# A 7.96 gram sample of silver reacts with oxygen to form 8.55 gram of the metal oxide. What is the formula of the oxide?

May 2, 2016

The empirical formula is $\text{Ag"_2"O}$.

#### Explanation:

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of $\text{Ag}$ to $\text{O}$.

$\text{Mass of Ag = 7.96 g}$

$\text{Mass of silver oxide = mass of Ag + mass of O}$

$\text{8.55 g = 7.96 g + mass of O}$

$\text{Mass of O = (8.55 – 7.96) g = 0.59 g}$

$\text{Moles of Ag" = 7.96 color(red)(cancel(color(black)("g Ag"))) × "1 mol Ag"/(107.9color(red)(cancel(color(black)( "g Ag")))) = "0.073 77 mol Ag}$

$\text{Moles of O "= 0.59 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "0.0369 mol O}$

To get this into an integer ratio, we divide both numbers by the smaller value.

From this point on, I like to summarize the calculations in a table.

$\text{Element"color(white)(Ag) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio"color(white)(mll)"Integers}$
stackrel(—————————————————-———)(color(white)(m)"Ag" color(white)(XXXm)7.96 color(white)(Xm)0.073 77 color(white)(Xll)2.00color(white)(mmm)2)
$\textcolor{w h i t e}{m l} \text{O" color(white)(XXXXl)0.59 color(white)(mm)} 0.0369 \textcolor{w h i t e}{X m l} 1 \textcolor{w h i t e}{m m m m l} 1$

There are 2 mol of $\text{Ag}$ for 1 mol of $\text{O}$.

The empirical formula of silver oxide is $\text{Ag"_2"O}$.

Here is a video that illustrates how to determine an empirical formula.