Laws Of Chemical Combination
Laws Of Chemical Combination
The Laws of Chemical Combination represent the cornerstone of quantitative chemistry, providing a framework for understanding how matter interacts at the atomic and molecular level. These laws, primarily formulated during the 18th and early 19th centuries through meticulous experimental work, laid the groundwork for Dalton's Atomic Theory and the subsequent development of modern chemical understanding. They describe the fundamental principles governing the formation of compounds and the progression of chemical reactions.
Law Of Conservation Of Mass
Also historically known as the Law of Conservation of Matter, this fundamental principle asserts that in any isolated or closed system, the total mass remains constant over time. This means that matter can neither be created nor destroyed during a chemical reaction or physical transformation, it can only change its form or arrangement. This law is a direct consequence of the fact that atoms themselves are not created or destroyed in chemical reactions; they are merely rearranged.
Historical Context and Proponents:
While the concept had been hinted at by earlier alchemists and scientists like Robert Boyle, it was definitively enunciated and experimentally validated by the French chemist Antoine Lavoisier in his treatise Traité Élémentaire de Chimie (Elementary Treatise of Chemistry) in 1789. Lavoisier's groundbreaking work involved precise measurements of reactants and products in sealed vessels, meticulously accounting for all substances, including gases. His experiments, such as the combustion of mercury in a sealed container, unequivocally demonstrated that the mass of the system remained unchanged.
Detailed Explanation and Derivation:
The law can be understood by considering the conservation of individual atoms. In a chemical reaction, chemical bonds are broken and new bonds are formed, leading to the rearrangement of atoms. However, the number of atoms of each element and their individual masses remain unchanged. Therefore, the sum of the masses of all atoms on the reactant side must equal the sum of the masses of all atoms on the product side.
Consider a generalized chemical reaction:
$aA + bB \rightarrow cC + dD$
Where:
- A, B, C, and D represent chemical species (elements or compounds).
- a, b, c, and d are the stoichiometric coefficients, representing the number of molecules or moles of each species involved in the reaction.
The Law of Conservation of Mass states:
Sum of masses of reactants = Sum of masses of products
Mathematically, this can be expressed as:
$a \times (\text{Molar mass of A}) + b \times (\text{Molar mass of B}) = c \times (\text{Molar mass of C}) + d \times (\text{Molar mass of D})$
This holds true at the molecular level as well, where the total mass of all atoms in the reactant molecules equals the total mass of all atoms in the product molecules.
Real-World Application:
This law is fundamental to stoichiometry, allowing chemists to predict the amount of product formed from a given amount of reactant, or vice versa. It's crucial in industrial chemical processes, where precise calculations of material usage and yield are necessary for efficiency and safety. For example, in the Haber-Bosch process for ammonia synthesis ($N_2 + 3H_2 \rightleftharpoons 2NH_3$), the mass of nitrogen and hydrogen consumed will exactly equal the mass of ammonia produced, plus any unreacted starting materials.
Example 1. A student burns 5 grams of magnesium ribbon in a sealed container containing excess oxygen. After the reaction, the product, magnesium oxide (MgO), is weighed. If the Law of Conservation of Mass holds true, what would be the mass of the magnesium oxide formed? (Assume atomic mass of Mg = 24.3 g/mol, O = 16.0 g/mol).
Answer:
The reaction is:
$2Mg_{(s)} + O_{2(g)} \rightarrow 2MgO_{(s)}$
According to the Law of Conservation of Mass, the total mass of reactants must equal the total mass of products. In this scenario, the reactants are magnesium (Mg) and oxygen ($O_2$). The product is magnesium oxide (MgO).
We are given the mass of magnesium = 5 grams.
To determine the mass of oxygen that reacts with 5g of Mg, we can use stoichiometry:
Moles of Mg = Mass of Mg / Molar mass of Mg = 5 g / 24.3 g/mol ≈ 0.206 mol
From the balanced equation, 2 moles of Mg react with 1 mole of $O_2$. Therefore, 0.206 moles of Mg will react with $0.206 / 2 = 0.103$ moles of $O_2$.
Mass of $O_2$ = Moles of $O_2$ × Molar mass of $O_2$ = 0.103 mol × (2 × 16.0 g/mol) = 0.103 mol × 32.0 g/mol ≈ 3.296 g
Total mass of reactants = Mass of Mg + Mass of $O_2$ = 5 g + 3.296 g = 8.296 g
By the Law of Conservation of Mass, the mass of the product (MgO) must equal the total mass of the reactants.
Mass of MgO = 8.296 grams
Note: If the container was not sealed and oxygen could escape, the measured mass of MgO would be less than the initial mass of Mg, seemingly violating the law. This highlights the importance of closed systems for experimental verification.
Law Of Constant Proportions
This law, also known as the Law of Definite Proportions, is a fundamental principle stating that a pure chemical compound always consists of the same elements combined together in the same proportion by mass, irrespective of its origin or method of preparation. This means that the elemental composition of a specific compound is fixed.
Historical Context and Proponents:
This law was first clearly stated and extensively supported by experimental evidence by the French chemist Joseph Proust in 1794, following his extensive work on the composition of various compounds like metal oxides and carbonates. He observed that a given compound, such as copper carbonate, always had the same ratio of copper, carbon, and oxygen by mass, whether it was synthesized in the laboratory or found in nature.
Detailed Explanation and Derivation:
The law is a direct consequence of the nature of chemical bonds and the discrete atomic nature of matter. When atoms of different elements combine to form a molecule of a specific compound, they do so in a fixed ratio determined by their valencies and the nature of the chemical bond. This fixed ratio of atoms translates directly into a fixed ratio of masses.
Consider the formation of water ($H_2O$). The molecule consists of two hydrogen atoms and one oxygen atom. The approximate atomic masses are: Hydrogen (H) = 1.008 u, Oxygen (O) = 15.999 u.
The ratio of the mass of hydrogen to the mass of oxygen in a water molecule is:
Ratio of masses = $\frac{\text{Mass of 2 Hydrogen atoms}}{\text{Mass of 1 Oxygen atom}}$
Ratio = $\frac{2 \times (\text{Atomic mass of H})}{1 \times (\text{Atomic mass of O})}$
Ratio = $\frac{2 \times 1.008 \text{ u}}{1 \times 15.999 \text{ u}} \approx \frac{2.016 \text{ u}}{15.999 \text{ u}} \approx \frac{1}{7.937}$
Rounded to simple whole numbers, this is approximately $1:8$. This means that for every 1 gram of hydrogen in pure water, there are approximately 8 grams of oxygen. This ratio remains constant regardless of whether the water is from a tap, a glacier, or synthesized from hydrogen and oxygen gas.
Example: Carbon Dioxide ($CO_2$)
Atomic mass of Carbon (C) ≈ 12.01 u
Atomic mass of Oxygen (O) ≈ 16.00 u
In $CO_2$, there is 1 Carbon atom and 2 Oxygen atoms.
Ratio of masses = $\frac{\text{Mass of 1 Carbon atom}}{\text{Mass of 2 Oxygen atoms}}$
Ratio = $\frac{1 \times (\text{Atomic mass of C})}{2 \times (\text{Atomic mass of O})}$
Ratio = $\frac{1 \times 12.01 \text{ u}}{2 \times 16.00 \text{ u}} = \frac{12.01 \text{ u}}{32.00 \text{ u}} \approx \frac{3}{8}$
Thus, the ratio of Carbon to Oxygen by mass in carbon dioxide is approximately $3:8$.
Significance:
This law was instrumental in establishing that compounds have a fixed composition. It helped differentiate between pure compounds and mixtures, where composition can vary. It also supports the idea that elements combine in specific, fixed ways.
Example 2. A sample of ammonia ($NH_3$) was prepared by reacting nitrogen gas with hydrogen gas in a laboratory. It was found to contain 82.35% nitrogen and 17.65% hydrogen by mass. Another sample of ammonia, obtained from a natural gas well, was analyzed and found to contain 82.35% nitrogen and 17.65% hydrogen by mass. This observation supports which law of chemical combination?
Answer:
This observation strongly supports the Law of Constant Proportions (or Law of Definite Proportions). The law states that a pure chemical compound always has the same proportion by mass of its constituent elements, regardless of its source or method of preparation. Since both the laboratory-prepared ammonia and the naturally obtained ammonia show the exact same percentage composition of nitrogen and hydrogen (82.35% N and 17.65% H), it confirms that ammonia is a definite compound with a fixed elemental ratio.
For verification, let's check the mass ratio in $NH_3$: Atomic mass of N ≈ 14.01 u, Atomic mass of H ≈ 1.008 u.
Mass ratio N:H = $14.01 : (3 \times 1.008) = 14.01 : 3.024 \approx 4.63 : 1$.
Percentage of Nitrogen = $\frac{14.01}{14.01 + 3.024} \times 100\% = \frac{14.01}{17.034} \times 100\% \approx 82.25\%$
Percentage of Hydrogen = $\frac{3.024}{17.034} \times 100\% \approx 17.75\%$
The experimental values (82.35% N, 17.65% H) are very close to these calculated values, confirming the consistent composition.
Law Of Multiple Proportions
This law states that when two elements, let's call them A and B, combine to form more than one chemical compound, then the different masses of element B that combine with a fixed mass of element A are in the ratio of small, whole numbers.
Historical Context and Proponents:
This law was proposed by the English chemist and physician John Dalton in 1803 as part of his atomic theory. Dalton noticed that elements could combine in different ratios to form distinct compounds, and he sought to explain this phenomenon. His atomic theory proposed that elements were composed of indivisible atoms, and that these atoms combined in simple, whole-number ratios to form compounds. This law was one of the first pieces of evidence supporting this atomic hypothesis.
Detailed Explanation and Derivation:
The law arises from the fact that atoms of elements combine in discrete, whole-number ratios to form molecules. If two elements can form multiple compounds, it means they can combine in different whole-number ratios of their atoms.
Let's consider two elements, A and B. Suppose they form two compounds, $A_x B_y$ and $A_p B_q$. According to the law, if we fix the mass of element A, the masses of element B that combine with it will be in a simple whole-number ratio.
To demonstrate this, we fix a certain mass of element A. Let's say we take 1 unit of mass for element A.
In the first compound ($A_x B_y$), let the atomic mass of A be $M_A$ and that of B be $M_B$. The mass of B combining with $M_A$ of A is $M_B \times (y/x)$.
In the second compound ($A_p B_q$), the mass of B combining with $M_A$ of A is $M_B \times (q/p)$.
The ratio of the masses of B in the two compounds, when combined with a fixed mass of A, is:
$\frac{\text{Mass of B in compound 1}}{\text{Mass of B in compound 2}} = \frac{M_B \times (y/x)}{M_B \times (q/p)} = \frac{y/x}{q/p} = \frac{yp}{xq}$
Since x, y, p, and q are whole numbers representing the number of atoms in the molecule, the ratio $(yp)/(xq)$ will also be a ratio of small whole numbers.
Example: Nitrogen Oxides
Let's re-examine the example of nitrogen and oxygen compounds:
1. Nitrous Oxide ($N_2O$): Atomic masses: N ≈ 14, O ≈ 16.
- For 28 g of N (2 atoms), 16 g of O combines.
- Mass of O per gram of N = $16 / 28 \approx 0.57$ g
2. Nitric Oxide (NO):
- For 14 g of N (1 atom), 16 g of O combines.
- For 28 g of N, $(28/14) \times 16 = 32$ g of O combines.
- Mass of O per gram of N = $32 / 28 \approx 1.14$ g
3. Nitrogen Dioxide ($NO_2$):
- For 14 g of N (1 atom), 32 g of O combines.
- For 28 g of N, $(28/14) \times 32 = 64$ g of O combines.
- Mass of O per gram of N = $64 / 28 \approx 2.29$ g
The masses of Oxygen that combine with a fixed mass (28 g) of Nitrogen are 16 g, 32 g, and 64 g.
The ratio of these masses of Oxygen is $16 : 32 : 64$, which simplifies to $1 : 2 : 4$. This is a ratio of small whole numbers.
Table for Nitrogen Oxides:
| Compound | Formula | Mass of N (g) | Mass of O (g) | Mass of O per 1g of N (g/g) |
|---|---|---|---|---|
| Nitrous Oxide | $N_2O$ | 28 | 16 | $16/28 \approx 0.57$ |
| Nitric Oxide | NO | 14 | 16 | $32/28 \approx 1.14$ |
| Nitrogen Dioxide | $NO_2$ | 14 | 32 | $64/28 \approx 2.29$ |
| Ratio of Oxygen masses (for fixed 28g N) | 16 : 32 : 64 | 1 : 2 : 4 | ||
Significance:
This law provided strong empirical support for Dalton's Atomic Theory, particularly the idea that atoms combine in simple, whole-number ratios. It helped to explain the existence of different compounds formed from the same elements, like $CO$ and $CO_2$, or water ($H_2O$) and hydrogen peroxide ($H_2O_2$).
Example 3. Carbon and Hydrogen combine to form methane ($CH_4$), ethane ($C_2H_6$), propane ($C_3H_8$), and butane ($C_4H_{10}$). If we consider a fixed mass of carbon, say 12 grams, determine the masses of hydrogen that combine with it in each of these compounds and show that they satisfy the Law of Multiple Proportions.
Answer:
We will determine the mass of hydrogen that combines with 12 grams of Carbon in each hydrocarbon, using the atomic masses: C ≈ 12 u, H ≈ 1 u.
1. Methane ($CH_4$):
- Formula: $CH_4$. This means 1 atom of Carbon combines with 4 atoms of Hydrogen.
- Atomic mass of C = 12 u. Atomic mass of H = 1 u.
- The ratio of masses of C to H is $12 : (4 \times 1) = 12 : 4$.
- If 12 g of Carbon combines with 4 g of Hydrogen, then for a fixed 12 g of Carbon, 4 g of Hydrogen combine.
2. Ethane ($C_2H_6$):
- Formula: $C_2H_6$. This means 2 atoms of Carbon combine with 6 atoms of Hydrogen.
- Mass of 2 Carbon atoms = $2 \times 12 = 24$ g. Mass of 6 Hydrogen atoms = $6 \times 1 = 6$ g.
- The ratio of masses of C to H is $24 : 6$.
- To find the mass of Hydrogen combining with 12 g of Carbon, we can set up a proportion:
- If 24 g of Carbon combines with 6 g of Hydrogen,
- Then 12 g of Carbon combines with $(12/24) \times 6 = 0.5 \times 6 = 3$ g of Hydrogen.
- So, for a fixed 12 g of Carbon, 3 g of Hydrogen combine.
3. Propane ($C_3H_8$):
- Formula: $C_3H_8$. This means 3 atoms of Carbon combine with 8 atoms of Hydrogen.
- Mass of 3 Carbon atoms = $3 \times 12 = 36$ g. Mass of 8 Hydrogen atoms = $8 \times 1 = 8$ g.
- The ratio of masses of C to H is $36 : 8$.
- To find the mass of Hydrogen combining with 12 g of Carbon:
- If 36 g of Carbon combines with 8 g of Hydrogen,
- Then 12 g of Carbon combines with $(12/36) \times 8 = (1/3) \times 8 = 8/3$ g of Hydrogen.
- So, for a fixed 12 g of Carbon, 8/3 g of Hydrogen combine.
4. Butane ($C_4H_{10}$):
- Formula: $C_4H_{10}$. This means 4 atoms of Carbon combine with 10 atoms of Hydrogen.
- Mass of 4 Carbon atoms = $4 \times 12 = 48$ g. Mass of 10 Hydrogen atoms = $10 \times 1 = 10$ g.
- The ratio of masses of C to H is $48 : 10$.
- To find the mass of Hydrogen combining with 12 g of Carbon:
- If 48 g of Carbon combines with 10 g of Hydrogen,
- Then 12 g of Carbon combines with $(12/48) \times 10 = (1/4) \times 10 = 10/4 = 5/2$ g of Hydrogen.
- So, for a fixed 12 g of Carbon, 5/2 g of Hydrogen combine.
Demonstration of the Law of Multiple Proportions:
The masses of Hydrogen that combine with a fixed mass (12 g) of Carbon in methane, ethane, propane, and butane are:
4 g, 3 g, 8/3 g, and 5/2 g.
To see the ratio of small whole numbers, let's express these with a common denominator or convert to integers by multiplying by a suitable factor (e.g., 6):
- 4 g = $4/1$ g
- 3 g = $3/1$ g
- 8/3 g
- 5/2 g
Multiply by 6:
- $4 \times 6 = 24$
- $3 \times 6 = 18$
- $(8/3) \times 6 = 16$
- $(5/2) \times 6 = 15$
The ratio of the masses of hydrogen is $24 : 18 : 16 : 15$. This is a ratio of small whole numbers, thereby demonstrating the Law of Multiple Proportions.