What is the empirical formula for ethyl fluoride if the compound contains 49.97% carbon,10.51% hydrogen,and 39.52% fluorine by mass? A)C₂H₅F B)C₄H₁₀F₂ C)C₄H₁₀F₄ D)C₂₅F₂

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Quizplus · Accepted Answer

The empirical formula is determined by converting the mass percentages to moles, then finding the simplest whole number ratio. For carbon (C), 49.97% corresponds to 49.97/12.01 moles (using the atomic mass of carbon, 12.01 g/mol). For hydrogen (H), 10.51% corresponds to 10.51/1.01 moles (using the atomic mass of hydrogen, 1.01 g/mol). For fluorine (F), 39.52% corresponds to 39.52/19.00 moles (using the atomic mass of fluorine, 19.00 g/mol). After calculating the moles, the smallest number of moles is used to divide all mole quantities to get the simplest whole number ratio, which results in CHF for the empirical formula.

What Is the Empirical Formula for Ethyl Fluoride If the Compound