Question

Butane (C4H10) is used as a fuel where natural gas is not available. How many grams of butane will fill a 3.50-liter container at 35.6 °C and 758 torr? I've put the values through n = PV/RT to get n = 0.0138 and then converted to grams to get 0.77423g but it's not the right answer on my practice midterm. Any help is gratefully appreciated!

Answer 1

We use the formula PV = nRT. P = 758 torr = 0.997 atm. V = 3.50 L. T = 35.6 C = 308.15 K. R = 0.0821. Rearranging the equation gives up n = PV/Rt and we get .0138 moles of butane. Mass of 0.0138 moles of butane = .0138 x 58.12 = 8.02g.