Teach me this content?
LEARNING OBJECTIVE
SPQ-2.B
Explain the quantitative
relationship between the
elemental composition by
mass and the composition of
substances in a mixture.
ESSENTIAL KNOWLEDGE
SPQ-2.B.1
While pure substances contain molecules or
formula units of a single type, mixtures contain
molecules or formula units of two or more
types, whose relative proportions can vary.
SPQ-2.B.2
Elemental analysis can be used to determine
the relative numbers of atoms in a substance
and to determine its purity.

user
<think>
We are going to cover the learning objective SPQ-2.B: "Explain the quantitative relationship between the elemental composition by mass and the composition of substances in a mixture."

We have two essential knowledge points to consider:
1. SPQ-2.B.1: Pure substances vs. mixtures in terms of molecular composition and variability.
2. SPQ-2.B.2: How elemental analysis helps determine atomic ratios and purity.

Let me break this down step by step.

### Step 1: Understanding Pure Substances and Mixtures
- **Pure Substance**: A single type of molecule or formula unit. Examples: pure water (H₂O), pure table salt (NaCl). Every sample has the same composition and properties.
- **Mixture**: Contains two or more different types of molecules or formula units. The relative amounts of these components can vary. Examples: salt water (mixture of H₂O and NaCl), air (mixture of N₂, O₂, etc.).

### Step 2: Elemental Composition by Mass
- **Elemental composition by mass** refers to the percentage (by mass) of each element in a sample.
  - For a pure compound, the elemental composition is fixed. For example, water (H₂O) always has 11.19% hydrogen and 88.81% oxygen by mass.
  - For a mixture, the elemental composition depends on the relative amounts of the components. For example, in a mixture of water and salt, the percentage of sodium, chlorine, hydrogen, and oxygen will change depending on how much salt is dissolved in the water.

### Step 3: Quantitative Relationship in Mixtures
- In a mixture, the overall elemental composition by mass is a weighted average of the compositions of the individual components.
- Example: Consider a mixture of two compounds, A and B.
  - Let the mass fraction of A in the mixture be \( w_A \) and that of B be \( w_B \) (so \( w_A + w_B = 1 \)).
  - The percentage of an element E in the mixture is:
    \[
    \%E = (\%E \text{ in } A) \times w_A + (\%E \text{ in } B) \times w_B
    \]
  - This formula can be extended to more components.

### Step 4: Elemental Analysis and Purity
- **Elemental analysis** is a technique used to determine the elemental composition of a sample. For example, combustion analysis for organic compounds (to find C, H, O, etc.).
- **Determining atomic ratios**: In a pure compound, the elemental analysis gives the mass percentages, which can be converted to the mole ratio (and hence the empirical formula). For example, if we have a compound that is 40% carbon, 6.7% hydrogen, and 53.3% oxygen, we can find the empirical formula (which would be CH₂O for this example).
- **Determining purity**: If we have a sample that is supposed to be a pure compound, but it is actually a mixture, the elemental analysis will not match the expected percentages. For instance:
  - Pure glucose (C₆H₁₂O₆) has 40.00% C, 6.71% H, and 53.29% O by mass.
  - If we analyze a sample and get 38% C, then it might be impure (mixed with something that has less carbon). The deviation from the expected composition indicates impurity.

### Example Problem:
**Problem:** A sample is labeled as pure glucose (C₆H₁₂O₆). Upon elemental analysis, it is found to contain 38.5% carbon by mass. Is the sample pure? If not, what might be the reason?

**Solution:**
1. Calculate the expected carbon percentage in pure glucose:
   - Molar mass of glucose = 6*12 + 12*1 + 6*16 = 180 g/mol
   - Mass of carbon = 72 g/mol
   - %C = (72/180) * 100% = 40.00%
2. The sample has 38.5% carbon, which is less than 40%. Therefore, the sample is not pure.
3. The impurity must be a substance that has a lower carbon percentage than glucose. For example, water (H₂O) has 0% carbon. If the sample is mixed with water, the overall carbon percentage would decrease.

### Summary:
- Pure substances have fixed elemental composition by mass.
- Mixtures have variable elemental composition that depends on the relative amounts of the components.
- Elemental analysis can be used to:
  - Determine the empirical formula of a pure compound.
  - Check the purity of a substance by comparing the measured elemental composition to the expected one.

By understanding these concepts, you can explain the quantitative relationship between the elemental composition by mass and the composition of substances in a mixture.
</think>
### Understanding SPQ-2.B: Elemental Composition in Pure Substances vs. Mixtures  
**Learning Objective:** Explain the quantitative relationship between elemental composition by mass and the composition of substances in a mixture.  

#### Key Concepts:  
1. **Pure Substances vs. Mixtures (SPQ-2.B.1):**  
   - **Pure substances** (e.g., distilled water, pure salt) contain **only one type of molecule or formula unit**.  
     - Example: Pure water (H₂O) always has **exactly 11.2% H and 88.8% O by mass**.  
   - **Mixtures** (e.g., saltwater, air) contain **two or more types of molecules/formula units**, with **varying proportions**.  
     - Example: Saltwater (NaCl + H₂O) can have different % Na, Cl, H, or O depending on salt concentration.  

2. **Elemental Analysis (SPQ-2.B.2):**  
   - **Purpose:**  
     - Determine **atomic ratios** (e.g., finding an empirical formula).  
     - Assess **purity** by comparing measured composition to a pure substance’s expected composition.  
   - **How it works:**  
     - Techniques like combustion analysis measure the **mass % of elements** (e.g., % C, H, O in a sample).  
     - Deviations from expected values indicate impurities.  

---

### Quantitative Relationship: Elemental Composition in Mixtures  
**Core Idea:** The elemental composition by mass in a mixture is a **weighted average** of its components.  

#### Formula:  
For a mixture with components A and B:  
\[
\% \text{ Element } = \left( \% \text{ Element in A} \times \frac{\text{Mass of A}}{\text{Total Mass}} \right) + \left( \% \text{ Element in B} \times \frac{\text{Mass of B}}{\text{Total Mass}} \right)
\]  

#### Example: Saltwater Mixture  
- **Components:**  
  - **Pure water (H₂O):** 11.2% H, 88.8% O.  
  - **Pure salt (NaCl):** 39.3% Na, 60.7% Cl.  
- **Scenario:** 100 g mixture with 70 g H₂O + 30 g NaCl.  
  - **% H:** \(\left(11.2\% \times \frac{70}{100}\right) + \left(0\% \times \frac{30}{100}\right) = 7.84\%\)  
  - **% Na:** \(\left(0\% \times \frac{70}{100}\right) + \left(39.3\% \times \frac{30}{100}\right) = 11.8\%\)  
- **Conclusion:**  
  - H% decreases (vs. pure water) due to NaCl dilution.  
  - Na% depends on salt amount.  

---

### Using Elemental Analysis to Determine Purity  
**Principle (SPQ-2.B.2):**  
- A **pure substance** has a **fixed elemental composition**.  
- A **mixture** deviates from this fixed composition.  

#### Example: Testing Glucose Purity  
- **Pure glucose (C₆H₁₂O₆):**  
  - Molar mass = 180 g/mol.  
  - Expected % C = \(\frac{72 \text{ g C}}{180 \text{ g/mol}} \times 100\% = 40.0\%\).  
- **Sample analysis:**  
  - **Measured % C = 38.0%** (less than 40.0%).  
  - **Conclusion:** Sample is **impure** (e.g., mixed with water, which has 0% C).  

---

### Summary:  
| **Aspect**               | **Pure Substance**                          | **Mixture**                                  |  
|---------------------------|---------------------------------------------|----------------------------------------------|  
| **Composition**           | Fixed molecules/formula units               | Variable molecules/formula units             |  
| **Elemental % by mass**   | Constant (e.g., H₂O always 11.2% H)         | Varies with component proportions           |  
| **Elemental analysis**    | Confirms purity or gives atomic ratios      | Reveals impurities via deviation from expected % |  

**Key Takeaway:** Elemental composition by mass directly reflects a mixture’s components and their ratios. Deviations in measured vs. expected values signal impurities.  

Let me know if you’d like a practice problem to test your understanding!
assistant
Teach me this content?
LEARNING OBJECTIVE
SPQ-2.B
Explain the quantitative
relationship between the
elemental composition by
mass and the composition of
substances in a mixture.
ESSENTIAL KNOWLEDGE
SPQ-2.B.1
While pure substances contain molecules or
formula units of a single type, mixtures contain
molecules or formula units of two or more
types, whose relative proportions can vary.
SPQ-2.B.2
Elemental analysis can be used to determine
the relative numbers of atoms in a substance
and to determine its purity.

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English
user
California
Certainly! Let's explore the concepts outlined in the learning objective SPQ-2.B and its essential knowledge statements SPQ-2.B.1 and SPQ-2.B.2.

### SPQ-2.B: Quantitative Relationship in Mixtures

Understanding the composition of substances in a mixture involves knowing how the mass of each component relates to the overall mass of the mixture. When you have a mixture, unlike pure substances, there is no fixed ratio of the elements present because the proportions of different substances can vary. The challenge then becomes to analyze and quantify the amounts of each substance within the mixture.

### SPQ-2.B.1: Pure Substances vs. Mixtures

- **Pure Substances**: These are materials made of only one type of particle, which can be either atoms of an element or molecules/formula units of a compound. For instance, pure water contains only H2O molecules, and each molecule has the same fixed composition: two hydrogen atoms bounded to one oxygen atom.

- **Mixtures**: Mixtures consist of two or more different particles, which may be elements, compounds, or both, mixed together without a fixed ratio. For example, seawater is a mixture containing mainly water (H2O) but also has various salts, such as sodium chloride (NaCl), dissolved within it, alongside other substances. The percentage of each component can vary from one sample to another.

### SPQ-2.B.2: Elemental Analysis for Composition and Purity

- **Elemental Analysis**: This is a process used in chemistry to determine the percentage composition of different elements in a compound or the percentages of substances in a mixture. It helps in identifying the relative numbers of atoms of each element within a substance. Techniques for elemental analysis include gravimetric analysis (measuring mass changes), spectroscopy methods (identifying atoms by their emission or absorption spectra), or chromatography (separating components based on their movement through a medium).

- **Determining Purity**: Elemental analysis can also be employed to assess the substance's purity. If a substance is supposed to be pure, any deviation in elemental composition from the expected can indicate impurity. For example, a sample of pure copper (Cu) subjected to elemental analysis should contain close to 100% copper atoms. If other elements are present, it suggests that the copper is not pure, and the analysis can quantify how much of each impurity is present.

#### Example

We'll look at an example to practically understand the application of these principles:

If you have a mixture of salt (NaCl) and sand (SiO2), and you want to determine the percentage by mass of each component, you could:

1. **Dissolve the Mixture**: Add water to the mixture. Salt dissolves in water, but sand does not.
2. **Filter the Sand**: Filter the mixture. The sand will remain on the filter, while the salt solution will pass through.
3. **Evaporate the Water**: Now, evaporate the water from the salt solution to obtain the dry salt.
4. **Weigh the Components**: By weighing the dry sand and dry salt obtained from these processes, you can calculate the percentage by mass of each component in the initial mixture.

With the given masses, let's say:
- Mass of salt (NaCl) recovered: 5 grams
- Mass of sand (SiO2) recovered: 15 grams
- Total mass of the mixture: 5g + 15g = 20 grams

Percentage by mass of NaCl in the mixture: (5g / 20g) * 100% = 25%
Percentage by mass of SiO2 in the mixture: (15g / 20g) * 100% = 75%

In this way, elemental and compound compositions by mass are crucial for understanding and manipulating mixtures in both science and industry.
English
assistant
2024-03-13T18:40:15+00:00