A10.00 g sample of a compound containing only carbon, hydrogen, and oxygen forms 23.98 g co2 and 4.91 g h2o upon complete combustion. what is the empirical formula of the compound?

(23.98 g CO2) / (44.00964 g CO2/mol) × (1 mol C / 1 mol CO2) = 0.544881 mol C 
(0.544881 mol C) × (12.01078 g C/mol) = 6.54445 g C 

(4.91 g H2O) / (18.01532 g H2O/mol) × (2 mol H / 1 mol H2O) = 0.54509 mol H 
(0.54509 mol H) × (1.007947 g H/mol) = 0.54942 g H 

10 g total - 6.54445 g C - 0.54942 g H = 2.90613 g O 
(2.90613 g O) / (15.99943 g O/mol) = 0.18164 mol O 

Divide by the smallest number of moles: 
(0.544881 mol C) / 0.18164 mol = 2.9998 
(0.54509 mol H) / 0.18164 mol = 3.0009 
(0.18164 mol O) / 0.18164 mol = 1.0000 
Round to the nearest whole number to find the empirical formula: 
C3H3O 

Gravity bc it’s gravity

    5.5 n======    assuming that      we have      m = 0.145 kg    a = 38 m/s²therefore    5.51 n, but two significant figures.final answer: 5.5 n