A 3.68 g sample of a compound consisting of carbon, hydrogen, oxygen, nitrogen, and sulfur was combusted in excess oxygen. This produced 2.07 g CO2 and 1.27 g H2O . A second sample of this compound with a mass of 6.46 g produced 6.62 g SO3 . A third sample of this compound with a mass of 8.70 g produced 3.51 g HNO3 . Determine the empirical formula of the compound. Enter the correct subscripts on the given chemical formula.
empirical formula C5H3NSO3
N/A
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Steps
=> Find out percentage of elements
=> Divide by atomic mass
=> Determine the least ratio
%C = 38.2
%N = 8.9
%O = 30.6
%H = 1.9
%S = 20.4
To determine the empirical formula of the compound, we need to find the ratio of each element present in the compound. We can do this by analyzing the masses of the elements in the given compounds produced during combustion.
1. Start by calculating the moles of carbon, hydrogen, oxygen, nitrogen, and sulfur in each product:
- For CO2:
- Carbon (C): 2.07 g CO2 / 44.01 g/mol = 0.047 mol C
- Oxygen (O): 2.07 g CO2 / 44.01 g/mol = 0.047 mol O
- For H2O:
- Hydrogen (H): 1.27 g H2O / 18.02 g/mol = 0.071 mol H
- Oxygen (O): 1.27 g H2O / 18.02 g/mol = 0.071 mol O
- For SO3:
- Sulfur (S): 6.62 g SO3 / 80.06 g/mol = 0.083 mol S
- Oxygen (O): 6.62 g SO3 / 80.06 g/mol = 0.083 mol O
- For HNO3:
- Hydrogen (H): 3.51 g HNO3 / 63.02 g/mol = 0.056 mol H
- Nitrogen (N): 3.51 g HNO3 / 63.02 g/mol = 0.056 mol N
- Oxygen (O): 3.51 g HNO3 / 63.02 g/mol = 0.056 mol O
2. Next, we need to find the lowest whole-number ratio of these elements. We can divide each mole value by the lowest number of moles calculated (which is 0.047 mol):
- Carbon (C): 0.047 mol C / 0.047 mol = 1
- Hydrogen (H): 0.071 mol H / 0.047 mol = 1.51 (approximately)
- Nitrogen (N): 0.056 mol N / 0.047 mol = 1.19 (approximately)
- Sulfur (S): 0.083 mol S / 0.047 mol = 1.77 (approximately)
- Oxygen (O): 0.083 mol O / 0.047 mol = 1.77 (approximately)
3. Now, we need to adjust the subscripts to achieve whole numbers. Since the ratios for hydrogen, nitrogen, and sulfur are slightly above 1, we can multiply all the subscripts by a factor of 2 to obtain whole numbers:
- Carbon (C): 1
- Hydrogen (H): 2
- Nitrogen (N): 2
- Sulfur (S): 4
- Oxygen (O): 4
Therefore, the empirical formula of the compound is C1H2N2S4O4, which can be simplified to C5H3NSO3 by multiplying all subscripts by 5.
Note: It's important to mention that the empirical formula represents the simplest whole-number ratio of atoms in a compound, but it may not necessarily represent the actual molecular formula of the compound.