# Chemistry

A particular metallic element, M, has a heat capacity of 0.36 J*g*K, and it forms an oxide that contains 2.90 grams of M per gram of oxygen.

(a) Using the law of Dulong and Petit (Cp x M = 25 J*mol*K), estimate the molar mass of the metal M [Answer: _____ g/mol]

(b) From the composition of the oxide (2.90 g M/g O) and the molar mass of oxygen (16 g/mol), determine the mass of M that combines with each mole of oxygen [Answer: _____ g M/mol O]

(c) Use your estimate of the atomic mass from question (a) and the information from question (b) to determine the empirical formula of the oxide (the moles of M that combine with each mole of O) [Answer: _____]

(d) Now that you know the composition of the oxide and its formula, what is the accurate value of the atomic mass of the metal M? What is the identity of M? [Answer: _____ g/mol]

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1. This problem looks simple enough. I would be interested in your thoughts before I go all out on a typing blitz for four questions. Part a is the easiest and you can start there. Cp x M = 25. You are given Cp and you can calculate M.
Tell us what you don't understand (in some detail) if you get stuck.

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2. Hi, DrBob222!

Here's what I got for:

(a) 69.44 g/mol
(b) 46.4 g M/mol O

I don't understand how to use those two information in order to determine the empirical formula (question (c)). I'm guessing it's some sort of ratio but I don't know how to use the answers that I got for (a) and (b) to solve for (c)

Thanks for you help!

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3. 46.4/69.4 = 0.668 = moles M
16/16 = 1 mole O

Now take the ratio. You want small whole numbers with the lowest number being 1. The easiest way to do that is to divide the smallest number by itself, then to keep things equal divide the other number(s) by that same small number.
0.668/0.668 = 1 mole M
1/0.668 = 1.5 moles O
so that is the ratio of 2:3 and the formula is xxxxx?

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4. OK, so the formula is:

(c) M2O3

...and the identity and atomic mass of the metal M is:

(d) Chromium which has a 51.9 amu

Did I get the answers right?

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5. I wouldn't have gone that route. Isn't the atomic mass of M something like 69.4 (from part a)? Cr isn't close to that although it does form Cr2O3 as the oxide.

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6. Eh, so what is the answer - Ga2O3 (basing it on the atomic mass of part (a))?

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7. That's what I would pick as the oxide and Ga as M.

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8. Great! Thank you so much for all the help, DrBob222!

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9. How did you get the answer on part (B)

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10. @LostChemist,

For part b:
(16.0g O/1 mol O) x (2.90g M/1g O)
= 46.4g M/1 mol O

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11. How did you get part d?

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