Fundamentals L39

Moderators: Chem_Mod, Chem_Admin

Nick Fiorentino 1E
Posts: 102
Joined: Wed Sep 18, 2019 12:16 am

Fundamentals L39

Postby Nick Fiorentino 1E » Thu Oct 03, 2019 2:51 pm

I'm a bit confused on how to go about this problem. Does anybody have any hints or ideas? How would we know what the oxide is?

Samuel Tzeng 1B
Posts: 103
Joined: Sat Aug 24, 2019 12:15 am

Re: Fundamentals L39

Postby Samuel Tzeng 1B » Thu Oct 03, 2019 4:53 pm

Start off by finding moles of tin and oxygen

Sn: 1.50g Sn / 118.71g/mol = .01264 mol Sn
O: (28.35g - 26.45g - 1.50g) / 16.00g/mol = 0.025 mol O

With this you can tell the empirical formula is SnO2

The oxide would be Tin(IV) Oxide since each oxygen has a -2 charge so the tin must have a +4 charge

Maya Gollamudi 1G
Posts: 100
Joined: Thu Jul 25, 2019 12:15 am

Re: Fundamentals L39

Postby Maya Gollamudi 1G » Thu Oct 03, 2019 4:57 pm

We find the identity of the oxide by using the known masses to determine the empirical formula of the oxide.

We know that the mass of the crucible and product together is 28.35 g, so the mass of the total product is 1.9 g. The mass of tin (Sn) in the product is 1.50 g so we know the mass of O in the product is 0.4 g.

Divide the known masses of Sn and O in the product by their molar masses in order to get the number of moles of each. We have 0.0126 moles of Sn and 0.025 moles of O. After dividing by the smallest number of moles (0.0126), we have a 1:2 ratio in the relative number of atoms of Sn and O in the oxide, making the empirical formula of the oxide SnO2.


Return to “Balancing Chemical Reactions”

Who is online

Users browsing this forum: No registered users and 1 guest