To find the empirical formula, we need to first find the moles of water, carbon dioxide, and ammonia produced in the combustion reaction:
Moles of H2O = 6.608 g / 18.015 g/mol = 0.367 moles
Moles of CO2 = 12.923 g / 44.01 g/mol = 0.294 moles
Moles of NH3 = 2.501 g / 17.031 g/mol = 0.147 moles
Next, we need to determine the simplest whole number ratio of these moles. We can divide each of the mole amounts by the smallest of the three, which is 0.147:
Moles of H2O / 0.147 = 2.50
Moles of CO2 / 0.147 = 2.00
Moles of NH3 / 0.147 = 1.00
Since these ratios are all close to whole numbers, we can round them to the nearest whole number to determine the empirical formula:
H5C2O5N (empirical formula)
To check our answer, we can calculate the molar mass of this empirical formula:
(5x1.01 g/mol) + (2x12.01 g/mol) + (5x16.00 g/mol) + (1x14.01 g/mol) = 117.18 g/mol
The experimental molar mass of the unknown compound can be determined by adding the molar masses of the products produced in the combustion reaction:
(2x1.01 g/mol) + (2x12.01 g/mol) + (5x16.00 g/mol) + (1x14.01 g/mol) = 117.19 g/mol
This is close to the molar mass of the empirical formula, indicating that our answer is reasonable. However, the unknown compound may have a molecular formula that is a multiple of the empirical formula (e.g. C10H10N5O10), so further analysis would be needed to determine the molecular formula.
An unknown compound was subjected to combustion.6.608 g of water together with 12.923 g of carbon dioxide and 2.501 g NH3 were produced. What is the compound’s empirical formula?
(Molar mass: N=14 g/mol; C=12 g/mol; O= 16 g/mol)
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