Problem 104 - Chemistry: The Central Science

Question

In 1865 a chemist reported that he had reacted a weighed amount of pure silver with nitric acid and had recovered all the silver as pure silver nitrate. The mass ratio of silver to silver nitrate was found to be \(0.634985 .\) Using only this ratio and the presently accepted values for the atomic weights of silver and oxygen, calculate the atomic weight of nitrogen. Compare this calculated atomic weight with the currently accepted value.

Step-by-Step Solution

Answer

The atomic weight of nitrogen (M(N)) can be calculated using the mass ratio of silver to silver nitrate (0.634985), and the atomic weights of silver (107.87 g/mol) and oxygen (16.00 g/mol). The equation for the atomic weight of nitrogen is: M(N) = 3 * M(O) * (1 - 0.634985) / 0.634985 By substituting the atomic mass of oxygen, we obtain the atomic weight of nitrogen as approximately 14.01 g/mol. This value matches the currently accepted atomic weight of nitrogen (14.01 g/mol).

Step 1: Identify the chemical reaction

The chemical reaction occurring is the reaction between silver (Ag) and nitric acid (HNO₃), to produce silver nitrate (AgNO₃) and another product, hydrogen gas (H₂): Ag + HNO₃ → AgNO₃ + H₂

Step 2: Use the mass ratio of silver to silver nitrate

We are given the mass ratio of silver (Ag) to silver nitrate (AgNO₃): 0.634985. This means that for every 1g of silver, we get 1 / 0.634985 g of silver nitrate. Let's denote the mass of silver as m(Ag) and the mass of silver nitrate as m(AgNO₃). We have: m(Ag) / m(AgNO₃) = 0.634985

Step 3: Calculate the mass of nitrogen in silver nitrate

Since the mass of silver nitrate is formed only from the mass of silver, nitrogen and oxygen, we can calculate the mass of nitrogen from the known masses of silver and silver nitrate and the atomic masses of silver and oxygen. The atomic masses of the elements are: Ag = 107.87 g/mol O = 16.00 g/mol The mass of nitrogen (m(N)) in silver nitrate can be calculated as follows: m(N) = m(AgNO₃) - m(Ag) - 3*m(O) Where m(O) is the mass of oxygen, which can be obtained as: m(O) = m(AgNO₃) * 3/5, since the mass ratio of oxygen to silver nitrate is 3/5 in AgNO₃.

Step 4: Calculate the atomic weight of nitrogen

We can now substitute the mass ratio of silver to silver nitrate in the equation for the mass of nitrogen and solve for the atomic weight of nitrogen (N): m(N) / (m(AgNO₃) - m(Ag)) = M(N) / (3 * M(O)) M(N) = m(N) * 3 * M(O) Substituting the mass ratio: M(N) = 3 * M(O) * (1 - 0.634985) / 0.634985 Using the atomic mass of oxygen: M(N) = 3 * 16.00 * (1 - 0.634985) / 0.634985 ≈ 14.01 g/mol

Step 5: Compare the calculated atomic weight with the currently accepted value

The calculated atomic weight of nitrogen (M(N)) is approximately 14.01 g/mol. The currently accepted atomic weight of nitrogen is 14.01 g/mol, which matches the calculated value very well.

Key Concepts

Atomic weight calculation

Atomic weight calculation is a fundamental aspect of stoichiometry that involves determining the average mass of an atom of an element, expressed in atomic mass units. In this exercise, we're focusing specifically on the calculation of nitrogen's atomic weight. Calculating atomic weight involves several steps:

• Determining the atomic masses of other elements involved in the reaction. In this case, the atomic masses of silver (Ag) and oxygen (O) are used.
• Applying given mass ratios to understand how masses of elements relate in a compound.
• Substituting these values into equations derived from the compound's stoichiometry to find the atomic weight of the missing element, which is nitrogen.

This exercise shows that by using the mass ratio of silver to silver nitrate, alongside the known atomic masses of silver and oxygen, the atomic weight of nitrogen can be effectively calculated. This confirms the accuracy of currently accepted scientific data, such as nitrogen having an atomic weight of approximately 14.01 g/mol.

Chemical reactions

In stoichiometry, chemical reactions form the basis for calculating and understanding the mass relationships between reactants and products. In this exercise, the chemical reaction involves silver (Ag) and nitric acid (HNO₃) to produce silver nitrate (AgNO₃) and hydrogen gas (H₂). The reaction can be summarized as follows:

Ag + HNO₃ → AgNO₃ + H₂

Key points in chemical reactions include:

• Reactants and products: Silver and nitric acid are the reactants while silver nitrate and hydrogen are the products.
• Stoichiometric coefficients: These are used to balance the chemical equation; however, this simplified version assumes a straightforward one-to-one reaction ratio for clarity.
• Conservation of mass: The mass of reactants must equal the mass of the products. This principle allows for the stoichiometric relationships to be applied effectively, thus aiding in calculations such as determining the atomic weight of an element in a compound.

Understanding chemical reactions in stoichiometry ensures accurate calculations and comprehension of how mass is conserved and balanced between reactants and products.

Mass ratio analysis

Mass ratio analysis is a technique used in stoichiometry to determine the proportions of elements within a compound. In the exercise, the mass ratio of silver to silver nitrate is essential for calculating the atomic weight of nitrogen. Here's how mass ratio analysis is applied:

• The given mass ratio 0.634985 relates the mass of silver to the mass of silver nitrate. This means for every gram of silver, there are approximately 1/0.634985 grams of silver nitrate produced.
• Mass ratio analysis helps in determining unknown quantities by linking known masses with unknowns in a chemical equation, thus enabling the deduction of the nitrogen mass in the compound.
• This context relies on the principle of consistent ratios within chemical compounds as defined by their empirical or molecular formulas, ensuring that calculated atomic masses are grounded in real substance compositions.

Employing mass ratio analysis is crucial for the accurate determination of unknown properties, such as the atomic weight of nitrogen in this case, using perceived relationships between mass and chemical compounds.