Question

Quinone, a chemical used in the dye industry and in photography, is an organic compound containing only \(\mathrm{C}\) H, and O. What is the empirical formula of the compound if \(0.105 \mathrm{g}\) of the compound gives \(0.257 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(0.0350 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}\) when burned completely in oxygen?

Short answer

The empirical formula is \( \mathrm{C}_3 \mathrm{H}_2 \mathrm{O} \).

Step-by-step solution

Determine Moles of CO2 and H2O

First, calculate the number of moles of \( \mathrm{CO}_2 \) and \( \mathrm{H}_2\mathrm{O} \) produced from the combustion. \[\text{Moles of } \mathrm{CO}_2 = \frac{0.257 \text{ g}}{44.01 \text{ g/mol}} = 0.00584 \text{ mol}\] \[\text{Moles of } \mathrm{H}_2\mathrm{O} = \frac{0.0350 \text{ g}}{18.02 \text{ g/mol}} = 0.00194 \text{ mol}\]

Calculate Moles of Carbon and Hydrogen

Next, use the moles of \( \mathrm{CO}_2 \) and \( \mathrm{H}_2\mathrm{O} \) to find the moles of carbon and hydrogen in the sample. Each mole of \( \mathrm{CO}_2 \) contains 1 mole of carbon and each mole of \( \mathrm{H}_2\mathrm{O} \) contains 2 moles of hydrogen. \[\text{Moles of C} = 0.00584 \text{ mol}\] \[\text{Moles of H} = 2 \times 0.00194 \text{ mol} = 0.00388 \text{ mol}\]

Calculate Mass of Carbon and Hydrogen

Now, calculate the mass of carbon and hydrogen using their respective atomic masses. The atomic mass of carbon is 12.01 g/mol and hydrogen is 1.008 g/mol. \[\text{Mass of C} = 0.00584 \times 12.01 = 0.0701 \text{ g}\] \[\text{Mass of H} = 0.00388 \times 1.008 = 0.0039 \text{ g}\]

Determine Mass of Oxygen

Subtract the combined mass of carbon and hydrogen from the total mass of the compound to find the mass of oxygen. \[\text{Mass of O} = 0.105 \text{ g} - (0.0701 \text{ g} + 0.0039 \text{ g}) = 0.031 \text{ g}\]

Calculate Moles of Oxygen

Find the moles of oxygen by dividing the mass of oxygen by its atomic mass (16.00 g/mol). \[\text{Moles of O} = \frac{0.031}{16.00} = 0.00194 \text{ mol}\]

Determine the Empirical Formula

Divide the moles of C, H, and O by the smallest number of moles calculated in the previous steps, which is 0.00194 mol. \[\text{Ratio of C} = \frac{0.00584}{0.00194} \approx 3\] \[\text{Ratio of H} = \frac{0.00388}{0.00194} \approx 2\] \[\text{Ratio of O} = \frac{0.00194}{0.00194} = 1\] Thus, the empirical formula is \( \mathrm{C}_3 \mathrm{H}_2 \mathrm{O} \).

Key concepts

Organic Compounds

Organic compounds are chemical substances that contain carbon atoms bonded to other elements, primarily hydrogen, oxygen, nitrogen, and more. These compounds form the basis of all known life on Earth. Carbon's ability to form long chains and stable rings allows it to create complex and diverse molecules. These molecules are the building blocks of living organisms and are prevalent in applications such as fuels, plastics, and medicines.
Understanding the composition of organic compounds is crucial in chemistry. Each compound has a unique combination of elements in particular ratios, leading to specific structures and properties. For example, quinone is a type of organic compound used in dyes and photography. By analyzing its combustion products, we can determine its empirical formula — a representation showing the simplest whole-number ratio of atoms in the compound. Learning how organic compounds are structured and how their molecular compositions affect their functions is vital in numerous scientific fields.

Combustion Analysis

Combustion analysis is a common laboratory technique used to determine the composition of organic compounds. By burning the compound in the presence of oxygen, it breaks down into simpler substances, typically carbon dioxide ( CO_2 ) and water ( H_2O ).
This method allows chemists to deduce the amounts of carbon and hydrogen present in the original compound. For instance, if a sample of quinone is burned, collecting and measuring the resultant CO_2 and H_2O helps us ascertain the quantities of carbon and hydrogen. Knowing these amounts, along with the initial mass of the sample, allows for determining the oxygen content by subtraction.
The precision of combustion analysis makes it an indispensable tool in chemistry. It not only helps in verifying the purity of organic substances but also in identifying unknown compounds by deriving their empirical formulas. Understanding this process is essential for chemists who work on synthesizing new compounds or analyzing existing ones.

Mole Calculations

Mole calculations are fundamental in chemistry as they allow scientists to relate the mass of a substance to the number of atoms or molecules it contains. The mole is a convenient unit because it is based on Avogadro's number, which is 6.022 imes 10^{23} entities per mole.
In the combustion analysis of quinone, calculating moles involves determining how many moles of CO_2 and H_2O are produced. From these moles, we can directly find the moles of carbon and hydrogen present in the original compound. Each molecule of CO_2 corresponds to a single carbon atom, and each molecule of H_2O corresponds to two hydrogen atoms.
Using mole calculations allows for precise conversion between mass and molecular quantity. It is an essential skill for any chemistry student, enabling better understanding of empirical formulas, reaction stoichiometry, and the overall reactions taking place in an experiment. By mastering mole calculations, students can solve complex chemical problems with confidence.