An organic compound has the empirical formula C3H3O if the molar mass of the compound is 110.15 gmol-1. The molecular formula of this organic compound is: (A, of C=12, H=1.008and O=16)
Correct answer: A. C6H6O2
- A. C6H6O2
- B. C3H3O
- C. C9H9O3
- D. C6H6O3
Explanation
The empirical formula mass can be calculated as follows:Empirical formula mass = (3 x atomic mass of C) + (3 x atomic mass of H) + (1 x atomic mass of O) Empirical formula mass = (3 x 12.011) + (3 x 1.008) + 16.00 Empirical formula mass = 55.05 g/molThe molecular formula mass can be calculated using the following equation:Molecular formula mass = n x empirical formula masswhere n is the number of empirical formula units in the molecular formula.n can be calculated as follows:n = Molecular formula mass / Empirical formula massn = 110.15 / 55.05 n = 2Therefore, the molecular formula is 2 times the empirical formula, and the molecular formula is:2(C3H3O) = C6H6O2So, the correct option is A) C6H6O2.
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Mole calculations connect mass, number of particles and Avogadro's number, while balanced equations provide the mole ratios used in stoichiometry. Questions cover limiting and excess reactants, theoretical yield and percentage yield, including identifying which reactant is consumed first and comparing the actual product with the maximum possible product.
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