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?

  • $\begingroup$ 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, 2014 at 17:43

1 Answer 1


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.

  • $\begingroup$ Yes, I thought an oxide was always an ionic compound, that is, nonmetal with a metal. Ok, thanks. $\endgroup$
    – user137452
    Sep 24, 2014 at 16:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.