Answer to Question #279354 in General Chemistry for jhon

Question

An unknown compound was discovered in a laboratory by a chemist. To be able to

find out the molecular formula of this compound, 1.13 g of the unknown compound was

placed into a combustion chamber, and end of the analysis 1.252 g CO2 and 0.3404 g

H2O were obtained. Later, the chemist analyzed the molar mass of the compound as 240

g/mol. What is the molecular formula of the unknown compound?

Answer 1

Assuming the compound contains C, H and O

And taking molar masses as:

$CO_2=44.011\\H_2O=18.016\\C=12.011\\H=1.008\\O=15.999$

Mass of Carbon in the compound:

$= \frac{1.252}{44.011}×12.011=0.34168$

Mass of Hydrogen in the compound:

$= \frac{0.3404}{18.016}×2×1.008=0.03809g$

Mass of Oxygen in the compound:

$1.13-(0.34168+0.03809)\\=0.75023g$

Moles of $C=\frac{0.34168}{12.011}=0.02845$

Moles of $H=\frac{0.03809}{1.008}=0.03779$

Moles of O$=\frac{0.75023}{15.999}=0.04689$

Simplest ratio:

$\frac{0.02845}{0.02845}:\frac{0.03779}{0.02845}:\frac{0.04689}{0.02845}$

$=1:1.3283:1.64815\\=3(1:1.3283:1.64815)\\\approx3:4:5$

Empirical formula $=C_3H_4O_5$

Mass =$120g$

Molecular formula $\frac{240}{120} =2$ times Empirical formula

Molecular formula is:

$C_6H_8O_{10}$