Answer:
Explanation:
mass of carbon in 10.27 g of CO₂ = 12 x 10.27 / 44 = 2.80 g
mass of hydrogen ( H ) in 3.363 g of H₂O = 2 x 3.363 / 18
= .373 g
These masses would have come from the sample of 6.165 g .
Rest of 6.165 g of sample is oxygen .
So oxygen in the sample = 6.165 - ( 2.8 + .373 ) = 2.992 g
Ratio of C , H , O in the sample
2.8 : .373 : 2.992
C: H : O : : 2.8 : .373 : 2.992
Ratio of moles
C: H : O : : 2.8/12 : .373/1 : 2.992 / 16
C: H : O : : .2333 : .373 : .187
C: H : O : : .2333/.187 : .373/.187 : .187/.187
C: H : O : : 1.247 : 1.99 : 1
C: H : O : : 5 : 8 : 4 ( after multiplying by 4 )
Hence empirical formula
C₅H₈O₄
Molecular formula ( C₅H₈O₄ )n
n ( 5 x 12 + 8 x 1 + 4 x 16 ) = 132
n x ( 60 + 8 + 64 ) = 132
n = 1
Molecular formula = C₅H₈O₄.
