Covalently bonded compounds are characterized by a molecular formula, because such compounds exist in the form of discrete molecules, all of the atoms of which move together as kinetic-molecular units. Ionically bonded compounds do not have molecules in this sense: Their compounds are made up of at least two kinds of ions, one positive and the other negative, and the ions in these compounds can move separately as kinetic-molecular units: If a positive ion is separated from the particular negative ion with which it was most closely associated initially in a fluid mixture, usually a solution in an ionizing solvent, of both kinds of ions, another negative ion with the same properties is always close by.
What you write for an ionic compound is called the formula unit, but the formula unit is almost always the same as the empirical formula. The answer to your question could not be the molecular formula because an ionic compound is not a molecule.
The empirical formula C2H3 has a molecular mass of 27 (C: 12, H: 1). To determine the molecular formula with a molecular mass of 54, the molecular formula would simply be double the empirical formula, so the molecular formula would be C4H6.
ionic molecules dissolve the most. but some polar covalent molecules also do dissolve in water.
The molecular formula for iodine pentafluoride is IF5. No prefix in front of iodine is understood to be one, but mono- is not used for the first element in a binary covalent compount. The prefix penta- means five, so the subscript for fluoride is 5. Unfortunately, there is no way to write the 5 as a subscript.
A molecular formula is identical to the empirical formula, and is based on quantity of atoms of each type in the compound.The relationship between empirical and molecular formula is that the empirical formula is the simplest formula, and the molecular can be the same as the empirical, or some multiple of it. An example might be an empirical formula of C3H8. Its molecular formula may be C3H8 , C6H16, C9H24, etc. Looking at it the other way, if the molecular formula is C6H12O6, the empirical formula would be CH2O.
What you write for an ionic compound is called the formula unit, but the formula unit is almost always the same as the empirical formula. The answer to your question could not be the molecular formula because an ionic compound is not a molecule.
I would expect the molecular formula to be FX4. This is because fluorine typically forms a single bond resulting in a fluoride with a 1- charge, and X is a halogen that could potentially form four covalent bonds in a molecule.
Molecular. The empirical formula would simply be S.
The empirical formula C2H3 has a molecular mass of 27 (C: 12, H: 1). To determine the molecular formula with a molecular mass of 54, the molecular formula would simply be double the empirical formula, so the molecular formula would be C4H6.
Basically Nicotine is bonding of 2 molecules. The bond in between these two molecule is very weak and so it is IONIC.
That's a very good question and there are arguments for both. I would class it more as a simple molecular substance because a sample will contain many separate molecules, with forces other than covalent bonds between them. A giant covalent structure like diamond is wholly held together by covalent bonds.
molecular
ionic molecules dissolve the most. but some polar covalent molecules also do dissolve in water.
The molecular formula for iodine pentafluoride is IF5. No prefix in front of iodine is understood to be one, but mono- is not used for the first element in a binary covalent compount. The prefix penta- means five, so the subscript for fluoride is 5. Unfortunately, there is no way to write the 5 as a subscript.
To find the molecular formula, you first need to calculate the empirical formula mass of C3H4. C3H4 has an empirical formula weight of 40 g/mol. If the molecular weight is 120 g/mol, then the molecular formula would be 3 times the empirical formula, so the molecular formula would be C9H12.
Ammonia and nitrogen are two distinct compounds, not a single compound that would have a molecular formula.
Yes, it is possible for different covalent compounds to have the same empirical formula. This occurs when compounds have different arrangements of atoms but the same ratio of elements. An example is ethyne (C2H2) and benzene (C6H6), both of which have an empirical formula of CH.