It was found that compound contained 68.1% carbon, 13.7% hydrogen, and 18.2% oxygen by mass.
a. find empirical formula for the compound.
b. if the molar mass of the compounds is 176.34 g/mol, what it’s the molecular formula?

A. The empirical formula is the simplest ratio of components making up a compound.

In 100 g of the compound

                             C                            H                               O

mass               68.1 g                         13.7 g                         18.2 g

number of moles

                    68.1 g / 12 g/mol        13.7 g / 1 g/mol         18.2 g / 16 g/mol

                     = 5.68 mol                 = 13.7 mol                 1.14 mol

divide by the least number of moles - which is 1.14 mol

                  5.68 / 1.14 = 5.0             13.7 / 1.14 = 12.0     1.14 / 1.14 = 1.0

the ratio of the components C : H : O is 5: 12 : 1

therefore empirical formula is C₅H₁₂O

B. Molecular formula is the actual ratio of the components making up the compound. We need to find the mass of the empirical formula / empirical unit and then find how many empirical units make up the molecular formula

mass of 1 empirical unit - C₅H₁₂O  - 12 g/mol x 5 + 1 g/mol x 12 + 16 g/mol x 1

                                       = 60 + 12 + 16 = 88 g

number of empirical units = molecular formula mass / mass of 1 empirical unit

                                           = 176.34 g/mol  / 88 g = 2.00 = 2

there are 2 empirical formulas in the molecular formula

therefore molecular formula  = 2 x (C₅H₁₂O) = C₁₀H₂₄O₂ ,

therefore molecular formula is C₁₀H₂₄O₂

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

fdvwsa gfwsargebgeebbfwfwffw gwf   vgwgw gw f gwgw fgw v v dsv dwfff