# If something is made of 1.05 g "Ni" and 0.58 g "O", what is the empirical formula?

Aug 20, 2017

The empirical formula is ${\text{NiO}}_{2}$, which is nickel(IV) oxide.

#### Explanation:

The empirical formula of a compound is the lowest whole number ratio of the elements in the compound. These whole numbers are the subscripts.

To determine an empirical formula, convert the given mass to moles. Then divide the number of moles of each element by the lowest number of moles. If you get whole numbers, those are the lowest whole number ratio. If you don't get whole numbers, you will need to multiply by a number that will give whole numbers.

Moles

Determine the moles of each element by dividing its given mass by its molar mass. The molar mass of each element is its atomic weight on the periodic table in g/mol. Since molar mass is a fraction, $\text{g"/"mol}$, multiply the given mass of each element by the inverse of its molar mass.

$\text{Ni:}$ 1.05color(red)cancel(color(black)("g Ni"))xx(1"mol Ni")/(58.6934color(red)cancel(color(black)("g Ni")))="0.0179 mol Ni"

$\text{O:}$ 0.58color(red)cancel(color(black)("g O"))xx(1"mol O")/(15.999color(red)cancel(color(black)("g O")))="0.036 mol O"

Mole Ratios

Determine the mole ratio between the two elements by dividing the mol of each element by the lowest number of moles.

$\text{Ni:}$ $\left(0.0179 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mol")))/(0.0179color(red)cancel(color(black)("mol}}}}\right) = 1.00$

$\text{O:}$ $\left(0.036 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mol")))/(0.0179color(red)cancel(color(black)("mol}}}}\right) = 2.0$

The empirical formula is ${\text{NiO}}_{2}$.