When 0.273 g of Mg is heated strongly in a nitrogen (N2) atmosphere, a chemical reaction occurs

When 0.273 g of Mg is heated strongly in a nitrogen (N2)...

When $0.273 \mathrm{~g}$ of $\mathrm{Mg}$ is heated strongly in a nitrogen $\left(\mathrm{N}_{2}\right)$ atmosphere, a chemical reaction occurs. The product of the reaction weighs 0.378 g. Calculate the empirical formula of the compound containing $\mathrm{Mg}$ and $\mathrm{N}$. Name the compound.

Key Concepts

Empirical Formula Determination

This concept involves finding the simplest whole-number ratio of atoms in a compound, based on the mass or percentage composition of its elements. It requires converting the masses of individual elements to moles, finding the ratio between them, and then adjusting to the simplest integral relationship.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It uses the concept of moles as a bridge between the mass of substances and the number of particles, enabling calculations of product yields and reactant consumption based on balanced chemical equations.

Conversion of Mass to Moles

This fundamental concept involves using the molar mass of a substance to convert a given mass into moles. It is essential for linking the measurable mass of a substance with the number of particles, which is crucial for performing stoichiometric calculations in chemical reactions.

Chemical Nomenclature

Chemical nomenclature is the systematic method of naming chemical compounds. It ensures that each compound has a unique and standardized name, which is especially important for clearly communicating the identity and composition of compounds formed in chemical reactions.

Transcript

00:01 So for this problem, we know that some magnesium reacts with nitrogen gas, and we want to find the empirical formulas compound and name it.

00:09 So we are given that there's 0 .273 grams of magnesium, and that the entire compound that contains both magnesium and nitrogen, the mass of that compound is 0 .378 grams.

00:29 So now we want to convert both the mass of magnesium to moles and then find the mass of the nitrogen and convert that to moles.

00:38 So the first thing that we're going to want to do is find moles of magnesium.

00:42 Well, we know that we have .273 grams of magnesium.

00:48 And if we divide that by the molar mass of magnesium, which is 24 .31, and you'll get moles of magnesium.

00:58 But because this is going to be such a small number as we can already see to multiply this by a thousand to get millimoles.

01:12 And if we do that, we will get 11 .23 millimoles of magnesium.

01:22 Now we want to do the same thing for nitrogen.

01:25 Well, first we need to define the mass of the nitrogen, but we know that that's just going to be the mass of the compound minus the mass of the magnesium, because the compound contains only magnesium and nitrogen.

01:38 So if you subtract this tool, we'll get only the nitrogen.

01:42 And when you do that, we get 0 .105 grams of nitrogen gas.

01:50 So now you want to convert grams of nitrogen into moles of nitrogen.

01:58 So we have 0 .105 grams of nitrogen.

02:07 And if i want to find moles, i divide by the molar mass, which is 14 .01...

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