# What is the name of the compound with 77.3% silver, 7.4% phosphorous and 15.3% oxygen?

Jan 8, 2017

$\text{Silver phosphate}$, $A {g}_{3} P {O}_{4}$.

#### Explanation:

As with all of these problems, we assume there are $100 \cdot g$ of compound, and proceeed on this basis to derive the empirical formula by calculating the individual molar quantites. And given $100 \cdot g$ of stuff, there are:

$\text{Moles of silver} = \frac{77.3 \cdot g}{107.9 \cdot g \cdot m o {l}^{-} 1} = 0.716 \cdot m o l$

$\text{Moles of phosphorus} = \frac{7.4 \cdot g}{31.00 \cdot g \cdot m o {l}^{-} 1} = 0.239 \cdot m o l$

$\text{Moles of oxygen} = \frac{15.3 \cdot g}{16.00 \cdot g \cdot m o {l}^{-} 1} = 0.956 \cdot m o l$

For each element we have divided by the atomic mass of the given element to give a molar quantity of that element.

We now divide thru by the SMALLEST molar quantity, that of phosphorus:

$\text{Equivs of silver}$ $=$ $\frac{0.716 \cdot m o l}{0.239 \cdot m o l}$ $=$ $3$

$\text{Equivs of phosphorus}$ $=$ $\frac{0.239 \cdot m o l}{0.239 \cdot m o l}$ $=$ $1$

$\text{Equivs of oxygen}$ $=$ $\frac{0.956 \cdot m o l}{0.239 \cdot m o l}$ $=$ $4$

And thus the empirical formula, the simplest whole number ratio representing equivalent quantities in a species is:

$A {g}_{3} P {O}_{4}$