The combustion analysis of an organic compound with relative molar mass 99 g shows that it contains 4.04% hydrogen, 24.24% carbon and 71.72% chlorine. (Note: Atomic mass of H= 1 amu, C = 12 amu and Cl = 35.5 amu) What are the (i) empirical formula and (ii) molecular formula of the given compound? Show the steps of working clearly.​

To find the empirical and molecular formula, we first need to convert the percentages into moles.

(i) Empirical Formula:

1. Convert the percentages to grams (since the percentages are based on 100g, the percentage is the same as the mass in grams for this calculation):

   • Hydrogen (H) = 4.04g
   • Carbon (C) = 24.24g
   • Chlorine (Cl) = 71.72g

2. Convert the grams to moles using the atomic masses:

   • Moles of H = 4.04g / 1g/mol = 4.04 moles
   • Moles of C = 24.24g / 12g/mol = 2.02 moles
   • Moles of Cl = 71.72g / 35.5g/mol = 2.02 moles

3. Divide each mole quantity by the smallest number of moles calculated (in this case, 2.02) to get the ratio of atoms:

   • Ratio of H = 4.04 / 2.02 = 2
   • Ratio of C = 2.02 / 2.02 = 1
   • Ratio of Cl = 2.02 / 2.02 = 1

So, the empirical formula is CH2Cl.

(ii) Molecular Formula:

1. Calculate the molar mass of the empirical formula:

   • Molar mass of CH2Cl = (112.01) + (21.01) + (1*35.45) = 49.48 g/mol

2. Divide the given molar mass by the empirical formula molar mass and round to the nearest whole number:

   • n = 99 g/mol / 49.48 g/mol = 2

3. Multiply the subscripts in the empirical formula by this number to get the molecular formula:

   • Molecular formula = C(12)H(22)Cl(1*2) = C2H4Cl2

So, the molecular formula of the compound is C2H4Cl2.