Empirical Equation Textbook Problem

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Empirical Equation Textbook Problem

Postby Rose_Malki_3G » Mon Oct 26, 2020 5:27 pm

A .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

When I was converting the amount of oxygen from grams to mols I divided by 32g/mol because I thought that if the oxygen were in the air it would be O2. However, when I looked at the solution, it was only divided by 16g/mol, meaning that the fact that oxygen is diatomic was not taken into account. Can someone please explain why the oxygen in the air isn't O2

Hannah Lechtzin 1K
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Re: Empirical Equation Textbook Problem

Postby Hannah Lechtzin 1K » Mon Oct 26, 2020 6:02 pm

They aren't using the molar mass of the oxygen in the air. In that solution they are using the molar mass of oxygen to find the mass percentage of oxygen that is within the oxide in order to find the empirical formula for the oxide, so they use 16g/mol instead of 32g/mol. I hope this helps!

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