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Published on January 16, 2008

Author: Tomasina

Source: authorstream.com

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Calculations with Chemical Formulas and Equations:  Calculations with Chemical Formulas and Equations Chapter 3 Molecular and Formula Weight:  Molecular and Formula Weight The molecular weight is the sum of the masses of all atoms in a molecule. C2H5OH has a molecular mass of 46.0688. The formula weight is the sum of the masses of all atoms in a formula unit of any compound, molecular or ionic. CaCO3 has a formula mass of 100.0892 Mole and Molar Mass:  Mole and Molar Mass A mole of any formula unit is the amount whose mass is equal to the molecular or formula weight in grams. A mole of any formula unit is 6.022142 x 1023 of those formula units. The molar mass of a substance described by a given formula unit is the total mass of a mole of that formula unit. Slide4:  1 mole C8H17OH 1 mole HgI2 1 mole CH3OH 1 mole S8 Molar Mass Example:  Molar Mass Example CaCO3 has a formula mass of 100.0892 amu. One formula unit of CaCO3 weighs 100.0892 amu. CaCO3 has a molar mass of 100.0892 grams. One mole of CaCO3 formula units weighs 100.0892 grams. 6.022142 x 1023 amu = 1 gram Conversion of Grams & Moles:  Conversion of Grams & Moles The conversion factor between grams and moles of any formula unit is 1 mole A = (formula mass A) grams Example: 1 mole C2H5OH = 46.0688 grams Example:  Example Aspirin has the formula C9H8O4. How many moles of aspirin are in a tablet weighing 500 mg? Formula mass = 180.1598 g/mole Percent Composition:  Percent Composition The percent composition of a compound is the percent by mass of each substituent element. Percent composition can be calculated from the chemical formula by converting moles to grams. Example:  Example What is the percent composition of ammonium nitrate, NH4NO2? Solution: 2 moles N = 2 (14.0067) g = 28.0134 g 4 moles H = 4 (1.00794) g = 4.03176 g 2 moles O = 2 (15.9994) g = 31.9988 g Total mass = 64.0440 g Elemental Analysis:  Elemental Analysis In elemental analysis, a known weight of a compound containing only, C, H, and O is burned and the products CO2 and H2O are collected and weighed. From the weights of these products, the % composition may be found. mass CO2  mass C mass H2O  mass H mass O by difference Elemental Analysis:  Elemental Analysis Combustion in Oxygen:  Combustion in Oxygen CxHyOz + w O2  x CO2 + ½ y H2O CO2 is 27.29% C H2O is 11.19% H mass C in sample = 0.2729  mass CO2 mass H in sample = 0.1119  mass H2O mass O in sample = mass sample – mass C – mass H Example:  Example A 505 mg sample of an organic substance gives 855 mg CO2 and 624 mg H2O. What is the % composition? Solution: 855 mg CO2 = 233 mg C 624 mg H2O = 69.8 mg H Therefore 505 – 233 – 70 = 202 mg O %C = 233/505 = 46.1 % %H = 69.8/505 = 13.8% %O = 202/505 = 40.0% Mass Spectrometry Methylene Chloride:  Mass Spectrometry Methylene Chloride Combustion in Oxygen:  Combustion in Oxygen CxHyOz + w O2  x CO2 + ½ y H2O CO2 is 27.29% C H2O is 11.19% H mass C in sample = 0.2729  mass CO2 mass H in sample = 0.1119  mass H2O mass O in sample = mass sample – mass C – mass H Chloride Analysis:  Chloride Analysis A 7.35 g sample of methylene chloride is treated with sodium peroxide and the liberated chloride ion yields 24.80 g AgCl. Calculate the % chlorine in methylene chloride. Hint: AgCl is 24.74% Cl by weight. Is your result consistent with the chemical formula CH2Cl2? Example:  Example Combusion of a 8.23 mg sample of a compound gave 9.62 mg carbon dioxide and 3.94 mg water. Analysis of a separate 5.32 mg sample of the same compound gave 13.49 mg Cl. What is the % composition of the compound? Empirical Formula:  Empirical Formula The empirical formula is the smallest formula unit consistent with a given elemental composition. Example: Hydrogen peroxide has the molecular formula H2O2. Its empirical formula is HO. Example: Oxalic acid is H2C2O4. Its empirical formula is HCO2. Empirical vs. Molecular Formula:  Empirical vs. Molecular Formula Empirical Formula from %’s:  Empirical Formula from %’s Convert mass percentages to moles of each element. Convert the moles of each element to mole ratios relative to the least abundant element. Convert the mole ratios to equivalent whole number ratios. These ratios are the formula subscripts. Example:  Example A substance has 14.25% C, 56.93% O, and 28.83% Mg. What is the empirical formula? Solution 14.25 g C = 1.19 mole C 56.93 g O = 3.56 mole O 28.83 g Mg = 1.19 mole Mg Ratios: Mg/C = 1 and O/C = 3 Therefore, the substance is MgCO3 Example:  Example 1.00 g of menthol on combustion yields 1.161 g of H2O and 2.818 g of CO2. What is the empirical formula? Solution: 1.161 g H2O  0.1298 g H = 0.1288 mol H 2.818 g CO2  0.7690 g C = 0.0640 mol C difference = 0.101 g O = 0.00632 mol O ratios: H/O = 20.4 and C/O =10.1 Therefore, C10H20O Conversion of Moles to Moles:  Conversion of Moles to Moles The conversion factor between moles A, B, C and D in the reaction a A + b B  c D + d D is a moles A = b moles B = c moles C = … Example: CH4 + 2O2  CO2 + 2H2 1 mole CH4 = 2 moles O2 = 1 mole CO2 = … Example:  Example The combustion of propane occurs according to C3H8 + 5O2  3CO2 + 4H2O How many moles of H2O are produced if 12 moles of oxygen are consumed? Solution Mass Stoichiometry:  Mass Stoichiometry Goal: Given a quantity of one substance in a reaction, calculate the quantity of any other substance required or produced in the reaction. Example: CH4 + 2O2  CO2 + 2H2O Suppose that 20.0 g CH4 burned. How many grams of H2O are produced? Solution Strategy:  Solution Strategy Convert grams of A to mole A using the formula weight of A Convert moles A to moles B using the balanced chemical equation Convert moles B to grams B using the formula weight of B Example:  Example In the reaction: 2 NaOH + Cl2  NaOCl + NaCl + H2O How many grams of NaOH are needed to react with 25.0 g of Cl2? Solution: 25.0 g Cl2  0.3526 mole Cl2  0.7052 mole NaOH  28.2 g NaOH Limiting Reagent Example:  Limiting Reagent Example Which is the limiting reagent – cheese or bread? Limiting Reactants:  Limiting Reactants  is the limiting reactant. is in excess. Limiting Reactant Problems:  Limiting Reactant Problems Determine which reactant is limiting. Use the limiting reactant to calculate the quantities of products produced and excess reactants used. Never use the quantity of excess reactant to calculate product produced. Limiting Reactant Example:  Limiting Reactant Example CH4 + 2 Cl2  CH2Cl2 + 2 HCl Suppose 1.85 kg of CH4 is allowed to react with 12.62 kg of Cl2. What is the theoretical yield of CH2Cl2? Solution: 1.85 kg CH4 = 115.3 mol CH4 12.62 kg Cl2 = 178.0 mol Cl2 Cl2 is limiting reactant  7.559 kg CH2Cl2 Theoretical Yield:  Theoretical Yield The theoretical yield of a product is the mass that would be obtained from a given quantity of starting material if it were all converted to product (and recovered as such). If the reaction is a A  b B, and the mass of A is given, find the mass of B. g A  moles A  moles B  g B. Actual and Percent Yield:  Actual and Percent Yield The actual yield is the mass of product actually obtained and recovered in a laboratory experiment. The percent yield is the ratio of the actual to the theoretical yield, expressed as a percent. % yield = (actual / theoretical) x 100 Example:  Example Dichloromethane is produced by: CH4 + 2 Cl2  CH2Cl2 + 2 HCl How many grams of CH2Cl2 result from reaction of 1.85 kg of CH4 if the % yield is 43.1%? Solution: 1850 g CH4  115.3 mol  9790 g CH2Cl2 actual yield = .431 x theoretical yield = 4220 g CH2Cl2 = 4.22 kg

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