What is the empirical formula for a compound that is 17.15 % carbon, 1.44 % hydrogen, and 81.41 % fluorine?

Answer 1

Regards for the question on the empirical formula.

The lowest whole-number ratio of moles (or atoms) in a chemical formula is known as an empirical formula.

Since the percents are already by mass, we will use the percents as if they are grams and convert to moles; the denominators of each element are taken from the periodic table. The first step in determining an empirical formula is to ascertain the relative number of moles in the compound.

carbon, or 17.15/12.01, equals 1.428.

= 1.44/1.008 = 1.429 for hydrogen

= 81.41/19.00 = 4.285 for fluorine

The next step is to divide all of the mole ratios by the SMALLEST MOLE RATIO in an attempt to reduce the ratios to the smallest whole numbers, since it is now evident that we cannot have partial atoms in our chemical formula.

Carbon: 1 / 1.428 = 1.428

Water: 1.429 x 1.428 = 1.

Fluorine: 3.22 (4.285 / 1.428)

Fortunately, these are all whole numbers, and these ratios provide us with our empirical formula.

The empirical formula is: CH #F_3#.

I hope this is useful.

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Answer 2

The empirical formula is CF4.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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