L.39 Question: Empirical Formula of Oxide

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Nhi Vo 3A
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L.39 Question: Empirical Formula of Oxide

Postby Nhi Vo 3A » Tue Sep 27, 2016 11:55 am

L.39. 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 in air to form an oxide. The crucible and product together were found to weigh 28.35 g.
a). What is the empirical formula of the oxide?
b). Write the name of the oxide.

I found the moles of Sn to be 0.001264 mol Sn and moles of oxygen to be 0.025 mol O, but how do I find the mole ratio of Sn: O ?
Last edited by Nhi Vo 3A on Tue Sep 27, 2016 4:38 pm, edited 1 time in total.

Chem_Mod
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Re: L39 Question: Empirical Formula of Oxide

Postby Chem_Mod » Tue Sep 27, 2016 12:52 pm

The moles of tin you found is off by a decimal point. You simply take the larger number, divided by the smaller number. In this case, we find out there are two moles of O for every one mole of Sn. Therefore, the empirical formula is SnO2, meaning that tin has a +4 charge. So the name would be tin (IV) oxide.

Nhi Vo 3A
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Joined: Sat Jul 09, 2016 3:00 am

Re: L39 Question: Empirical Formula of Oxide

Postby Nhi Vo 3A » Tue Sep 27, 2016 4:33 pm

Thank you!

Aleena_Sorf_2A
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Re: L.39 Question: Empirical Formula of Oxide

Postby Aleena_Sorf_2A » Wed Oct 12, 2016 5:42 pm

L.39
Given:
mass of Sn= 1.50 g
mass of crucible= 26.45 g
mass of crucible and oxide= 28.35 g
mass of oxide= 28.35g-26.45g= 1.90 g
mass of oxygen= 1.90-1.50 g= .4 g

1.50 g Sn x (1 mol Sn/118.71 g Sn)= 1.264 x 10^-2 mol Sn

.4 g x (1mol O/16.00 g O)= .025 mol O

You would then divide the moles by the smallest number of moles, which in this case is 1.264 x 10^-2 mol.
1.264 x 10^-2 / 1.264 x 10^-2= 1
.025 / 1.264 x 10^-2 = 1.977= 2
Mole Ratio of Sn:O is 1:2
Empirical Formula- SnO2

b) The name of this oxide is tin(IV) oxide.


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