Calculating Empirical Formulas and Waters of Hydration from Percent Composition, Summaries of Chemistry

Download Calculating Empirical Formulas and Waters of Hydration from Percent Composition and more Summaries Chemistry in PDF only on Docsity! Empirical Formulas The compound disilicon hexafluoride has the formula Si2F6. The percent composition would be calculated from the atomic masses (atomic weights) as follows: Si 28.092 = 56.18  56.18 100% 170.2  = 33.01 % Si F 19.006 = 114.00  114.0 100% 170.2  = 66.99 % F  170.18  molecular mass (molecular weight) To obtain the formula from the percent composition one can convert masses to moles and find the mole ratio. If masses are not given, it is convenient to (arbitrarily) start with 100. g of compound to obtain elemental masses. So in this case 100.0 g would contain 33.01 g Si and 66.99 g F. Convert these masses to moles and scale to simple whole numbers. 33.01 g Si 1 mol Si 28.09 g Si = 1.175  1.175 1.175 = 1 Divide both numbers by the smallest number. 66.99 g F  1 mol F 19.00 g F = 3.526  3.526 1.175 = 3.00  Si1F3  SiF3 SiF3 is the empirical formula. The empirical formula is the molecular formula reduced to the simplest whole number ratio of atoms. It is not possible to get back to the molecular formula, Si2F6, from percent composition without further information. A silicon bromide compound is 11.65% Si and 88.35% Br. Calculate the empirical formula. 11.65 g Si 1 mol Si 28.09 g Si = 0.415  0.415 0.415 = 1   3  3 88.35 g Br  1 mol Br 79.90 g Br = 1.106  1.106 0.415 = 2.667   3  8.00  Si3Br8 Notice that the division by the smallest number (0.415) did not give whole numbers in this case, so a further step is required. 2.667 cannot be rounded to 3.0 to give the compound SiBr3, because SiBr3 will not agree with the percent composition . ( SiBr3 is 10.5% Si and 89.5%Br) Since 2.667 = 2 2 /3 it is appropriate to multiply both numbers by 3. If you are not certain what to multiply by, try multiplying by integers 2, 3, 4, 5… until both subscripts reach integer values or nearly integer values—say within 0.05. If the final numbers seem unusually large, double-check all calculations. If the numbers came directly from an experiment, it may be necessary to try the experiment again to get better data. Determination of Waters of Hydration from Percentage Data. A hydrated compound like CuCl22H2O has a percentage of the compound that is H2O and a percentage that is the anhydrous (without water) CuCl2. This is easily calculated from the molecular weight: MW of CuCl22H2O is 134.5 (= 63.55 + 235.45) + 218.016 = 170.5 % CuCl2 = 2 2 2 134.5 g CuCl 170.5 g CuCl 2H O   100% = 78.9 % CuCl2 % H2O = 2 2 2 2 18.016 g H O 170.5 g CuCl 2H O    100% = 21.1 % H2O Problem: The hydrated compound CuCl2xH2O is 21.1 % water. Determine the value of x. The problem can be solved by methods used for determination of an empirical formula. Determine the mole ratio of CuCl2 and H2O in the compound. Calculate moles of CuCl2 and H2O in 100.0 g of the hydrate, CuCl2xH2O. 78.9 g CuCl2 2 2 1 mol CuCl 134.5 g CuCl = 0.587 mol CuCl2 21.1 g H2O 2 2 1 mol H O 18.016 g H O = 1.17 mol H2O so 2 2 1.17 mol H O 0.587 mol CuCl = 1.99 so x must equal 2, i.e, in the hydrate there are 2 moles of water per every mole of CuCl2. The formula for the hydrate is: CuCl22H2O Try these problems: 1. VSO4∙xH2O is 46.175 % water; calculate the value of x. 2. Mg2Fe(CN)6∙xH2O is 45.346 % water; calculate the value of x 3. A compound of neptunium (Np) and oxygen is 15.25% oxygen; determine the empirical formula of the compound. Answers: 1. x=7 VSO4∙7H2O 2. x= 12 Mg2Fe(CN)6∙12H2O 3. Np3O8