Question

A gaseous compound is composed of \(85.7 \%\) by mass carbon and \(14.3 \%\) by mass hydrogen. It's density is \(2.28 \mathrm{~g}\) /litre at \(300 \mathrm{~K}\) and \(1.0\) atm pressure. Determine the molecular formula of the compound: (a) \(\mathrm{C}_{2} \mathrm{H}_{2}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (c) \(\mathrm{C}_{4} \mathrm{H}_{8}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{10}\)

Short answer

The molecular formula of the compound is \text{(c) } \text{C}_4\text{H}_8.

Step-by-step solution

Calculate the molar mass

The molar mass (M) can be determined using the ideal gas law equation, PV = nRT. Rearrange the equation to n = PV/RT where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Since density (d) is mass (m) over volume (V), we can also say that d = m/V. Here, d = 2.28 g/L, P = 1.0 atm, and T = 300 K. We will use the gas constant R = 0.0821 L·atm/(K·mol). Substituting these values, the molar mass M can be calculated by M = (dRT)/P.

Determine the Molar Mass

Using the values from Step 1: M = (2.28 g/L * 0.0821 L·atm/(K·mol) * 300 K) / 1.0 atm = 56.0892 g/mol. This value is the molar mass of the gaseous compound.

Calculate the empirical formula mass

Assuming 100 g of the compound, we would have 85.7 g of carbon and 14.3 g of hydrogen according to the percentage composition. The molar mass of carbon (C) is 12 g/mol, and hydrogen (H) is 1 g/mol. Calculate the moles of each: Moles of C = 85.7 g / 12 g/mol = 7.1417 mol; Moles of H = 14.3 g / 1 g/mol = 14.3 mol. These are the ratios of the elements in the empirical formula, and they can be simplified to integers.

Determine the empirical formula

As the ratio of C to H is 7.1417 mol to 14.3 mol, or approximately 1 to 2 when simplified, the empirical formula of the compound is CH2.

Deduce the molecular formula

Divide the molar mass of the compound (obtained in Step 2) by the molar mass of the empirical formula (CH2 has a molar mass of 12 g/mol for C and 2*1 g/mol for H, totaling 14 g/mol). Molar mass of compound / Empirical formula mass = 56.0892 g/mol / 14 g/mol = 4.0064 which rounds to 4. This means the molecular formula is four times the empirical formula, resulting in C4H8.

Match the molecular formula with the given options

The calculated molecular formula C4H8 corresponds to option (c) C4H8, which is the correct answer.

Key concepts

Empirical Formula Calculation

Understanding the empirical formula of a compound is a fundamental step in analyzing its composition. The empirical formula represents the simplest whole number ratio of the elements in the compound. To calculate it, you must first convert the mass percentages of each component to moles by using the atomic masses of the elements, a kind of stoichiometry in itself.

For instance, if you have a compound that’s 85.7% carbon by mass and 14.3% hydrogen, you would assume a sample size of 100 grams for simplicity. With carbon's atomic mass approximately 12 g/mol and hydrogen's 1 g/mol, the moles of each can be calculated. The moles of carbon would be approximately 7.14 (85.7 g / 12 g/mol) and hydrogen would be 14.3 (14.3 g / 1 g/mol). These values give you a direct mole ratio that can then be simplified to the smallest whole numbers, representing the empirical formula.

However, empirical formulas can be deceiving because they don't tell you about the actual numbers of atoms in a molecule—only the ratio. This is where molecular formula determination comes into play, which can be found by comparing the empirical formula mass to the molar mass of the compound.

Molar Mass Calculation

The molar mass calculation is crucial for converting between grams and moles, enabling us to ease through stoichiometric calculations and use the ideal gas law. Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol).

To determine the molar mass of a compound, as shown in the exercise, you can use the ideal gas law, which connects the pressure (P), volume (V), temperature (T), and the number of moles (n) of a gas. When the pressure, volume, and temperature are known, the number of moles can be deduced, and from there, the molar mass, since molar mass is the mass of one mole of the gas. By rearranging the ideal gas law to solve for molar mass, and using the density to link mass and volume, one can find the molar mass of the gas at the given conditions.

Ideal Gas Law Application

The ideal gas law is a powerful equation that relates the four critical physical properties of a gaseous sample: pressure (P), volume (V), number of moles (n), and temperature (T). Expressed as PV = nRT, this relationship allows scientists and students alike to calculate any one of these variables if the others are known.

As demonstrated in solving the textbook problem, the ideal gas law is rearranged to find the molar mass by setting n (number of moles) equal to the mass (m) divided by the molar mass (M): that is, n = m/M. Applying the density (d), which is mass per unit volume (m/V), leads to the expression M = (dRT)/P when density, the gas constant (R), temperature, and pressure are given. This application of the ideal gas law is indispensable in finding the molar mass and ties directly into the stoichiometry and empirical formula concepts when solving for a compound's molecular formula.

Stoichiometry

Stoichiometry is the section of chemistry that involves using the relationships between reactants and products in a chemical reaction to determine desired quantitative data. It heavily relies on the balanced chemical equation and the concept of the mole to let us predict the amounts of substances consumed and produced in a given reaction.

In the context of finding the molecular formula, stoichiometry is used to scale the empirical formula up to the actual molecular formula. You do this by dividing the molar mass of the entire compound by the mass of the empirical formula to find out how many times larger the actual molecule is compared to the empirical formula. This allows you to multiply the ratio of atoms in the empirical formula by this number to get the precise number of atoms in each element that the molecular formula has.

Such stoichiometric calculations are integral for chemists when synthesizing materials, formulating products, or just understanding the nature of the substances they are working with.