Laws of Chemical Combination - Ninth Grade Science
Before the late eighteenth century, chemistry was largely qualitative and empirical, dominated by alchemical traditions that lacked rigorous quantitative foundations. When substances burned, bubbled, or corroded, observers had no reliable mathematical framework to predict how much matter was consumed or created. This transformed into a quantitative physical science through the pioneering work of French chemist Antoine Lavoisier and his contemporaries. By introducing analytical chemical balances to measure the precise masses of reactants and products before and after chemical reactions, Lavoisier and Joseph Proust established the Laws of Chemical Combination. These foundational laws demonstrated that chemical reactions are governed by conservation and mathematical proportions. In this chapter, you will investigate the Law of Conservation of Mass, the Law of Definite Proportions, the Law of Multiple Proportions, and the empirical evidence that laid the groundwork for modern atomic theory.
The Law of Conservation of Mass
Formulated by Antoine Lavoisier in 1774, the Law of Conservation of Mass is the cornerstone of modern chemistry.
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| Law of Conservation of Mass |
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| Mass can neither be created nor destroyed in a chemical reaction. |
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| Total Mass of Reactants = Total Mass of Products |
| Sigma m_reactants = Sigma m_products |
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Lavoisier's Closed-System Reaction
Before Reaction (Unmixed): After Reaction (Precipitate Formed):
Flask contains Barium Chloride; White Barium Sulfate precipitate forms
Suspended tube has Sodium Sulfate. with Sodium Chloride solution.
+---------------+ +---------------+
| [Flask A+B] | | [Flask C+D] |
+---------------+ +---------------+
=============+============= =============+=============
[ Balance: 300.25 g ] [ Balance: 300.25 g ]
Mass Before Reaction = Mass After Reaction!
The Closed-System Proof
Lavoisier conducted combustion experiments inside sealed, airtight glass retorts to prevent volatile gases from entering or escaping. When he burned tin or phosphorus inside a sealed flask, the metal reacted vigorously with the enclosed oxygen, forming a powdery oxide crust. When placed back on his balance, the total mass of the sealed system remained identical to the milligram. Mass appeared to change in open-air burning only because invisible gases—such as oxygen being absorbed or carbon dioxide escaping—were omitted from the calculation. In any chemical reaction, atoms are neither manufactured out of nothing nor obliterated into nonexistence; they are rearranged into new chemical configurations.
Classic Laboratory Demonstration: Precipitation Reactions
The conservation of mass is demonstrated using precipitation reactions in a conical flask fitted with an internal test tube:
1. Place aqueous barium chloride solution ($ ext{BaCl}_2$) in the flask.
2. Suspend a small test tube containing aqueous sodium sulfate solution ($ ext{Na}_2 ext{SO}_4$) inside the flask without mixing, seal with a rubber cork, and weigh on a digital balance (e.g., registering 250.50 g).
3. Tilt and swirl the flask to mix the two liquids. An instant double displacement reaction occurs, producing a dense white precipitate of barium sulfate ($ ext{BaSO}_4$) suspended in aqueous sodium chloride ($ ext{NaCl}$):
BaCl2 (aq) + Na2SO4 (aq) ---> BaSO4 (s) [Precipitate] + 2 NaCl (aq)
- Reweigh the sealed flask. The balance registers exactly 250.50 g. Despite the dramatic visual phase change, total mass is conserved.
The Law of Definite Proportions (Law of Constant Composition)
Formulated by French chemist Joseph Louis Proust in 1799, the Law of Definite Proportions describes the elemental composition of pure chemical compounds.
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| Law of Definite Proportions |
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| In a pure chemical compound, the constituent elements are always |
| combined together in a fixed, definite proportion by mass, regardless |
| of the source or method of preparation. |
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Definite Proportions in Water (H_2O)
Source A: Natural Rainwater Source B: Deep Well Spring Source C: Laboratory Synthesis
[ H_2O ] [ H_2O ] [ 2H_2 + O_2 ]
| | |
v v v
Mass Ratio of H : O = 1 : 8 Mass Ratio of H : O = 1 : 8 Mass Ratio of H : O = 1 : 8
(1 g Hydrogen per 8 g Oxygen) (1 g Hydrogen per 8 g Oxygen) (1 g Hydrogen per 8 g Oxygen)
Analysis of Pure Water and Carbon Dioxide
Consider pure water ($ ext{H}_2 ext{O}$). Chemical analysis reveals that every 9.0 grams of pure water decomposes into 1.0 gram of hydrogen gas and 8.0 grams of oxygen gas. Therefore, the mass ratio of hydrogen to oxygen in water is always:
\text{Mass Ratio of } H : O = 1 : 8
If you attempt to synthesize water by reacting 3.0 grams of hydrogen with 16.0 grams of oxygen, you will not obtain 19.0 grams of water. Because oxygen reacts in an exact $8:1$ mass ratio with hydrogen, 16.0 grams of oxygen can combine with only 2.0 grams of hydrogen, producing 18.0 grams of water ($ ext{H}_2 ext{O}$). The remaining 1.0 gram of unreacted hydrogen gas is left behind.
Similarly, in carbon dioxide ($ ext{CO}_2$), carbon and oxygen combine in a constant mass ratio of $12:32$, which simplifies to:
\text{Mass Ratio of } C : O = 3 : 8
Whether carbon dioxide is generated by burning wood, decomposing limestone ($ ext{CaCO}_3$), exhaling from human lungs, or reacting vinegar with baking soda, it contains carbon and oxygen in an exact $3:8$ mass ratio.
The Law of Multiple Proportions
Formulated by John Dalton in 1803, the Law of Multiple Proportions addresses cases where two elements can combine to form more than one distinct chemical compound.
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| Law of Multiple Proportions |
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| When two elements combine to form more than one compound, the different |
| masses of one element that combine with a fixed mass of the other |
| element are in the ratio of small whole numbers. |
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Carbon Oxides Illustrating Multiple Proportions
Compound 1: Carbon Monoxide (CO) Compound 2: Carbon Dioxide (CO_2)
Mass of Carbon = 12 g (Fixed Mass) Mass of Carbon = 12 g (Fixed Mass)
Mass of Oxygen = 16 g Mass of Oxygen = 32 g
Ratio of Oxygen Masses combining with 12 g Carbon:
16 g : 32 g ===> 1 : 2
(A Simple Whole-Number Ratio!)
Analysis of Carbon Oxides
Carbon and oxygen combine to form two distinct gaseous compounds: carbon monoxide ($ ext{CO}$) and carbon dioxide ($ ext{CO}_2$).
1. In carbon monoxide ($ ext{CO}$), 12 grams of carbon combine with 16 grams of oxygen.
2. In carbon dioxide ($ ext{CO}_2$), 12 grams of carbon combine with 32 grams of oxygen.
Holding the mass of carbon fixed at 12 grams, the masses of oxygen that combine with it are 16 grams and 32 grams. The ratio of these oxygen masses is:
\frac{16\text{ g}}{32\text{ g}} = \frac{1}{2} = 1 : 2
This $1:2$ ratio is a simple small whole number. Dalton realized that this small integer ratio could only occur if oxygen attached as discrete, indivisible atomic units: one atom of oxygen in carbon monoxide, and two atoms of oxygen in carbon dioxide.
Chapter Exercises
Exercise 1: In an experimental investigation, 5.0 grams of calcium carbonate ($ ext{CaCO}_3$) are heated strongly in a closed test tube, decomposing completely into solid calcium oxide ($ ext{CaO}$) and gaseous carbon dioxide ($ ext{CO}_2$). The mass of the solid calcium oxide residue remaining in the tube is measured as 2.8 grams. Calculate the mass of carbon dioxide gas produced during the reaction, stating the fundamental chemical law that enables this calculation.
Exercise 2: When 3.0 grams of pure carbon are burned in 8.0 grams of oxygen, 11.0 grams of carbon dioxide are produced. Calculate the mass of carbon dioxide that will be formed if 3.0 grams of carbon are burned in 50.0 grams of oxygen. Name the chemical law governing your prediction and identify any unreacted mass remaining.
Exercise 3: Hydrogen and oxygen react to form two distinct liquid compounds: water ($ ext{H}_2 ext{O}$) and hydrogen peroxide ($ ext{H}_2 ext{O}_2$). In water, 2.0 grams of hydrogen combine with 16.0 grams of oxygen. In hydrogen peroxide, 2.0 grams of hydrogen combine with 32.0 grams of oxygen. Demonstrate that these experimental data illustrate the Law of Multiple Proportions.
Solutions and Explanations
Solution 1: We are given the initial mass of the reactant: calcium carbonate mass is $m_{ ext{reactant}} = 5.0 ext{ g}$. The reaction decomposes into two products: calcium oxide ($m_{ ext{CaO}} = 2.8 ext{ g}$) and carbon dioxide ($m_{ ext{CO}*2}$). According to the Law of Conservation of Mass, the total mass of the reactants must equal the total mass of the products: $m*{ ext{CaCO}*3} = m*{ ext{CaO}} + m_{ ext{CO}*2}$. Substituting the known values: $5.0 ext{ g} = 2.8 ext{ g} + m*{ ext{CO}*2}$. Solving for the mass of carbon dioxide: $m*{ ext{CO}_2} = 5.0 - 2.8 = 2.2$ grams. Exactly 2.2 grams of carbon dioxide gas are generated during the reaction.
Solution 2: According to the Law of Definite Proportions, carbon and oxygen always combine in a fixed mass ratio of $3:8$ to form carbon dioxide. In the first reaction, 3.0 grams of carbon combine with 8.0 grams of oxygen to produce $3.0 + 8.0 = 11.0$ grams of carbon dioxide, with zero excess. When 3.0 grams of carbon are placed in 50.0 grams of oxygen, the 3.0 grams of carbon can still combine with only 8.0 grams of oxygen, because the stoichiometric proportion is fixed by nature. Consequently, exactly 11.0 grams of carbon dioxide will be formed. The remaining oxygen does not react; the unreacted mass of oxygen is $50.0 - 8.0 = 42.0$ grams. This illustrates the Law of Definite Proportions and the concept of a limiting reactant.
Solution 3: To demonstrate the Law of Multiple Proportions, we keep the mass of one element constant and examine the ratio of the masses of the second element. In water, the mass of oxygen combining with 2.0 grams of hydrogen is 16.0 grams. In hydrogen peroxide, the mass of oxygen combining with the exact same 2.0 grams of hydrogen is 32.0 grams. The ratio of the masses of oxygen combining with the fixed 2.0-gram mass of hydrogen is: $16.0 ext{ g} : 32.0 ext{ g} = 16 : 32 = 1 : 2$. Because $1:2$ is a ratio of small whole numbers, this experimental data illustrates the Law of Multiple Proportions.