Compound Analysis: Empirical Formula Calculation

What is the empirical formula for a compound containing 9.0231 g of calcium, 6.9722 g of phosphorus, and 12.6072 g of oxygen?

Choose one:

A. $Ca_5P_5O_{18}$

B. $Ca_3P_2O_7$

C. $Ca_3P_3O_10$

Answer:

The empirical formula for the compound is $Ca_5P_5O_{18}$.

Hello! In this case, we need to calculate the moles of each element to determine the empirical formula of the compound. Firstly, let's convert the given masses of calcium, phosphorus, and oxygen into moles:

Calcium (Ca): $\displaystyle n_{Ca} = \frac{9.0231 \, g}{40.08 \, \frac{g}{mol}} = 0.225 \, mol$

Phosphorus (P): $\displaystyle n_{P} = \frac{6.9722 \, g}{31.97 \, \frac{g}{mol}} = 0.218 \, mol$

Oxygen (O): $\displaystyle n_{O} = \frac{12.6072 \, g}{16.00 \, \frac{g}{mol}} = 0.788 \, mol$

Next, we need to find the ratio of moles between each element. The lowest moles belong to phosphorus (P). Divide the moles of calcium, phosphorus, and oxygen by the moles of phosphorus:

Calcium (Ca): $\displaystyle \frac{0.225}{0.218} \approx 1$

Phosphorus (P): $\displaystyle \frac{0.218}{0.218} = 1$

Oxygen (O): $\displaystyle \frac{0.788}{0.218} \approx 3.61$

Finally, multiply these ratios by 5 to obtain whole numbers and determine the empirical formula:

Empirical Formula: $Ca_5P_5O_{18}$

Feel free to ask if you have any more questions!