## Fundamental L.39

Leslie Cheng 4B
Posts: 35
Joined: Fri Sep 28, 2018 12:29 am

### Fundamental L.39

Hi! The problem says that you have a 1.50g sample on tin placed in a 26.45g crucible that is heated until the tin reacts to the oxygen in the air to form and oxide. The product and the crucible together weigh 28.35g. How do you figure out the empirical formula of the oxide, and how do you name it? Also as a side note, what do the roman italics in for example, copper (II) oxide mean?

ThomasLai1D
Posts: 61
Joined: Fri Sep 28, 2018 12:17 am

### Re: Fundamental L.39

Hi,

The roman numerals following the metals name are an indication the number of positive cations of that metal. Some transition metals may have multiple cation forms, therefore the number of positive cations of the ionic compound are notated with roman numerals. So copper (II) oxide would have two copper cations.

Albert Duong 4C
Posts: 66
Joined: Fri Sep 28, 2018 12:17 am
Been upvoted: 1 time

### Re: Fundamental L.39

Okay, so you want to start by finding how much product of tin oxide you've made. We do this by subtracting the crucible's mass (26.45 g) from the total mass of the product AND the crucible (28.35 g). That should get you 1.9 g of tin oxide. And since we know we started with 1.5 g of tin, we can subtract that from the mass of the tin oxide to get the mass of oxygen which would be 0.4 g. Then we find the mass % composition for tin and oxygen: (1.5 g Sn)/(1.9 g tin oxide) = 78.9% and for oxygen it's 21.1 %. Imagine both as 78.9 g Sn and 21.1 g O then using their molar masses (118.71 g/mol Sn and 16.00 g/mol O), we get .665 mol Sn and 1.32 mol O. Divide both by the smallest number of moles (.665) and we get 1 mol Sn and 2 mol O. Thus the empirical formula is SnO2.

The roman italics for a metal represent its charge (which is always positive in my experience). In the case of copper(II), it has a charge of 2+, which makes sense since a single oxygen atom tends to have a charge of 2- anyways. Can you guess why SnO2 is called tin(IV) oxide now?

005115864
Posts: 64
Joined: Fri Sep 28, 2018 12:15 am

### Re: Fundamental L.39

Hi, I just wanted to share something interesting I found. Another way to solve this problem, oddly enough, is by simply creating a balanced equation. Because they explicitly tell us the reactants and ultimate product, I simply combined Tin and Oxygen to make my product. In this case, the molecular and the empirical formula are the same which Professor Lavelle has said is possible. Just to make sure though, I would do the math since this is not always the case. Also, all the transition metals will work in this matter because they have no set charge assigned to them and so their charge can be determined by seeing what other element they're paired to, since there is two oxygens in this case with a -2 charge, that means in total it has a -4 charge, so Sn will have a +4 charge.

Noh_Jasmine_1J
Posts: 71
Joined: Fri Sep 28, 2018 12:15 am

### Re: Fundamental L.39

Hi. I have a question regarding this problem as well: why is the oxygen written as "O" without the subscript 2 until the formula is written? I though oxygen was always O2

Lucy Agnew 3E
Posts: 30
Joined: Fri Sep 28, 2018 12:26 am

### Re: Fundamental L.39

@AlbertDuong Thank you for that explanation! Completely forgot how to do it and so that was very helpful

Aidan Ryan 1B
Posts: 60
Joined: Fri Sep 28, 2018 12:17 am

### Re: Fundamental L.39

@Noh_Jasmine_1G Oxygen in its gaseous state is always O2 however if bonded to a substance it does not. I think when others wrote about oxygen they may have been talking about the molar mass of oxygen which is needed not about oxygen gas (which has a mass of 32g). Hope that helps