# A compound containing Na, S, H, and O where, Na =14.28 %, S = 9.92%, H = 6.2% and the rest are O. The molecular mass of the compound is 322. Calculate the molecular formula of compound considering all H atoms are present as water of crystallization?

## The last part which explains that H is present as the water of crystallization is a bit confusing for me.

Jun 1, 2018

Warning! Long Answer. The molecular formula is $\text{Na"_2"SO"_4·10"H"_2"O}$.

#### Explanation:

Step 1. Calculate the empirical formula

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 $\text{Na}$ to $\text{S}$ to $\text{H}$ to $\text{O}$.

Your compound contains 14.28 % $\text{Na}$, 9.92 % $\text{S}$, and 6.2 % $\text{H}$.

Assume that you have 100 g of sample.

Then it contains 14.28 g of $\text{Na}$, 9.92 g of $\text{S}$, and 6.2 g of $\text{H}$.

$\text{Mass of O = (100 - 14.28 - 9.92 - 6.2) g = 69.6 g}$

$\text{Moles of Na" = 14.28 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)( "g C")))) = "1.189 mol B}$

$\text{Moles of S" = 9.92 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(32.06 color(red)(cancel(color(black)( "g C")))) = "0.3094 mol B}$

$\text{Moles of H" = 6.2 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "6.15 mol H}$

$\text{Moles of O" = 69.6 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)( "g O")))) = "4.350 mol O}$

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

$\underline{\boldsymbol{\text{Element"color(white)(m) "Mass/g"color(white)(Xl) "Moles"color(white)(Xll) "Ratio"color(white)(m)"Integers}}}$
color(white)(m)"Na" color(white)(XXXml)14.28 color(white)(Xml)0.6211 color(white)(mm)2.007color(white)(mmmm)2
$\textcolor{w h i t e}{m} \text{S} \textcolor{w h i t e}{X X X m m l l} 9.92 \textcolor{w h i t e}{X m l} 0.3094 \textcolor{w h i t e}{m m l} 1 \textcolor{w h i t e}{m m m m m l l} 1$
$\textcolor{w h i t e}{m} \text{H} \textcolor{w h i t e}{X X X X m l l} 6.2 \textcolor{w h i t e}{m m l l} 6.15 \textcolor{w h i t e}{X m m} 19.9 \textcolor{w h i t e}{m m m m l} 20$
$\textcolor{w h i t e}{m} \text{O} \textcolor{w h i t e}{X X X m m} 69.6 \textcolor{w h i t e}{X m l l} 4.350 \textcolor{w h i t e}{m m l} 14.06 \textcolor{w h i t e}{m m m m} 14$

The empirical formula is ${\text{Na"_2"SH"_20"O}}_{14}$.

Step 2. Calculate the molecular formula of the compound

The empirical formula mass of ${\text{Na"_2"SH"_20"O}}_{14}$ is 322.19 u.

The molecular mass is 322 u.

The molecular mass must be an integral multiple of the empirical formula mass.

"MM"/"EFM" = (322color(red)(cancel(color(black)("u"))))/(322.19 color(red)(cancel(color(black)("u")))) = 0.999 ≈ 1

The molecular formula must be the same as the empirical formula.

${\text{MF" = "EF" = "Na"_2"SH"_20"O}}_{14}$

Step 3. Calculate the water of crystallization

If all the $\text{H}$ atoms are present as water of crystallization,

$\text{20H = 10H"_2"O}$

The molecular formula is $\text{Na"_2"SO"_4·10"H"_2"O}$.