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

Answer:

$Ca_5P_5O_{18}$

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:

$n_{Ca}=9.0231g*\frac{1mol}{40.08g}= 0.225mol\\\\n_{P}=6.9722g*\frac{1mol}{31.97g}=0.218mol\\\\n_{O}=12.6072g*\frac{1mol}{16.00g} =0.788mol$

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:

$Ca:\frac{0.225}{0.218}= 1\\\\P:\frac{0.218}{0.218}=1\\\\O:\frac{0.788}{0.218}=3.6$

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

$Ca_5P_5O_{18}$

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