HW Question L.39 - What in the world is a crucible you guys?

Julia Mazzucato 4D
Posts: 64
Joined: Fri Aug 30, 2019 12:17 am

HW Question L.39 - What in the world is a crucible you guys?

L.39 A 1.50-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.

This might be a dumb question but I don't know what a crucible is and Google results aren't super helpful. Is it part of the reaction or is it just a vessel for the reaction?

DarrenKim_1H
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Joined: Fri Sep 20, 2019 12:17 am
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Re: HW Question L.39 - What in the world is a crucible you guys?

To my knowledge, it is solely a container that houses chemical reactions. For this question, I think you are given its mass (26.45g) so that you can subtract it from the combined mass of the crucible and product (28.35g) in order to find the mass percentage of the oxide.

TL;DR 28.35g-26.45g = molar mass of post-reaction oxide (1.9g)

Alexa Mugol 3I
Posts: 54
Joined: Sat Aug 17, 2019 12:17 am

Re: HW Question L.39 - What in the world is a crucible you guys?

A crucible is a container that you can heat to high temperatures.

In this question, tin was put into a crucible and heated so it can react with oxygen to make a tin oxide.

The problem gives you the mass of the crucible and then the mass of the crucible with the product inside it. Therefore, you subtract the mass of crucible from the mass of the crucible + the product to get the mass of the product.
28.35-26.45= 1.90g product

Use that mass for the product to solve for the empirical formula. Hope this helps!

Brittany Tran 3I
Posts: 50
Joined: Sat Jul 20, 2019 12:16 am

Re: HW Question L.39 - What in the world is a crucible you guys?

A crucible is a vessel that you can use to perform chemical reactions. To find the empirical formula, you have to find the mass of the two reactants (tin and oxygen). To find the mass of oxygen, you can find the mass of the product (28.35 g - 26.45 g = 1.90 g) and subtract the mass of tin from that (1.90 g - 1.50 g = 0.40 g oxygen) because of the law of conservation of mass. You're given the mass of tin so you can find the empirical formula from there!