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?

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.

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