Balancing Chemical Equations and Calculating Molar Masses, Study notes of Chemistry

Download Balancing Chemical Equations and Calculating Molar Masses and more Study notes Chemistry in PDF only on Docsity! Chapter 3 Chapter 3. Stoichiometry: Calculations with Chemical Formulas and Equations Chapter 3 Objective 1 • Write a balanced chemical equation given a word description or the formulas of the reactants and products. Balancing Chemical Equations • The number of atoms of each element on the reactant side equals the number of atoms of each element on the product side – CH4(g) + O2(g)  CO2(g) + H2O(l) Not balanced – CH4(g) + 2 O2(g)  CO2(aq) + 2 H2O(l) Balanced Strategies: • Balance the equation by inspection, starting with the most complicated molecule(s). Balance free elements last. • Determine what coefficient is necessary so that the same number of each type of atoms appears on both reactant and product sides. • If fractional coefficients, multiply thru by denominator • Finally, check that smallest whole number coefficients are used. • Subscripts should never be changed when trying to balance a chemical equation. Changing a subscript changes the actual identity of a product or reactant Examples: Balancing Chemical Equations Balance the following equation: C3H7OH(l) + O2(g)  CO2(g) + H2O(l) Balance the following equation: _ AgNO3(aq) + _ CaCl2 (aq)  _ AgCl(s) + _ Ca (NO3)2(aq) Note – you can count the nitrate (NO3 -) ion as a whole rather than splitting it Practice: Balanced Chemical Equations Write a balanced Chemical Equation to correspond to each of the following descriptions. 1. Solid lithium hydroxide is used in space vehicles to remove the carbon dioxide gas exhaled by astronauts. Solid lithium hydroxide reacts with gaseous carbon dioxide to form solid lithium carbonate and liquid water. 2. Balance the following chemical equation. a) Ca(OH)2(aq) + H3PO4 (aq)  Ca3(PO4) 2(s) + H2O(l) Combination reaction • Two elements and/or compounds combine to form ONE compound GENERAL FORM: A + B → AB • A metal element reacts with a nonmetal element to form an ionic compound – 2 Na(s) + Cl2(g)  2 NaCl(s) • A nonmetal element reacts with another nonmetal element to form a molecular compound – N2(g) + 3 H2(g)  2 NH3(g) • A nonmetal oxide compound reacts with water to form an acid – SO3(g) + H2O(l) → H2SO4 (aq) • A a metal oxide compound reacts with water to form a base – CaO (s) + H2O (l) → Ca(OH)2 (aq) Decomposition reactions • A compound is decomposed into two or more substances (elements and/or compounds) GENERAL FORM: AB → A + B • The decomposition of a compound to form its elements – 2 H2O(l) → 2H2(g) + O2(g) – 2 NaN3 (s) → 2 Na (s) + 3 N2 (g) • Decomposition of carbonates to yield oxides and CO2 – CaCO3 (s)  CaO (s) + CO2 (g • Decomposition of chlorates to yield chlorides and O2 – 2 KClO3 (s)  2 KCl (s) + O2 (g) Decomposition reaction of NaN3(s) Is used in inflating automobile airbags . Combustion reaction • Reactions of substances with oxygen to produce CO2, H2O and heat • General Form: CxHy + O2 (g) → CO2 + H2O (g) or CxHyOz + O2 (g) → CO2 (g) + H2O (g) • Hydrocarbons react with oxygen in the air to produce CO2 and H2O. – C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) • Combustion of oxygen-containing derivatives of hydrocarbons such as alcohols, sugars also produce CO2 and H2O – 2 C3H7OH(l) + 9 O2(g) → 6 CO2(g) + 8 H2O(g) – C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(g) Practice: Predict the products and balance the following reactions 1. Combination reaction of Mg(s) + N2(g)  CaO(s) + H2O(l)  2. Combustion reaction of octane C8H18 3. Decomposition of BaF2(s)  CaCO3(s)  Chapter 3 Objectives 4 and 5 4. Given a chemical formula or name, convert among moles, grams, amu, molecules and atoms. 5. Given the name or formula of an element or compound, determine its molar mass. The mole • If you understand what a dozen is, you understand the mole • One dozen = 12 • One mole 602200000000000000000000= 6.022 X 1023 • This number is called Avogadro’s number. • Why so many? Atoms and molecules are so small that we can only deal with large quantities of them. Formula Weights and Molar Mass • The weight of an individual molecule (covalent) or formula unit (ionic) • Is calculated by adding the masses of the atoms in a single molecule or formula unit – Formula weight of NaCl = 1(23.0 amu) + 1(35.5 amu) = 58.5 amu • The molar mass of a compound is the mass in grams of 1 mole of the compound. It is numerically equal to formula weight – Molar mass of NaCl = 1 (23.0 g/mol) + 1(35.5 g/mol) = 58.5 g/mol • The term molecular weight is sometimes used for formula weight and molar mass Molar Mass • Is the mass of 1 mole of a substance (i.e. g/mol) • The molar mass of an element is the atomic mass in g/mole for the element • Atomic mass of O is 16.0 amu • Molar mass of O is 16.0 g/mole • The molar mass of a substance is found by adding the molar mass of its component atoms • The molar mass (in g/mol) of any substance is always numerically equal to its formula weight (in amu) – Formula weight of NaCl = 58.5 amu – Molar mass of NaCl = 58.5 g/mole Figure 03.09 One mole each of a solid, a liquid, and a gas. One mole of NaCl, the solid, has a mass of 58.45 g. One mole of H2O, the liquid, has a mass of 18.0 g and occupies a volume of 18.0 mL. One mole of O2, the gas, has a mass of 32.0 g and occupies a balloon whose diameter is 35 cm. Practice: Interconvert among moles, grams and molecules (or atoms) 1. Calculate the mass in grams of 6.05 mol of NaHCO3. Strategy: mol NaHCO3  g NaHCO3 (Ans: 508 g) 2. Calculate the mass of 450. atoms of Copper (Cu). Strategy: atoms Cu  Mole Cu  g Cu (Ans: 4.75 x 10-20 g) Chapter 3 Objectives 6-8 6. Given the chemical formula or name of a substance, calculate its percent composition. 7. Determine the empirical formula of a compound from a molecular formula or percent composition. 8. Given the empirical formula and molar mass of a compound determine its molecular formula. Steps to Calculate Percent Composition of Elements 1. Compute the molar mass of the compound 2. Calculate how much mass comes from each element (number of atoms of that element in the molecule x its atomic mass) 3. Divide step 2 by step 1 and multiply by 100 to convert to percent 4. Sum the individual mass percent values – they should total to 100% within round-off error. Empirical Formulas From Analyses Empirical formula can be calculated from the percent composition of elements 1. Assume the compound has a mass of exactly 100 grams. – You can therefore convert percentage to grams 2. Calculate moles of each kind of atoms present 3. Divide each value of moles by the smallest of the values. 4. (If needed) Multiply each number by an integer to obtain all whole numbers. (if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3; if ratio 0.25 or 0.75, multiply all by 4; etc.) Practice: Calculating Empirical Formula Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is its empirical formula? • Given Information: 40.92 g C, 4.58 g H, 54.50 g O Find: Empirical Formula, CxHyOz Strategy: g --------> mole -------> mole ratio -----> empirical formula C,H,O C,H,O divide all moles multiply with an by the smallest integer if necessary mole Answer:C3H4O3 Example: Molecular Formula from Empirical Formula and Molar mass What is the molecular formula of a compound with empirical formula C2H4O and molar mass = 88 g/mol? Molecular formula = n (Empirical formula) Empirical formula (C2H4O) mass = 2(12.0 g/mol) + 4 (1.0 g/mol) + 1(16.0 g/mol) = 44.0 g/mol Molecular Formula = 2 (C2H4O) = C4H8O2 n = 88 g/mol 44.0 g/mol = 2 Quantitative Information From Balanced Equations • The coefficients in the balanced equation give the ratio of moles of reactants and products 2 mol H2 1 mol O2 2 mol H2 2 mol H2O 1 mol O2 2 mol H2O Equation: 2H2(g) + O2 (g) 2H2O(l) Stoichiometry • Deals with the quantities of substances consumed and produced in chemical reactions • We’ll first look at the stoichiometry of sandwiches Stoichiometry & Balanced Equations • Our sandwiches will have one and only one slices of bologna between two (no more, no less) pieces of bread • Our “sandwich reaction” is then: 2B + M MB2 • Any reaction, such as 2H2 + O2 2H2O, has the same constraints Grams of Substance A Moles of Substance A Moles of Substance B Grams of Substance B Use molar mass of A Use molar mass of B Use coefficients of A & B from balanced equation Examples: Stoichiometry • For the reaction C + O2 CO2 – What mass C is burned to CO2 in the presence of 320.g O2 ? – What mass CO2 is produced? • For the reaction C3H8 + 5O2 3CO2 + 4 H2O – What mass of oxygen is consumed in the combustion of 1.00 g C3H8 ? Chapter 3 Objectives 10 and 11 10.Given the amounts of two or more reactants and the chemical equation, determine the limiting reagent and the theoretical yield of the product. 11.Given the theoretical yield and the actual yield, calculate the percent yield. Theoretical Yield, Actual Yield and Percent Yield • The amount of product predicted from stoichiometry, taking into account limiting reagents, is called the theoretical yield. • Actual yield—the amount of product actually obtained in the reaction. • The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical yield: Percent yield = Actual yield Theoretical yield X 100% Percent yield will be always less than 100%. Aluminum and oxygen react according to the following equation:4Al(s) + 3O2(g)  2Al2O3(s) In a certain experiment, 4.6 g Al was reacted with excess oxygen; 6.8 g of product was obtained. What was the theoretical yield, if the percent yield of the reaction was 78%? Theoretical yield = 8.7 g Example: Calculating the theoretical yield for a Reaction given percent yield and the actual yield Percent yield = Actual yield Theoretical yield X 100% 78% = 6.8 g Theoretical yield X 100% Example : Theoretical and Percent Yield In the previous problem, 330.g carbon dioxide is the theoretical yield. It is the amount of product formed when the limiting reactant is completely consumed. If 300.g carbon dioxide is actually produced, what is the percent yield of carbon dioxide ?