Question

When 3.537 grams of a hydrocarbon, C,Hy, were burned in a combustion analysis apparatus, 11.96 grams of CO2 and 2.448 grams of H2O were produced In a separate experiment, the molar mass of the compound was found to be 78.11 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon. Enter the elements in the order presented in the question. empirical formula = molecular formula =

Answer 1

UILS alinmamin anofficients in a chemical equation can CxHy = 3.537g CO₂ = 11.96 g H₂O = 2.4489 No. 95 7. Of carbon can be calculated as CO2 = 449 6 12g 44g of co₂ contain a = 12 g 11.969 of co, contain c= 12 x 11.96 44 % of c = x 100 weight of c. weight of compound 7 of 12 * 1.96 x100 44 3.537 C = 92.21% similarly % of t = 7 * 2.44 x 100 x 2.44 x100 3-537 H = 7.66%

meanino maneira Saraswali No. 96 Dt. calculation of molecular formula - Element % Al masc moles | mule ratio I simplest while no, ratio 192.21 92.21 I 12 7.68 1.002 12 = 7.68 4 7.661 7.66 7.66 = 1 Empirical formula = CH Empirical formula mass = 13 2. Molecular mass = 78.11 Imol. na molecular mass Empirical formula mass C n- 78.11 13 formula - Cho Molecular