Answer:
CH₃O
Step-by-step explanation:
1. Calculate the mass of each element
Mass of C = 1.45 mg CO₂ × (12.01 mg C/44.01 mg CO₂) = 0.3957 mg C
Mass of H = 0.89 mg H₂O × (2.016 mg H/18.02 mg H₂O) = 0.0996 mg H
Mass of O = Mass of compound - Mass of C - Mass of H = (1.55 – 0.3957 – 0.0996) mg = 1.055 mg
2. Calculate the moles of each element
Moles of C = 0.3957 mg C × 1mmol C/12.01 mg C = 0.03295 mol C
Moles of H = 0.0996 mg H × 1 mmol H/1.008 mg H = 0.0988 mol H
Moles of O = 1.055 mg O × 1 mmol O/ 16.00 mg O = 0.06592 mol O
3. Calculate the molar ratios
Divide all moles by the smallest number of moles.
C: 0.032 95/0.032 95 = 1
H: 0.0988/0.032 95 = 2.998
O: 0.065 92/0.032 95 = 2.001
4. Round the ratios to the nearest integer
C:H:O = 1:3:2
5. Write the empirical formula
The empirical formula is CH₃O.
