Balancing Chemical Equations and Calculating Empirical Formulas - Prof. Scott A. Miller, Study notes of Chemistry

Download Balancing Chemical Equations and Calculating Empirical Formulas - Prof. Scott A. Miller and more Study notes Chemistry in PDF only on Docsity! Balancing Chemical Equations Ex. 1. Fe + O2  Fe2O3 1 Fe, 2 O  2 Fe, 3 O Balance oxygen… Fe + 3O2  2Fe2O3 1 Fe, 6 O  4 Fe, 6 O …Then Iron… 4Fe + 3O2  2Fe2O3 4 Fe, 6 O  4 Fe, 6 O Balancing Chemical Equations Ex. 2. C2H4 + O2  CO2 + H2O2 C, 4 H, 2 O  1 C, 2 H, 3 O Balance carbon… C2H4 + O2  2CO2 + H2O 2 C, 4 H, 2 O  2 C, 2 H, 5 O …Then hydrogen… C2H4 + O2  2CO2 + 2H2O 2 C, 4 H, 2 O  2 C, 4 H, 6 O …Then oxygen… C2H4 + 3O2  2CO2 + 2H2O 2 C, 4 H, 6 O  2 C, 4 H, 6 O Formula Weights • Ex. 1. Formula weight of Neon – Exists as individual Ne atoms – Formula is an atomic symbol, so FW=atomic weight – Formula weight of Ne is 20.18 amu. • Ex. 2. Formula weight of molecular oxygen – Exists as two bonded oxygen atoms, molecular formula is O2. – Formula weight is 2 x 16.0 = 32 amu. – Since this is a molecule • We may also say “molecular weight” Formula Weight • Ex. 3. – Formula weight of glucose, C6H12O6 – (6 x 12.0 amu) + (12 x 1.01 amu) + (6 x 16.0 amu) = 180. amu – Again, a molecule, so we can say molecular weight • Ex. 4. – Formula weight of magnesium chloride, MgCl2 – (1 x 24.3 amu) + (2 x 35.5 amu) = 95.3 amu – This is an ionic compound—so we don’t say molecular weight, only formula weight Percentage Composition of an Element in a Compound • Percentage composition is – The weight percent of each element in a compound • Calculation is straightforward – Know chemical formula and formula weight of the compound 100% Compound) of Wt.(Formula element) that of wt.micatoms)(Ato of(Number % xElement  Class Work: in 5.380 g Glucose • Determine FW of Glucose: C6H12O6 • Calculate Moles of Glucose: • Calculate Molecules of Glucose: • Calculate number of atoms of C 6126 6126 61266126 02989.0 0.180 1380.5 OHmolC OHgC OHmolCOHgC             Now that you’ve got the number of moles, use NA to convert to number of molecules: 6126 22 6126 6126 23 6126 OHC molecules1080.1 OHC mol 1 OHC molecules106.02OHC mol 0.02989 x x             The molecular formula can be used to convert number of molecules to number of atoms: C atoms1008.1 OHC molecule 1 C atom 6OHC molecules1080.1 23 6126 6126 22 x x             Calculating Empirical Formulas • Combination of the percent composition and atomic weight gives the number of moles in a sample, if the amount of sample is known. • In calculating empirical formulas, when a percentage composition is given, assume a sample size of 100.0 g to get the units to work out. Ex. A sample is found to be 73.9% mercury (Hg) and 26.1% chlorine (Cl) by mass. Find the empirical formula. 1. Assume 100.0 g of sample—so you’ve got 73.9 g of Hg and 26.1 g of Cl 2. Convert these masses to number of moles: Calculating Empirical Formulas Hg mol 0.368 Hg g 200.6 Hg mol 1Hg g 73.9 Hg Moles              Cl mol 735.0 Cl g 35.5 Cl mol 1Cl g 26.1 Cl Moles              Calculating Empirical Formulas 3. Divide each number of moles by the smallest number of moles. This gives the mole ratio:                   Hg mol 1 Cl mol 99.1 Hg mol 368.0 Cl mol 735.0 Hg of moles Cl of moles In this case, we get (close enough to) whole numbers. So, we can say that the ratio of chlorine to mercury is 2:1. This ratio becomes the subscripts in the empirical formula: HgCl2 Calculating Empirical Formulas Ex. Ascorbic acid is 40.92 % C, 4.58 % H, and 54.50 % O by mass. What is the empirical formula of ascorbic acid? Given this information: • Step 1 = assume a 100.0 g sample size • 40.92 g C • 4.58 g H, and • 54.50 g O Calculating Empirical Formulas A fraction cannot be a subscript. So, we multiply these values by a number to make them all “whole”. In this case, 3: C : H : O = 3 : 4 : 3 These number then become the subscripts in the empirical formula for ascorbic acid: C3H4O3 Molecular Formula from Empirical Formula The procedure for finding the empirical formula only finds the relative amount of each element in a substance. To find the actual molecular formula, you need the molecular mass of the substance. Dividing the molecular mass of a substance by the molecular mass of its empirical formula results in a whole number. To find the molecular formula, multiply each of the subscripts by that number. Molecular Formula from Empirical Formula In the previous example we found the empirical formula of ascorbic acid to be C3H4O3. The mass of the empirical formula is: (3)(12.01 g/mol) + (4)(1.01 g/mol) + (3)(16.0 g/mol) = 88.0 g/mol Known molar mass of ascorbic acid is 176.0 g/mol. Dividing the molar mass by the empirical mass gives the factor by which the subscripts are multiplied to get the molecular formula. Combustion Analysis A sample of isopropyl alcohol, which contains C, H, and O, is tested by combustion analysis. Combustion of 0.255 g of isopropyl alcohol yields 0.561 g of CO2 and 0.306 g H2O. Find the empirical formula for isopropyl alcohol. To find an empirical formula, we need the number of moles of each element. We first find the number of grams of each element in the original sample, and then convert that to moles to find the empirical formula. Combustion Analysis C g 153.0 C mol 1 C g 12.0 CO mol 1 C mol 1 CO g 44.0 CO mol 1CO g 561.0 22 22                         Step 1. Grams of Carbon (from CO2): Step 2. Grams of Hydrogen (from H2O): H g 0343.0 H mol 1 H g 1.01 OH mol 1 H mol 2 OH g 18.0 OH mol 1OH g 306.0 22 22                         Combustion Analysis Step 3. We now need the mass of oxygen in the original sample. We’ve found the masses of the other two elements, and we know the original mass, so we can find the mass of oxygen by subtraction: Mass of O = Mass of sample – mass of H – Mass of C = 0.255 g – 0.153 g – 0.0343 g = 0.068 g O Now, we have the masses of each of the elements in the sample. Limiting Reactants 2C4H10 + 13O2  8CO2 + 10 H2O • In a butane combustion example, we observ that 2 moles of butane reacts with 13 moles of oxygen. But what if only 1 mole of butane were present, and you still had 13 moles of oxygen? • The reaction would proceed, but not all of the oxygen would be used because we would run out of butane. Butane would be the limiting reactant. • “Limiting Reactant” = “Limiting Reagent” Ex. How many moles of oxygen are used in the combustion of 1.00 mole of butane? If 10.0 moles of oxygen are present, how much oxygen is left over? How many moles of CO2 are produced? Use stoichiometry: Limiting Reactants used areO mol 50.6 HC mol 2 O mol 13HC mol 1.00 2 104 2104             2C4H10 + 13O2  8CO2 + 10 H2O overleft O mol 3.5 0 mol 6.5 - O mol 0.10 222  Limiting Reactants 2C4H10 + 13O2  8CO2 + 10 H2O produced areCO mol 00.4 HC mol 2 CO mol 8HC mol 1.00 2 104 2104             2C4H10 + 13O2  8CO2 + 10 H2O Compare quantities (moles) of each reactant and product: Starting 1 10 0 0 Change -1 -6.5 4 5 Final 0 3.5 4 5 Limiting Reactants 4 43 4343 PONa mol 0.0213 PONa g 164 PONa mol 1PONa g 3.50 3            2Na3PO4 + 3Ba(NO3)2  Ba3(PO4)2 + 6NaNO3 23 23 2323 )Ba(NO mol 0.0245 )Ba(NO g 612 )Ba(NO mol 1)Ba(NO g 6.40             • Need 1.5 times as many moles Ba(NO3)2 as Na3PO4 • But we only have 1.15 times as much. • So, barium nitrate is the limiting reactant. • For complete reaction of the 3.50 g of sodium phosphate, we would need 0.0320 mole, or 8.34 g, of barium nitrate. Limiting Reactants Because barium nitrate is the limiting reactant, we use its mass to find the mass of barium phosphate. Now it’s a stoichiometry problem: 2Na3PO4 + 3Ba(NO3)2  Ba3(PO4)2 + 6NaNO3 243 243 243 23 243 23 2323 )(POBa g 92.4 )(POBa mol 1 )(POBa g 602 )Ba(NO mol 3 )(POBa mol 1 )Ba(NO g 612 )Ba(NO mol 1)Ba(NO g 6.40                         Similar calculations can be used to find the mass of sodium nitrate produced, or the mass of sodium phosphate used up in the reaction. Theoretical Yield * Most reactions don’t actually go all the way to completion. This can be due to impurities, side reactions or other considerations. The quantity of product produced if the reaction went to completion is called the theoretical yield. The actual yield* is what is actually produced. They are related through the percent yield*. 100% x yield lTheoretica yield Actual YieldPercent        * Aspirin Yield in Lab