Example (PageIndex{4}): Combustion of Naphalene
Naphthalene, the active ingredient in one variety of mothballs, is an organic compound that contains carbon and hydrogen only. Complete combustion of a 20.10 mg sample of naphthalene in oxygen yielded 69.00 mg of CO2 and 11.30 mg of H2O. Determine the empirical formula of naphthalene.
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Given: mass of sample and mass of combustion products
Asked for: empirical formula
Strategy:
- Use the masses and molar masses of the combustion products, CO2 and H2O, to calculate the masses of carbon and hydrogen present in the original sample of naphthalene.
- Use those masses and the molar masses of the elements to calculate the empirical formula of naphthalene.
Solution:
A Upon combustion, 1 mol of (ce{CO2}) is produced for each mole of carbon atoms in the original sample. Similarly, 1 mol of H2O is produced for every 2 mol of hydrogen atoms present in the sample. The masses of carbon and hydrogen in the original sample can be calculated from these ratios, the masses of CO2 and H2O, and their molar masses. Because the units of molar mass are grams per mole, we must first convert the masses from milligrams to grams:
[ mass , of , C = 69.00 , mg , CO_2 times {1 , g over 1000 , mg } times {1 , mol , CO_2 over 44.010 , g , CO_2} times {1 , mol C over 1 , mol , CO_2 } times {12.011 ,g over 1 , mol , C} ]
[ = 1.883 times 10^{-2} , g , C ]
[ mass , of , H = 11.30 , mg , H_2O times {1 , g over 1000 , mg } times {1 , mol , H_2O over 18.015 , g , H_2O} times {2 , mol H over 1 , mol , H_2O } times {1.0079 ,g over 1 , mol , H} ]
[ = 1.264 times 10^{-3} , g , H ]
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B To obtain the relative numbers of atoms of both elements present, we need to calculate the number of moles of each and divide by the number of moles of the element present in the smallest amount:
[ moles , C = 1.883 times 10^{-2} ,g , C times {1 , mol , C over 12.011 , g , C} = 1.568 times 10^{-3} , mol C ]
[ moles , H = 1.264 times 10^{-3} ,g , H times {1 , mol , H over 1.0079 , g , H} = 1.254 times 10^{-3} , mol H ]
Dividing each number by the number of moles of the element present in the smaller amount gives
[H: {1.254times 10^{−3} over 1.254 times 10^{−3}} = 1.000 , , , C: {1.568 times 10^{−3} over 1.254 times 10^{−3}}= 1.250]
Thus naphthalene contains a 1.25:1 ratio of moles of carbon to moles of hydrogen: C1.25H1.0. Because the ratios of the elements in the empirical formula must be expressed as small whole numbers, multiply both subscripts by 4, which gives C5H4 as the empirical formula of naphthalene. In fact, the molecular formula of naphthalene is C10H8, which is consistent with our results.
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