Texts a 165010 4 mol of an unidentified gaseous substance...

Texts: A) 1.650×10^-4 mol of an unidentified gaseous substance effuses through a tiny hole in 91.8 s. Under identical conditions, 1.528×10^-4 mol of argon gas takes 84.9 s to effuse. What is the molar mass of the unidentified substance (in g/mol)? B) The percentage C by mass in the above unidentified substance is 89.9%. It may also contain H and/or O, but no other element. 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.) C) Under identical conditions, how many moles of cyclopropane (C3H6) gas would effuse in 78.2 s?

Transcript

00:09 I'm confident i can do this one, but let's give it a try here.

00:12 I'm told that i have 1 .115 times 10 to the minus 4th moles of an unknown gas, and it diffuses in 79 .9 seconds.

00:31 I'm also told that under identical conditions, i have 1 .48 -478 times 10 to the minus 4th moles of argon effuses in 81 .2 seconds.

00:50 Since this took longer to effuse, takes longer.

00:59 Thus, it has a greater molar mass.

01:07 So in my formula at the top, i'm going to have this one be gas b.

01:13 And my molar mass for argon is 39 .95 grams per mole.

01:25 Okay, so let's go ahead and figure this out.

01:28 I'm going to take my rate of a.

01:30 This is going to be gas a.

01:33 So i will take 1 .115 times 10 to the minus 4th divided by 79 .95 seconds, moles, over 1 .478 times 10 to the minus 4th moles per 81 .2 seconds.

01:57 I'll abbreviate my s like this.

01:59 And then i'm going to take my molar mass of 39 .95, and whatever i want to call this, i'll go grams per moles over x.

02:09 Very good.

02:10 So first of all, let me get my, i'm going to solve for my equation on the left, 1 .115 times 10 to the minus 4th, divided by 79 .95, enter.

02:24 Enter.

02:26 And then i'm going to divide this by close 1 .478 times 10 to the minus 4, divided by 81 .2, close parentheses, enter.

02:39 And i get 0 .76619 equals the square root of 39 .95 over x.

02:51 So if i take, if i square the square root of 39 .95 over x.

02:55 This whole thing, i get 0 .08705 equals 39 .95 over x.

03:11 So now i can take my x.

03:13 I'll take 39 .95, and i will divide this by 0 .58705.

03:19 And i will get 68 .05.