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By the way, there are three more examples of this problem at the end of the alternate solution just below.

A 4.000 g sample of M2S3 is converted to MO2 and loses 0.277 g. What is the atomic weight of M?

A different solution path from the worksheet:

1) Some facts that I can't think of a good title for:

a) The moles of M in M2S3 equals the moles of M in 2MO2. (Notice the inclusion of the coefficient.)

b) Let x = the atomic weight of M.

2) More facts that I can't think of a good title for:

molar mass of M2S3 ⇒ 2x + 96 g mol¯1

"molar mass" of 2MO2 ⇒ 2x + 64 g mol¯1

3) Determine moles of M2S3 and moles of 2MO2:

M2S3 ⇒ 4.000 / (2x + 96)

2MO2 ⇒ 3.723 / (2x + 64)

4) Set moles of M in M2S3 equal to moles of M in 2MO2:

[4.000 / (2x + 96)] = [3.723 / (2x + 64)]

x = 183 g mol¯1

For a bit of clarity with the numbers, I didn't pay close attention to putting units on numbers. I hope this didn't mess you up too much.


First Extra Problem: When M2S3(s) is heated in air, it is converted to MO2(s). A 3.000 g sample of M2S3(s) shows a decrease in mass of 0.353 g when it is heated in air. What is the average atomic mass of M?

[3.000 / (2x + 96)] = [2.647 / (2x + 64)]

(3.000) (2x + 64) = (2.647) (2x + 96)

x = 87.98 g mol¯1


Second Extra Problem: A 3.500-g sample of M2S3(s) (when converted to MO2) shows a decrease in mass of 0.295 g when it is heated in air. What is the average atomic mass of M?

[3.500 / (2x + 96)] = [3.205 / (2x + 64)]

x = 141.8 g mol¯1


Third Extra Problem: When M2S3(s) is heated in air, it is converted to MO2(s). A 5.000 g sample of M2S3(s) shows a decrease in mass of 0.347 g when it is heated in air. What is the average atomic mass of M?

No solution provided.

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