# How do we use the molar mass of Oxygen to solve this stoichiometry problem?

## For the reaction represented by the equation $C {X}_{4} + 2 {O}_{2} \implies C {O}_{2} + 2 {X}_{2} O$ where $\text{X}$ is an unknown element 9g of $C {X}_{4}$ completely reacts with 1.74g of Oxygen. What is the approximate molar mass of $X$?

Apr 25, 2018

The approximate molar mass of $\text{X}$ is 80 g/mol.

#### Explanation:

Let $x =$ the molar mass of $\text{X}$.

The balanced equation is

${M}_{\textrm{r}} : \textcolor{w h i t e}{m m} \text{12.01+4} x$
$\textcolor{w h i t e}{m m m l m m} \text{CX"_4 + "2O"_2 → "CO"_2 + "2X"_2"O}$
$\text{Mass/g} : \textcolor{w h i t e}{m m} 9 \textcolor{w h i t e}{m m} 1.74$

Step 1. Calculate the moles of ${\text{O}}_{2}$

${\text{Moles of O"_2 = 1.74 color(red)(cancel(color(black)("g O"_2))) × "1 mol O"_2/(32.00 color(red)(cancel(color(black)("g O"_2)))) = "0.0544 mol O}}_{2}$

Step 2. Calculate the moles of ${\text{CX}}_{4}$

${\text{Moles of CX"_4 = 0.0544 color(red)(cancel(color(black)("mol O"_2))) × "1 mol CX"_4/(2 color(red)(cancel(color(black)("mol O"_2)))) = "0.0272 mol CX}}_{4}$

Step 3. Calculate the molar mass of ${\text{CX}}_{4}$

$\text{Molar mass of CX"_4 = ("9 g CX"_4)/("0.0272 mol CX"_4) = "330 g/mol}$

Step 5. Calculate the molar mass of $\text{X}$

$12.01 + 4 x = 330$

$4 x = 330 - 12.01 = 320$

${A}_{\textrm{r}} = x = \frac{320}{4} = 80$