Answer:
Empirical formula is C3H3O
Molecular formula C9H9O3
Explanation:
From the question given, we obtained the following data:
Carbon = 63.15%
Hydrogen = 5.30%
Oxygen = 31.55%
We can obtain the empirical and molecular formula by doing the following as illustrated in the attached file. Please see attachment for explanation.
CH4 + 2O2 → CO2 + 2H2O + 890 kJ
MM of CH4 = (12.01 + 4x1.008) g/mol = 16.04 g/mol
Moles of CH4 = 45.5 g CH4 x (1 mol CH4/16.04 g CH4) = 2.837 mol CH4
q = 2.837 mol CH4 x (890 kJ/1 mol CH4) = 2520 kJ
The standard atomic weight of a C is 12, and the standard atomic weight of a H is 1. So to find molar ratio of C and H in the compound: 60.0/12=5, 5.05/1=5. This means the molar ratio of C and H is 5:5, thus 1:1. Assuming the molecular formula is CnHn, to find molar mass: 12n + 1n = 78.12. n=78.12/(12+1) = 6. So the compound's molecular formula is C6H6, benzene.
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