Question

A 1.87 g sample of a hydrocarbon gas occupies a volume of 672.0 mL at 35°C and 741 mmHg. What is the molecular formula of the gas in this sample? MC= 12.01 g/mol, MH = 1.01 g/mol, R = 0.0821 L∙atm/mol∙K

Answer 1

Answer:

C₅H₁₂

Explanation:

Using general law of gases we can determine the moles of the sample and, as we know the mass the molecular formula is the ratio between mass in grams per mole of susbtance. With this, we can find the molecular formula, thus:

Moles:

PV = nRT -General gas law-

Where P is pressure: 741mmHg * (1atm / 760mmHg) = 0.975atm

V is volume in liters: 672.0mL = 0.672L

n are moles -Our incognite-

R is gas constant: 0.0821atmL/molK

And T is absolute temperature: 273.15 + 35°C = 308.15K

Replacing:

n = PV/RT

n = 0.975atm*0.672L / 0.0821atmL/molK*308.15K

n = 0.0259moles

And molar mass is:

1.87g / 0.0259moles =

72.2g/mol.

In 72.2g/mol we can have just 5 moles of C:

5 moles of C will weigh: 12.01g/mol * 5 = 60.05g/mol

And the other mass must come from hydrogen:

72.2g/mol - 60.05g/mol = 12.15g/mol / 1.01g/mol = 12 moles of hydrogen

And molecular formula is:

C₅H₁₂