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#?

1 Answer
Apr 25, 2018

Answer:

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#