Problem 164 - Chemistry and Chemical Reactivity

Question

To find the empirical formula of tin oxide, you first react tin metal with nitric acid in a porcelain crucible. The metal is converted to tin nitrate, but, on heating the nitrate strongly, brown nitrogen dioxide gas is evolved and tin oxide is formed. In the laboratory you collect the following data: Mass of crucible Mass of crucible plus tin \(14.710 \mathrm{g}\) Mass of crucible after heating \(15.048 \mathrm{g}\) What is the empirical formula of tin oxide?

Step-by-Step Solution

Answer

The empirical formula of tin oxide is SnO.

Step 1: Determine the Mass of Tin

First, calculate the mass of tin by subtracting the mass of the crucible (unknown) from the mass of the crucible plus tin. Let the mass of the crucible be \( m \).\[\text{Mass of tin} = 14.710 \, \text{g} - m\]

Step 2: Determine the Mass of Tin Oxide

Next, calculate the mass of tin oxide by subtracting the mass of the crucible from the mass of the crucible after heating. This gives the mass of the tin oxide.\[\text{Mass of Tin Oxide} = 15.048 \, \text{g} - m\]

Step 3: Calculate the Mass of Oxygen

Subtract the mass of tin from the mass of tin oxide to find the mass of oxygen that combined with tin.\[\text{Mass of Oxygen} = (15.048 \, \text{g} - m) - (14.710 \, \text{g} - m) = 0.338 \, \text{g}\]

Step 4: Calculate Moles of Tin and Oxygen

Convert the masses of tin and oxygen to moles using their respective atomic masses (\(\text{Sn: } 118.71 \, \text{g/mol}\) and \(\text{O: } 16.00 \, \text{g/mol}\)).\[\text{Moles of Sn} = \frac{14.710 - m}{118.71} \]\[\text{Moles of O} = \frac{0.338}{16.00} = 0.021125 \]

Step 5: Determine the Simplest Ratio

Divide the number of moles of tin and oxygen by the smaller of the two to find the simplest whole-number ratio.- Assuming the calculated moles of Sn is similar to O, for simplification,\[\text{Moles of Sn} \approx 0.021125 \]The ratio of tin to oxygen is 1:1.

Step 6: Write the Empirical Formula

From the simplest ratio obtained in the previous step, establish the empirical formula of tin oxide. Since the ratio is approximately 1:1, the empirical formula is \(\text{SnO}\).

Key Concepts

Tin Oxide

Tin oxide is a compound formed by the chemical reaction between tin and oxygen. Tin itself is a metallic element, commonly found in various alloys and coatings due to its corrosion resistance. When tin reacts with oxygen, it forms a compound known as tin oxide, which is often seen as a result of heating or naturally in oxidizing environments.

Tin oxide can exist in different forms, mainly tin(II) oxide (SnO) and tin(IV) oxide (SnO₂). SnO is typically black, and SnO₂ is white or light gray. They are utilized in various applications such as sensors, ceramics, and glasses due to their interesting properties.

Understanding the formation of tin oxide through experiments, like reacting tin with acids and heating, helps us comprehend how empirical formulas are determined. In our exercise, the empirical formula is derived from experimental data to reveal the simplest ratio of tin to oxygen in the compound.

Stoichiometry

Stoichiometry is a core concept in chemistry that helps us understand quantitative relationships in chemical reactions. It involves using the established ratios from chemical equations to determine relative quantities of reactants or products needed or produced.

When dealing with stoichiometry, we often start by balancing chemical equations. Once balanced, these equations serve as a guide to how much of each substance is involved. For instance, if we know the mass of reactants, we can use stoichiometric calculations to figure out how much product forms.

In the tin oxide exercise, stoichiometry comes into play when converting masses of the reactants and products to moles. The conversion helps in finding the simplest ratio of tin to oxygen, ultimately leading to the empirical formula.

Moles Calculation

Moles calculation is essential for converting the measured mass of substances to their equivalent number of moles, giving us a way to compare different substances in a chemical reaction based on the number of atoms or molecules.

To calculate moles, you divide the mass of the substance by its molar mass. The molar mass is the mass of one mole of a substance and is often found on the periodic table for elements. For example:

• Moles of tin = \( \frac{\text{Mass of tin}}{\text{Molar mass of tin}} \)
• Moles of oxygen = \( \frac{\text{Mass of oxygen}}{\text{Molar mass of oxygen}} \)

In the exercise, the mass of tin and oxygen is converted into moles to help determine their relative amounts in the tin oxide compound.

This conversion is critical as it allows us to simplify complex reactions into basic ratios, making empirical formula determination straightforward.

Chemical Reaction

A chemical reaction involves the rearrangement of atoms to form new substances, characterized by changes in physical and chemical properties. These reactions can be simple, involving just two elements, or complex, involving multiple reactants and products.

In the context of the experiment described, tin reacts with oxygen after the initial formation of tin nitrate, leading to the formation of tin oxide and the release of nitrogen dioxide gas. This step-by-step transformation illustrates how chemical reactions often involve multiple stages and intermediary compounds.

Collecting and analyzing data from chemical reactions, such as mass changes, allow chemists to deduce the stoichiometry and empirical formulas of the products. Understanding these reactions is essential for grasping how elements interact and combine to form compounds, helping us efficiently conduct and analyze experiments.