A certain anesthetic contains 64.9% C, 13.5% H, and 21.6% O by mass. At 120°C an

Question

A certain anesthetic contains 64.9% C, 13.5% H, and 21.6% O by mass. At 120°C and 0.987 atm, 1.00 L of the gaseous compound weighs 2.30 g. What is the molecular formula of the compound?
A certain anesthetic contains 64.9% C, 13.5% H, and 21.6% O by mass. At 120°C and 0.987 atm, 1.00 L of the gaseous compound weighs 2.30 g. What is the molecular formula of the compound?
A certain anesthetic contains 64.9% C, 13.5% H, and 21.6% O by mass. At 120°C and 0.987 atm, 1.00 L of the gaseous compound weighs 2.30 g. What is the molecular formula of the compound?

Explanation / Answer

Frist we will calculate the emperical formula

so emperical formula is (C4H10O)k now we need to calculate k .

let n mole of gas of temperature T = (120+273 ) = 393 K of volume 1 liter have pressure P = .987 atm

PV = nRT

n = (.987 * 1 ) / ( 393 * .082 ) = .0306

weight of .0306 mole gas is 2.30

so molecular weight = 2.30 / .0306 = 75 gm

now moleweght of emperical formula = (C4H10O)k = 74 k = 75

so k = 1

molecular formula is (C4H10O)1 = C4H10O

parcentage 64.9 13.5 21.6 molecular weight of element 12 1 16 weight ratio 5.4 13.5 1.3 divide all by 1.3 smallest number 4 10 1

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