for a compound from the mass percent (or mass fraction). The empirical formula is the simplest formula for a compound that agrees with the elemental analysis results for that compound. It shows the lowest whole number of moles and gives the relative number of atoms of each element present. It is derived from mass analysis of the compound. – The empirical formula for hydrogen peroxide (H2O2) is HO. – Mass analysis of hydrogen peroxide gives 1:16 mass ratio of H:O, yielding a 1:1 molar ratio of H:O, hence the HO empirical formula. The molecular formula shows the actual number of atoms of each element in a molecule of the compound. – The molecular formula for hydrogen peroxide is H2O2. ©McGraw-Hill Education. Empirical Formulas There are 3 steps to find the empirical formula: 1) Determine the mass (g) of each element in the compound. This is done experimentally by quantitative chemical analysis of elements; 2) Convert each mass (g) to amount (mol). The ratio of # of moles of each element gives the preliminary formula. The preliminary formula may have fractional (rational) subscripts. 3) Convert the preliminary formula to empirical formula by finding mathematically the smallest integers (whole numbers). To do this conversion: – Divide each fractional subscript by the smallest of the subscripts – If necessary, multiply through by the smallest integer that turns all subscripts into integers. ©McGraw-Hill Education. Sample Problem 3.7A – Problem and Plan Determining an Empirical Formula from Amounts of Elements PROBLEM: A sample of an unknown compound contains 0.21 mol of zinc, 0.14 mol of phosphorus, and 0.56 mol of oxygen. What is its empirical formula? PLAN: Find the relative number of moles of each element. Divide by the lowest mol amount to find the relative mol ratios (empirical formula). ©McGraw-Hill Education. Sample Problem 3.7A - Solution SOLUTION: Using the numbers of moles of each element given, we write the preliminary formula Zn0.21P0.14O0.56. Next we divide each fraction by the smallest one; in this case 0.14: 0.21/0.14 = 1.5 0.14/0.14 = 1.0 0.56/0.14 = 4.0 This gives Zn1.5P1.0O4.0 We convert to whole numbers by multiplying by the smallest integer that gives whole numbers; in this case 2: 1.5 x 2 = 3 1.0 x 2 = 2 4.0 x 2 = 8 This gives us the empirical formula Zn3P2O8 ©McGraw-Hill Education. Sample Problem 3.7B – Problem and Plan Determining an Empirical Formula from Masses of Elements PROBLEM: Analysis of a sample of an ionic compound yields 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What is the empirical formula and the name of the compound? PLAN: Find the relative number of moles of each element. Divide by the lowest mol amount to find the relative mol ratios (empirical formula). ©McGraw-Hill Education. Sample Problem 3.7B - Solution SOLUTION: 1 mol Na 2.82 g Na x = 0.123 mol Na 22.99 g Na 1 mol Cl 4.35 g Cl x = 0.123 mol Cl 35.45 g Cl 1 mol O 7.83 g O x = 0.489 mol O 16.00 g O 0.123 0.489 Na and Cl = = 1 and O = = 3.98 0.123 0.123 The empirical formula is Na1Cl1O3.98 or NaClO4; this compound is named sodium perchlorate. ©McGraw-Hill Education. Determining the Molecular Formula The molecular formula gives the actual numbers of moles of each element present in 1 mol of compound. The molecular formula is a whole-number multiple of the empirical formula: 𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 = 𝑧 × (𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎) molar mass (g/mol) = whole−number multiple=z empirical formula mass (g/mol) Hydrogen Peroxide has the empirical formula HO. The molar mass is 34.02 g/mol. What is the molecular formula? (𝐻𝑂)𝑧 = 𝐻𝑧 𝑂𝑧 𝑔 𝑔 34.02 34.02 𝑚𝑜𝑙 𝑚𝑜𝑙 = 2.000 = 2 𝑧= = 𝑀 𝐻𝑂 17.01 𝑔/𝑚𝑜𝑙 𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎: 𝐻2 𝑂2 ©McGraw-Hill Education. Data for the Molecular Formula Molar Mass (g/mol) of an unknown compound is measured by Mass Spectrometry in the form of Molecular Mass (amu). Analytical laboratories do not report compositional data as absolute masses (g). Instead, they provide mass percents. To get the masses of elements, we assume 100 g of sample, therefore percent masses are the absolute masses in 100 g sample. ©McGraw-Hill Education. Sample Problem 3.8 – Problem and Plan Determining a Molecular Formula from Elemental Analysis and Molar Mass PROBLEM: During excessive physical activity, lactic acid ( M = 90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental analysis shows that this compound has 40.0 mass % C, 6.71 mass % H, and 53.3 mass % O. a) Determine the empirical formula of lactic acid. b) Determine the molecular formula. PLAN: Assume 100g of sample. Percent masses become absolute masses. ©McGraw-Hill Education. Sample Problem 3.10 - Solution SOLUTION: 1 mol C 40.0 g C x = 3.33 mol C 12.01 g C 1 mol O 53.3 g O x = 3.33 mol O 16.00 g O 1 mol H 6.71 g H x = 6.66 mol H 1.008 g H Empirical formula = CH2O molar mass of lactic acid 90.08 g/mol = ≅3 molar mass of CH2O 30.03 g/mol C3H6O3 is the molecular formula ©McGraw-Hill Education. Combustion Analysis of Organic Compounds Organic compounds contain C, H and possibly other elements, O, N, S, etc. Burning the organic compound in O2(g) totally converts C to CO2 and H to H2O: 1 mol C yields 1 mol CO2, 1 mol H yields 0.5 mol H2O. Figure 3.4. A sample of an organic compound containing C, H and other elements is burned in a stream of O2(g). The resulting H2O(g) is absorbed by Mg(ClO4)2, and the resulting CO2 is absorbed by NaOH(s) on asbestos. ©McGraw-Hill Education. Sample Problem 3.11 – Problem and Plan Determining a Molecular Formula from Combustion Analysis PROBLEM: When a 1.000 g sample of vitamin C (M = 176.12 g/mol) is placed in a combustion chamber and burned, the following data are obtained: mass of CO2 absorber after combustion = 85.35 g mass of CO2 absorber before combustion = 83.85 g mass of H2O absorber after combustion = 37.96 g mass of H2O absorber before combustion = 37.55 g PLAN: The masses of CO2 and H2O produced will give us the masses of C and H present in the original sample. From this we can determine the mass of O. ©McGraw-Hill Education. Sample Problem 3.11 - Solution SOLUTION: m CO2 = 85.35 g - 83.85 g = 1.50 g 12.01 g C = 0.409 g C 44.01 g CO2 m H2O = 37.96 g - 37.55 g = 0.41 g 2×1.008 g H 0.41 g H2O x = 0.046 g H 18.02 g H2O 1.50 g CO2 x mass of O = mass of vitamin C sample – (mass of C + mass of H) = 1.000 g - (0.409 + 0.046) g = 0.545 g O ©McGraw-Hill Education. Sample Problem 3.11 – Solution, Cont’d Convert mass to moles: 0.409 g C = 0.0341 mol C 12.01 g/mol C 0.046 g H = 0.046 mol H 1.008 g/mol H 0.545 g O = 0.0341 mol O 16.00 g/mol O Divide by smallest to get the preliminary formula: 0.0341 0.046 0.0341 C =1 H = 1.3 O =1 0.0341 0.0341 0.0341 C1H1.3O1 = C3H3.9O3 → C3H4O3 Divide molar mass by mass of empirical formula: 176.12 g/mol = 2.000 88.06 g/mol ©McGraw-Hill Education. C6H8O6 Table 3.2 Compounds with Empirical Formula CH2O Name Molecular Formula Whole-Number Multiple M (g/mol) Use or Function Formaldehyde CH2O 1 30.03 Disinfectant; biological preservative Acetic acid C2H4O2 2 60.05 Acetate polymers; vinegar (5% solution) Lactic acid C3H6O3 3 90.08 Causes milk to sour; forms in muscles during exercise Erythrose C4H8O4 4 120.10 Forms during sugar metabolism Ribose C5H10O5 5 150.13 Component of many nucleic acids and vitamin B2 Glucose C6H12O6 6 180.16 Major nutrient for energy in cells ©McGraw-Hill Education. 3.3 Chemical Equations A chemical equation is a statement of chemical or physical change that uses chemical formulas to express the identities and quantities of substances involved. A chemical equation shows the change from the initial substances (reactants) to the final substances (products). The direction of change is indicated by a straight arrow. The arrow points to the products. Figure 3.6 ©McGraw-Hill Education. Features of Chemical Equations To present a chemical change quantitatively, an equation must be balanced: the same number and type of each atom must appear on both sides. The equation below is NOT balanced: ©McGraw-Hill Education. Balancing a Chemical Equation 1) We first translate the chemical statement into an unbalanced “skeleton” equation: 𝑀𝑔 + 𝑂2 → 𝑀𝑔𝑂 2) Balancing the atoms. We match the numbers of each type of atom on the left and the right of the yield arrow: 1 𝑀𝑔 + 𝑂2 → 𝑀𝑔𝑂 2 3) Adjusting the coefficients. In most cases, the smallest wholenumber coefficients are preferred: 2𝑀𝑔 + 𝑂2 → 2𝑀𝑔𝑂 4) Checking. After balancing and adjusting the coefficients, we always check that the equation is balanced: 2𝑀𝑔 + 2𝑂 = 2 × 𝑀𝑔𝑂 5) Specifying the states of matter. The final equation also indicates the physical state of each substance or whether it is dissolved in water: ©McGraw-Hill Education. 2𝑀𝑔(𝑠) + 𝑂2 (𝑔) → 2𝑀𝑔𝑂(𝑠) A Three-level View of a Reaction Figure 3.7 ©McGraw-Hill Education. © McGraw-Hill Education/Charles Winters/Timeframe Photography, Inc. Sample Problem 3.10 – Problem and Solution Balancing a Chemical Equation PROBLEM: Within the cylinders of a car’s engine, the liquid hydrocarbon octane (C8H18), one of many components of gasoline, mixes with oxygen from the air and burns to form carbon dioxide and water vapor. Write a balanced equation for this reaction. SOLUTION: C8H18 + O2 → CO2 + H2O 25 C8H18 + ( ) O2 → 8 CO2 + 9 H2O 2 2C8H18 + 25O2 → 16CO2 + 18H2O 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠: 16𝐶, 36𝐻, 50𝑂 → 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠: 16𝐶, 36𝐻, 50𝑂 2C8H18(l) + 25O2 (g) → 16CO2 (g) + 18H2O(g) ©McGraw-Hill Education. Visualizing a Reaction with a Molecular Scene Combustion of Octane ©McGraw-Hill Education. Sample Problem 3.11 – Problem and Plan Balancing an Equation from a Molecular Scene PROBLEM: The following molecular scenes depict an important reaction in nitrogen chemistry. The blue spheres represent nitrogen while the red spheres represent oxygen. Write a balanced equation for this reaction. PLAN: Determine the formulas of the reactants and products from their composition. Arrange this information in the correct equation format and balance correctly, including the states of matter. ©McGraw-Hill Education. Sample Problem 3.11 - Solution SOLUTION: The reactant circle shows only one type of molecule, composed of 2 N and 5 O atoms. The formula is thus N2O5. There are 4 N2O5 molecules depicted. The product circle shows two types of molecule; one has 1 N and 2 O atoms while the other has 2 O atoms. The products are NO2 and O2. There are 8 NO2 molecules and 2 O2 molecules shown. The reaction depicted is 4N2O5 → 8NO2 + 2O2. Writing the equation with the smallest whole-number coefficients and states of matter included; 2N2O5 (g) → 4NO2 (g) + O2 (g) ©McGraw-Hill Education. 3.4 Stoichiometric Calculations In a balanced equation, the amounts (mol) of substances are stoichiometrically equivalent to each other, which means that a specific amount of one substance is formed from, produces, or reacts with a specific amount of the other. The quantitative relationships are expressed as stoichiometrically equivalent molar ratios that we use as conversion factors to calculate the amounts. ©McGraw-Hill Education. Table 3.4 Information Contained in a Balanced Equation 𝐶3 𝐻8 𝑔 + 5𝑂2 𝑔 → 3𝐶𝑂2 + 4𝐻2 𝑂(𝑔) Viewed in Terms of Reactants C3H8(g) + 5O2(g) → Products 3CO2(g) + 4H2O(g) Molecules 1 molecule C3H8 + 5 molecules O2 → 3 molecules CO2 + 4 molecules H2O 1 mol C3H8 + 5 mol O2 → 3 mol CO2 + 4 mol H2O 44.09 amu C3H8 + 160.00 amu O2 → 132.03 amu CO2 + 72.06 amu H2O 44.09 g C3H8 + 160.00 g O2 → 132.03 g CO2 + 72.06 g H2O Amount (mol) Mass (amu) Mass (g) Total mass (g) ©McGraw-Hill Education. 204.09 g → 204.09 g Stoichiometrically Equivalent Molar Ratios 𝐶3 𝐻8 𝑔 + 5𝑂2 𝑔 → 3𝐶𝑂2 + 4𝐻2 𝑂(𝑔) Possible stoichiometrically equivalent molar ratios (s.e.m.r.): 1 𝑚𝑜𝑙 𝐶3 𝐻8 1 𝑚𝑜𝑙 𝐶3 𝐻8 1 𝑚𝑜𝑙 𝐶3 𝐻8 5 𝑚𝑜𝑙 𝑂2 𝑂𝑅 𝑂𝑅 𝑂𝑅 5 𝑚𝑜𝑙 𝑂2 3 𝑚𝑜𝑙 𝐶𝑂2 4 𝑚𝑜𝑙 𝐻2 𝑂 3 𝑚𝑜𝑙 𝐶𝑂2 5 𝑚𝑜𝑙 𝑂2 3 𝑚𝑜𝑙 𝐶𝑂2 𝑂𝑅 𝑂𝑅 4 𝑚𝑜𝑙 𝐻2 𝑂 4 𝑚𝑜𝑙 𝐻2 𝑂 ©McGraw-Hill Education. Stoichiometric Calculations – Cont’d The quantitative relationships are expressed as stoichiometrically equivalent molar ratios (s.e.m.r.) that we use as conversion factors to calculate the amounts. 𝐶3 𝐻8 𝑔 + 5𝑂2 𝑔 → 3𝐶𝑂2 + 4𝐻2 𝑂(𝑔) How many moles of O2 are consumed when 10.0 mol of H2O is produced? 5 𝑚𝑜𝑙 𝑂2 𝑛 𝑂2 , 𝑚𝑜𝑙 = 10.0 𝑚𝑜𝑙 𝐻2 𝑂 × = 12.5 𝑚𝑜𝑙 𝑂2 4 𝑚𝑜𝑙𝐻2 𝑂 NOTE THAT: Stoichiometry only works in mol. It never works in mass (g). The conversion factor (s.e.m.r.) is always a ratio of moles of 2 reaction participants: either 2 reactants, or 2 products, or 1 reactant and 1 product. Correctly balancing the chemical equation is critical! ©McGraw-Hill Education. Amount-mass-number Relationships Figure 3.8 Start at any box and move to any other by using the conversion factor on the arrow. Always convert to amount (mol) first. ©McGraw-Hill Education. Sample Problem 3.12a – Problem and Plan Calculating Quantities of Reactants and Products: Amount (mol) to Amount (mol) PROBLEM: Copper is obtained from copper(I) sulfide by roasting it in the presence of oxygen gas to form powdered copper(I) oxide and gaseous sulfur dioxide. How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide? PLAN: Step 1 – set up the balanced chemical equation. The first statement (period) gives us the reaction participants. – Reactants (formulas, states): ? – Products (formulas, states): ? 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 (𝑓𝑜𝑟𝑚𝑢𝑙𝑎𝑠, 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠, 𝑠𝑡𝑎𝑡𝑒𝑠) → 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 ©McGraw-Hill Education. Sample Problem 3.12a –Solution PLAN: Step 2 – decide the roadmap for stoichiometric calculation SOLUTION: 2Cu2S (s) + 3O2 (g) → 2Cu2O (s) + 2SO2 (g) 3 mol O2 n(mol 𝑂2 ) = 10.0 mol Cu2S x = 𝟏𝟓. 𝟎 𝒎𝒐𝒍 𝑶𝟐 2 mol Cu2S ©McGraw-Hill Education. Sample Problem 3.12b – Problem and Plan Calculating Quantities of Reactants and Products: Amount (mol) to Mass (g) PROBLEM: During the process of roasting copper(I) sulfide, how many grams of sulfur dioxide form when 10.0 mol of copper(I) sulfide reacts? PLAN: Using the balanced equation from the previous problem, we again use the mole ratio as a conversion factor. ©McGraw-Hill Education. Sample Problem 3.12b - Solution SOLUTION: 2Cu2S (s) + 3O2 (g) → 2Cu2O (s) + 2SO2 (g) M(SO2)= 32.06 g S mol + 2 × 16.00 𝑔 𝑂 𝑚𝑜𝑙 = 64.06 𝑔 𝑆𝑂2 𝑚𝑜𝑙 2 mol SO2 64.06 g SO2 m (𝑆𝑂2 )= 10.0 mol Cu2S x × = 𝟔𝟒𝟏 𝒈 𝑺𝑶𝟐 2 mol Cu2S 1 mol SO2 (3 sig. figs) ©McGraw-Hill Education. Sample Problem 3.13 – Problem and Plan Calculating Quantities of Reactants and Products: Mass to Mass PROBLEM: During the roasting of copper(I) sulfide, how many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide? PLAN: We convert the mass of Cu2O from kg to g and then to amount (mol). Then, we use the molar ratio (3 mol O2/2 mol Cu2O) to find the amount (mol) of O2 required. Finally, we convert the amount of O2 to g and then kg. ©McGraw-Hill Education. Sample Problem 3.13 - Solution SOLUTION: 2Cu2S (s) + 3O2 (g) → 2Cu2O (s) + 2SO2 (g) M(Cu2O) = 143.10 g/mol M(O2) = 32.00 g/mol 103 g 1 mol Cu2O 2.86 kg Cu2O × × = 20.0 𝑚𝑜𝑙 Cu2O 1 kg 143.10 g Cu2O 3 mol O2 32.00 g O2 1 kg 20.0 mol Cu2O × × × 3 = 𝟎. 𝟗𝟓𝟗 𝒌𝒈 𝑶𝟐 2 mol Cu2O 1 mol O2 10 g ©McGraw-Hill Education. Overall (Net, Combined) Reactions For stoichiometric purposes, when the same substance forms in one reaction and reacts in the next (it is common to both reactions), we eliminate that substance in an overall (net) equation. The steps in writing the overall equation are: 1) Write 2 or more balanced chemical equations in the sequence these occur: 1 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 → 𝐼𝑛𝑡𝑒𝑟𝑚𝑒𝑑𝑖𝑎𝑡𝑒1 + 𝑠𝑖𝑑𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑖𝑓 𝑎𝑛𝑦 𝐼𝑛𝑡𝑒𝑟𝑚𝑒𝑑𝑖𝑎𝑡𝑒1 + 𝑜𝑡ℎ𝑒𝑟 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠(𝑖𝑓 𝑎𝑛𝑦) → 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 2) Adjust the equations arithmetically to cancel the common substance(s). You must cancel out the intermediate(s). 3) Add the adjusted equations together to obtain 1 overall balanced equation, known as the net equation. ©McGraw-Hill Education. Sample Problem 3.14 - Problem and Plan Writing an Overall Equation for a Reaction Sequence PROBLEM: Roasting is the first step in extracting copper from chalcocite, the ore used in the previous problem. In the next step, copper(I) oxide reacts with powdered carbon to yield copper metal and carbon monoxide gas. Write a balanced overall equation for the two-step process. PLAN: Write individual balanced equations for each step. Adjust the coefficients so that any common substances can be canceled. Add the adjusted equations together to obtain the overall equation. ©McGraw-Hill Education. Sample Problem 3.14 - Solution SOLUTION: Write individual balanced equations for each step: Roasting (1): 2Cu2S (s) + 3O2 (g) → 2Cu2O (s) + 2SO2 (g) Copper formation (2): Cu2O (s) + C (s) → 2Cu (s) + CO (g) Identify the intermediate to be canceled out: Cu2O (s) Adjust the coefficients so that the 2 moles of Cu2O formed in reaction 1 are used up in reaction 2: 2Cu2S (s) + 3O2 (g) → 2Cu2O (s) + 2SO2 (g) 2Cu2O (s) + 2C (s) → 4Cu (s) + 2CO (g) Add the equations together to generate the net equation: 2Cu2S (s) + 3O2 (g) + 2C (s) → 2SO2 (g) + 4Cu (s) + 2CO (g) ©McGraw-Hill Education. Limiting Reactants So far we have assumed that reactants are present in the correct amounts to react completely. In reality, one reactant may be present in a smaller amount than what is required by the stoichiometry of the chemical equation and will limit the amount of product that can form. The limiting reactant will be completely used up in the reaction, while there will be unreacted amounts of the other reactants. The reactant that is not limiting is in excess – some of this reactant will be left over. ©McGraw-Hill Education. Analogy for Limiting Reactions ©McGraw-Hill Education. Figure 3.10 Sample Problem 3.15 - Problem Using Molecular Depictions in a Limiting-Reactant Problem PROBLEM: Chlorine trifluoride, an extremely reactive substance, is formed as a gas by the reaction of elemental chlorine and fluorine. The molecular scene shows a representative portion of the reaction mixture before the reaction starts. (Chlorine is green, and fluorine is yellow.) (a)Find the limiting reactant by stoichiometric calculations. (b)Track the budget of molecules in a reaction table. (c)Draw a representative portion of the mixture after the reaction is complete. ©McGraw-Hill Education. Sample Problem 3.15 – Plan and Solution (a) PLAN: Write a balanced chemical equation. To determine the limiting reactant, find the number of molecules of product that would form from the given numbers of molecules of each reactant. Use these numbers to write a reaction table and use the reaction table to draw the final reaction scene. SOLUTION: a) The balanced equation is Cl2 (g) + 3F2 (g) → 2ClF3 (g) There are 3 molecules of Cl2 and 6 molecules of F2 depicted: 2 molecules ClF3 3 molecules Cl2 x = 6 molecules ClF3 1 molecule Cl2 2 molecules ClF3 6 molecules F2 x = 4 molecules ClF3 3 molecule F2 Since the given amount of F2 can form less product, it is the limiting reactant. ©McGraw-Hill Education. Sample Problem 3.15 – Solution (b) and (c) (b) We use the amount of F2 to determine the “change” in the reaction table, since F2 is the limiting reactant: Molecules Cl2 (g) + 3F2 (g) → 2ClF3 (g) Initial Change 3 1𝐶𝑙2 −6𝐹2 × = −2 3𝐹2 6 -6 0 2𝐶𝑙𝐹3 +6𝐹2 × = +4 3𝐹2 Final 3–2=1 0 0+4=4 (c) The final reaction scene shows that all the F2 has reacted and that there is Cl2 left over. 4 molecules of ClF3 have formed: ©McGraw-Hill Education. Sample Problem 3.16 – Problem and Plan Calculating Quantities in a Limiting-Reactant Problem: Amount to Amount PROBLEM: In another preparation of ClF3, 0.750 mol of Cl2 reacts with 3.00 mol of F2. (a) Find the limiting reactant. (b) Write a reaction table. PLAN: Find the limiting reactant by calculating the amount (mol) of ClF3 that can be formed from each given amount of reactant. Use this information to construct a reaction table. ©McGraw-Hill Education. Sample Problem 3.16 –Solution (a) and (b) SOLUTION: (a) The balanced equation is Cl2 (g) + 3F2 (g) → 2ClF3 (g) 2 mol ClF3 0.750 mol Cl2 x = 1.50 mol ClF3 1 mol Cl2 2 mol ClF3 3.00 mol F2 x = 2.00 mol ClF3 3 mol F2 Cl2 is limiting, because the given amount yields less ClF3. (b) All the Cl2 reacts since this is the limiting reactant. For every 1 Cl2 that reacts, 3 F2 will react, so 3(0.750) or 2.25 moles of F2 reacts. Amount (mol) Cl2(g) Initial 0.750 3.00 0 Change −0.750 −2.25 +1.50 0.75 1.50 Final ©McGraw-Hill Education. 0 + 3F2(g) → 2ClF3(g) Sample Problem 3.17 – Problem and Plan Calculating Quantities in a Limiting-Reactant Problem: Mass to Mass PROBLEM: A fuel mixture used in the early days of rocketry consisted of two liquids, hydrazine (N2H4) and dinitrogen tetroxide (N2O4), which ignite on contact to form nitrogen gas and water vapor. (a) How many grams of nitrogen gas form when 1.00 x 102 g of N2H4 and 2.00 x 102 g of N2O4 are mixed? (b) Write a reaction table for this process. PLAN: Find the limiting reactant by calculating the amount (mol) of product that can be formed from each given mass of reactant. Make sure you calculate the mass of the SAME product for each given amount of reactant. Use this information to construct a reaction table. ©McGraw-Hill Education. Sample Problem 3.17 – Plan, Cont’d ©McGraw-Hill Education. Sample Problem 3.17 – Solution (a) SOLUTION: (a) 2N2H4 (l) + N2O4 (l) → 3N2 (g) + 4H2O (g) 1 mol N2H4 = 3.12 mol N2H4 32.05 g N2H4 3 mol N2 3.12 mol N2H4 x = 4.68 mol N2 2 mol N2H4 For N2H4: 1.00x 102 g N2H4 x 1 mol N2O4 For N2O4: 2.00x g N2O4 x = 2.17 mol N2O4 92.02 g N2O4 3 mol N2 2.17 mol N2O4 x = 6.51 mol N2 1 mol N2O4 102 N2H4 is limiting and only 4.68 mol of N2 can be produced: 28.02 g N2 4.68 mol N2 x = 131 g N2 1 mol N2 ©McGraw-Hill Education. Sample Problem 3.17 – Solution (b) (b) All the N2H4 reacts since it is the limiting reactant. For every 2 moles of N2H4 that react 1 mol of N2O4 reacts and 3 mol of N2 form: 1 mol N2O4 3.12 mol N2H4 x = 1.56 mol N2O4 reacts 2 mol N2H4 4 mol H2O 3.12 mol N2H4 x = 6.24 mol 𝐻2 𝑂 2 mol N2H4 Amount (mol) 2N2H4(l) + N2O4(l) → 3N2(g) + 4H2O(g) Initial 3.12 2.17 0 0 Change −3.12 −1.56 +4.68 +6.24 0.61 4.68 6.24 Final ©McGraw-Hill Education. 0 Stoichiometric Relationships Figure 3.11 ©McGraw-Hill Education. Reaction Yields The theoretical yield is the amount (mol) or mass (g) of a product calculated (predicted) using the molar ratios from the balanced equation. The actual yield is the amount (mol) or mass (g) of a product actually obtained by running the reaction in a laboratory. The actual yield is usually less than the theoretical yield and can never exceed it. See why on the next slide. The percent yield is the actual yield expressed as a percentage of the theoretical yield. actual yield % yield = × 𝟏𝟎𝟎 theoretical yield NOTE: units must cancel out in %-yield (mol/mol, g/g, kg/kg) ©McGraw-Hill Education. Reaction Yields – cont’d What causes the actual yield to be smaller than the theoretical yield? Common reasons: 1) Losses associated with isolating and purifying the desired product from the reaction mixture: other products, solvents, catalysts, excess reactants 2) Losses during synthesis (running the reaction) from the reaction vessel: evaporation or leaking of reactants, products 3) Side (competing, parallel) reactions. 4) Incomplete reaction: a) Incomplete reaction time for slow reactions (see Chemical Kinetics in Gen Chem II) b) Reaction cannot complete 100% to products due to entropy (see Thermodynamics and Chemical Equilibrium in Gen Chem II) ©McGraw-Hill Education. Effect of Side Reactions on Yield of Main Product Example: 𝑀𝑎𝑖𝑛 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛: 2𝑁2 𝐻4 + 𝑁2 𝑂4 → 3𝑁2 + 4𝐻2 𝑂 𝑆𝑖𝑑𝑒 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛: 𝑁2 𝐻4 + 2𝑁2 𝑂4 → 6𝑁𝑂 + 2𝐻2 𝑂 ©McGraw-Hill Education. Sample Problem 3.21 – Problem and Plan Calculating Percent Yield PROBLEM: Silicon carbide (SiC) is made by reacting sand (silicon dioxide) with powdered carbon at high temperature. Carbon monoxide is also formed. What is the percent yield if 51.4 kg of SiC is recovered from processing 100.0 kg of sand? PLAN: After writing the balanced equation, we convert the given mass of SiO2 (100.0 kg) to amount (mol). We use the molar ratio to find the amount of SiC formed and convert it to mass (kg) to obtain the theoretical yield. ©McGraw-Hill Education. Sample Problem 3.21 - Solution SOLUTION: SiO2(s) + 3C(s) → SiC(s) + 2CO(g) M(SiO2) = 60.09 g/mol M(SiC) = 40.10 g/mol 103 g 1 mol SiO2 100.0 kg SiO2 x x = 1 664 mol SiO2 1 kg 60.09 g SiO2 mol SiO2 = mol SiC = 1664 mol SiC 40.10 g SiC 1 kg 1664 mol SiC x x 3 = 66.73 kg 1 mol SiC 10 g %-yield = ©McGraw-Hill Education. 51.4 kg × 100 = 𝟕𝟕. 𝟎% 66.73 kg