calculate the percent composition & the empirical formula of the following compound combustion of 3.02 milligram sample of compound give 8.86 milligram CO2 & 5.43 milligram of water.
....... please answer
Answers
Answer:
Empirical and Molecular Formulas
CHEM 1A
Empirical Formula: The lowest whole number ratio between the elements in a compound (not
necessarily the actual formula of the compound).
Molecular Formula: The actual formula of a molecular compound (the fixed ratio between the
elements in the molecule).
Example: glucose
molecular formula empirical formula
C6H12O6 CH2O
The empirical formula is useful because it can be determined experimentally from the percent
composition by mass or from the combustion products (see following pages).
The molecular formula can be found from the empirical formula using the scaling factor if the
molar mass of the compound is known (the molar mass can also be determined experimentally).
Example: The molar mass of a compound with the empirical formula CH2O is 180.156 g/mol.
What is the molecular formula of the compound?
CH2O empirical formula
x 6 multiply subscripts by scaling factor
C6H12O6 molecular formula
A. Romero 2009
CH2O = 1(12.01 g/mol) + 2(1.008 g/mol) + 1(16.00 g/mol)
Scaling Factor =
molar mass of compound
molar mass of empirical formula
mass compound
mass CH2O
180.156 g/mol
30.026 g/mol
Scaling Factor = = = 6
Molecular formula = C6H12O6
Explanation:
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