What is the simplest formula of compound which has 80% carbon and 20% nitrogen?
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Explanation:
As with all these problems, it is usually assumed that we have a 100⋅g mass of unknown compound, and we work out the molar quantities:
And thus moles of carbon≡80⋅g12.011⋅g⋅mol−1=6.66⋅mol.
And thus moles of hydrogen≡20⋅g1.008⋅g⋅mol−1=19.8⋅mol.
We divide the molar quantities thru by the SMALLER molar quantity:
C:6.66⋅mol6.66⋅mol=1
H:19.8⋅mol6.66⋅mol=2.97
And thus the empirical formula = CH3
And we know that the molecular formula is always a multiple of the empirical formula, and thus is terms of mass:
i.e. molecular formula = n×empirical formula
But we have a molecular mass of 30⋅amu
So 30⋅amu = n×(12.01+3×1.01)⋅amu
Clearly, n=2, and the MOLECULAR FORMULA is C2H6.
This is not a realistic problem, as few analysts would perform combustion on a liquid, and no analyst could perform combustion on a gas.
As with all these problems, it is usually assumed that we have a 100⋅g mass of unknown compound, and we work out the molar quantities:
And thus moles of carbon≡80⋅g12.011⋅g⋅mol−1=6.66⋅mol.
And thus moles of hydrogen≡20⋅g1.008⋅g⋅mol−1=19.8⋅mol.
We divide the molar quantities thru by the SMALLER molar quantity:
C:6.66⋅mol6.66⋅mol=1
H:19.8⋅mol6.66⋅mol=2.97
And thus the empirical formula = CH3
And we know that the molecular formula is always a multiple of the empirical formula, and thus is terms of mass:
i.e. molecular formula = n×empirical formula
But we have a molecular mass of 30⋅amu
So 30⋅amu = n×(12.01+3×1.01)⋅amu
Clearly, n=2, and the MOLECULAR FORMULA is C2H6.
This is not a realistic problem, as few analysts would perform combustion on a liquid, and no analyst could perform combustion on a gas.
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