1) Assume 100 g of the compound is present. This changes the percents to grams:
C ⇒ 40 g
H ⇒ 6.67g
O ⇒ 53.33 g
2) Convert the masses to moles:
C ⇒ 40 g g / 12 = 3.33
H ⇒ 6.67g / 1 = 6.67
O ⇒ 53.33g / 16 = 3.33
3) Divide by the lowest, seeking the smallest whole-number ratio:
C ⇒ 3.33 / 3.33 = 1
H ⇒ 6.67 / 3.33 = 2.003
O ⇒ 3.33 / 3.33 = 1
4) Write the empirical formula:
CH2O
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