A certain alcohol contains only carbon, oxygen and hydrogen. When 50
grams are combusted in air, 26.05 grams of C and 6.630 grams of H20 are
produced. If the molecular mass of this alcohol is 189 amu, what is its
molecular formula?

C12H208 is the molecular formula of the alcohol.

Explanation:

The weight of carbon given is 26.05 grams

The weight of H20 = 6.630 grams

first let us find the number of moles of:

number of moles = weight given ÷ mass of 1 mole of the element

                              = 26.05 ÷ 12

                               =  2.170 moles of carbon

number of moles of hydrogen present in H2O

    =  6.630 ÷ 18

     = 0.368

1 mole of water contains 2 mole of hydrogen so, 2 × 0.368 grams

= 0.7366 moles

Number of moles of oxygen cannot be calculated as of now.

So from the data obtained,

50 gms C2H2Ox is combusted.

Lets, calculate the gms from moles using the same equation

For carbon, 2.170 × 12

             = 26.04 grams

For hydrogen, 0.368 × 1.01

              =  0.3716 grams

adding the carbon and hydrogen content, 26.04 grams

now the oxygen content can be obtained by

50 - 26.04

= 23.96 grams

Now moles of oxygen calculated as

n = 23.96 ÷ 16

  = 1.49 moles of oxygen

The emperical formula for the alcohol is C6H04

now the molar mass is divided by emperial formula mass.

molar mass = 189 amu

emperial formula mass = 137

the division would give= 1.37

it can be taken as 2

The subscript in the formula is multiplied by 2 to get molecular formula from emperial formula.

So, C12H208 is the molecular formula of the alcohol.