Given Empirical formula =CH2O Mole-ratio of C, H and O=1:2:1 Also 0.0833 moles contains 1g of hydrogen ∴0.0833×n=1g of hydrogen where n= number of moles of hydrogen (a) For C5H10O5(n=10) ∴0.0833×10=0.833≠1 (b) For C3H4O3(n=4) ∴0.0833×4=0.333≠1 (c) For C12H22O11(n=22) ∴0.0833×22=1.8326≠1 (d) For C6H12O6(n=12) ∴0.0833×12=0.9996≈1=1