# A 10 g sample of a compound contains 4.00g C, 0.667g H, and 5.33g O. What is the molecular formula?

## The molecular mass is 180.156 g/mol.

Nov 29, 2016

The molecular formula is ${\text{C"_6"H"_12"O}}_{6}$.

#### Explanation:

We must first determine the empirical formula of the compound.

We must figure out the moles of $\text{C}$, $\text{H}$, and $\text{O}$ and then calculate their ratio.

Step 1. Calculate the moles of each element.

$\text{Moles of C" = 4.00 color(red)(cancel(color(black)("g C"))) × "1 mol C"/(12.01 color(red)(cancel(color(black)("g C")))) = "0.3331 mol C}$

$\text{Moles of H" = 0.667 color(red)(cancel(color(black)("g H"))) × "1 mol H"/(1.008 color(red)(cancel(color(black)("g H")))) = "0.6617 mol C}$

$\text{Moles of O" = 5.33 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)("g O")))) = "0.3331 mol O}$

Step 2. Calculate the empirical formula.

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

$\text{Element"color(white)(Agll) "Mass/g"color(white)(Xm) "Moles"color(white)(Xm) "Ratio"color(white)(m)color(white)(l)"Integers}$
stackrel(———————————————————)(color(white)(m)"C" color(white)(XXXmm)4.00 color(white)(Xmm)0.3331 color(white)(Xm)1color(white)(mmmmml)1)
$\textcolor{w h i t e}{m} \text{H} \textcolor{w h i t e}{X X X X m} 0.667 \textcolor{w h i t e}{m m l} 0.6617 \textcolor{w h i t e}{X m} 1.987 \textcolor{w h i t e}{m m m l} 2$
$\textcolor{w h i t e}{m} \text{O} \textcolor{w h i t e}{X X X X m} 5.33 \textcolor{w h i t e}{m m m} 0.3331 \textcolor{w h i t e}{X m} 1.000 \textcolor{w h i t e}{m m m l} 1$

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

Step 3. Calculate the empirical formula mass

The empirical formula mass of $\text{CH"_2"O}$ is 30.03 u.

Step 4. Calculate the molecular mass.

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

$\text{MM" = n × "EFM}$

n = "MM"/"EFM" = (180.156 color(red)(cancel(color(black)("u"))))/(30.03 color(red)(cancel(color(black)("u")))) = 5.999 ≈ 6

Step 5. Calculate the Molecular Formula

The molecular formula is 6 times the empirical formula.

The molecular formula is ("CH"_2"O")_6 = "C"_6"H"_12"O"_6.