Given:
- 74 gm of sample on complete combustion gives 132 gm CO2 and 54 gm of H2O.

To find:
- The molecular formula of the compound.

Solution:

Step 1: Calculate the number of moles of CO2 and H2O produced.

Moles of CO2 = 132 gm / 44 gm/mol = 3 moles
Moles of H2O = 54 gm / 18 gm/mol = 3 moles

Step 2: Calculate the number of moles of carbon and hydrogen present in the sample.

The carbon in the sample combines with oxygen to form CO2, so the number of moles of carbon in the sample is equal to the number of moles of CO2.
Moles of carbon = 3 moles

The hydrogen in the sample combines with oxygen to form H2O, so the number of moles of hydrogen in the sample is equal to the number of moles of H2O.
Moles of hydrogen = 3 moles

Step 3: Determine the empirical formula of the compound.

The empirical formula is the simplest whole number ratio of atoms in a compound.
The empirical formula of the compound is CH.

Step 4: Determine the molecular formula of the compound.

The molecular formula is a multiple of the empirical formula.
To find the molecular formula, we need to know the molecular weight of the compound.

Molecular weight = empirical formula weight x n
where n is the number of empirical formula units in the molecule.

Empirical formula weight = 12 (for 1 carbon atom) + 1 (for 1 hydrogen atom) = 13 g/mol
Molecular weight = 74 g/mol
n = molecular weight / empirical formula weight = 74 g/mol / 13 g/mol = 5.7

The molecular formula is the empirical formula multiplied by n, rounded to the nearest whole number.
Molecular formula = (CH) x 6 = C6H6

However, C6H6 is not a possible molecular formula based on the given information.

We need to consider the fact that oxygen is also present in the compound.
We can calculate the amount of oxygen in the compound by subtracting the mass of carbon and hydrogen from the total mass of the sample.

Mass of oxygen = 74 g - (12 g + 3 g) = 59 g

The molecular formula should be such that the ratio of carbon, hydrogen, and oxygen atoms is in the simplest whole number ratio.

Divide the number of moles of each element by the smallest number of moles to get the simplest ratio.

Moles of carbon = 3
Moles of hydrogen = 3
Moles of oxygen = 3.69

Divide by 3 (the smallest number of moles) to get the simplest ratio.

Moles of carbon = 1
Moles of hydrogen = 1
Moles of oxygen = 1.23

Round off to the nearest whole number to get the empirical formula:
Empirical formula = C1H1O1 = CHO

The molecular formula is a multiple of the empirical formula.
Molecular weight = 74 g/mol
Empirical formula weight = 12 + 1 + 16 = 29 g/mol
n = molecular weight / empirical