# Question 87de9

Oct 1, 2016

$\textcolor{red}{{C}_{2} {H}_{4} O}$

#### Explanation:

Atomic mass of $H = 1 \text{ g/mol}$
Atomic mass of $C = 12 \text{ g/mol}$
Atomic mass of $O = 16 \text{ g/mol}$
Molar mass of $C {O}_{2} = 44 \text{ g/mol}$
Molar mass of ${H}_{2} O = 18 \text{ g/mol}$

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

$C : H : O = \frac{2.07}{12} : \frac{0.345}{1} : \frac{1.38}{16}$
$= 0.1725 : 0.345 : 0.08625$

$= \left(\frac{0.1725}{0.08625}\right) : \left(\frac{0.345}{0.08625}\right) : \left(\frac{0.08625}{0.08625}\right)$
$= 2 : 4 : 1$

Hence empirical formula of the compound is $\textcolor{red}{{C}_{2} {H}_{4} O}$