Respuesta :
Answer:
The empirical formula is the formula with a minimum whole number ratio of the constituent atoms present in a molecule. While the molecular formula has the actual number of atoms of various elements present in the molecule of the substance.
Empirical formula is C₃H₄O₃
Molecular formula is C₆H₈O₆
- This question involves finding the empirical and molecular formulas. Empirical formula is the one where minimum whole number ratio of the constituent atoms present in a molecule is shown whereas molecular formula is the one that shows the number of each type of atom that are present in a molecule.
- The compound contains only carbon, oxygen and hydrogen and undergoes combustion. Thus;
[tex]C_{x}[/tex][tex]H_{y} O_{z}[/tex] + O₂ ⟶ xCO₂ + [tex]\frac{y}{z} H_{2} O[/tex]
- The combustion yields 12.81 mg of CO₂.
From online sources, molar mass of CO₂ = 44 g/mol
number of moles of CO₂ = 12.81/44 = 0.291m mol
Mass of carbon in it is; C = 12 × 0.291 = 3.492 mg
- Similarly the combustion yields 3.50 mg H2O.
molar mass of water = 18 g/mol
number of moles of water in it = 3.5/18 = 0.194m mol
Hydrogen has 2 nos in water and so;
number of moles of hydrogen = 2 × 0.194 = 0.388m mol
Mass of hydrogen = 1 × 0.388 = 0.388 mg
- Compound is 8.544 mg and so;
Oxygen mass = 8.544 - (3.492 + 0.388) = 4.664 mg
Similarly, moles of oxygen that has molar mass of 16 is;
no of moles of oxygen = 4.664/16 = 0.2915m mol
- Thus;
Ratio of C:H:O is; 0.291:0.388;0.2915
Simplifying this gives us the empirical formula ratio as;
3:4:3
Thus;
Empirical formula is C₃H₄O₃
- Molar mass of C₃H₄O₃ = (12 × 3) + (1 × 4) + (16 × 3) = 88 g/mol
We are told that the molar mass of the compound is 176.1 g/mol.
Thus;
Molecular formula is; (176.1/88)C₃H₄O₃ = C₆H₈O₆
Read more at; brainly.com/question/14853529
