L.39

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FionaGriffin-1K
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Joined: Fri Sep 29, 2017 7:04 am

L.39

Postby FionaGriffin-1K » Wed Oct 04, 2017 4:46 pm

Hi everyone!
I'm trying to solve L.29 but having trouble understanding how the empirical formula comes into play.
"A 1.50 g sample of metallic tin was placed in a 26.45 g crucible and heated until all the tin had reacted with the oxygen I air to form an oxide. The crucible and product together were found to weigh 28.35g. (a) What is the empirical formula of the oxide? (b) Write the name of the oxide.

I added 1.50g to 26.45g to get 27.95, then subtracted that from 28.35g. From that I had .4g, which I used for my oxygen and converted that to moles, but then I get confused with where to go from there.

Clara Hu 1G
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Re: L.39

Postby Clara Hu 1G » Wed Oct 04, 2017 5:01 pm

After converting the amount of oxygen to moles (0.025mol), also convert the 1.50g of tin to moles, which is 0.0126 mol. Then because you have the amount of oxygen and tin in moles you can get the empirical formula because you know the ratio of tin to oxygen. You have Sn0.0126O0.025, and because 0.0126 is the smaller number divide both the subscripts by 0.0126 to get SnO2, which is the empirical formula.

Anna Goldberg 2I
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Re: L.39

Postby Anna Goldberg 2I » Wed Oct 04, 2017 5:06 pm

I first subtracted 26.45g (mass of crucible) from 28.35(the mass of the crucible and product), to get 1.9 g, which is the mass of the tin oxide. I then subtracted 1.5g (mass of the metallic tin) from 1.9g (the mass of the tin oxide) to get .4g, which is the mass of the oxygen in the air that the tin reacted with.
To obtain the empirical formula, I first calculated the amount of moles for both the metallic tin and the oxygen. For the metallic tin, I did grams divided by molar mass, so 1.5g/118.71(g/mol) to get 0.0124 moles of tin. For oxygen, using grams divided by molar mass, I did 0.4g/16.00*(g/mol) to get 0.025 moles of Oxygen.
I then compared the moles of tin with the moles of oxygen. If you multiply the moles of tin (0.0124) by two, you get 0.025, which is the number of moles of oxygen. Thus, the mole ratio of tin to oxygen is 1:2. The empirical formula of the oxide product is thus SnO2.
The name of the oxide is tin(IV)oxide

Humza_Khan_2J
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Joined: Thu Jul 13, 2017 3:00 am

Re: L.39

Postby Humza_Khan_2J » Wed Oct 04, 2017 5:33 pm

To provide a more conceptual view on the topic, we can see that the oxygen used in the reaction accounts for the added mass(.4 g) in the crucible along with the metallic tin. You can then find how many moles of oxygen that is, and compare it to how many moles are in 1.5 g of tin. By seeing this ratio, we can find the empirical formula of the oxide.

FionaGriffin-1K
Posts: 4
Joined: Fri Sep 29, 2017 7:04 am

Re: L.39

Postby FionaGriffin-1K » Thu Oct 05, 2017 4:47 pm

thanks so much! thats super helpful :)

Esin Gumustekin 2J
Posts: 57
Joined: Thu Jul 27, 2017 3:01 am

Re: L.39

Postby Esin Gumustekin 2J » Thu Oct 05, 2017 7:24 pm

For this problem, does anyone understand why we assume that the oxygen in the air is O and not O2? I thought oxygen in the air is always in the form of O2.


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