yes i can answer it
While both formaldehyde (CH2O) and acetic acid (C2H4O2) share the same empirical formula (CH2O), they are different compounds with distinct chemical properties. The difference lies in their molecular structures - formaldehyde is a simple aldehyde whereas acetic acid is a carboxylic acid. This structural difference leads to variations in their physical and chemical properties.
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).
CH3- C(=O)OH Acetic acid is the archaic name for 'Ethanoic Acid'.
C12h24
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.
While both formaldehyde (CH2O) and acetic acid (C2H4O2) share the same empirical formula (CH2O), they are different compounds with distinct chemical properties. The difference lies in their molecular structures - formaldehyde is a simple aldehyde whereas acetic acid is a carboxylic acid. This structural difference leads to variations in their physical and chemical properties.
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).
C2h4o2 is the molecular formula for CH2O.
CH3- C(=O)OH Acetic acid is the archaic name for 'Ethanoic Acid'.
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
C12h24
Vinegar is a mixture and does not have a molecular formula as such. The "active" ingredient in vinegar is acetic acid, which has the formula CH3COOH, in one common method of writing it.
In one molecule of C2H4O2, there are 4 atoms of hydrogen.
5.00 moles H x 1 mole C2H4O2/4 moles H = 1.25 moles of C2H4O2 present.