Question

In an experiment \(0.500 \mathrm{~g}\) of tin reacted with nitric acid to give tin oxide. If the oxide had a mass of \(0.635 \mathrm{~g}\), what is the empirical formula of tin oxide?

Short answer

The empirical formula of tin oxide is \( \text{SnO}_2 \).

Step-by-step solution

Determine the mass of oxygen

Subtract the initial mass of tin from the final mass of tin oxide to find the mass of oxygen. The mass of oxygen is given by the difference: \[ \text{Mass of oxygen} = 0.635 \text{ g} - 0.500 \text{ g} = 0.135 \text{ g} \]

Convert masses to moles

Convert the masses of tin and oxygen to moles using their molar masses. The molar mass of tin (Sn) is approximately 118.71 g/mol, and oxygen (O) is 16.00 g/mol.For tin:\[ \text{Moles of Sn} = \frac{0.500 \text{ g}}{118.71 \text{ g/mol}} \approx 0.00421 \text{ mol} \] For oxygen:\[ \text{Moles of O} = \frac{0.135 \text{ g}}{16.00 \text{ g/mol}} = 0.00844 \text{ mol} \]

Determine the mole ratio

To find the empirical formula, we need the simplest whole number ratio of moles of tin to moles of oxygen. Divide the moles of each element by the smallest number of moles:For tin:\[ \text{Ratio of Sn} = \frac{0.00421}{0.00421} = 1 \]For oxygen:\[ \text{Ratio of O} = \frac{0.00844}{0.00421} \approx 2 \]

Write the empirical formula

Using the mole ratios, the empirical formula is determined by assigning the smallest whole number ratios to the elements involved. Since the mole ratio of Sn to O is 1:2, the empirical formula of tin oxide is \( \text{SnO}_2 \).

Key concepts

Mole Ratio

In chemistry, the concept of mole ratio is fundamental when it comes to understanding the proportions of elements in compounds and reactions. A mole ratio tells us how many moles of one substance is related to a certain number of moles of another. In a chemical reaction or formula, the mole ratio allows us to predict the quantities of reactants and products involved.

For example, if you're looking for the empirical formula of a compound, the mole ratio will give you insight into the number of atoms of each element present in the simplest form of the compound. By dividing the moles of each constituent by the smallest number of moles calculated, you ensure that you are using the smallest possible integers to describe how the elements are combined. This is critical because chemistry often uses these ratios to further calculate real-world implications like yields or predicting product formation.

The mole ratio is particularly important in the context of stoichiometry, as it helps relate quantities from among different parts of a reaction or compound.

Chemical Reactions

Chemical reactions occur when substances interact to form new products through breaking and forming chemical bonds. Understanding chemical reactions is crucial for determining the relationship between reactants and products, involving changes in the arrangement of atoms.

In an experiment where tin reacts with nitric acid to form tin oxide, you start with reactants (tin and nitric acid) and end with products (tin oxide and possibly other by-products). The reaction involves a transformation at the molecular level, where tin and oxygen atoms rearrange.

To make sense of what amounts these atoms need, mole ratios are applied to predict how much tin is required to react with determined amounts of oxygen to produce tin oxide efficiently. This understanding not only helps in determining the empirical formula but also in controlling the reaction’s conditions to maximize yield and minimize waste in practical applications.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It uses the principle of conservation of mass and the ratios of molecules participating in the reactions.

Let's say you have a certain mass of tin, and you want to find out how much of a product such as tin oxide it will form when reacting with nitric acid. Stoichiometry lets you calculate the masses or volumes of reactants needed or the products that will result.

In the provided exercise, stoichiometry was used to convert the masses of tin and oxygen to moles using their respective molar masses. By determining the mole ratio of tin to oxygen, it’s possible to derive the empirical formula of tin oxide as SnO₂. This conversion from mass to moles is a fundamental stoichiometric calculation that helps bridge the gap between the experimental data and theoretical chemistry.

Stoichiometry is essential not just for calculating empirical formulas, but also for practical applications in lab settings and industrial reactions where precise material usage is crucial to efficiency and safety.