A \(2.000 \mathrm{~g}\) sample of a liquid compound that contains only carbon and hydrogen in its formula is subjected to combustion analysis. From the result, the lab determines that \(0.2874 \mathrm{~g}\) of the original sample's mass is due to hydrogen. (a) What is the percent by mass hydrogen for this sample? (b) What is the empirical formula for this compound? (c) The molar mass of this compound is determined to be about \(71 \mathrm{~g} / \mathrm{mol}\). What is the molecular formula for this compound?

Understanding the concept of percent by mass is crucial when dealing with chemical compounds. It is a ratio that compares the mass of a single element to the total mass of a compound, expressed as a percentage. Often used in chemistry, this shows how much of a compound's mass is made up by a specific element. To calculate it, one must divide the mass of the element of interest by the mass of the entire compound and then multiply the result by 100.

For example, if we have a 2.000 g sample of a compound, and we find that 0.2874 g of that sample is hydrogen, we would calculate the percent by mass of hydrogen like this:
Percent by mass hydrogen = (Mass of hydrogen / Mass of the sample) \times 100 = (0.2874 g / 2.000 g) \times 100 \[\approx 14.37 \%\]This calculation is fundamental in stoichiometry and helps chemists understand the composition of substances.

The empirical formula of a compound gives the simplest whole-number ratio of each type of atom in the compound. It may not represent the actual number of atoms in a molecule but does provide the proportional relationship. To determine an empirical formula, one must first convert the masses of each element in a compound to moles by dividing by their respective molar masses. Then, these mole values are compared by dividing them by the smallest mole value among them to obtain the simplest ratio.

In the case of the compound from our problem, we calculated the moles of carbon and hydrogen and then divided these by the smallest number of moles (in this case, moles of carbon) to find a ratio of 1:2. This resulted in an empirical formula of CH\(_2\), signifying that for every carbon atom, there are two hydrogen atoms.

While the empirical formula indicates the simplest ratio of atoms in a compound, the molecular formula shows the actual number of each type of atom in a molecule. To find it, one needs to know the compound's empirical formula and its molar mass (the mass of one mole of the compound).

The molar mass of the empirical formula is calculated by adding up the molar masses of all atoms according to the empirical formula. Then, the molar mass of the compound is divided by the empirical formula's molar mass to find a ratio. This ratio or multiplier is then used to scale up the empirical formula to get the molecular formula. For example, our compound's empirical formula is CH\(_2\) with a molar mass of approximately 14.03 g/mol, and the given molar mass of the compound is 71 g/mol. Dividing these gives us a multiplier of about 5, which when applied to the empirical formula, gives C\(_5\)H\(_{10}\), the molecular formula.

The concept of molar mass is another cornerstone of chemistry. It represents the mass of one mole of a substance (usually expressed in grams per mole), which is numerically equivalent to the average mass of one molecule of the substance (expressed in atomic mass units, amu). To calculate the molar mass, you add up the atomic masses (found on the periodic table) of all the atoms in a molecule.

For carbon (C), the atomic mass is 12.01 amu, and for hydrogen (H), it is 1.01 amu. Hence, for our empirical formula CH\(_2\), the calculation would be the sum of one carbon atom and two hydrogen atoms: \[Molar mass of CH_2 = (1 \times 12.01 \text{ g/mol}) \ + (2 \times 1.01 \text{ g/mol}) = 14.03 \text{ g/mol}\]When dealing with molecular formulas, the molar mass is a key element to transition between empirical formulas and molecular formulas, as seen in solving the exercise.