# What is the empirical formula of a compound composed of 3.25% hydrogen (H), 19.36% carbon (C), and 77.39% oxygen (O) by mass?

Feb 27, 2018

CH2O3

#### Explanation:

i used their percentage masses divided by each element's relative atomic mass then divided thru by the smallest mole ratio.

moles of carbon= $\frac{19.36}{12} = 1.613$
moles of hydrogen= $\frac{3.25}{1} = 3.25$
moles of oxygen= $\frac{77.39}{16} = 4.8369$

divide thru with the smallest ratio

C $\frac{1.613}{1.613}$H $\frac{3.25}{1.613}$O $\frac{4.8369}{1.613}$

=CH2O3

Feb 27, 2018

The empirical formula is the SIMPLEST whole number ratio defining constituent elements in a species...

#### Explanation:

As with all these problems, the usual rigmarole is to assume a $100 \cdot g$ mass and then interrogate the molar quantities of each element.

$\text{Moles of hydrogen,}$ $\frac{3.25 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 3.220 \cdot m o l .$

$\text{Moles of carbon,}$ $\frac{19.36 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 1.612 \cdot m o l .$

$\text{Moles of oxygen,}$ $\frac{77.39 \cdot g}{16.00 \cdot g \cdot m o {l}^{-} 1} = 4.84 \cdot m o l .$

And we divide thru by the SMALLEST molar quantity to get the empirical formula...

$\text{Empirical formula} \equiv {C}_{\frac{1.612 \cdot m o l}{1.612 \cdot m o l}} {H}_{\frac{3.220 \cdot m o l}{1.612 \cdot m o l}} {O}_{\frac{4.84 \cdot m o l}{1.612 \cdot m o l}} \equiv C {H}_{2} {O}_{3}$

Feb 27, 2018

The empirical formula is $\text{CH"_2"O"_3}$.

#### Explanation:

An empirical formula represents the lowest whole number ratio of elements in a compound. There are several steps involved.

1. Determine the moles of each element by dividing its given mass by its molar mass. When percentages add up to 100%, we can directly convert percentage to mass in grams.

2. Determine the mole ratios between each element and the lowest number of moles by dividing the moles of each element by the lowest number of moles. This step determines the subscripts for the elements.

3. If all mole ratios are whole numbers, you have the lowest whole number ratio and can write them as subscripts for the compound.

4. If one or more mole ratio is not a whole number, then all ratios must be multiplied by a factor that will make all ratios whole numbers.

Moles of elements

Since molar mass is a fraction (g/mol), I prefer to divide by the molar mass by multiplying by its reciprocal (mol/g). I think it makes what happens with the units more clear.

$\text{H} :$ 3.25color(red)cancel(color(black)("g H"))xx(1"mol H")/(1.008color(red)cancel(color(black)("g H")))="3.22 mol H"

$\text{C} :$ 19.36color(red)cancel(color(black)("g C"))xx(1"mol C")/(12.011color(red)cancel(color(black)("g C")))="1.61 mol C"

$\text{O} :$ 77.39color(red)cancel(color(black)("g O"))xx(1"mol O")/(15.999color(red)cancel(color(black)("g O")))="4.84 mol O"

Mole ratios

Because moles cancel and mole ratios are dimensionless, I am not going to label the moles.

$\text{H} :$ $\frac{3.22}{1.61} = 2$

$\text{C} :$ $\frac{1.61}{1.61} = 1$

$\text{O} :$ $\frac{4.84}{1.61} = 3$

The empirical formula is $\text{CH"_2"O"_3}$.