What is the empirical formula of a compound with 1.20% H, 42.00% Cl, and 56.80% O?

Take a hypothetical sample of exactly 100. grams of the compound:
(1.20 g H) / (1.007947 g H/mol) = 1.19054 mol H

(42.00 g Cl) / (35.4532 g Cl/mol) = 1.184660 mol Cl

(56.80 g O) / (15.99943 g O/mol) = 3.550126 mol O

Divide by the smallest number of moles:
(1.19054 mol H) / 1.184660 mol = 1.00496
(1.184660 mol Cl) / 1.184660 mol = 1.0000
(3.550126 mol O) / 1.184660 mol = 2.9967

Round to the nearest whole numbers to find the empirical formula:
HClO3

Mole ratio H : Cl : O
= (1.20/1.008) : (42.00/35.45) : (56.80/16.00)
= 1.19 : 1.18 : 3.55
= (1.19/1.18) : (1.18/1.18) : (3.55/1.18)
= 1 : 1 : 3

Empirical formula = HClO₃

For simplicity, assume you have 100g
1.2 g H (1 mole/xx g) = ___ moles
repeat for each
Only you have moles, find multipliers to get formula