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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 M_{2}S_{3} is converted to MO_{2} 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 M_{2}S_{3}equals the moles of M in 2MO_{2}. (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 M_{2}S_{3}⇒ 2x + 96 g mol¯^{1}"molar mass" of 2MO

_{2}⇒ 2x + 64 g mol¯^{1}

3) Determine moles of M_{2}S_{3} and moles of 2MO_{2}:

M_{2}S_{3}⇒ 4.000 / (2x + 96)2MO

_{2}⇒ 3.723 / (2x + 64)

4) Set moles of M in M_{2}S_{3} equal to moles of M in 2MO_{2}:

[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 M_{2}S_{3}(s) is heated in air, it is converted to MO_{2}(s). A 3.000 g sample of M_{2}S_{3}(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¯

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**Second Extra Problem:** A 3.500-g sample of M_{2}S_{3}(s) (when converted to MO_{2}) 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¯

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**Third Extra Problem:** When M_{2}S_{3}(s) is heated in air, it is converted to MO_{2}(s). A 5.000 g sample of M_{2}S_{3}(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.