Determining Chemical Formulas and Percent Composition, Lecture notes of Chemistry

Download Determining Chemical Formulas and Percent Composition and more Lecture notes Chemistry in PDF only on Docsity! 29 Chemical Formulae Chemical formulae are used as shorthand to indicate how many atoms of one element combine with another element to form a compound. C2H6, 2 atoms of carbon combine with 6 atoms of hydrogen to form ethane. Mn(Cr2O7)2, one Mn combines with 4 Cr and 14 O to form manganese(II) dichromate. If we know the formula of a compound, determining the molar mass is simple. Since we know the atomic masses of the elements, we can sum the atomic masses to obtain the molecular masses. 1 atom of C2H6 contains 2 atoms of C 12.01 amu = 24.02 1 atom of C2H6 contains 6 atoms of H 1.0079 amu = 6.0474 30.0674 = 30.07 amu We also know that 1 mole of C atoms weighs 12.01 g and 1 mol of H atoms weighs 1.0079 g. So, we know the molar mass of ethane also. 1 mole C2H6, 2 moles of C 12.01 g x 2 24.02 g 1 mole C2H6, 6 moles of H 1.0079 g x 6 6.0474 g 30.0674 = 30.07 g/mole Or, 30.07 g in 1 mole C2H6. So, we say the molar mass of ethane is 30.07 g/mol (that’s grams per mole). 1 mole Na2Cr2O7, 2 mole of Na 22.989 g x 2 47.978 g 1 mole Na2Cr2O7, 2 mole of Cr 51.996 g x 2 103.992 g 1 mole Na2Cr2O7, 7 moles of O 15.9994 g x 7 111.9958 g 263.9658 = 263.97 g/mole 29 30 MW (which means molar mass) of (C5(CH3)5)Re(NO)(CO)2BF4. 12 mol C x 12.01 g C = 144.12 g C 1 mol C 15 mol H x 1.0079 g H = 15.1185 g H 1 mol H 1 mol B x 10.81 g B = 10.81 g B 1 mol B 4 mol F x 18.998 g F = 75.992 g F 1 mol F 1 mol N x 14.01 g N = 14.01 g N 1 mol N 3 mol O x 15.9994 g O = 47.9982 g O 1 mol O 1 mol Re x 186.207 g Re = 186.207 g Re 1 mol Re _____________ MW= 494.26 g/mol 30 33 molar mass = (a number) x (empirical mass) Since, empirical mass CH2O = 30.25 g/mole 180.16 g/mol = (a number) x 30.25 g/mol (a number) = 5.956 or 6 molecular formula = (6) x (CH2O) So, we did have C6H12O6 Essentially we found how many empirical formula units are in the whole. A bottle of improperly labeled “iron oxide” is found in a laboratory. Elemental analysis reveals that the sample is iron and oxygen, and the sample is 69.94 % iron by mass. What is the formula of the material, and what should the bottle be labeled; that is, name the compound? To find the formula we need to know the % mass of each of the elements. Since, there are only 2 different elements in iron oxide we know... %Fe + %O = 100 % 100 % - %Fe = %O 100 % - 69.94 % = %O %O = 30.06 % now that we have % mass of both elements let’s find the formula assume 100 g sample 30.06 g O x 1 mol O = 1.8788 mol O 15.9994 g O 69.94 g Fe x 1 mol Fe = 1.2523 mol Fe 55.85 g Fe 33 34 reduce to lowest ratio by dividing each number by the smallest number... 1.8788 mol O = 1.500 mol O 1.2523 1.2523 mol Fe = 1 mol Fe 1.2523 do not want fractions in a formula so we must find the lowest common whole number multiple Multiplying 1.5 by 2 gives a whole number; i.e., 3 So, multiply each number by 2 1.5 x 2 = 3 mol O and 1 Fe x 2 = 2 mol Fe So, the formula is Fe2O3 The name is iron(III) oxide 34 35 To find a formula we do not need to have % composition. If we are told the mass of each element present in a compound we can find the formula. The mass of the elements can be converted to moles of the elements. The mole ratio tells us the empirical formula For example, A compound containing only Mn and Cl contains 1.9228 g Mn and 2.4817 g Cl. Determine the empirical formula. We simply need to convert the masses of the elements to moles. The empirical formula is determined by the mole ratio of Mn to Cl. 1.9228 g Mn 54.938 g Mn 1 mol Mnx = 0.034999 mol Mn 2.4817g Cl 35.453 g Cl 1 mol Clx = 0.070000 mol Cl reducing to 1 by dividing by 0.034999 1.9228 g Mn 54.938 g Mn 1 mol Mnx = 0.034999 mol Mn 2.4817g Cl 35.453 g Cl 1 mol Clx = 0.070000 mol Cl = 1 = 2 The formula is 35