About Chemistry

For the 9.11 g sample
no. of mole of 14.67g CO2 = 14.67 / (12+16X2) = 0.3334
mass of 0.3334 mole carbon = 0.3334 X 12 = 4 g
% of carbon in the carbohydrate = (4 / 9.11)X100 = 43.9%
mass of H2O in the carbohydrate = 9.11 - 4 = 5.11 g
no. of mole of H2O = 5.11 / 18 = 0.2839
mole ratio of carbon to water = 0.3334 : 0.2839 = 1.1744 : 1 = 12 : 10
empirical formulae = C12H20O10
mass % of hydrogen = ((0.2839X2)/9.11)X100 = 6.23%

For the 2.93g sample
no. of mole of water = 1.93/18 = 0.107
mass of hydrogen = 0.107X2 = 0.2144 g
mass % of hydrogen = (0.2144/2.93)X100 = 7.32%
As this is a carbohydrate, mass of oxygen in sample = 0.107X16 = 1.712 g
mass of carbon = 2.93 - 1.712 - 0.2144 = 1.0036 g

no. of mole of carbon = 1.0036/12 = 0.08363
no. of mole of water = 0.107
ratio of carbon to water = 0.08363 : 0.107 = 1:1.28 = 10 : 13
empirical formulae = C10H26O13

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