A 0.325 g sample of an organic compound containing carbon, hydrogen, and nitrogen is burned in oxygen to yield 0.562 g of carbon dioxide and 0.217 g of water. Determine the empirical formula of the compound.
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To determine the empirical formula of the compound, we need to calculate the moles of each element present in the compound.
First, we need to calculate the moles of carbon dioxide. The molar mass of carbon dioxide (CO2) is 44.01 g/mol.
moles of CO2 = mass of CO2 / molar mass of CO2
= 0.562 g / 44.01 g/mol
= 0.01274 mol
Next, we need to calculate the moles of water. The molar mass of water (H2O) is 18.015 g/mol.
moles of H2O = mass of H2O / molar mass of H2O
= 0.217 g / 18.015 g/mol
= 0.01203 mol
Since there is no nitrogen mentioned in the products, we can assume that all the nitrogen in the original sample is converted to nitrogen gas during combustion.
Now, we need to calculate the moles of carbon, hydrogen, and nitrogen in the original sample by using the mole ratios from the balanced combustion reaction:
CxHyNz + (x + y/4 - z/2)O2 -> xCO2 + y/2H2O + z/2N2
From the balanced reaction equation, we can see that 1 mole of CxHyNz corresponds to x moles of CO2, y/2 moles of H2O, and z/2 moles of N2.
moles of C = moles of CO2 = 0.01274 mol
moles of H = 2 * moles of H2O = 2 * 0.01203 mol = 0.02406 mol
moles of N = (1/2) * moles of N2 = 0.5 * 0.325 g / (14.01 g/mol)
= 0.11586 mol
To find the empirical formula, we need to determine the simplest whole number ratio of moles for each element. We can divide the moles of each element by the smallest number of moles (moles of N in this case).
moles of C = 0.01274 mol / 0.01274 mol = 1
moles of H = 0.02406 mol / 0.01274 mol = 1.89 (approximated to 2)
moles of N = 0.11586 mol / 0.01274 mol = 9.11 (approximated to 9)
Therefore, the empirical formula of the compound is C1H2N9, which can be further simplified to CH2N9.
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