Question

A certain anesthetic contains 64.9 percent C, 13.5 percent H, and 21.6 percent O by mass. At 110°C and 721 mmHg, 3.00 L of the gaseous compound weighs 6.66 g. What is the molecular formula of the compound?

Answer 1

1st calculate the molar mass
P*V = n*R*T
P*V = (mass/MM)*R*T
(721/760) atm*3.00 L = (6.66/MM)*0.0821 atmL/mol-k * (110+273)k
MM = 73.6 g/mol

Now lets find the empirical formula:
Let total mass be 100 gm.
C : 64.9 g
H : 13.5 g
O: 21.6 g

Divide by molar mass to get number of moles of each:
C: 64.9/12 = 5.41
H: 13.5/1 = 13.5
O:21.6/16 = 1.35

Divide by smallest to get simplest whole number ratio:
C: 5.41/1.35 = 4
H: 13.5/1.35 = 10
O:1.35/1.35 = 1

So empirical formula is:
C4H10O

emprical formula mass = 4*12 + 10 + 16 = 74 g

same is the molar mass approximately

hence molecular formula is
C4H10O