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Chemistry 11 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.

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