Question

A certain anesthetic contains 64.9 percent C, 13.5 percent H, and 21.6 percent O by mass. At 144°C and 755 mmHg, 2.60 L of the gaseous compound weighs 5.93 g. What is the molecular formula of the compound?

Answer 1

Let's first of all calculate the molar ratio of atoms, assuming that we have % = g (if we take 100 g of a compound):

n (C) = 64.9 g / 12.01 g/mol = 5.40
n (H) = 13.5 g / 1.01 g/mol = 13.37
n (O) = 21.6 g / 16.00 g/mol = 1.35

n (C) : n (H) : n (O) = 5.40 : 13.37 : 1.35, divide everything by the smallest number (1.35)
n (C) : n (H) : n (O) = 4.00 : 9.90 : 1, i. e. 4 : 10 : 1, so the empirical formula would be (C4H10O)n.

Now, let's use pV = nRT formula. Let's rearrange it to get the molecular mass of our compound:

pV = (m/M) RT => M = m / (pV / RT) = mRT / pV = 5.93 g * 8.314 J K^-1 mol^-1 * 393 K / (101325 Pa x (755 mm Hg / 760 mm Hg) * 0.0026 m^3) = 78.55 g/mol, which yields n = 1.

The molecular formula is C4H10O.