Question

• A compound used in glue has a composition of 54.5% C, 9.1 % H and 36.4% O. • At 100 °C and 1.00 atm, the compound is a gas and 0.345 g occupies a volume of 120.0 mL. • What is the molecular formula of the compound?

Answer 1

Given:

P = 1.0 atm

V = 120.0 mL

= (120.0/1000) L

= 0.12 L

T = 100.0 oC

= (100.0+273) K

= 373 K

find number of moles using:

P * V = n*R*T

1 atm * 0.12 L = n * 0.08206 atm.L/mol.K * 373 K

n = 3.92*10^-3 mol

mass(solute)= 0.345 g

use:

number of mol = mass / molar mass

3.92*10^-3 mol = (0.345 g)/molar mass

molar mass = 88 g/mol

we have mass of each elements as:

C: 54.5 g

H: 9.1 g

O: 36.4 g

Divide by molar mass to get number of moles of each:

C: 54.5/12.01 = 4.5379

H: 9.1/1.008 = 9.0278

O: 36.4/16.0 = 2.275

Divide by smallest to get simplest whole number ratio:

C: 4.5379/2.275 = 2

H: 9.0278/2.275 = 4

O: 2.275/2.275 = 1

  

So empirical formula is:C2H4O

Molar mass of C2H4O,

MM = 2*MM(C) + 4*MM(H) + 1*MM(O)

= 2*12.01 + 4*1.008 + 1*16.0

= 44.052 g/mol

Now we have:

Molar mass = 88.0 g/mol

Empirical formula mass = 44.052 g/mol

Multiplying factor = molar mass / empirical formula mass

= 88.0/44.052

= 2

So molecular formula is:C4H8O2

Answer: C4H8O2