# What is the empirical formula for a compound which contains 0.0134 g of iron, 0.00769 g of sulfur and 0.0115 g of oxygen?

Jul 2, 2017

$F e S {O}_{3}$

#### Explanation:

Convert to % per 100 wt and follow the scheme...

Given $0.0134 g F e + 0.00769 g S + 0.0115 g O = 0.03259 g$

=> %Fe = ((0.0134g)/(0.03259g))100% = 41.1169% per 100 wt

=> %S = ((0.00769g)/(0.03259g))100% = 23.5962% per 100 wt

=> %O = ((0.0115g)/(0.03259g))100% = 35.2669% per 100wt

%/100wt => g/100wt => moles => normalize (divide by smallest moles) => Empirical Ratio => Empirical Formula Jul 2, 2017

The empirical formula is ${\text{FeSO}}_{3}$.

#### Explanation:

The empirical formula is the simplest whole number molar ratio of the elements in the compound.

We must convert the masses of $\text{Fe}$, $\text{S}$, and $\text{O}$ to moles and then find the ratio.

$\text{Moles of Fe" = 0.0134 color(red)(cancel(color(black)("g Fe"))) × "1 mol Fe"/(55.84 color(red)(cancel(color(black)("g Fe")))) = 2.400 × 10^"-4"color(white)(l) "mol Fe}$

$\text{Moles of S" = "0.007 69" color(red)(cancel(color(black)("g S"))) × "1 mol S"/(32.06 color(red)(cancel(color(black)("g S")))) = 2.399 × 10^"-4"color(white)(l)"mol S}$

$\text{Moles of O" = 0.0115 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = 7.188 × 10^"-4"color(white)(l) "mol O}$

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

$\boldsymbol{\text{Element"color(white)(Ag) "Mass/g"color(white)(Xm) "Moles"color(white)(Xmm) "Ratio"color(white)(m)color(white)(l)"Integers}}$
$\textcolor{w h i t e}{m} \text{Fe" color(white)(XXXml)0.0134 color(white)(Xml)2.400 × 10^"-4} \textcolor{w h i t e}{X l l} 1.000 \textcolor{w h i t e}{m m m l l} 1$
$\textcolor{w h i t e}{m} \text{S" color(white)(XXXXm)0.00769 color(white)(mll)2.399 × 10^"-4} \textcolor{w h i t e}{X l l} 1 \textcolor{w h i t e}{m m m m m l l} 1$
$\textcolor{w h i t e}{m} \text{O" color(white)(XXXXll)0.0115 color(white)(mml)7.188 × 10^"-4} \textcolor{w h i t e}{X l l} 2.996 \textcolor{w h i t e}{m m m l l} 3$

The molar ratios are $\text{Fe:S:O = 1:1:3}$.

The empirical formula is ${\text{FeSO}}_{3}$.

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