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# 2.511×10-4 mol of an unidentified gaseous substance effuses through a tiny hole in 88.4 s. Under identical co?

2.511×10-4 mol of an unidentified gaseous substance effuses through a tiny hole in 88.4 s.

Under identical conditions, 1.593×10-4 mol of argon gas takes 88.5 s to effuse.

1. What is the molar mass of the unidentified substance (in g/mol)?

The percentage C by mass in the above unidentified substance is 74.9%. It may also contain H and/or O, but no other element.

2. What is the molecular formula of the substance?

(In typing this formula use the convention of C first, H second and O last. For example, C6H8O2 would be C6H8O2.)

3. Under identical conditions, how many moles of cyclopropane (C3H6) gas would effuse in 90.1 s?

mol

Help Please !!!

### 1 Answer

- BobbyLv 79 years agoFavorite Answer
Graham's law says

Rate 1 / Rate 2 = square root ( M2 / M1 )

if 1 is our compound rate1 = 2.511 x 10 ^-4 / 88.4 = 2.8405 x 10^-06 moles / second

rate 2 = 1.593 x 10 ^ -4 / 88.5 = 1.80x 10^ -06 moles / second

2.8405 x 10 ^ -06 /1.80 x 10^ -06 = square root ( 40 / M1)

1.58 ^2 = 40 / M1

M1 = 40 / 2.49

molecular mass of the unidentified gas = 16 g / mole

carbon = 74.9 % of 16 = 12 g or 1 mole

H = 25.1 % of 16 = 4 g and 4 moles of H

molecular formula is CH4 methane

rate 1 is C3H6 rate 2 is argon

rate 1 / 1.80x 10^ -06 = square root (40 / 42)

rate 1 / 1,8 x 10 ^-6 = 0.9759

rate 1 = 2.77 x 10^-06 moles / sec

or 2.77 x 10 ^-06 x 90.1 sec = 2.50 x 10 ^ -04 moles in 90.1 sec