An organic compound contains \(40 \%\) carbon and \(6.67 \%\) hydrogen by mass. Which of the following represents the empirical formula of the compound? (a) \(\mathrm{CH}_{2}\) (b) \(\mathrm{CH}_{2} \mathrm{O}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) (d) \(\mathrm{CH}_{3} \mathrm{O}\)

Understanding the mass percentage composition of a compound is fundamental in chemistry. It tells us the fraction of a compound's mass that each element contributes. This information is expressed as a percentage and is calculated by dividing the mass of the specific element by the total mass of the compound, then multiplying by 100.

For example, in the given exercise, an organic compound is said to contain 40% carbon and 6.67% hydrogen by mass. This implies that if we consider a sample of 100 grams of this compound, 40 grams would be carbon, and 6.67 grams would be hydrogen. The remaining mass, which is calculated by subtracting the mass of carbon and hydrogen from the total mass, can be attributed to oxygen since we are dealing with an organic compound.

It's crucial to note that getting the masses from the percentage composition is the first practical step towards finding the empirical formula, which is the ratio of the elements in the simplest form.

After determining the composition by mass, chemists often need to relate these masses to the number of atoms, which requires converting mass into moles. The mole is a key concept in stoichiometry that acts as a bridge between the atomic scale and the macroscopic scale. To convert grams to moles, one must use the molecular weight of each element, which is the weight in grams of one mole of that element.

In our exercise, we convert the mass of carbon, hydrogen, and oxygen into moles. For example, for carbon with a mass of 40 grams, we divide by its atomic weight, 12.01 gram per mole, which gives us 3.33 moles of carbon. Performing similar calculations, we find the moles of hydrogen and oxygen in our sample, providing us the basis to deduce the empirical formula.

Stoichiometry is the portion of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. In the context of finding empirical formulas, stoichiometry allows us to convert between masses, moles, and atomic ratios.

Once we have the moles for each element, stoichiometry comes into play to find the ratio that will give us the empirical formula. By dividing the number of moles of each element by the smallest number obtained, we're able to find the simplest whole number ratio of the elements. This ratio provides a clear picture of the elemental composition of the compound, which is crucial for understanding its properties and reactions it may undergo.

In stoichiometry, precision is key. It's vital to carry out all calculations accurately to avoid errors that can lead to incorrect empirical formulas and thus, misunderstandings about the compound.

Analysis of organic compounds involves determining their composition and structure. The empirical formula is a crucial part of the analysis as it provides information about the simplest whole number ratio of the atoms of each element in the compound. This exercise demonstrates the procedure in a straightforward context.

In practice, complex analytical techniques such as mass spectrometry, infrared spectroscopy, and nuclear magnetic resonance are used to determine the empirical formula of organic compounds. However, the principles remain the same: finding out the elemental makeup and their ratios.

Note that empirical formulas do not give any information about the structure of the molecule - that is, how the atoms are bonded or arranged. For example, the empirical formula CH2 could correspond to many different compounds with distinct structures and properties, highlighting the importance of structural formulas in organic chemistry.