What is the empirical formula of a compound that is 48.6% C,8.2% H,and 43.2% O by mass?

Question

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1. Assume a 100 g sample of the compound. 2. Convert the percentages to grams for each element: - C: 48.6 g - H: 8.2 g - O: 43.2 g 3. Convert each element's grams to moles: - C: 48.6 g / 12.01 g/mol = 4.049 mol - H: 8.2 g / 1.008 g/mol = 8.132 mol - O: 43.2 g / 16 g/mol = 2.7 mol 4. Find the smallest whole number mole ratio between the elements. To do that, divide each element's moles by the smallest number of moles: - C: 4.049 mol / 2.7 mol = 1.5 - H: 8.132 mol / 2.7 mol = 3 - O: 2.7 mol / 2.7 mol = 1 5. Round the ratios to the nearest whole number (or a simple ratio, if the numbers are already close). The empirical formula is CH1.5O1, but we can simplify it further by multiplying everything by 2: C2H3O2. 6. The answer choice E, C₃H₆O₂, matches the empirical formula we found, C2H3O2.

What Is the Empirical Formula of a Compound That Is