Ask your own question, for FREE!
Chemistry 7 Online
OpenStudy (anonymous):

a 0.04356 g sample of gas occupies 10.0 mL at 291.5K and 1.10 atm upon further analysis the compound is found to be 25.305 percent C and 74.695 percent C what is the molecular formula

OpenStudy (wolfe8):

First, use PV=nRT to find the number of moles. But you put both as percentages of C. Is that right?

OpenStudy (anonymous):

what do you mean by both as percentages of C?

OpenStudy (wolfe8):

"25.305 percent C and 74.695 percent C" I meant that.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!