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Sagot :
Answer:
C2H3O3
Explanation:
Empirical formula is the simplest whole number ratio of moles of atoms that you can find in a molecule.
In combustion analysis all Carbon reacts producing CO2 and all hydrogen reacts producing H2O. With the differences in masses we can find the mass of oxygen and their moles:
Moles CO2 = Moles C:
14.08g * (1mol/44.01g) = 0.3199 moles C * (12.01g/mol) = 3.8423g C
Moles H2O:
4.32g H2O * (1mol/18.01g) = 0.2399 moles H2O * (2mol H / 1molH2O) = 0.4797moles H = 0.4797g H
Mass O:
12.01g = Mass O + 3.8423g C + 0.4797g H
Mass O = 7.688g O
Moles O:
7.688g O * (1mol/16g) = 0.48 moles O
The ratio of atoms (Dividing in the moles of C that are the lower number of moles):
O: 0.48moles O / 0.3199 moles C = 1.50
C: 0.3199 moles C / 0.3199 moles C = 1
H: 0.4797 moles H / 0.3199 moles C = 1.50
As empirical formula requires whole numbers:
O: 1.50* 2 = 3
C: 1*2 = 2
H: 1.50*2 = 3
The empirical formula is:
C2H3O3
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