Chapter 2 Empirical and Molecular Formulas 2 1 Chemical Formula and Composition 2 2 Balancing Chemical Equations Mass Relationships in Chemical Eqns Limiting Reagent and Yield 2 5 2 3 2 4 1 A mole of XmYn contains Review Moles of Compounds m moles of atom X and n moles of atom Y 1 mol of H2O contains 2 mol of H atoms and 1 mol of O atoms Molar mass sum of the atomic masses Mass of 1 water molecule 2 1 008 amu 1 15 999 amu 18 015 amu Molar mass of water 2 1 008 g mol 1 15 999 g mol 18 015 g mol 2 2 Percent Composition Two names used percent composition by mass or mass percent of the compound Example What is the mass percent of each element in sodium chlorite NaClO2 molar mass 22 990 35 453 2 15 999 90 441 g mol 1 2 2Percent Composition mass of Na in 1 mol NaClO2 x 100 Na mass of NaClO2 in 1 mol NaClO2 x 100 25 42 22 990 g 90 441 g O mass of O mass of NaClO2 x 100 2 15 999 g 90 441 g x 100 35 38 2 2 Percent Composition Cl mass of Cl mass of NaClO2 x 100 35 453 g 90 441 g x 100 39 20 Check your work Na O Cl 25 42 35 38 39 20 100 6 2 1 Empirical Molecular Formulas Last example molecular formula percent composition The process can be reversed percent composition empirical formula Not molecular formula Empirical formula the simplest ratio of atoms in a compound 2 1 Empirical Molecular Formulas Compound Mol formula Emp formula Hydrogen peroxide H2O2 HO Borane boron trihydride BH3 BH3 Diborane diboron hexahydride B2H6 BH3 Octene C8H16 CH2 Butene C4H8 CH2 2 1 Empirical Molecular Formulas Example An orange compound is 26 6 K 35 4 Cr and 38 0 O Determine its empirical formula Assume a 100 0 g sample becomes mass in grams Divide each mass by its atomic mass Gives the number of moles of each in 100 g Divide each by the smallest answer found The smallest integer ratio empirical formula 2 1 Empirical Molecular Formulas Unknown 26 6 K 35 4 Cr 38 0 O In 100 0 g 26 6 g K 1 mol K 39 10 g K 0 6803 mol K 35 4 g Cr 1 mol Cr 52 00 g Cr 0 6808 mol Cr 38 0 g O 1 mol O 16 00 g O 2 375 mol O 2 1 Empirical Molecular Formulas Empirical formula smallest integer ratio Divide by the smallest ratios stay the same K Cr O 0 6803 mol 0 6803 mol 0 6808 mol 0 6803 mol 2 375 mol 0 6803 mol 1 000 x2 2 1 001 x2 2 3 491 x2 7 Choose a multiplier to make integer Empirical formula K2Cr2O7 2 1 Empirical Molecular Formulas The molecular formula can be determined if the molecular mass is known Example Vitamin C has the empirical formula C3H4O3 and molecular mass 175 g mol Empirical mass 3 12 01 4 1 008 3 15 99 88 03 g mol Empirical mass molecular mass Mol formula 2 emp formula C6H8O6 13 2 1 Empirical Formula Problems Empirical Formula can be determined from 1 Percent Composition 2 Measured Mass Composition 3 Combustion Analysis 14 Percent Composition Empirical Formulas Empirical formula simplest ratio of atoms of each element in a compound Found for organic compounds by combustion analysis O2 a furnace Sample in H2O absorber CO2 absorber Mg ClO4 2 NaOH C and H are converted to CO2 H2O Both are trapped and the weight gain measured Other elements N O with other traps or by mass difference Combustion of Organic Species CxHy O2 CO2 H2O CxHyOz O2 CO2 H2O CxHy Nz O2 CO2 H2O N2 CxHyNzOa O2 CO2 H2O N2 16 Combustion Analysis C and H A 1 50 g sample of hydrocarbon undergoes complete combustion to produce 4 40 g of CO2and 2 70 g of H2O What is the empirical formula of this compound Step 1 determine the grams of carbon in 4 40 g CO2and the grams of hydrogen in 2 70 g H2O Step 2 convert grams of C and H to their respective amount of moles Step 3 divide each molar amount by the lowest value seeking to modify the molar amounts into small whole numbers 17 Step 1 determine the grams of carbon in 4 40 g CO2and the grams of hydrogen in 2 70 g H2O carbon 4 40 g x 12 011 g 44 0098 g 1 201 g hydrogen 2 70 g x 2 0158 g 18 0152 g 0 3021 g Step 2 convert grams of C and H to their respective amount of moles carbon 1 201 g 12 011 g mol 0 0999 mol hydrogen 0 3021 g 1 0079 g mol 0 2997 mol Step 3 divide each molar amount by the lowest value into small whole numbers carbon 0 0999 mol 0 0999 mol 1 hydrogen 0 2997 mol 0 0999 mol 3 18 Percent Composition Empirical Formulas Vitamin C 176 12 g mol contains C H O only If 1 000 g is burned in O2 1 502 g CO2 and 0 409 g H2O form Determine its empirical molecular formula Mass of C 1 502 g CO2 1 mol CO2 44 009 g CO2 1 mol C 1 mol CO2 12 011 g C 1 mol C or mC 0 4099 g C mC 1 502 g CO2 12 011 g C 44 009 g CO2 0 4099 g C Percent Composition Empirical Formulas Combustion of 1 000 g of vitamin C gave1 502 g CO2 and 0 409 g H2O 2 H in H2O mH 0 409 g H2O 2 0158 g H 18 015 g H2O 0 04577 g H Mass of H Mass of O mO sample mass mC mH 1 000 g 0 4099 g 0 04577 g 0 544 g Percent Composition Empirical Formulas Convert to moles 0 03413 mol C 0 4099 g C 12 011 g mol 0 04577 g H 1 0079 g mol 0 04541 mol H 0 544 g O 15 999 g mol 0 0340 mol O Percent Composition Empirical Formulas Find the mole ratio divide by smallest C 0 03413 0 0340 1 00 H 0 04541 0 0340 1 34 O 0 0340 0 0340 1 00 Close to 1 1 1 C H O Multiply by 3 to get an integer ratio 3 4 3 Empirical formula is C3H4O3 Percent Composition Empirical Formulas Vitamin C contains C H O only molar mass of vitamin C is 176 12 g mol find its empirical and molecular formula Empirical formula C3H4O3 Empirical mass 3 12 4 1 3 16 g 88 g Molar mass 176 12 g Molar mass 2 x empirical mass Vitamin C has the molecular formula C6H8O6 Combustion Analysis Caffeine a stimulant found in coffee tea and certain soft drinks contains C H O and N Combustion of 1 000 mg of …