Question

Complete combustion of a 17.12mg sample of xylene In oxygen yielded 56.77mg

Answer 1

Xylene moles =\frac{17.12}{106.16×1000}=0.00016moles=

106.16×1000

17.12

=0.00016moles

Moles of CO_2 =\frac{56.77}{44.01×1000}=0.0013CO

2

=

44.01×1000

56.77

=0.0013

Moles of H_2O= =\frac{14.53}{18.02×1000}=0.0008H

2

O==

18.02×1000

14.53

=0.0008

Moles ratios

\frac{0.0013}{0.0008}=1.625

0.0008

0.0013

=1.625

\frac{0.0008}{0.0008}=1

0.0008

0.0008

=1

Hence molecular fomula

The empirical formula is C 4H 5.

The molecular formula C8H10