# Question 34aeb

Jun 14, 2017

The empirical formula of the compound is $\text{Li"_2"O}$.

#### Explanation:

In order to determine the empirical formula for a compound, you must first determine the number of moles of each element from its molar mass, then divide the number number of moles of each element by the least number of moles.

color(blue)("Molar Masses"

The molar mass of an element is its atomic weight on the periodic table in grams/mol, or g/mol.

Molar mass Li: $\text{6.941 g/mol}$

Molar mass: $\text{15.999 g/mol}$

color(blue)("Moles Li"

Divide the given mass of lithium by its molar mass by multiplying by the inverse of its molar mass.

197.5color(red)cancel(color(black)("g Li"))xx(1"mol Li")/(6.941color(red)cancel(color(black)("g Li")))="28.45 mol Li"

color(blue)("Moles O"

Divide the given mass of oxygen by its molar mass by multiplying by the inverse of its molar mass.

227.5color(red)cancel(color(black)("g O"))xx(1"mol O")/(15.999color(red)cancel(color(black)("g O")))="14.22 mol O"

Divide each number of moles by the lowest number of moles.

$\text{Li:}$ (28.45color(red)cancel(color(black)("mol")))/(14.22 color(red)cancel(color(black)("mol")))="2.001# $\approx 2$

$\text{O} :$$\left(14.22 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mol")))/(14.22color(red)cancel(color(black)("mol}}}}\right) = 1.000$

The mole ratios, if they are whole numbers, are the subscripts for each element in the compound.

Empirical formula of the compound:

$\text{Li"_2"O}$

This formula represents the ionic compound lithium oxide, which is known to have the formula unit $\text{Li"_2"O}$. The vast majority of ionic compounds represent the lowest whole number ratio of the elements of which they are made.