Question

Calculate the empirical formula of each compound from the percent compositions given: (a) \(63.6 \% \mathrm{~N}, 36.4 \% \mathrm{O}\) (d) \(43.4 \% \mathrm{Na}, 11.3 \% \mathrm{C}, 45.3 \% \mathrm{O}\) (b) \(46.7 \% \mathrm{~N}, 53.3 \% \mathrm{O}\) (e) \(18.8 \% \mathrm{Na}, 29.0 \% \mathrm{Cl}, 52.3 \% \mathrm{O}\) (c) \(25.9 \% \mathrm{~N}, 74.1 \% \mathrm{O}\) (f) \(72.02 \% \mathrm{Mn}, 27.98 \% \mathrm{O}\)

Short answer

(a) NO (b) NO (c) NO2 (d) Na2CO3 (e) NaClO3 (f) MnO2

Step-by-step solution

- Convert Percentages to Grams

Assume you have 100 grams of the compound. This simplifies the conversion since each percentage can be directly interpreted as grams. For example, 63.6% of N translates to 63.6 grams of N.

- Convert Grams to Moles

Use the molar mass of each element to convert the mass in grams to moles. The formula is: \[ \text{moles} = \frac{\text{mass in grams}}{\text{molar mass of the element}} \]

- Calculate Moles for Each Compound

(a) For N: \[ \frac{63.6 \text{ grams}}{14.01 \text{ g/mol}} = 4.54 \text{ moles} \] For O: \[ \frac{36.4 \text{ grams}}{16.00 \text{ g/mol}} = 2.28 \text{ moles} \]

- Divide by the Smallest Number of Moles

Identify the smallest number of moles calculated and divide all the moles by this number to get a simple whole-number ratio.

- Determine the Empirical Formula

Use the mole ratios from the previous step to write the empirical formula, ensuring all subscripts are in the smallest possible whole numbers.

Key concepts

percent composition

Percent composition tells us the percentage by mass of each element in a compound. It's typically expressed in percentages. To calculate this, you follow the steps below:

• Firstly, assume you have 100 grams of the compound. That way, each percentage can be easily converted to grams.
• For example, if a compound is 63.6% nitrogen, you would assume it has 63.6 grams of nitrogen per 100 grams of the compound.

This simplifies things for later steps, such as molar mass and mole conversion. Once you have the grams, you're ready to move on to the next step.

molar mass

Molar mass is the mass of one mole of a substance, measured in g/mol. Every element has a unique molar mass, which you can find on the periodic table. For nitrogen, it's 14.01 g/mol, whereas for oxygen, it's 16.00 g/mol.

• To find the molar mass of an element, look at its atomic weight on the periodic table.
• Express this mass in grams per mole (g/mol).

For instance, if you have 63.6 grams of nitrogen, you convert it to moles by dividing by its molar mass (14.01 g/mol), resulting in 4.54 moles of nitrogen. This conversion is crucial for determining the empirical formula of a compound.

mole conversion

Mole conversion involves converting mass to moles or vice versa using the molar mass. To perform this, use the formula:
\[ \text{moles} = \frac{\text{mass in grams}}{\text{molar mass}} \] For example:

• Convert 36.4 grams of oxygen to moles by dividing by its molar mass (16.00 g/mol), giving 2.28 moles of oxygen.

This conversion allows you to work with moles, which is essential for comparing ratios of different elements in the compound. Always remember that the mole is a fundamental unit in chemistry, making it easier to balance equations and find empirical formulas.

chemistry problem solving

Chemistry problem solving often follows a structured method to ensure accuracy. For empirical formulas, the steps include:

• Converting percentages to grams, and then grams to moles.
• Next, divide all moles by the smallest number of moles obtained to find the simplest ratio.
• Finally, use these ratios to write the empirical formula.

For example, after converting grams to moles for nitrogen and oxygen in a compound and dividing by the smallest number of moles, you might find the ratio of nitrogen to oxygen to be 2:1. This ratio helps in writing the empirical formula, such as \(\text{N}_2\text{O}\). Sticking to these steps helps solve complex problems systematically.

scientific method

The scientific method is crucial in solving chemistry problems like empirical formulas. It involves systematic steps:

• Observation: Identifying what you need to find (empirical formula).
• Hypothesis: Assuming 100 grams to simplify percentage conversion.
• Experiment: Converting grams to moles using molar mass.
• Analysis: Dividing moles by the smallest number to get ratios.
• Conclusion: Determining the empirical formula.

This method provides a clear, logical pathway to follow, ensuring all steps are covered and no part of the problem is overlooked. It helps in organizing thoughts and procedures, making complex problems more manageable.