00:01
Cl and i'll react with excess cl.
00:08
And it gives 0 .233 grams of hcl and 0 .402 grams of h2 .0.
00:23
Find the empirical formula, which will be cx, clx, o, y.
00:40
Okay.
00:41
Okay.
00:43
So let's first figure out our grams of hcl by doing the following.
00:52
I like to just figure out my grams of each.
00:55
So hcl is 36 .0 .36 .46.
01:10
And my h2 contributes 1 .008 grams.
01:18
And that'll give me a mass.
01:19
Oh, i want to do cl.
01:21
Sorry about that.
01:25
I'm used to doing combustion analysis, 35 .45, and that will equal 0 .23, 0 .23, times 35 .45 divided by 36 .46, is 0 .260 grams of cl...