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An oxide of chromium used in chrome plating has a formula mass of 100.0 u and contains four atoms per formula unit. Establish the formula of this compound, with a minimum of calculation.

Express your answer as a chemical formula.

So the question tells me that the mass of the Chromium Oxide is 100.0 amu and has 4 atoms for very unit. Thus, the question wants to know a formula that has 4 atoms? Right?

Hence, if $\ce{Cr}$'s (Chromium) molecular mass is 51.9961 and since the question is stating that we are dealing with an oxide, which would mean that it is a ionic compound (Chromium gives two electrons to Oxygen [15.999 amu]), wouldn't the answer be $\ce{Cr2O2}$; but that would equal 135.9902 grams.

On the other hand, if I am supposed to view a formula unit as an empirical formula, wouldn't that require me to write the simplest formula for a compound?

Yet, $\ce{CrO}$ would give me (51.9961 + 15.999) 67.9951 g/mol and that is not correct, or is it?

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  • $\begingroup$ Welcome to Chemistry.SE! Please have a look at this tutorial to acquaint yourself with the way math and chemical formulae can be nicely formatted on this site. $\endgroup$ – Philipp Sep 24 '14 at 17:43
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Rounding judiciously, the mass of chromium is 52 amu and oxygen is 16 amu.

Our compound is CrxOy and has 4 atoms. The sum of those 4 atoms' mass is 100 amu.

x + y = 4

52x + 16y = 100

Solve the system of equations to find the answer. Alternatively, you can just guess and check.

Any species with 2 chromium atoms must have a mass greater than 100 amu, since Cr2 = 104 amu and we haven't included any oxygens. Thus we know for certain that there is only 1 chromium atom in the compound. The remaining three must be oxygen. I leave it to you to check that the answer is correct.

I think your confusion stems from assuming that something called "chromium oxide" must have 1 chromium atom for every 1 oxygen atom, and that isn't the case.

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  • $\begingroup$ Yes, I thought an oxide was always an ionic compound, that is, nonmetal with a metal. Ok, thanks. $\endgroup$ – user137452 Sep 24 '14 at 16:55

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