Question

Aspirin has a molar mass of \(180 \mathrm{~g} / \mathrm{mol}\). If the empirical formula is \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\), what is the molecular formula of aspirin?

Short answer

The molecular formula of aspirin is the same as the empirical formula: \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\).

Step-by-step solution

Determining Empirical Formula Molar Mass

To find the molar mass of the empirical formula, calculate the total mass of all atoms in the formula: - Carbon (C): 9 atoms, each with a molar mass of 12 g/mol, contributing \( 9 \times 12 = 108 \mathrm{~g/mol} \).- Hydrogen (H): 8 atoms, each with a molar mass of 1 g/mol, contributing \( 8 \times 1 = 8 \mathrm{~g/mol} \).- Oxygen (O): 4 atoms, each with a molar mass of 16 g/mol, contributing \( 4 \times 16 = 64 \mathrm{~g/mol} \).Sum these to get: \( 108 + 8 + 64 = 180 \mathrm{~g/mol} \). Thus, the empirical formula molar mass is \(180 \mathrm{~g/mol} \).

Comparing with Molecular Molar Mass

Compare the empirical formula molar mass with the given molecular molar mass of aspirin, which is also \(180 \mathrm{~g/mol} \). They are equal, which means the molecular formula is the same as the empirical formula as it contains all parts of the molecule at stoichiometric ratios.

Key concepts

Understanding Molar Mass

Molar mass is an essential concept in chemistry that refers to the mass of one mole of a substance. Molecules are made up of atoms, and each atom contributes to the overall weight based on its atomic mass. In our case, we are investigating aspirin's molar mass, which is given as 180 g/mol.
To calculate the molar mass of a compound, you need to sum the molar masses of all the individual atoms present in the compound's formula. For example, if we have a compound with a formula such as C\(_7\)H\(_5\)O\(_2\), we would:

• Multiply the number of carbon (C) atoms by its molar mass (12 g/mol) and add this to the total.
• Multiply the number of hydrogen (H) atoms by its molar mass (1 g/mol) and add this to the total.
• Multiply the number of oxygen (O) atoms by its molar mass (16 g/mol) and add this to the total.

Once added up, the result is the total molar mass of the compound.

In the case of aspirin, its given molar mass is already 180 g/mol, matching the calculation from its empirical formula.

Empirical Formula Explained

The empirical formula of a compound gives us the simplest whole number ratio of atoms of each element present. It does not necessarily tell us about the actual number of atoms, but rather the "shortened" version of the atomic arrangement in the compound.

For instance, in the case of aspirin, the empirical formula provided is C\(_9\)H\(_8\)O\(_4\). This compound indicates that for every 9 carbon (C) atoms, there are 8 hydrogen (H) atoms and 4 oxygen (O) atoms.
This ratio can provide valuable insight into the compound’s chemical properties.

• The empirical formula is useful in reactions where proportions, rather than actual quantities, are important.
• It is often a stepping stone to finding the molecular formula, especially when paired with molar mass information.

In our aspirin example, it turns out the empirical formula perfectly matches the molecular formula since the calculated empirical molar mass is identical to the known molar mass of aspirin.

Stoichiometry in Chemistry

Stoichiometry is a key concept in chemistry that involves calculating the quantities of reactants and products in chemical reactions. It allows chemists to predict how much product will be formed from given amounts of reactants, or even determine how much of each reactant is necessary.

It's like a recipe, ensuring that each ingredient is measured accurately for the desired outcome. For instance:

• If you know the molar mass and empirical formula of a compound, you can determine the stoichiometric proportions in a reaction.
• Stoichiometry helps in balancing chemical equations, ensuring the conservation of mass in reactions.

When dealing with compounds like aspirin, stoichiometry can tell us how the empirical formula relates to the actual molecular structure.
This is especially true when comparing its empirical formula molar mass to the given molecular molar mass, as it determines whether any additional scaling is required for the molecular formula.
In aspirin’s case, no further scaling was necessary as the molar masses matched, simplifying the stoichiometric calculations needed.