# How many moles are in 160 g of methane?

Jun 10, 2017

$10.$ ${\text{mol CH}}_{4}$

(or $9.97$ ${\text{mol CH}}_{4}$)

#### 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 ($6.022 \times {10}^{23}$) of individual units of that substance.]

Methane has the chemical formula ${\text{CH}}_{4}$, 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 ($\text{C}$ and $\text{H}$) is present in the compound

• multiply the number of $\text{C}$ atoms by the relative atomic mass of $\text{C}$ (masses can be found on most periodic tables)

• multiply the number of $\text{H}$ atoms by the relative atomic mass of $\text{H}$

• 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

• $\text{C}$: $12.01$ $\text{amu}$

• $\text{H}$: $1.01$ $\text{amu}$

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

• $\text{C}$: $12.01$ $\text{g"/"mol}$

• $\text{H}$: $1.01$ $\text{g"/"mol}$

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

• $\text{C}$: (1)(12.01"g"/"mol") = 12.01"g"/"mol"

• $\text{H}$: (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:)