Answer:
1. Empirical formula => CH2O2
2. Molecular formula => C2H4O4
Explanation:
From the question given above, the following data were obtained:
Carbon (C) = 26.4 %
Hydrogen (H) = 3.3 %
Oxygen (O) = 70.3 %
Molar mass of compound = 91.0 g/mol
Empirical formula =..?
Molecular formula =..?
1. Determination of the empirical formula of the compound.
C = 26.4 %
H = 3.3 %
O = 70.3 %
Divide by their molar mass
C = 26.4 /12 = 2.2
H = 3.3 /1 = 3.3
O = 70.3 /16 = 4.39
Divide by the smallest
C = 2.2 /2.2 = 1
H = 3.3 /2.2 = 2
O = 4.39 /2.2 = 2
Empirical formula => CH2O2
2. Determination of the molecular formula of the compound.
Molar mass of compound = 91.0 g/mol
Empirical formula => CH2O2
Molecular formula => [CH2O2]n
We shall determine the value of n as follow:
[CH2O2]n = 91
[12 + (2×1) + (2×16)]n = 91
[12 + 2 + 32]n = 91
46n = 91
Divide both side by 46
n = 91/46
n = 2
Molecular formula => [CH2O2]n
Molecular formula => C2H4O4
In the given case where 26.4 % Carbon , 3.3 % Hydrogen , 70.3 % Oxygen , and Molar Mass: 91.0 g/mol the:
Given:
Carbon (C) = 26.4 %
Hydrogen (H) = 3.3 %
Oxygen (O) = 70.3 %
Molar mass of compound = 91.0 g/mol
Determination of the empirical formula of the compound:-
The ratio is CHO = 1 : 1.5 : 2
multiply with 2 to find correct and complete number ratio
C = 1 × 2 = 2
H = 1.5 × 2 = 3
O = 2 × 2 = 4
Thus, the Empirical formula => [tex]C_2H_3O_4[/tex]
Mass × n = molar mass
12 × 2 + 1 × 3 + 16 × 4 = 91
24 + 3 + 64 = 91
91 = 91
Thus moles are 1 which means
molecular formula = ([tex]C_2H_3O_4[/tex])[tex]_1[/tex]
Thus, here in the given data:
Molecular Formula: ([tex]C_2H_3O_4[/tex])[tex]_1[/tex]
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