What is the percent of nitrogen in 4.444 moles of ammonium sulfide?

Answer 1

It is important to know that the percent of nitrogen in 4.444 moles of ammonium sulfide is the same as the percent of nitrogen in 454 grams or 4843 moles or 96 kg, etc. Remember the law of definite proportions - chemical compounds always contain the same proportion of elements by mass. Perhaps you were asking how much nitrogen is in 4.444 moles of ammonium sulfide given the percent of nitrogen in any given mass. So we'll do that too: find the percent of nitrogen in any given sample and apply it specifically to 4.444 moles.

Before we go directly to the 4.444 moles, we have to figure out how much nitrogen is in any amount of ammonium sulfide by percent. To do this, we need the atomic weights of the elements and add them up to find the total molar mass of the compound.

Ammonium sulfide = (NH4)2S

Nitrogen = 14.0 grams × 2 = 28.0 grams

Hydrogen = 1.01 grams × 8 = 8.08 grams

Sulfur = 32.1 grams

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Ammonium sulfide = 68.2 grams

Now we take the mass of nitrogen and divide it by the total mass to get our percent.

Nitrogen ÷ Ammonium sulfide = % Nitrogen

28.0 grams ÷ 68.2 grams = 0.411 = 41.1% Nitrogen in Ammonium sulfide

Since we know that in any amount of Ammonium sulfide contains 41.1% of Nitrogen, we can apply it to the mass given.

41.1% of 4.444 moles = .411 × 4.444 = 1.83 moles of Nitrogen in 4.444 moles Ammonium sulfide

Answer 2

Ammonium sulfide has the formula (NH4)2S, which contains 2 nitrogen atoms per formula unit. One mole of (NH4)2S contains 2 moles of nitrogen. Therefore, in 4.444 moles of (NH4)2S, there are 8.88 moles of nitrogen. The percent of nitrogen in the compound would be (8.88 moles / 4.444 moles) * 100%, which is 200%.