Dr. Xander Reichert PhD
Dec 16, 2024
21.7% Carbon,9.6% Oxygen,68.7% Fluorine is what empirical formula?
What is the empirical formula for a compound that contains 21.7% carbon, 9.6% oxygen, and 68.7% fluorine?
4 Answers
I’m happy to bring the calculator out ..
in 100g of compound we have 21.7g of C or 21.7 / 12 moles of C = 1.808 moles
O = 9.6 / 16 = .6 moles
F = 68.7 / 19 = 3.62
molar ratio of C : O : F = 1.80 : .6 : 3.62 or 3:1:6
formula is C3OF6
Assume that it’s a 100 g sample; thus you have 21.7 g C, 9.6 g O, and 68.7 g F. Convert all those numbers to moles, so that you have X moles F, Y moles O, Z moles F. Divide all three of those numbers by the lowest number, which in this case would be Y moles of O. And then multiply all three of those numbers by a number so that all three moles of elements are integers. And then make sure that’s the empirical value (some cases it will be the molecular formula also).
It would be a lot easier to just do the work myself, but I’m not about to get my calculator out, and I don’t need to learn how to do it. I already do. I need to teach you how to do it.
F is 19g/mole, C is 12g/mole (from periodic table) assume which you have a hundred g of compound. then you definately would have seventy six g of F and 24 g of C. seventy six/19=4 for F, 24/12=2 for C So the formula would be C2F4
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