Question

Butane, C₄H_10, is used as a fuel for barbecues and as an aerosol propellant. If you have 108mL of butane at 715 mmHg and 25^oC, what is the mass, in grams, of butane? Show all work!

Answer 1

given:

P = 715 mmHg

= (715/760) atm

P = 0.9407 atm

T = 25 + 273 K = 298 K

V = 108 mL = 0.108 L , R = 0.0821

and

n = PV/RT
= 0.9407*0.108/0.0821*298

n = 0.004155 moles

now,

mass of butane = molar mass of butane*moles

= 58*0.004155

mass of butane = 0.241 gm ............. answer