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What is the empirical formula of a compound consisting of 29.6 percent oxygen and 70.4 percent fluorine by mass?
We assume 100 grams total and turn those percentages into grams. Get moles elements.
29.6 grams O (1 mole O/16.0 grams)
= 1.85 mole O
70.4 grams F (1 mole F/19.0 grams)
= 3.71 mole F
now divide both moles by smallest number to get mole ratio for empirical formula
1.85/1.85 = 1 for O
3.71/1.85 = 2 for F
so the empirical formula for this compound is........
OF2
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What is the empirical formula of the ionic compound that forms sodium and fluorine?
The empirical formula for the ionic compound formed by sodium and fluorine is NaF. Sodium is a metal that gives away one electron, becoming Na+, while fluorine is a non-metal that gains one electron, becoming F-. The resulting compound has a 1:1 ratio of sodium to fluorine ions, giving NaF as the empirical formula.
Predict the empirical formula of the ionic compound that forms from magnesium and fluorine?
The empirical formula of the compound formed between magnesium and fluorine is MgF2. This is because magnesium has a 2+ charge and fluorine has a 1- charge, so one magnesium ion will combine with two fluorine ions to achieve a neutral compound.
What is the empirical formula of magnesium and fluorine?
The empirical formula of magnesium fluoride is MgF2. This is because the ratio of magnesium atoms to fluorine atoms in the compound is 1:2.
Whats the empirical formula of a compound of uranium and fluorine that is composed of 67.6 percent uranium and 32.4 percent fluorine is?
The empirical formula of the compound is UF6 (uranium hexafluoride). This is because the ratio of uranium to fluorine in the compound is close to 1:6, indicating that there are six fluorine atoms for every one uranium atom in the compound.
Predict the empirical formula of the ionic compound that forms from sodium and fluorine?
The empirical formula of the ionic compound formed by sodium and fluorine is NaF, which is sodium fluoride. Sodium typically forms a +1 cation (Na+) and fluorine typically forms a -1 anion (F-), leading to a one-to-one ratio in the compound.
What is the empirical formula of the ionic compound that forms sodium and fluorine?
The empirical formula for the ionic compound formed by sodium and fluorine is NaF. Sodium is a metal that gives away one electron, becoming Na+, while fluorine is a non-metal that gains one electron, becoming F-. The resulting compound has a 1:1 ratio of sodium to fluorine ions, giving NaF as the empirical formula.
Predict the empirical formula of the ionic compound that forms from magnesium and fluorine?
The empirical formula of the compound formed between magnesium and fluorine is MgF2. This is because magnesium has a 2+ charge and fluorine has a 1- charge, so one magnesium ion will combine with two fluorine ions to achieve a neutral compound.
What is the empirical formula of magnesium and fluorine?
The empirical formula of magnesium fluoride is MgF2. This is because the ratio of magnesium atoms to fluorine atoms in the compound is 1:2.
Whats the empirical formula of a compound of uranium and fluorine that is composed of 67.6 percent uranium and 32.4 percent fluorine is?
The empirical formula of the compound is UF6 (uranium hexafluoride). This is because the ratio of uranium to fluorine in the compound is close to 1:6, indicating that there are six fluorine atoms for every one uranium atom in the compound.
What is the empirical formula for fluorine?
the empirical formula for fluorine is F. the chemical formula is F2.
Predict the empirical formula of the ionic compound that forms from sodium and fluorine?
The empirical formula of the ionic compound formed by sodium and fluorine is NaF, which is sodium fluoride. Sodium typically forms a +1 cation (Na+) and fluorine typically forms a -1 anion (F-), leading to a one-to-one ratio in the compound.
What is the empirical formula if a 100.0 g sample of a compound is composed of 16.3 g of carbon 32.1 g of chlorine and 51.6 g of fluorine?
To find the empirical formula, convert the masses of each element to moles. The molar ratio of carbon to chlorine to fluorine is 1:1:2. Therefore, the empirical formula is CClF2.
Is calcium fluoride a compound?
Yes, calcium fluoride is a compound. It is an inorganic compound consisting of the elements calcium and fluorine, with the chemical formula CaF2.
How do you know if a formula is an empirical formula?
An empirical formula has no data about the structure of a compound.
What compound does the formula KrF4 make?
The compound with the formula KrF4 is called krypton tetrafluoride. It is a chemical compound consisting of one krypton atom bonded to four fluorine atoms.
The lowest whole number ratio of the elements in a compound is called the?
An empirical formula refers to the chemical formula that indicates the simplest ratio of atoms in a compound. Two different compounds may have the same empirical formula.
What is carbon and fluorine molecular formular?
The molecular formula for a compound consisting of carbon and fluorine can vary depending on the specific compound. For example, the simplest binary compound is carbon tetrafluoride, which has the molecular formula CF₄. Another example is carbon difluoride, with the formula CF₂. The specific formula will depend on the ratio of carbon to fluorine in the compound being considered.
