Determine the molecular formula of a compound with an empirical formula of nh2 and a formula mass of 32.06 amu

N2H4

Explanation: Empirical Formula, EF= NH2

Mass of the EF = mass of Nitrogen + mass of Hydrogen

EF = 14.007 + 1.008

= 15.015

Then, divide the given formula mass by the EF. That is,

32.06/15.015 = 2.135 = 2 (to the nearest whole number) (1)

To get the molecular formula, multiple the empirical formula with the number from the equ (1) above.

Molecular formula = 2[NH2] = N2H4

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g