A sample of O2 of volume 4.56 L was collected over water at 25c and a total pressure of 1.00 atm. The partial pressure of water is 34.5 Torr. How many moles of O2molescules were collected?

0.178 moles of oxygen gas molecules were collected.

Explanation:

Pressure at which oxygen gas collected = p= 1.00 atm

Vapor pressure of the water

Pressure of the oxygen gas = P

p = p' + P

P = 1.00 atm -  0.0454 atm = 0.9546 atm

Volume of the oxygen gas = V = 4.56 L

Moles of oxygen gas = n

Temperature at which gas collected = T =

(ideal gas)

0.178 moles of oxygen gas molecules were collected.

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.