Question

A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of CO2 and 0.01961 mol of H2O. The empirical formula of the compound was found to be C3H6O2

Answer 1

Answer:

The essence including its given problem is outlined in the following segment on the context..

Explanation:

The given values are:

Moles of CO₂,

x = 0.01962

Moles of water,

$\frac{y}{2} =0.01961$

$y=2\times 0.01961$

  $=0.03922$

Compound's mass,

= 0.4647 g

Let the compound's formula will be:

$C_{x}H_{y}O_{z}$

Combustion's general equation will be:

⇒  $C_{x}H_{y}O_{z}+x+(\frac{y}{4}-\frac{z}{2}) O_{2}=xCO_{2}+\frac{y}{2H_{2}O}$

On putting the estimated values, we get

⇒  $12\times x=1\times y+16\times z=0.4647$

⇒  $12\times 0.01962+1\times 0.03922+16\times z=0.4647$

⇒  $0.27466+16z=0.4647$

⇒                     $z=0.01187$

Now,

x : y : z = $0.01962:0.03922:0.01187$

           = $\frac{0.01962}{0.0118}:\frac{0.03922}{0.0188}:\frac{0.0188}{0.0188}$

           = $1.6:3.3:1.0$

           = $3:6:2$

So that the empirical formula seems to be "C₃H₆O₂".