L.39 Textbook question

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Valerie Doan 3I
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L.39 Textbook question

Postby Valerie Doan 3I » Fri Oct 09, 2020 12:08 am

Could someone help me with L.39 from the textbook hw? I keep getting the wrong answer but I don't know why.

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.

Thanks :)

Madilyn Schindler 3E
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Re: L.39 Textbook question

Postby Madilyn Schindler 3E » Fri Oct 09, 2020 12:32 am

For part (a), I first found the mass of the oxygen to be 0.4g by using conservation of mass and subtracting 26.45g and 1.5g from 28.35g. I then found the number of moles of tin and oxygen to be 0.0126 and 0.025, respectively. Since the oxygen had twice as many moles as the tin, the ratio of oxygen to tin was 2:1. Therefore, the empirical formula of the oxide is SnO2. For part (b), the name of the oxide is tin oxide.

Yuelai Feng 3E
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Re: L.39 Textbook question

Postby Yuelai Feng 3E » Fri Oct 09, 2020 12:51 am

Hi! For this question, we can first calculate the masses of the product and reactants.
Given the crucible is 26.45 g, and the crucible + product are 28.35 g in total, the mass of the product (the oxide) is 28.35g-26.45g = 1.90g. And we also know that the tin sample is 1.50g. So we have:
Sn (1.50g) + O2 (?g) --> the oxide (1.90g)
According to the conservation of mass, we know that the mass of oxygen gas involved in the reaction is 1.90g-1.50g = 0.40g.

Then, convert masses to moles:
- The molar mass of Sn is 118.7g/mol, so the number of tin atoms is 1.50g / 118.7g/mol = 0.0126 mol
- The molar mass of O is 15.999g/mol, so the number of oxygen atoms in this reaction is: 0.40g / 15.999g/mol = 0.025 mol
Therefore, the Sn:O molar ratio is 0.0126 : 0.025 ≈ 1:2
So the empirical formula for the oxide (which is made up by all the atoms in the reactants) is SnO2. Name: tin(IV) oxide.

Hope this helps!

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Re: L.39 Textbook question

Postby rhettfarmer-3H » Fri Oct 09, 2020 11:54 am

I also found this problem troublesome. The start is tough and there is enough information that is extra and very irrelevant or can be discard after a quick subtraction. Furthermore, I attached my work that follows the steps first get the amount of product by subtracting the mass of the crucible and the combined mass of product. and then use converstation of matter hence mass must equal on both sides to find the oxygen. Then it's simple. A easy, convert the grams to moles. Divide by the lowest mole ratio and boom you have the elements ratios. See work for math.
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