Answer:
C₆H₄O₂
Explanation:
Given parameters:
Mass of the unknown compound = 0.315g
Atoms contained in the compound C, H and O
Mass of CO₂ produced = 0.771g
Mass of H₂O = 0.105g
Molar mass of compound = 108gmol⁻¹
The complete combustion of most hydrocarbon compounds like this would produce carbon dioxide and water only.
Solution
We first find the mass of the Carbon, hydrogen and Oxygen in the compounds given.
Mass of carbon in compound = [tex]\frac{Molar mass of Carbon}{Molar mass of CO_{2} }[/tex] x mass of CO₂
Molar mass of C = 12
Molar mass of CO₂ = 12 + (16x2) = 44
Mass of CO₂ produced = 0.771g
Mass of carbon in compound = [tex]\frac{12}{44 }[/tex] x 0.771=0.2103g
Mass of H in compound = [tex]\frac{Molar mass of H}{Molar mass of H_{2}O }[/tex] x mass of H₂O
Molar mass of H in H₂O = 2
Molar mass of H₂O = 2 + 16 = 18
Mass of H₂O = 0.105g
Mass of H in compound = [tex]\frac{2}{18}[/tex] x 0.105 = 0.012g
Now, to find the mass of oxygen in the compound, we sum the mass H and C and subtract from the mass of the compound given:
Mass of oxygen = 0.315 - (0.2103 + 0.012)= 0.0927g
We then continue to derive the empirical formula of the compound.
C H O
mass of
atoms 0.2103 0.012 0.0927
Moles 0.2103/12 0.012/1 0.0927/16
0.018 0.012 0.006
Dividing by
the smallest 0.018/0.006 0.012/0.006 0.006/0.006
3 2 1
The empirical formula of the compound is C₃H₂O
From the empirical formula, we can find the compound from the given molar mass:
Molecular formula = (Empirical formula)n
Where n is the number of repeating times of the empirical formula present in one mole of the molecule.
Therefore n = [tex]\frac{molar mass of the compound}{molar mass of the empirical formula of the compound}[/tex]
Molar mass of the empirical formula C₃H₂O = (12x3) + (2x1) + (16) = 54g/mol
n = [tex]\frac{108}{54}[/tex] = 2
The molecular formula of the compound is = 2(C₃H₂O) = C₆H₄O₂
