Question

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

Answer 1

`Ca_5P_5O_(18)`

Step-by-step explanation:

Hello!

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:

`n_(Ca)=9.0231g*(1mol)/(40.08g)= 0.225mol\\\\n_(P)=6.9722g*(1mol)/(31.97g)=0.218mol\\\\n_(O)=12.6072g*(1mol)/(16.00g) =0.788mol`

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:

`Ca:(0.225)/(0.218)= 1\\\\P:(0.218)/(0.218)=1\\\\O:(0.788)/(0.218)=3.6`

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

`Ca_5P_5O_(18)`

Best regards!