A sample of a compound used to polish dentures and as a nutrient and dietary supplement is analyzed and found to contain 9.0231 g of calcium, 6.9722 g of phosphorus, and 12.6072 g of oxygen. What is the empirical formula for this compound

Explanation:

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In this case, since the determination of empirical formulas requires the moles of the constituents, we first need to calculate the moles in the given grams of the listed elements:

Next, we divide each moles by the fewest moles, in this case, those of phosphorous, in order to determine their subscripts in the empirical formula:

Thus, we multiply these subscripts by 5 to get whole numbers:

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first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

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