## Fundamentals L.39:

Veronica_Lubera_2A
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### Fundamentals L.39:

A 1.5 g sample of metallic tin was placed in a 26.45g crucible and heated until all the tin had reacted with the oxygen 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 name of the oxide.

So I found the empirical formula of the oxide to be SnO2, but I am unsure of how to solve for the name of the oxide (part b), which is supposed to be Tin (IV). Any suggestions?

sarahsalama2E
Posts: 164
Joined: Fri Aug 30, 2019 12:16 am

### Re: Fundamentals L.39:

so we know that the equation for tin reacting with the oxygen to form an oxide is the following:
Sn + O2 --> SnO2 (the 2's are subscripts)

28.35g (total product mass which is the crucible and the oxide) - 26.45 g (which is the tin sample and crucible mass) = 1.9 g which must be the mass of the reactants
we know that the mass of the metallic tin is 1.5g (given in the question) so we must subtract 1.5 from 1.9 and we get 0.4. 0.4g must be the mass of the oxygen that the tin reacted with.

then 0.4 g of oxygen is converted into moles (this is done by dividing 0.4 g by the molar mass of oxygen)--> answer for this is .025 mol O2 (subscript 2)
then 1.5 g of Sn is converted into moles (this is done by dividing 1.5 g by the molar mass of tin)--> answer for this is .013 mol Sn

Next, to determine empirical formula you divide both moles values by the smallest one, which in this case is the moles value for tin which is .013 (.013 is smaller than .025). After this we get .025/2 which is 2 (for oxygen) and .013/.013 (for Sn) which is 1. These resultant values then give us the relative number of atoms in the oxide, AKA, the empirical formula of the tin oxide which is SnO2 (2 is subscript) (which you got the answer for).

by nomenclature principles, the name of this oxide is tin (IV) oxide. The roman numeral is used because tin (Sn) can either have oxidation states of +2 or +4. Oxygen can have a charge of either + or - 2. ( we know these according to the column that the specified element is in). In this case, the tin has a charge of +4 and the oxygen has a charge of -2. We "cross" the charges and get Sn2O4 (2 and 4 are subscripts) and then reduce to get SnO2.

* by the way, this is an oxidation reduction reaction. tin loses 4 electrons, while oxygen gains 4 electrons and gets reduced.

When writing the name of the oxide, you must write the roman numeral IV in parenthesis to indicate which oxidation state the tin is in, in this case. so the name of the product formed in this rxn is Tin (IV) Oxide.

hope that helps.

KNguyen_1I
Posts: 101
Joined: Sat Aug 17, 2019 12:16 am

### Re: Fundamentals L.39:

So if you recall, most metals have more than just one oxidation state. However, most nonmetals have very specific and defined oxidation states we can always remember, like oxygen always being -2 and flourine being -1 for example (this can be corroborated with them having the highest electronegativty and tendency to cling to electrons, therefore we can always count on them being -2 and -1). So, looking at the compound that being SnO2, if a single atom of oxygen is -2, and we have 2 oxygens, their net charge would have to be -4. And checking on tin's possible oxidation states, 4 is definitely a viable oxidation state as well as balancing the net charge of the molecule. Therefore, it would have to be tin (IV) oxide.

Posts: 51
Joined: Sat Aug 24, 2019 12:16 am

### Re: Fundamentals L.39:

Tin is one of the transition metals on the periodic table. This means that it has multiple oxidation states. For tin specifically, the oxidations states are +2 or +4.

When naming molecules containing transition metals, a roman numeral is often placed after the transitional metal to denote the oxidation state.

In SnO2, because we know that the O2 will have a total -4 charge and the molecule as a whole is neutral, we can deduce that Sn has a +4 charge. This means that the name of the molecule is tin (IV) oxide, where (IV) denotes the oxidation state of tin.