How many moles are in 160 g of methane?

1 Answer
Jun 10, 2017

Answer:

#10.# #"mol CH"_4#

(or #9.97# #"mol CH"_4#)

Explanation:

To convert between moles and grams of any substance, we need to know that's substance's molar mass.

The molar mass of a substance (in this case, methane) is the mass, most typically in grams, of one mole of that substance. [One mole of any substance is Avogadro's number (#6.022xx10^23#) of individual units of that substance.]

Methane has the chemical formula #"CH"_4#, meaning that one mole of methane contains one mole of carbon atoms and four moles of hydrogen atoms. To find the molar mass of methane, we must

  • find how many of each element (#"C"# and #"H"#) is present in the compound

  • multiply the number of #"C"# atoms by the relative atomic mass of #"C"# (masses can be found on most periodic tables)

  • multiply the number of #"H"# atoms by the relative atomic mass of #"H"#

  • sum the total

First, let's realize that there is

  • #1# atom of carbon, and

  • #4# atoms of hydrogen

per methane molecule. The atomic masses are

  • #"C"#: #12.01# #"amu"#

  • #"H"#: #1.01# #"amu"#

The mass of a substance in atomic mass units (#"amu"#) is the same number as the molar mass of that substance, in #"g"/"mol"#.
Therefore, the molar masses of #"C"# and #"H"# are

  • #"C"#: #12.01# #"g"/"mol"#

  • #"H"#: #1.01# #"g"/"mol"#

Now, multiplying each by the number of moles per mole of methane, we have

  • #"C"#: #(1)(12.01"g"/"mol") = 12.01"g"/"mol"#

  • #"H"#: (4)(1.01"g"/"mol") = 4.40"g"/"mol"#

Finally, summing the total, we have

#underbrace(12.01"g"/"mol")_ "C" + underbrace(4.04"g"/"mol")_ "H" = underbrace(color(red)(16.05"g"/"mol"))_ "molar mass"#

Lastly, using dimensional analysis (which you'll find to use quite often), let's use this and the given mass to find the number of moles of methane:

#160# #"g CH"_4((1"mol CH"_4)/(16.05"g CH"_4)) = color(blue)(10.# #color(blue)("mol CH"_4#

rounded to #2# significant figures, the (technical) amount given in the problem. If you count the last #0# as significant, the answer is #color(blue)(9.97# #color(blue)("mol CH"_4#

I hope this wasn't too confusing. This really is a straightforward process that you'll use endless times:)