How many moles are in 160 g of methane?

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:)