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The empirical formula is the simplest formula of a chemical compound.

To find the empirical formula, we take the following steps;

1) O - 88.10/16,      H - 11.190/1

  O - 5.5,               H - 11.19

  O - 5.5/5.5,        H - 11.19/5.5

  O -  1,                 H - 2

Empirical formula = OH2

2) C - 41.368/12  H - 8.101/1,   N - 32.162/14,   O - 18.369/16

   C - 3,               H - 8,           N - 2,                  O - 1

   C - 3/1,            H - 8/1          N - 2/1                 O - 1/1

    C - 3,             H - 8,           N - 2,                   O - 1

Empirical formula = C3H8N2O

To obtain the molecular formula where n = number of atoms of each element;

Molecular weight = 174.204 g/mol

[ 3(12) + 8(1) + 2(14) + 16]n = 174

n= 174/88

n = 2 (to the nearest whole number)

Hence, we have;

[C3H8N2O]2

The molecular formula is C6H16N4O2

3)  C - 19.999/12,  H - 6.713/1,   N - 46.646/14,   O - 26.641/16

    C - 2,                H - 7,            N -  3,                 O - 2

    C - 2/2,            H - 7/2,         N -   3/2,             O - 2/2

    C - 1,                H - 4,            N -  2,                  O - 1

Empirical formula - CH4N2O

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