Answer:
The answer to your question is empirical formula C₂.₆H₂.₆O₁
molecular formula C₈H₈O₃
Explanation:
Data
Carbon 63.15%
Hydrogen 5.30%
Oxygen 31.55 %
Molar mass = 152.15 g
Process
1.- Express the percents as grams
Carbon 63.15g
Hydrogen 5.30g
Oxygen 31.55 g
2.- Calculate the moles of these elements
Carbon 12 g ---------------- 1 mol
63.15 g ---------- x
x = (63.15 x 1) / 12
x = 5.2625 moles
Hydrogen 1 g ---------------- 1 mol
5.3 g --------------- x
x = (5.3 x 1)/1
x = 5.3 moles
Oxygen 16 g -------------- 1 mol
31.55 g ---------- x
x = (31.55 x 1)/16
x = 1.97 moles
2.- Divide by the lowest number of moles
Carbon 5.2625/1.97 = 2.6
Hydrogen 5.3/1.97 = 2.6
Oxygen 1.97/197 = 1
3.- Write the empirical formula
C₂.₆H₂.₆O₁
4.- Write the molecular formula
Molar mass of empirical formula = (12 x8) + (1 x 8) + (16 x 3) = 152
152.15/149.8 = 3
Multiply the empirical formula by 3
C₈H₈O₃
