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Answer:
C₃H₆O₃
Explanation:
To solve this question we need to find, as first, the moles of each atom in order to find empirical formula (Simplest whole-number ratio of atoms present in a molecule).
With the molar mass of the substance and the empirical formula we can find the molecular formula as follows:
Moles C -Molar mass:12.0g/mol-
0.24g * (1mol/12.0g) = 0.020 moles C
Moles H = Mass H because molar mass = 1g/mol:
0.040 moles H
Moles O -Molar mass: 16g/mol-
Mass O: 0.60g - 0.24g - 0.040g = 0.32g O
0.32g O * (1mol/16g) = 0.020 moles O
Ratio of atoms (Dividing in moles of C: Lower number of moles):
C = 0.020 moles C / 0.020 moles C = 1
H = 0.040 moles H / 0.020 moles C = 2
O = 0.020 moles O / 0.020 moles C = 1
Empirical formula:
CH₂O.
Molar mass CH2O:
12g/mol + 2*1g/mol + 16g/mol = 30g/mol
As molecular formula has a molar mass 3 times higher than empirical formula, the molecular formula is 3 times empirical formula:
The molecular formula of the organic acid would be C3H6O3
Molecular formula = [empirical formula]n
Where n = molar mass/mass of empirical formula
Empirical formula
C = 0.24/12 = 0.02
H = 0.040/1 = 0.04
O = 0.6 - (0.24+0.04) = 0.32/16 = 0.02
Divide by the smallest
C = 1
H = 2
O = 1
Empirical formula = CH2O
Empirical formula mass = 12 + 2 + 16 = 30
n = 90/30 = 3
Molecular formula = [CH2O]3
= C3H6O3
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