Question

Arrange the following substances in order of increasing mass percent of nitrogen. \(\begin{array}{ll}{\text { a. } \mathrm{NO}} & {\text { c. } \mathrm{NH}_{3}} \\\ {\text { b. } \mathrm{N}_{2} \mathrm{O}} & {\text { d. SNH }}\end{array}\)

Short answer

The correct order of the substances in increasing mass percent of nitrogen is: SNH (29.79%), NO (46.67%), N2O (63.64%), NH3 (82.35%).

Step-by-step solution

Calculate the molar mass of each compound

To find the mass percentage of nitrogen in each compound, we must first calculate the molar mass of each compound using the atomic masses of the elements (Nitrogen, Oxygen, and Sulfur). Nitrogen (N): 14 g/mol Oxygen (O): 16 g/mol Sulfur (S): 32 g/mol For each compound: 1. NO: Molar mass = 14 (N) + 16 (O) = 30 g/mol 2. N2O: Molar mass = 14x2 (N) + 16 (O) = 44 g/mol 3. NH3: Molar mass = 14 (N) + 3x1 (H) = 17 g/mol 4. SNH: Molar mass = 32 (S) + 14 (N) + 1 (H) = 47 g/mol

Calculate mass percentage of nitrogen in each compound

Now we can calculate the mass percentage of nitrogen in each compound: 1. NO: Mass % of N = \(\frac{14}{30}\) x 100 = 46.67% 2. N2O: Mass % of N = \(\frac{28}{44}\) x 100 = 63.64% 3. NH3: Mass % of N = \(\frac{14}{17}\) x 100 = 82.35% 4. SNH: Mass % of N = \(\frac{14}{47}\) x 100 = 29.79%

Arrange the compounds in increasing order of nitrogen mass percent

Now that we have found the mass percentage of nitrogen in each compound, we can arrange them in increasing order: 1. SNH: 29.79% 2. NO: 46.67% 3. N2O: 63.64% 4. NH3: 82.35% So, the correct order of the substances in increasing mass percent of nitrogen is: SNH, NO, N2O, NH3.

Key concepts

Molar Mass

When studying chemistry, understanding molar mass is crucial. Molar mass refers to the weight of one mole of a substance, measured in grams per mole (g/mol).
In order to determine the molar mass of a compound, we add together the atomic masses of all the elements present in the molecule.
For instance, let's look at nitrogen monoxide (NO):

• Nitrogen (N) has an atomic mass of 14 g/mol
• Oxygen (O), on the other hand, has an atomic mass of 16 g/mol

Thus, the molar mass of NO is the sum of these masses: \[ 14 ext{ g/mol (N)} + 16 ext{ g/mol (O)} = 30 ext{ g/mol} \]
By calculating the molar mass, we are able to proceed with further calculations, such as finding mass percentages, which is extremely useful in chemical analyses.

Nitrogen Compounds

Nitrogen is a versatile element that forms a variety of compounds. Here, we look at a few common nitrogen compounds: NO, \( ext{N}_2 ext{O} \), \( ext{NH}_3 \), and SNH.
Each compound features nitrogen combined with different elements, affecting their properties and usage.

• NO (Nitric oxide) is a reactive gas, crucial in environmental chemistry and biological processes.
• \( ext{N}_2 ext{O} \) (Nitrous oxide) is well-known as "laughing gas," used as an anesthetic and has environmental implications.
• \( ext{NH}_3 \) (Ammonia) serves as a fundamental compound in fertilizers and industrial applications, highly soluble in water.
• SNH is a less common compound, containing sulfur and showcasing unique chemistry.

Understanding these compounds' differing chemical structures helps us comprehend their behavior and significance in various fields of study.

Chemical Calculations

Chemical calculations are an integral part of chemistry, enabling us to quantify and understand the compositions of substances. One of the most common calculations is determining the mass percent of an element within a compound.
Here's how it works:

• First, calculate the molar mass of the compound, considering each element's contribution based on its atomic mass.
• Next, find the total mass of the desired element (e.g., nitrogen) within the compound.
• The mass percent is then calculated by dividing the element's mass by the compound's total molar mass and multiplying by 100 to convert to a percentage:

For example, in \( ext{NH}_3 \), to find the mass percent of nitrogen, divide 14 (mass of N) by 17 (molar mass of \( ext{NH}_3 \)) and multiply by 100:\[ \frac{14}{17} \times 100 = 82.35\% \]
This process allows scientists and students to understand the concentration of specific elements in various substances, guiding experiments and applications effectively.