An organic compound C9H9O4

Question

An organic compound C9H9O4. ) What is the molar mass? (must show work to obtain credit)b) Show balanced combustion chemical equation of this compound.

c) How many isomers of this compound including the names, functional groups and families?

d) From all the isomes from part c), can any isomers have chiral(s) centers? Give isomer's name to justify your answer

Answer 1

 

 

a)  Molar mass:

     9 C 9* 12.01 =  108.09

     9 H  9* 1.008=      9.072

     4 O  4*  16.00=   64.0

                               -------

                               181.16

 

b) Notes:

     First npte:

        C9H9O4 + O2    ->       CO2 + H2O

     We have odd number of H (9) so we will need to double it to form whole H2O.

     We have to double the equation:

       2C9H9O4 + O2   -->   CO2 + H2O

      Next we now know the number of CO2 formed. We also know the number of H that forms water.

       2C9H9O4 + O2  ->  18CO2 + 9H2O

       Note that on the right we need 45 O atom, which is an odd number, while on the left we have even number of O atoms, including the C9H9O4.   Therefore we will need to double the equation again, to have even number of O atoms on the right. 

       4C9H9O4 + O2 -->  36 CO2 + 18H2O

Now we can calculate the number of O2 we need on the left:  On the right we have 72 + 18 'O' atoms, a total of 90.

On the left we have 16 'O" atoms from the organic compound so we subtract this number from the needed 90: 90 - 16 = 74.  We divide 74 by 2 to calculate the O2 molecules, which is 37.  The balanced equation is:

     4 C9H904 +37 O2  ->  36 CO2 + 18 H2O

Note that all the atoms must be balanced on both side of the equation:

        left     right

C      36       36

H      36       36

O     90        90

 

c)    In saturated organic compounds the formula is  CnH2n+2  where n can be any number.

       The presents of Oxygen does not change this formula:  CnH2n+2Ox  where n is as above and x can be any number (within reason x < n )  So for a 9 carbon compound we should have 20  hydrogens. When we introduce "unsaturation"  - a double or triple bond we take away a pair of hydrogens for each ( 2 for a double and 4 for a triple bond). Similarly if it is a cyclical compound we take away two hydrogens for the unsaturation.  Thus C9H9O4 must be a cation, with one hydrogen missing.

One can draw a large number of isomers for this formula(too many for me) with 6 "unsaturation", including several heterocyclical compounds, ethers, ketones, acid groups.  Some of these compounds may have chiral carbons.  

 

I hope this helps.