Elemental analysis of 0.2511g of a compound produced 0.372g of CO2, 0.178g of H2O and 0.039g of N2. What is the empirical formula of the compound?
Please give a detailed answer thanks.
Please give a detailed answer thanks.
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Mass of C in 0.372g CO2 = 12/44*0.372 = 0.1015g
Mass of H in 0.178g H2O = 2/18*0.178 = 0.0198g
Mass of N = 0.039g
Divide each by atomic mass
C = 0.1015/12 = 0.008458
H = 0.0198/1= 0.0198
N = 0.039/14 = 0.002786
divide by smallest:
C = 3.036
H = 7.10
N = 1
The closest I can make this is C3H5NH2 = cyclopropylamine.
But there is something not quite correct with this - the masses of the 3 elements making up the compound do not add up to the mass of the original compound. I am particularly suspicious of the statement "and 0.039g of N2"are you certain that this is correct? .
Mass of H in 0.178g H2O = 2/18*0.178 = 0.0198g
Mass of N = 0.039g
Divide each by atomic mass
C = 0.1015/12 = 0.008458
H = 0.0198/1= 0.0198
N = 0.039/14 = 0.002786
divide by smallest:
C = 3.036
H = 7.10
N = 1
The closest I can make this is C3H5NH2 = cyclopropylamine.
But there is something not quite correct with this - the masses of the 3 elements making up the compound do not add up to the mass of the original compound. I am particularly suspicious of the statement "and 0.039g of N2"are you certain that this is correct? .
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You might want to check the numbers. If you have a 0.2511g compound, you can't have 0.372g of CO2 as part of the compound. Part of the compound can't exceed the mass of the entire compound. Is it 2.511g by chance?