Learning Goal: To understand what an empirical formula is and to learn the procedure for finding empirical formulas. An empirical formula expresses the simplest ratio of the elements volved. The compound P4010 has a P:O ratio of 4:10. However, this ratio can be simplified to 2:5. Therefore the empirical formula for P4O10 is P2O5. Another example is hydrogen peroxide, which has the formula H₂O2 and an empirical formula of HO. One way that chemists analyze new or unknown compounds is to determine the percentage composition of elements experimentally, which yields the empirical formula. A compound is 80.0% carbon and 20.0% hydrogen by mass. Assume you have a 100.-g sample of this compound. Part A How many grams of each element are in this sample? Enter the number of grams of carbon followed by the number of grams of hydrogen, separated by a comma (e.g., 30.0,70.0). mass of C, mass of H = VO ΑΣΦ C ? 6.0 g

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### Understanding Empirical Formulas

**Learning Goal:**
To understand what an empirical formula is and to learn the procedure for finding empirical formulas.

#### Key Concepts:
1. **Definition:**
   - An empirical formula expresses the simplest ratio of the elements involved in a compound. 

2. **Example 1:**
   - The compound \( P_4O_{10} \) has a phosphorus to oxygen (P:O) ratio of 4:10. However, this ratio can be simplified to 2:5. Therefore, the empirical formula for \( P_4O_{10} \) is \( P_2O_5 \).

3. **Example 2:**
   - Hydrogen peroxide has the molecular formula \( H_2O_2 \) and its empirical formula is \( HO \).

#### Procedure:
- One way that chemists analyze new or unknown compounds is by determining the percentage composition of elements experimentally. This data is then used to derive the empirical formula.

### Example Problem:

**Problem Statement:**
A compound is made up of 80.0% carbon and 20.0% hydrogen by mass. Assume you have a 100-g sample of this compound.

**Part A:**
How many grams of each element are in this sample?

- **Instructions:**
  - Enter the number of grams of carbon followed by the number of grams of hydrogen, separated by a comma (e.g., 30.0, 70.0).

**Input Box:**
- Label: `mass of C, mass of H =`
- Units: grams (g)

### Visual Aid:

- The left pane includes the explanation of empirical formulas and examples.
- The right pane includes an example problem with a sample input box for calculating the mass of each element in a 100-g sample.

By breaking down the compound’s mass percentage, we are able to find practical real-world applications of empirical formulas. This foundational knowledge is critical for processes in chemistry and material sciences.
Transcribed Image Text:### Understanding Empirical Formulas **Learning Goal:** To understand what an empirical formula is and to learn the procedure for finding empirical formulas. #### Key Concepts: 1. **Definition:** - An empirical formula expresses the simplest ratio of the elements involved in a compound. 2. **Example 1:** - The compound \( P_4O_{10} \) has a phosphorus to oxygen (P:O) ratio of 4:10. However, this ratio can be simplified to 2:5. Therefore, the empirical formula for \( P_4O_{10} \) is \( P_2O_5 \). 3. **Example 2:** - Hydrogen peroxide has the molecular formula \( H_2O_2 \) and its empirical formula is \( HO \). #### Procedure: - One way that chemists analyze new or unknown compounds is by determining the percentage composition of elements experimentally. This data is then used to derive the empirical formula. ### Example Problem: **Problem Statement:** A compound is made up of 80.0% carbon and 20.0% hydrogen by mass. Assume you have a 100-g sample of this compound. **Part A:** How many grams of each element are in this sample? - **Instructions:** - Enter the number of grams of carbon followed by the number of grams of hydrogen, separated by a comma (e.g., 30.0, 70.0). **Input Box:** - Label: `mass of C, mass of H =` - Units: grams (g) ### Visual Aid: - The left pane includes the explanation of empirical formulas and examples. - The right pane includes an example problem with a sample input box for calculating the mass of each element in a 100-g sample. By breaking down the compound’s mass percentage, we are able to find practical real-world applications of empirical formulas. This foundational knowledge is critical for processes in chemistry and material sciences.
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