Question

A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products have cooled to 20.0 °C at 1 bar, there are 4.41 L of CO2 and 3.26 mL of H2O . The density of water at 20.0 °C is 0.998 g/mL.

Answer 1

Step-by-step explanation:

First, calculate the moles of
`CO_(2)` using ideal gas equation as follows.

PV = nRT

or, n =
`(PV)/(RT)`

=
`(1 atm * 4.41 ml)/(0.0821 Latm/mol K * 293 K)` (as 1 bar = 1 atm (approx))

= 0.183 mol

As, Density =
`(mass)/(volume)`

Hence, mass of water will be as follows.

Density =
`(mass)/(volume)`

0.998 g/ml =
`(mass)/(3.26 ml)`

mass = 3.25 g

Similarly, calculate the moles of water as follows.

No. of moles =
`\frac{mass}{\text{molar mass}}`

=
`(3.25 g)/(18.02 g/mol)`

= 0.180 mol

Moles of hydrogen =
`0.180 * 2` = 0.36 mol

Now, mass of carbon will be as follows.

No. of moles =
`\frac{mass}{\text{molar mass}}`

0.183 mol =
`(mass)/(12 g/mol)`

= 2.19 g

Therefore, mass of oxygen will be as follows.

Mass of O = mass of sample - (mass of C + mass of H)

= 3.50 g - (2.19 g + 0.36 g)

= 0.95 g

Therefore, moles of oxygen will be as follows.

No. of moles =
`\frac{mass}{\text{molar mass}}`

=
`(0.95 g)/(16 g/mol)`

= 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

C H O

No. of moles: 0.183 0.36 0.059

On dividing: 3.1 6.1 1

Therefore, empirical formula of the given compound is
`C_(3)H_(6)O`.

Thus, we can conclude that empirical formula of the given compound is
`C_(3)H_(6)O`.