Page 1 of 1

L.39 Question: Empirical Formula of Oxide

Posted: Tue Sep 27, 2016 11:55 am
by Nhi Vo 3A
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 ?

Re: L39 Question: Empirical Formula of Oxide

Posted: Tue Sep 27, 2016 12:52 pm
by Chem_Mod
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.

Re: L39 Question: Empirical Formula of Oxide

Posted: Tue Sep 27, 2016 4:33 pm
by Nhi Vo 3A
Thank you!

Re: L.39 Question: Empirical Formula of Oxide

Posted: Wed Oct 12, 2016 5:42 pm
by Aleena_Sorf_2A
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.