Question

Question #87de9

Answer 1

`color(red)(C_2H_4O)`

Explanation:

Atomic mass of `H=1" g/mol"`
Atomic mass of `C=12" g/mol"`
Atomic mass of `O=16" g/mol"`
Molar mass of `CO_2=44" g/mol"`
Molar mass of `H_2O=18" g/mol"`

Mass of C in 7.59g `CO_2=(7.59g)/(44"g/mol")xx12"g/mol"=2.07g`

Mass of H in 3.11g `H_2O=(3.11g)/(18"g/mol")xx2"g/mol"=0.345g`

These two amounts of C and H have been obtained from 3.8g compound containing C ,H and O.

So the remaining (3.80-2.07-0.35)g=1.38g is the mass of Oxygen present in 3.8 g compound.

Hence dividing these masses of elements with respective atomic masses of elements and taking their ratio we get the ratio of number of atoms of C, H and O in the molecule of the compound as below

`C:H:O=2.07/12:0.345/1:1.38/16`
`=0.1725:0.345:0.08625`

`=(0.1725/0.08625):(0.345/0.08625):(0.08625/0.08625)`
`=2:4:1`

Hence empirical formula of the compound is `color(red)(C_2H_4O)`