Question

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

For the reaction represented by the equation `CX_4+2O_2=>CO_2+2X_2O`
where `"X"` is an unknown element

9g of `CX_4` completely reacts with 1.74g of Oxygen. What is the approximate molar mass of `X`?

Answer 1

The approximate molar mass of `"X"` is 80 g/mol.

Explanation:

Let `x =` the molar mass of `"X"`.

The balanced equation is

`M_text(r):color(white)(mm)"12.01+4"x`
`color(white)(mmmlmm)"CX"_4 + "2O"_2 → "CO"_2 + "2X"_2"O"`
`"Mass/g":color(white)(mm)9color(white)(mm)1.74`

Step 1. Calculate the moles of `"O"_2`

`"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 `"CX"_4`

`"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 `"CX"_4`

`"Molar mass of CX"_4 = ("9 g CX"_4)/("0.0272 mol CX"_4) = "330 g/mol"`

Step 5. Calculate the molar mass of `"X"`

`12.01 + 4x = 330`

`4x = 330 - 12.01 = 320`

`A_text(r) = x = 320/4 = 80`