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Imagine you buy pure water from two different sources: one from a mountain spring and another from a city tap. Despite their different origins, the water molecules in both samples have the same chemical composition. This means the elements hydrogen and oxygen are always combined in the same fixed ratio by mass. This observation is the foundation of the Law of Constant Proportion, a fundamental principle in chemistry.
The Law of Constant Proportion states that a chemical compound always contains the same elements combined together in a fixed mass ratio, regardless of the source or method of preparation. This law helps us understand that compounds are pure substances with definite composition, unlike mixtures where proportions can vary.
Formally, the Law of Constant Proportion (also called the Law of Definite Proportions) can be stated as:
"A chemical compound is composed of elements combined in a definite proportion by mass, which remains constant for any sample of that compound."
This means that if you take any sample of a pure compound, the ratio of the masses of its constituent elements will always be the same.
For example, water (H2O) always contains hydrogen and oxygen in a mass ratio of approximately 1:8, no matter where the water is taken from or how it was prepared.
This is different from mixtures, where the components can be combined in any proportion. For instance, air is a mixture of gases like nitrogen, oxygen, and carbon dioxide, but their relative amounts can vary depending on location and conditions.
The law can be expressed mathematically by calculating the mass ratio of the elements in a compound. Suppose a compound contains two elements, A and B, with masses \( m_A \) and \( m_B \) respectively. Then, the mass ratio is:
This ratio remains constant for any pure sample of the compound.
Another useful way to express composition is the mass percent of an element in a compound, which tells us what fraction of the total mass is due to that element. It is calculated as:
| Compound | Element 1 | Mass (g) | Element 2 | Mass (g) | Mass Ratio (Element 1 : Element 2) |
|---|---|---|---|---|---|
| Water (H2O) | Hydrogen (H) | 2 | Oxygen (O) | 16 | 2 : 16 = 1 : 8 |
| Carbon Dioxide (CO2) | Carbon (C) | 12 | Oxygen (O) | 32 | 12 : 32 = 3 : 8 |
| Ammonia (NH3) | Nitrogen (N) | 14 | Hydrogen (H) | 3 | 14 : 3 |
Step 1: Find the molar mass of water.
Water has 2 hydrogen atoms and 1 oxygen atom.
Molar mass = (2 x 1) + (1 x 16) = 2 + 16 = 18 g/mol
Step 2: Calculate the mass percent of oxygen.
\[ \text{Mass \% of O} = \left( \frac{16}{18} \right) \times 100 = 88.89\% \]
Answer: Oxygen makes up 88.89% of the mass of water.
Step 1: Recall the fixed mass ratio of carbon to oxygen in CO2 is 12:32 or 3:8.
Step 2: For the first sample with 6 g oxygen, calculate carbon mass:
\[ \text{Mass of C} = \frac{3}{8} \times 6 = 2.25 \text{ g} \]
Step 3: For the second sample with 8 g oxygen, calculate carbon mass:
\[ \text{Mass of C} = \frac{3}{8} \times 8 = 3 \text{ g} \]
Step 4: Calculate mass ratios:
Sample 1: \( \frac{2.25}{6} = 0.375 \)
Sample 2: \( \frac{3}{8} = 0.375 \)
Since the mass ratios are equal, the law of constant proportion is verified.
Answer: Both samples have the same fixed mass ratio of carbon to oxygen, confirming the law.
Step 1: Use the mass percent of hydrogen in water.
From atomic masses, hydrogen mass = 2 g, total water mass = 18 g.
Mass percent of hydrogen = \( \frac{2}{18} \times 100 = 11.11\% \)
Step 2: Calculate hydrogen mass in 90 g water:
\[ \text{Mass of H} = \frac{11.11}{100} \times 90 = 10 \text{ g} \]
Answer: 90 g of water contains 10 g of hydrogen.
Step 1: Calculate the mass ratio of nitrogen to oxygen in both samples.
Sample 1: \( \frac{40}{60} = \frac{2}{3} = 0.6667 \)
Sample 2: \( \frac{20}{30} = \frac{2}{3} = 0.6667 \)
Step 2: Since the mass ratios are the same, the samples could represent a compound.
Step 3: However, nitrogen and oxygen do not form a stable compound with this ratio; they form a mixture (air). So, further chemical analysis is needed.
Answer: Equal mass ratios suggest a compound, but knowledge of chemistry tells us this is a mixture. This shows that constant mass ratio alone is not always sufficient; chemical bonding must be considered.
Step 1: Recall the mass ratio of hydrogen to oxygen in water is 2:16 or 1:8.
Step 2: Total parts = 2 + 16 = 18 parts.
Step 3: Find mass of oxygen in 54 g water:
\[ \text{Mass of O} = \frac{16}{18} \times 54 = 48 \text{ g} \]
Answer: 54 g of water yields 48 g of oxygen upon decomposition.
When to use: When given a sample to check if the law of constant proportion applies.
When to use: During mass percent and ratio calculations.
When to use: To quickly differentiate between mixtures and compounds in problems.
When to use: To avoid unit mismatch errors in calculations.
When to use: When verifying the law from experimental data.
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