Homework L39

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Johnathan Smith 1D
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Joined: Wed Sep 11, 2019 12:16 am

Homework L39

Postby Johnathan Smith 1D » Thu Oct 03, 2019 11:42 am

Can somebody explain to me why Sn02 is Tin(IV) Oxide?

Michelle Chan 1J
Posts: 50
Joined: Thu Jul 25, 2019 12:16 am

Re: Homework L39

Postby Michelle Chan 1J » Thu Oct 03, 2019 11:56 am

The roman numeral in the parenthesis represent the charge since those metals have various charges. Since 02 is -4, tin must be +4 (IV).

Audrie Chan-3B
Posts: 50
Joined: Thu Jul 25, 2019 12:17 am

Re: Homework L39

Postby Audrie Chan-3B » Thu Oct 03, 2019 11:57 am

Not quite sure, but I think that because oxygen has a charge of -2 on the periodic table and there are 2 oxygen atoms, you need tin to have a charge of 4 in order to balance it out. Tin is a transitional metal, meaning it does not have a set charge, so you need to look at the charge and number of oxygen to determine tin's charge.

Vicki Liu 2L
Posts: 101
Joined: Sat Aug 24, 2019 12:15 am

Re: Homework L39

Postby Vicki Liu 2L » Thu Oct 03, 2019 11:58 am

Tin is an element that exists in more than one oxidation state, meaning its ions could carry different charges. For tin specifically, it could be Sn2+or Sn4+. But since oxygen's oxidation number is -2 and there are two oxygen ions present in SnO2, for the net charge to equal 0, the tin ion must carry a charge of +4. Therefore, by following nomenclature rules, SnO2 should be written as Tin(IV) Oxide, with the Roman numeral IV representing that the tin ion has a charge of +4.

Johnathan Smith 1D
Posts: 108
Joined: Wed Sep 11, 2019 12:16 am

Re: Homework L39

Postby Johnathan Smith 1D » Thu Oct 03, 2019 12:04 pm

What about for cases like Cu(NO3)2? How do we get to know the charge of copper?

This is for M9.

Dakota Walker 1L
Posts: 58
Joined: Wed Sep 18, 2019 12:17 am

Re: Homework L39

Postby Dakota Walker 1L » Thu Oct 03, 2019 12:07 pm

For Cu the charge would be -2 because you pull up the number from the bottom of the opposite element which in this case is (NO3)2, making the charge of Cu -2.


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