Question

A 0.02887 g sample of gas occupies 10.0 mL at 288.0 K and 1.10 atm. Upon further analysis, the compound is found to be 38.734 % C and 61.266 % F . What is the molecular formula of the compound?

Answer 1

Answer: The molecular formula of the compound is
`C_2F_2`

Step-by-step explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 38.734 g

Mass of F = 61.266 g

Step 1 : convert given masses into moles.

Moles of C =

Moles of F =
`\frac{\text{ given mass of F}}{\text{ molar mass of F}}= (61.266g)/(19g/mole)=3.2245moles`

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =
`(3.2278)/(3.2245)=1`

For F =
`(3.2245)/(3.2245)=1`

The ratio of C : F = 1:1

Hence the empirical formula is
`CF`

The empirical weight of
`CF` = 1(12)+1(19)= 31 g.

Using ideal gas equation :

`PV=nRT`

where,

n = number of moles of gas = ?

P = pressure of the gas = 1.10 atm

T = temperature of the gas = 288.0 K

R = gas constant = 0.0821 L.atm/mole.K

V = volume of gas = 10.0 mL =0.01 L

`1.10* 0.01=n* 0.0821* 288.0`

`n=4.65* 10^(-4)`

`n=\frac{\text {given mass}}{\text {Molar mass}}`

`4.65* 10^(-4)=\frac{0.02887}{\text {Molar mass}}`

`{\text {Molar mass}}=62.0`

Now we have to calculate the molecular formula.

`n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=(62.0)/(31.0)=2`

The molecular formula will be=
`2* CF=C_2F_2`