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Upon heating, a 5.41 g sample of a compound decomposes into 2.37 g n 2 ​ and 3.04 g h 2 ​ o. if the molar mass of the compound is 64.06 g/mol, what is the chemical formula of the compound?

Respuesta :

Dejanras
Answer is: empirical formula is H₄N₂O₂.
m(unknown compound) = 8,5 g.
m(N₂) = 2,37 g.m(H₂O) = 3,04 g.
n(N₂) = m(N₂) ÷ M(N₂).
n(N₂) = 2,37 g ÷ 28 g/mol.
n(N₂) = 0,0846 mol.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 3,04 g ÷ 18 g/mol.
n(H₂O) = 0,168 mol.
n(H₂O) : n(N₂) = 0,168 mol : 0,0846 mol.
n(H₂O) : n(N₂) = 2 : 1.
Compound has four hydrogen, two oxygen and two oxygen.
W0lf93
At first note 2.37 g + 3.04 g = 5.41 g, which implies that the example was a compound of just N, H and O. Change over the majority of N2 and H2O into number of moles by utilizing the molar masses of each.  
1) N 
 moles = mass of N2/molar mass of N2 = 2.37 g/( 14.0 g/mol) = 0.169 mol of
N 
 2) H2O 
 moles = mass of H2)/molar mass of H2O = 3.04 g/(18.0 g/mol) = 0.169 mol 
 0.169 moles of H2O => 2 * 0.169 moles of H and 0.169 moles of O 
 3) Emipirical equation 
 N: 0.169/0.169 = 1 H: 2* 0.169/0.169 = 2 O: 0.169/0.169 = 1 
 => NH2O 
 4) Molar mass of the observational equation: 14.0 g/mol + 18 g/mol = 32
g/mol 
 5) Number of times that the observational equation is contained in the sub-atomic recipe: 64 g/mol/32 g/mol = 2 
 6) Molecular equation = 2 * Empirical formula = N2H4O2 
 Finally The chemical formula of the compound is N2H4O2

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