# Chlorine reacts with hot sodium hydroxide solution. A sample of a compound formed in the reaction was found to contain 10.8 g of sodium atoms, 16.7 g of chlorine atoms and 22.5 g of oxygen atoms. What is the empirical formula of the compound formed?

Jul 29, 2017

$N a C l {O}_{3}$

#### Explanation:

$\text{Moles of chlorine} = \frac{16.7 \cdot g}{35.45 \cdot g \cdot m o {l}^{-} 1} = 0.47 \cdot m o l$

$\text{Moles of sodium} = \frac{10.8 \cdot g}{22.99 \cdot g \cdot m o {l}^{-} 1} = 0.47 \cdot m o l$

$\text{Moles of oxygen} = \frac{22.5 \cdot g}{16.00 \cdot g \cdot m o {l}^{-} 1} = 1.41 \cdot m o l$

We divide thru by the smallest quantity, that of sodium, to give an empirical formula.....of $N a C l {O}_{3}$, i.e. $\text{sodium chlorate}$.

Jul 30, 2017

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

#### Explanation:

The empirical formula for a compound represents the lowest whole number ratio of elements. In order to determine the empirical formula of a compound, you must find the number of moles of each element. Once you know the moles, you divide the moles of each element by the smallest number of moles. This will give you the empirical formula, unless you do not have whole numbers. If you don't have all whole numbers, you need to multiply by a number that will result in all whole numbers.

Moles of Each Element

Divide the given mass of each element by its molar mass (atomic weight on the periodic table in g/mol) by multiplying by its inverse.

$\text{Na} :$10.8color(red)cancel(color(black)("g Na"))xx(1"mol Na")/(22.990color(red)cancel(color(black)("g Na")))="0.470 mol Na"

$\text{Cl} :$16.7color(red)cancel(color(black)("g Cl"))xx(1"mol Cl")/(35.453color(red)cancel(color(black)("g Cl")))="0.471 mol Cl"

$\text{O} :$22.5color(red)cancel(color(black)("g O"))xx(1"mol O")/(15.999color(red)cancel(color(black)("g O")))="1.41 mol O"

Mole Ratios (Subscripts)

Divide the moles of each element by the least number of moles.

$\text{Na} :$$\left(0.470 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mol")))/(0.470color(red)cancel(color(black)("mol}}}}\right) = 1.00$

$\text{Cl} :$$\left(0.471 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mol")))/(0.470color(red)cancel(color(black)("mol}}}}\right) = 1.00$

$\text{O} :$$\left(1.41 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mol")))/(0.470color(red)cancel(color(black)("mol}}}}\right) = 3.00$

Since the mole ratios are all whole numbers, you do not need to do anything else except write the formula.

The empirical formula is ${\text{NaClO}}_{3}$, which is the ionic compound sodium chlorate, composed of the ${\text{Na}}^{2 +}$ sodium cation and the ${\text{ClO"_3}}^{1 -}$ chlorate anion.