Empirical formula is a useless notion; important is the molecular formula.
C2H4O2 normally written as CH3COOH to show the presence of the carboxyic acid group
The empirical formula of acetic acid (CH3COOH) is CH2O. This is derived by dividing all the subscripts in the molecular formula by the greatest common factor (in this case, 2).
C2H4O2 represents the molecular formula for ethanoic acid, which is the compound acetic acid. Since this formula provides the actual number of each atom in a molecule, it is a molecular formula. To determine the empirical formula, you would need to simplify the ratio of the atoms present in the compound to the smallest whole number ratio, which in this case is CH2O.
C12h24
CH3- C(=O)OH Acetic acid is the archaic name for 'Ethanoic Acid'.
The empirical formula is similar.
C2H4O2 normally written as CH3COOH to show the presence of the carboxyic acid group
The empirical formula of acetic acid (CH3COOH) is CH2O. This is derived by dividing all the subscripts in the molecular formula by the greatest common factor (in this case, 2).
Yes, it LOOKS like an empirical formula BUT it is NOT a correct one:Either C9H20 or C8H18 are correct (both are saturated alkanes) but not C8H20
C2H4O2 represents the molecular formula for ethanoic acid, which is the compound acetic acid. Since this formula provides the actual number of each atom in a molecule, it is a molecular formula. To determine the empirical formula, you would need to simplify the ratio of the atoms present in the compound to the smallest whole number ratio, which in this case is CH2O.
C12h24
C2h4o2 is the molecular formula for CH2O.
To determine the molecular formula from the empirical formula CH2O and given molecular mass of 60.0 amu, calculate the empirical formula mass: (12.01 g/mol for C) + 2(1.01 g/mol for H) + 16.00 g/mol for O = 30.02 g/mol. Then divide the given molecular mass by the empirical formula mass to find the factor by which the empirical formula must be multiplied to get the molecular formula: 60.0 amu / 30.02 g/mol ≈ 2. Next, multiply the subscripts in the empirical formula by this factor to find the molecular formula: 2(C)2(H)2(O) = C4H4O2, giving the molecular formula as C4H4O2.
CH3- C(=O)OH Acetic acid is the archaic name for 'Ethanoic Acid'.
expanded Formula: CH3COOH, condensed Formula: C2H4O2
There are numerous compounds with this ratio. The simplest of them is formaldehyde with the formula CH2O. More complex molecules with more atoms but the same ratio include acetic acid (C2H4O2), lactic acid (C3H6O3), and glucose and its isomers (C6H12O6).
It's a better representation of its molecular structure and helps to distinguish it from its isomer methyl formate HCOOCH3.