I think I did the problem correctly but there is no answer key to check the revi

Question

I think I did the problem correctly but there is no answer key to check the review on so I'm not sure. I just need to verify my results. Fall 20 Final Examination 151 1) 0.1101 g of an organic compound containing C, H, and O was analyzed by combustion. The amount of CO2 produced was 0.2503 g and the amount of H2O produced was 0.1025 g. The molar mass was determined to be approximately 115 g/mol Determine the empirical and molecular formulas for the compound. Hint: use four significant figures. .2529- 1025 44 CAI 2. (Dl 0050813438

Explanation / Answer

moles of CO2 = 0.2503 / 44 = 0.005689

moles of C= 0.005689

mass of Carbon = 12 x 0.005689 = 0.0683 g

moles of H2O = 0.1025 / 18 = 0.00569

moles of H = 2 x 0.00569 = 0.0114

mass of hydroegen = 0.0114 g

Oxygen mass = 0.1101 - ( 0.0683 + 0.0114 ) = 0.0304 g

moles of O = 0.0304 /16 = 0.0019

C                      H           O    

0.005689       0.0114        0.0019

3                 15/2              1

    6                   15               2

emperical formula = C6H15O2

molar mass = 115 g/mol

emperical mass = 119 g/mol

n = 119 / 115 = 1

molecular formula = C6H15O2

emperical formula = C6H15O2

CH3O -----------------------> empirical formula

empirical formula mass = 48 + 32 + 2 = 82

n = molar mass / empirical formula mass

    = 166.13/82

     = 2 (nearly)

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