Answer in General Chemistry for jhon #279354

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?

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}u00d712.011=0.34168”

Mass of Hydrogen in the compound:

“= frac{0.3404}{18.016}u00d72u00d71.008=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}”

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