# A 0.967*g mass of an oxide of phosphorus contains 0.422*g of phosphorus. What is its empirical formula?

Sep 11, 2017

We gots ${P}_{2} {O}_{5}$.....

#### Explanation:

We calculate the empirical formula....

$\text{Moles of phosphorus} = \frac{0.422 \cdot g}{30.9737 \cdot g \cdot m o {l}^{-} 1} = 0.01361 \cdot m o l$

$\text{Moles of oxygen} = \frac{0.967 \cdot g - 0.422 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 0.0341 \cdot m o l$

We divide thru by the SMALLEST MOLAR quantity to give an empirical formula of ${P}_{\frac{0.01361}{0.01361}} {O}_{\frac{0.0341}{0.01361}}$

$\equiv P {O}_{2.50}$....

Now the empirical formula is by definition the simplest WHOLE number ratio, so we double the trial formula to give.....

${P}_{2} {O}_{5}$

The actual molecule is ${P}_{4} {O}_{10}$ but we would need further data to assess this.