You have an unknown substance that contains three elements: carbon, hydrogen, nitrogen. Your burn 7.114 grams of the unknown formula in excess oxygen gas to get 10.03 g of CO2 and 10.26 g of H20. What is the empirical formula of the unknown? what is the molecular formula if the molar mass of the unknown is 248.04 g/mol?
C% = 12*wt of CO2*100/44*wt of substance
= 12*10.03*100/(44*7.114) = 38.45%
H% = 2*wt of H2O*100/18*wt of substance
= 2*10.26*100/(18*7.114) = 16.025%
N% = 100-(C% + H%)
= 100-(38.45 + 16.025)
= 45.525%
Element % A. Wt Relative number simple ratio
C 38.45 12 38.45/12 = 3.2 3.2/3.2 = 1
H 16.025 1 16.025/1 = 16.025 16.025/3.2 = 5
N 45.525 14 45.525/14 = 3.25 3.25/3.2 = 1
The empirical formula = CH5N
empirical formula mass = 12 + 5 + 14 = 31g/mole
molecular formula = (empirical formula)n
n = molecular mass/empirical formula mass
= 248.04/31 = 8
The molecular formula = (CH5N)8
= C8H40N8
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