Analysis of a volatile liquid shows that it contains 62.04% carbon, 10.41% hydrogen, and 27.54% oxygen by mass. At 150.°C and 1.00 atm, 500. mL of the vapor has a mass of 0.8365 g. What is the molecular formula of the compound?

Molecular formula of the compound is C₃H₆O

Explanation:

Firstly let's determine the moles of the vapor (gas) with the Ideal Gases Law, so it can give us the molar mass with the mass and afterwards we can work with the percent composition.

Pressure. Volume = moles. Ideal Constant Gases. Temperature in K

Temperature in K = T°C + 273 → 150°C + 273 = 423K

P. V = n. R. T

n = (P. V) / (R. T)

n = 1 atm. 0.5L / (0.082. 423K)

n = 0.0144 moles

These are the moles for 0.8365 g, so let's determine the molar mass

Molar mass (g/mol) = 0.8365 g / 0.0144 mol → 58.02 g/mol

Percent composition means:

100 g of compound have 62.04 g of C

100 g of compound have 10.41 g of H

100 g of compound have 27.54 g of O

Let's make the rule of three:

100 g of compound have __ 62.04 g of C __ 10.41 g of H __ 27.54 g of O

The 58.02 g of compound must have:

(58.02 g. 62.04 g) / 100 g = 36 g of C

(58.02 g. 10.41 g) / 100 g = 6 g of H

(58.02 g. 27.54 g) / 100 g = 16 g of O

Let's find out the moles of each

Mass / Molar mass

36 g / 12 g/mol = 3 C

6 g / 1 g/mol = 6 H

16g / 16 g/mol = 1 O