Pon heating, a 5.41 g sample of a compound decomposes into 2.37 g n 22 and 3.04 g h 22 o. if the molar mass of the compound is 64.06 g/mol, what is the chemical formula of the compound

Initially note that 2.37 g + 3.04 g = 5.41 g, which means that the sample was a compound of only N, H and O.

Convert the masses of N2 and H2O into number of moles by using 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 formula

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 empirical formula: 14.0 g/mol + 18 g/mol = 32 g/mol

5) Number of times that the empirical formula is contained in the molecular formula: 64 g/mol / 32 g /mol = 2

6) Molecular formula = 2 * Empirical formula = N2H4O2

N2H4O2

Ibelieve your answer is true