Heating a 6.862 g sample of an ore containing a metal sulfide in excess oxygen produces 1053 mL....?

Question

of SO2 gas measured at 66 degrees C and 739 mmHg. Calculate the percentage by mass of sulfur in the ore. Help please!

Answer 1

find moles of SO2:

PV= nRT

(739Torr)(1.053L) = n (62.36Torr-Litres/mol-K)(339K)

n= 0.03681 moles of SO3

0.03681 moles of SO3  =  0.03681 moles of S

find grams:

0.03681 moles  @ 32.066 g S / mole = 1.180 grams of sulfur

Find %:

1.180 g S / 6.862 g sample times 100 = 17.20 %

your answer is 17.20 % sulfur

( but since the pressure only had 3 sigfigs , you might consider your answer to be 17.7%)