In this problem, we're given the number of molecules of NH3, and asked to find it's mass. Anytime we're dealing with a problem dealing with the number of something, and it's mass, we're going to have to use the molar mass in there, in order to get that calculation completed. And so, the first thing we're going to is actually find the molar mass of our NH three. So, nitrogen has a mass of 14.01 and there are one of those. Hydrogen is 1.008 and there are three of those, and so our molar mass as a compound is 17.03 grams per mol of our NH3. So now, we know the molar mass but we need to get to the moles of NH3. And so, we see our molecules, 5.00 times 10 to the 24th molecules, and we can abbreviate this molec, we want to avoid mol for molecules because that represents moles. So, we have f5 times 10 to the 24th molecules, and here I'm going to use Avogadro's number, I'm going to convert to Mols. Because then once I get to mols, then I can use my molar mass to find the mass. And so, here we have 6.022 times 10 to the 23rd molecules per mole. Remember, that any time we're dealing with the number of something we know it's going to be 6.022 times 10 to the 23rd, whatever it is per mole. It doesn't matter whether it's molecules, atoms, ions, whatever, this relationship will always be true for one mole. Now, we can use our molar mass, which is 17.03 grams per mole of NH3, because I need to make sure all my units are cancelling out. So, I see molecule cancelling out with molecules, moles cancels with moles, and the units I'm left with are grams of NH3, and so when I do the calculation, what I end up with is 141 grams of NH3. The key thing in doing these problems is always making sure your units cancel out correctly, and you get to the units that you want for your final answer.