Combustion analysis of 63.8 mg of a c, h and o containing compound produced 145.0 mg of co2 and 59.38 mg of h2o. what is the empirical formula for the compound?

: The empirical formula for the compound is C3H60 (see below) CO2 is the only product containing C, C produced = 145.0 mg CO2 x (1 g / 1000 mg) x (1 mole CO2 / 44.0 g CO2) x (1 mole C / 1 mole CO2) = 0.00330 moles C. H2O is the only product containing H, H produced = 59.38 mg H2O x (1 g / 1000 mg) x (1 mole H2O / 18.0 g H2O) x (2 moles H / 1 mole H2O) = 0.00660 moles H. Oxygen is in both and the unknown reacts with oxygen(in the air) 0.00330 moles C x (12.0 g C / 1 mole C) = 0.0396 g C = 39.6 mg C 0.00660 moles H x (1.01 g H / 1 mole H) = 0.00667 g H = 6.7 mg H Because the unknown weighed 63.8 mg and consists off justC, H, and O, then mass O = g unknown - g C - g H = 63.8 mg - 39.6 mg - 6.7 mg = 17.5 mg = 0.0175 g 0.0175 g O x (1 mole O / 16.0 g O) = 0.00109 moles O The mole ratio of C:H:O is: C = 0.00330 H = 0.00660 O = 0.00109 Divide by the smallest you get: C = 0.00330 / 0.00109 = 3.03 H = 0.00660 / 0.00109 = 6.06 O = 0.00109 / 0.00109 = 1.00

60.0 - there are 3 significant numbers

B"hf and f-"  is the correct answer took the test on edgenuity and got it right