# Question 2dda1

Aug 31, 2017

The formula of the second oxide is ${\text{X"_2"O}}_{3}$.

#### Explanation:

Calculate the molar mass of $\text{X}$

1 mol of ${\text{X"_3"O}}_{4}$ contains 3 mol of $\text{X}$ and 4 mol of $\text{O}$.

Assume that you have 1 mol of ${\text{X"_3"O}}_{4}$.

Then

$\text{Mass of O" = 4 color(red)(cancel(color(black)("mol O"))) × "16.00 g O"/(1 color(red)(cancel(color(black)("mol O")))) = "64.00 g O}$

and

$\text{Mass of X"_3"O"_4 = 64.00 color(red)(cancel(color(black)("g O"))) × ("100 g X"_3"O"_4)/(27.6 color(red)(cancel(color(black)("g O")))) = "232 g}$

$\text{Mass of X" = "Mass of X"_3"O"_4 - "Mass of O" = "232 g - 64.00 g" = "168 g}$

This is the mass of 3 mol of $\text{X}$.

$\text{Molar mass of X" = "168 g"/"3 mol" = "56.0 g/mol}$

Calculate the empirical formula of the second oxide

Assume that you have 100 g of the second oxide.

Then it contains 30 g of $\text{O}$ and 70 g of $\text{X}$.

$\text{Moles of X" = 70 color(red)(cancel(color(black)("g X"))) × "1 mol X"/(56.0 color(red)(cancel(color(black)( "g X")))) = "1.25 mol X}$

$\text{Moles of O" = 30 color(red)(cancel(color(black)("g O"))) × "1 mol O"/(16.00 color(red)(cancel(color(black)( "g O")))) = "1.88 mol O}$

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

$\boldsymbol{\underline{\text{Element"color(white)(m) "Mass/g"color(white)(X) "Moles"color(white)(Xll) "Ratio"color(white)(mm)"×2"color(white)(mm)"Integers}}}$
color(white)(mm)"X" color(white)(XXXmm)70 color(white)(Xmm)1.25 color(white)(mmm)1color(white)(mmmm)2color(white)(mmmmml)2#
$\textcolor{w h i t e}{m m} \text{O} \textcolor{w h i t e}{X X X m m} 30 \textcolor{w h i t e}{X m m} 1.88 \textcolor{w h i t e}{m m m} 1.50 \textcolor{w h i t e}{m m l l} 3.00 \textcolor{w h i t e}{m m m l l} 3$

The empirical formula is ${\text{X"_2"O}}_{3}$.