# Calculate the number of molecules of CH4 in 48g CH4?

Feb 22, 2018

There are $1.8021 \cdot {10}^{24}$ molecules of $C {H}_{4}$ in 48 grams of $C {H}_{4}$.

#### Explanation:

To answer this question, you must understand how to convert grams of a molecule into the number of molecules. To do this, you have to utilize the concepts of moles and molar mass.

A mole is just a unit of measurement. Avogadro's number is equal to $6.022 \cdot {10}^{23}$ molecules/mole. This number is the number of molecules of a specific compound in which when you multiply the compound by it, it converts atomic mass into grams.

For example, one mole of hydrogen gas (${H}_{2}$) or $6.022 \cdot {10}^{23}$ molecules of ${H}_{2}$ weighs 2.016 grams because one molecule of ${H}_{2}$ has an atomic weight of 2.016.

So the overall solution for this problem is to use molar mass of $C {H}_{4}$ (methane) to convert grams of methane into moles of methane. Then, use Avogadro's number to convert moles of methane into molecules of methane.

One mole of methane equals 16.04 grams because a molecule of methane has an atomic weight of 16.04. The conversion factor will be 16.04 grams/mole.

$48 g C {H}_{4} \cdot \frac{1 m o l C {H}_{4}}{16.04 g C {H}_{4}} \cdot \frac{6.022 \cdot {10}^{23} m c l s C {H}_{4}}{1 m o l C {H}_{4}}$

When you multiple and divide everything out, you get $1.8021 \cdot {10}^{24}$ molecules of $C {H}_{4}$

Notice this is a modified T-chart so the grams $C {H}_{4}$ cancels out when you do the first conversion, and the moles $C {H}_{4}$ cancels out when you do the second conversion. This leaves you with the unit molecules of $C {H}_{4}$ which corresponds to what the question asks.