S2i Thursday, September 27th Agenda: 1. Balancing Equations 2. Empirical and Molecular Formulas Balancing Equations Balance the following equations. Hints: 1. Write out word problems into symbols first 2. Combustion means adding O2 3. Don’t start with an element that is in more than one compound on the same side of the equation 1. The combustion of toluene (C7H8) to form carbon dioxide gas and water. C7H8 + 9O2è 7CO2 + 4H2O 2. Calcium hydride reacts with water to form calcium hydroxide (aqueous) and hydrogen gas. (Problem 3.66 on homework 3 #3) CaH2 + 2H2O è Ca(OH)2 + 2H2 3. An iron ore sample contains Fe2O3 together with other substances. Reaction of the ore with CO produces iron metal. Balance the equation: (Problem 3.64) Fe2O3 + 3CO è 2Fe + 3CO2 Empirical and Molecular Formulas Circle the correct word: The empirical / molecular formula is the empirical / molecular formula times an integer Ex) The empirical formula of pentane is C5H12 but an example of a molecular formula of pentane is (C5H12)3 or C15H36. Fill in the 4 steps to find the empirical formula from your notes: 1. Base calculation on 100 grams of compound 2. Determine moles of each element in 100 grams of compound 3. Divide each value of moles by the smallest of the value 4. Multiply each number by an integer to obtain whole numbers If you have a compound that is 34% Carbon, 20% Hydrogen, and 46% Oxygen, what is the empirical formula? = 2.83 mol C/2.88= 1 34 g C 1 mol C 12.01 g C = 19.8 mol H/2.88= 7 20 g H 1 mol H 1.008 g H = 2.88 mol O/ 2.88= 1 46 g O 1 mol O 16 g O CH7O If the compound has a molar mass of 245.5 g/mol what is the molecular formula? Molar mass of CH7O: 35.07 g/mol 245.5/35.07= 7 Molecular Formula= (CH7O)7= C7H49O7 The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of 2.50 mg of ethyl butyrate produces 5.69 mg of CO2 and 2.32 mg of H2O. What is the empirical formula of the compound? CxHyOz + O2 è CO2 + H2O 5.69 mg CO2 1 g CO2 1 mol CO2 1 mol C 1000 mg CO2 44.01 g CO2 1 mol CO2 = 1.29 x 10-‐4 mol C = 1.55 x 10-‐3 g C (times by molar mass) 2.32 mg H2O 1 g 1000 mg 1 mol H2O 18.016 g H2O 2 mol H 1 mol H2O = 2.58 x 10-‐4 mol H = 2.60 x 10-‐4 g H -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ Mass of O= mass of compound – mass (C+H) = .0025 g -‐ 1.55 x 10-‐3 g C -‐ 2.60 x 10-‐4 g H= 6.9 x 10-‐4 g O 6.9 x 10-‐4 g O /(16 g/mol)= 4.3 x 10-‐5 mol O -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 1.29 x 10-‐4 mol C ÷4.3 x 10-‐5 = 3 2.58 x 10-‐4 mol H ÷ 4.3 x 10-‐5= 6 C3H6O = empirical formula 4.3 x 10-‐5 mol O ÷ 4.3 x 10-‐5= 1 If the substance has a molar mass of 174.23 g/ mol, what is the molecular formula? Molar Mass of empirical formula: 58.078 g/mol 174.23/58.078= 3 (C3H6O)3= C9H18O3 Washing soda, a compound used to prepare hard water for washing laundry, is a hydrate, which means that a certain number of water molecules are included in the solid structure. Its formula can be written as Na2CO3xH2O , where x is the number of moles of H2O per mole of Na2CO3. When a 3.070 g sample of washing soda is heated at 25 °C, all the water of hydration is lost, leaving 1.14 g of Na2CO3. What is the value of x? 3.070 g sample-‐1.14 g Na2CO3= 1.93 g H2O 1.93 g H2O 1.14 g Na2CO3 !.!! !"# !!! !.!"" !"# !"! !"! 1 mol H2O 18.016 g H2O = 0.11 mol H2O 1 mol Na2CO3 105.99 g Na2CO3 = 0.011 mol Na2CO3 = 10=x Calculating the masses of products and reactants Automotive air bags inflate when sodium azide, NaN3, rapidly decomposes to its component elements: How many moles of N2 are produced by the decomposition of 1.60 mol of NaN3? = 2.40 mol N2 1.60 mol NaN3 3 mol N2 2 mol NaN3 How many grams of NaN3 are required to form 11.0 g of nitrogen gas? 11.0 g N2 1 mol N2 2 mol NaN3 65.02 g NaN3 = 17.0 g NaN3 28.02 g N2 3 mol N2 1 mol NaN3 How many grams of NaN3 are required to produce 13.0 ft3 of nitrogen gas if the gas has a density of 1.25 g/L? = 460 g N2 13 ft3 N2 (12 in)3 (2.54 cm)3 1 L 1.25 g N2 3 3 1 ft 1 in 1000 mL 1 L 460 g N2 1 mol N2 2 mol NaN3 65.02 g NaN3 = 712 g NaN3 28.02 g N2 3 mol N2 1 mol NaN3
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