Question

A compound made of only carbon, hydrogen and oxygen was analyzed by combustion in an excess of oxygen. When 520 mg of compound was used for the analysis, 780 mg of carbon dioxide and 213 mg of water were obtained. What is the empirical formula of the compound? Select one: o a. C,H,03 a b C6H8% 0 C. CHÁO O d. C₂H₂O₂ o e C H04 Next p

Answer 1

Solution :- The value provided in the question are, mass cy unknown Compound - 520 mg = 0.529 mass of CO2 = 780 mg = 0.78 g mass of H₂O = 213 mg = 0.2139 The Reaction is, 0.52 g Compound & g 0.78 g CO2 +0.213 g H₂O mass of onygen (0.78 g +0.213 g ) - 0.52 g = 0.4479 In coa, grams of Carbon = 0.78 g x 12 g 44 g = 0.21 g Carbon In H2O grams of Hydrogen = 0.2139x 29 189 = 0.0236 g Hydrogen In H₂o, coa 0.78 x 32 grams of onygon = box 0.213X16 - + _ 4 4 = 0.189 +0.567 = 0.756 g

mass of snygen in given compound 007586 = 0.756 g 0.4479 = 0.309 g mass of Carbon = 0.21 g mass of hydrogen = 0.0236g. mass of any gen = 6.309g Find the number of mole of each atom number of moles of carbon = 0.21 g 12 g/mol = 0.0175 mol number of moles of Hydrogen = 0.0236 g 19/mol = 0.0236 mol number of moles of onygen = 0.309 9 16 g/mol = 0.0193 mol

- 0.0175 H-0.0236 o - 0.0193 divide by least no of moles C - 1 0.0175 0.0175 0.0236 -1,34 0.0175 - 0.0193 -1-10 0.0175 to get a whole number. multiply each with 3 1.10x3 = 3.3 1x 3 =4.02 = 3 , 1.34 x 3 34 : 44 : 30 (3 Hao Hence the emperical Formula of the Compound is ( Ha 3