CH2 is the empirical formula for C4H8 because it is an alkene and the empirical fomula for ALL alkenes are C(n)H(2n) n being the number of molecules!!! (^-^)
CH2 The empirical formula of c4h8 is ch2. An empirical formula shows the ratio of elements present but not the actual number or arrangement.
The empirical formula for C4H8 would be CH2. For C2H4 it would be CH2 and for C2H2 it would be CH.
2av²x
Imperical fomula is C2H4.Molecular fomula is C4H8.
C12h24
C = 12H = 1 x 2 = 2-----------------total = 14 g/mole for empirical56.11/14 = 4.00 so there are 4 empirical formulae in 56.11 gChemical formula = C4H8
Well you know that Butanoic Acid's Molecular formula is C3H7COOH, and Empirical formula is a compound showig the simplest ratio of numbers of atoms of each element in the compound. Now the question is, can you simply C3H7COOH ? Nope! Then the Empirical formula is also C3H7COOH
Well its actually called cyclobutane. Erm, it's actually called butene - with an e
Imperical fomula is C2H4.Molecular fomula is C4H8.
Empirical formula (lowest whole number) for C4H8 is CH2, obtained by dividing by 4.
CH2 is the empirical formula for C4H8 because it is an alkene and the empirical fomula for ALL alkenes are C(n)H(2n) n being the number of molecules!!! (^-^)
C12h24
Yes, if you have some additional information, such as the molecular weight. For instance, the molecules C2H4 and C4H8 have exactly the same percent composition, but they are very different molecules. So you need some other information to tell them apart than the percent composition.Answer ExpandedThis is kind of a trick question. By knowing the percent composition, you would easily be able to determine its empirical formula, but molecular formula is a bit different. The molecular formula is the actual number of atoms in a molecule, so in order to find the specific molecular formula of a substance, you would also need to know how many grams there is of that substance.(This explains the difference between C2H4 and C4H8)
C2h4, c3h6, c4h8
C = 12H = 1 x 2 = 2-----------------total = 14 g/mole for empirical56.11/14 = 4.00 so there are 4 empirical formulae in 56.11 gChemical formula = C4H8
Well you know that Butanoic Acid's Molecular formula is C3H7COOH, and Empirical formula is a compound showig the simplest ratio of numbers of atoms of each element in the compound. Now the question is, can you simply C3H7COOH ? Nope! Then the Empirical formula is also C3H7COOH
C4H8 may be an alkene, Butene or methyl propene and may be a cycloalkane, cyclobutane or methyl cyclopropane.
* For the alkanes, the general formula is CnH2n+2 EXAMPLE: CH4, C4H10 * For the alkenes, the equation is CnH2n EXAMPLE: C2H4 where n is the number of atoms in one molecule of the hydrocarbon.
Cycloalkene is represented by a molecular formula. This formula is CnH (2n - 2) wherein C and H are the compositions of the cyclic and 2H is removed from the end C to connect to the other C.
If the C4H8 compound is butene-1 or -2 or 2-methylpropene, the product is one of the chlorobutanes with general formula C4H9Cl. If the C4H8 compound is cyclobutane, there is no reaction at standard temperature and pressure.