How do you calculate the enthalpy of formation of butaine

 "Calculate the enthalpy of formation of butane, C4H10, using the balanced chemical equation and tables displaying information about enthalpy of formation/combustion. Write out the solution according to Hess's law." That's all I'm given. The key says that the answer is -125.4 kJ. C4H10   +  6.5 O2  =   4CO2  +   5 H20      The heat of combustion for one mole of butane is -2877 K.  Heats of formation from table   C02 =  -393.5 Kj/mole, H20 (g)   -241.8 Kj/mole. Heat of combustion of butane =  sum heats of formation products minus heats of formation reactants -2877 Kj =   4(-393.5Kj) + 5(-241.8Kj)    minus heat or enthalpy of formation butane.

The enthalpy formation oxygen an element is assigned a value of zero.

            -2877 Kj  =   -1582 Kj  +  -1209 Kj   minus   X

                  -2877 =  - 2791 Kj minus X

                      X =   -86 Kj.       The heats of formation for various compounds vary a little from table to table,   so   this does not quite agree with your answer.    

This is just an example

a compound is a molecule made of atoms from different elements.

a. by the combination of an unlimited number of elements

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i