# What is the empirical formula for a compound whose analysis is 74.97% aluminum and 25.03% carbon?

Feb 10, 2016

The empirical formula is ${\text{Al"_4"C}}_{3}$.

#### Explanation:

Assume that you have 100 g of the compound.

Then you have 74.97 g of $\text{Al and}$25.03 g of $\text{C}$.

Our job is to calculate the ratio of the moles of each element.

$\text{Moles of Al" =74.97 color(red)(cancel(color(black)("g Al"))) × "1 mol Al"/(26.98 color(red)(cancel(color(black)("g Al")))) = "2.7787 mol Al}$

$\text{Moles of C" = 25.03 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01color(red)(cancel(color(black)("g C")))) = "2.0841 mol C}$

To get the molar ratio, we divide each number of moles by the smaller number ($2.0841$).

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

$\text{Element"color(white)(X) "Mass/g"color(white)(X) "Moles"color(white)(XXl) "Ratio" color(white)(XX)"×2"color(white)(mm)"×3"color(white)(mll)"Integers}$
stackrel(—————————————————-———)(color(white)(l)"Al" color(white)(XXXX)74.97 color(white)(Xlll)2.7787 color(white)(Xll)1.3333color(white)(X)2.6666)color(white)(l)3.9999color(white)(mm)4
$\textcolor{w h i t e}{l l} \text{C} \textcolor{w h i t e}{X X X X} 25.03 \textcolor{w h i t e}{X X} 2.0841 \textcolor{w h i t e}{X l l} 1 \textcolor{w h i t e}{X X X l l} 2 \textcolor{w h i t e}{m m m} 3 \textcolor{w h i t e}{m m m m l l} 3$

The ratio is not in small whole numbers, so we multiply by integers until the ratio comes close to small whole numbers.

The ratio comes out as $\text{Al:C} = 4 : 3$.

Thus, the empirical formula is ${\text{Al"_4"C}}_{3}$.