Question

The percent by mass of nitrogen is \(46.7 \%\) for a species containing only nitrogen and oxygen. Which of the following could be this species? a. \(\mathrm{N}_{2} \mathrm{O}_{5}\) b. \(\mathrm{N}_{2} \mathrm{O}\) c. \(\mathrm{NO}_{2}\) d. NO e. \(\mathrm{NO}_{3}\)

Short answer

The correct answer is d) NO, which has a percent by mass of nitrogen approximately equal to the given value (\(46.7 \%\)).

Step-by-step solution

Calculate molar mass of nitrogen and oxygen

To calculate the percent of nitrogen by mass, we need the molar masses of nitrogen and oxygen. In the periodic table, the molar mass of nitrogen (N) is \(14.01 \, g/mol\), and that of oxygen (O) is \(16.00 \, g/mol\).

Calculate percent by mass for each species

For each species, calculate the percent by mass of nitrogen using the formula: Percent by mass of nitrogen = \(\frac{\text{mass of nitrogen in the species}}{\text{total mass of the species}} × 100\) Let's do this for each species: a) N2O5 Mass of nitrogen in the species = 2 × 14.01 = 28.02 g Total mass of the species = 28.02 g (for nitrogen) + 5 × 16.00 g (for oxygen) = 108.02 g Percent by mass of nitrogen = \(\frac{28.02}{108.02}\) × 100 = 25.91% b) N2O Mass of nitrogen in the species = 2 × 14.01 = 28.02 g Total mass of the species = 28.02 g (for nitrogen) + 16.00 g (for oxygen) = 44.02 g Percent by mass of nitrogen = \(\frac{28.02}{44.02}\) × 100 = 63.62% c) NO2 Mass of nitrogen in the species = 14.01 g Total mass of the species = 14.01 g (for nitrogen) + 2 × 16.00 g (for oxygen) = 46.01 g Percent by mass of nitrogen = \(\frac{14.01}{46.01}\) × 100 = 30.44% d) NO Mass of nitrogen in the species = 14.01 g Total mass of the species= 14.01 g (for nitrogen) + 16.00 g (for oxygen) = 30.01 g Percent by mass of nitrogen = \(\frac{14.01}{30.01}\) × 100 = 46.68% (approximately equal to 46.7%) e) NO3 Mass of nitrogen in the species = 14.01 g Total mass of the species = 14.01 g (for nitrogen) + 3 × 16.00 g (for oxygen) = 62.01 g Percent by mass of nitrogen = \(\frac{14.01}{62.01}\) × 100 = 22.58%

Compare values with the given percent by mass of nitrogen

Now we will compare the calculated values with the given percent by mass of nitrogen (\(46.7 \%\)) for each species: a) N2O5: 25.91% b) N2O: 63.62% c) NO2: 30.44% d) NO: 46.68% (approximately equal to 46.7%) e) NO3: 22.58% Only species d) NO has a percent by mass of nitrogen approximately equal to the given value (\(46.7 \%\)). Therefore, the correct answer is d).

Key concepts

Molar Mass

When studying chemistry, you'll frequently hear the term 'molar mass'. It's quite similar to the concept of weight, but specific to chemistry. Molar mass is essentially the weight of one mole of a substance (molecular or atomic). What's a mole, you ask? Think of it as a standard number of particles – much like a 'dozen' means 12 items, a 'mole' means approximately 6.022 x 10²³ particles (Avogadro's number).

This number helps chemists relate the weight of substances to how many particles are present, which is invaluable when preparing chemical solutions and conducting reactions. The units for molar mass are grams per mole (g/mol).

To find the molar mass, you would look at the chemical formula of a substance and sum the average atomic masses of all the elements in a compound, multiplied by their respective number of atoms. This leads us directly to understanding the chemical formula itself, which gives us the ratios of elements within a compound.

Chemical Formula

The chemical formula is like a recipe for molecules. It tells you which elements make up a compound and in what ratios. Take for example water (H₂O), its chemical formula tells us that each molecule consists of 2 hydrogen (H) atoms and 1 oxygen (O) atom.

If you change the numbers, you're 'cooking up' a whole new chemical. The chemical formula is crucial because it allows us to calculate the molar mass of the substance. In the context of our original exercise, understanding chemical formulas is key for calculating percent by mass composition.

For instance, to calculate the percent by mass of nitrogen in N₂O, we add together the molar masses of the elements based on the quantity indicated by the chemical formula. This calculation tells us how much of the compound's weight comes from each element.

Stoichiometry

Stoichiometry is a section of chemistry that deals with the quantitative relationships of the elements in chemical reactions. Essentially, it's the art of balancing and predicting the outcomes of chemical equations.

One of the most common stoichiometric calculations involves using a balanced chemical equation to determine the amount of reactants needed or products formed. Percent by mass composition, which is central to our exercise, is one aspect of stoichiometry, as it requires understanding the proportions of elements within a compound to calculate how much of the mass comes from each element.

The term 'percent by mass' refers to the percentage of the total mass of a compound that is attributable to a given component. In our exercise, we examined various nitrogen-oxygen compounds to find which one had a nitrogen percent by mass that matched the given data. This is an excellent example of stoichiometry at work, as we use molar masses and the chemical formulas to solve the problem systematically.