Question

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of \(0.157 \mathrm{~g}\) of the compound produced \(0.213 \mathrm{~g} \mathrm{CO}_{2}\) and \(0.0310 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) In another experiment, it is found that \(0.103 \mathrm{~g}\) of the compound produces \(0.0230 \mathrm{~g} \mathrm{NH}_{3} .\) What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\). Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\). Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

Short answer

The empirical formula of the compound is C4H3NO2.

Step-by-step solution

Find the moles of carbon and hydrogen from the combustion experiment.

We know from the problem that the combustion produces carbon dioxide (CO2) and water (H2O), with all the carbon in CO2 and all the hydrogen in H2O. Using the given masses of CO2 and H2O produced and the molar masses of CO2 (44.01 g/mol) and H2O (18.02 g/mol), we can find the moles of carbon and hydrogen. Moles of CO2 = Mass of CO2 / Molar mass of CO2 = 0.213 g / 44.01 g/mol ≈ 0.00484 mol CO2 Moles of H2O = Mass of H2O / Molar mass of H2O = 0.0310 g / 18.02 g/mol ≈ 0.00172 mol H2O Since there is 1 carbon atom in 1 CO2 molecule and 2 hydrogen atoms in 1 H2O molecule, we can find the moles of carbon and hydrogen. Moles of C = 0.00484 mol CO2 × (1 mol C / 1 mol CO2) ≈ 0.00484 mol C Moles of H = 0.00172 mol H2O × (2 mol H / 1 mol H2O) ≈ 0.00344 mol H

Find the moles of nitrogen from the second experiment.

In the second experiment, we are given that the compound produces ammonia (NH3) and the mass of NH3 formed when reacting with nitrogen. Since all the nitrogen in the compound ends up in NH3, we can find the moles of nitrogen using the given mass of NH3 (0.0230 g) and the molar mass of NH3 (17.03 g/mol). Moles of NH3 = Mass of NH3 / Molar mass of NH3 = 0.0230 g / 17.03 g/mol ≈ 0.00135 mol NH3 Since there is 1 nitrogen atom in 1 NH3 molecule, we can find the moles of nitrogen. Moles of N = 0.00135 mol NH3 × (1 mol N / 1 mol NH3) ≈ 0.00135 mol N

Determine the moles of oxygen.

Now that we have the moles of carbon, hydrogen, and nitrogen, we can find the moles of oxygen. We know the total mass of the compound is 0.157g, and the molar masses of the elements: carbon (12.01 g/mol), hydrogen (1.01 g/mol), nitrogen (14.01 g/mol), and oxygen (16.00 g/mol). First, we find the total mass of C, H, and N. Total mass of C, H, and N = Moles of C × Molar mass of C + Moles of H × Molar mass of H + Moles of N × Molar mass of N Total mass of C, H, and N = (0.00484 mol × 12.01 g/mol) + (0.00344 mol × 1.01 g/mol) + (0.00135 mol × 14.01 g/mol) = 0.121 g  Now, we can find the mass of oxygen (O) in the compound by subtracting the mass of C, H, and N from the total mass of the compound. Mass of O = Total mass of the compound - Total mass of C, H, and N Mass of O = 0.157 g - 0.121 g = 0.036 g Finally, we can find the moles of oxygen. Moles of O = Mass of O / Molar mass of O = 0.036 g / 16.00 g/mol ≈ 0.00225 mol O

Determine the empirical formula.

To find the empirical formula, we need to determine the smallest whole-number ratio of moles of the elements. First, find the smallest number of moles (use the smallest number of moles from the calculations): Smallest no. of moles = 0.00135 mol N Divide the number of moles of each element by the smallest number of moles: C: 0.00484 mol / 0.00135 mol ≈ 3.58 H: 0.00344 mol / 0.00135 mol ≈ 2.55 N: 0.00135 mol / 0.00135 mol = 1 O: 0.00225 mol / 0.00135 mol ≈ 1.67 To obtain whole numbers, we can round the ratios to the nearest whole number. The empirical formula of the compound is C4H3NO2.

Key concepts

Combustion Analysis

Combustion analysis is a powerful technique used to determine the composition of an organic compound. In this method, the compound is burned in excess oxygen.
This results in the formation of products such as carbon dioxide (CO2) and water (H2O). By measuring the amounts of these combustion products, we can deduce the quantities of carbon and hydrogen initially present in the compound.
- **Carbon Analysis**: Since all carbon in the compound ends up in CO2, we can simply calculate the moles of CO2 and thus determine the moles of carbon. - **Hydrogen Analysis**: Similarly, all hydrogen in the compound appears in water (H2O), allowing for the calculation of hydrogen moles. Using the mass and molar mass of these combustion products, we convert the amounts to moles to backtrack to the composition of the original compound.

Organic Compound

An organic compound primarily consists of carbon and hydrogen atoms, but it can also contain other elements like nitrogen and oxygen, which are crucial for the compound's characteristics.
These elements form the backbone of the compound's structure and determine its properties and reactions.
During combustion analysis, such compounds are typically burned to yield CO2, H2O, and sometimes other products, which aids in revealing the elemental make-up of the substance.
Organic chemistry, which studies these compounds, uses combustion analysis to infer empirical formulas, which are the simplest representation of the compound's elemental ratio.

Stoichiometry

Stoichiometry is the science of measuring the quantitative relationships and ratios between reactants and products in a chemical reaction. It is essential for determining the empirical formula from combustion analysis.
This concept allows us to relate the mass of a substance to the number of moles through its molar mass. Using stoichiometry, you can predict the amount of products formed from certain reactants and vice-versa.

• **Balancing Chemical Equations**: This involves ensuring that the number of atoms for each element is equal on both sides of an equation.
• **Mole Ratios**: Using known reactions, like the combustion reaction, to find how much of each element corresponds to the products like CO2 and H2O.

In the context of the given problem, stoichiometry connects the masses with moles, helping establish the composition of the compound.

Mole Calculation

Mole calculations are central to stoichiometry and combustion analysis, allowing us to bridge the gap between mass and the number of particles. The mole is a fundamental unit in chemistry that helps quantify and compare amounts of substances in a reaction.
Using the formula \( \text{Moles} = \frac{\text{mass}}{\text{molar mass}} \), we convert grams of a substance to moles, enabling us to compare different elements' roles in a chemical reaction.
In the problem scenario, calculating moles for CO2, H2O, and NH3 provided the basis for determining the amounts of carbon, hydrogen, and nitrogen in the original organic compound.

• **Consistency in Units**: Always ensure the use of correct units and precise molar masses to achieve accurate calculations.
• **Calculation Steps**: Solving step-by-step makes it easier to trace errors and ensures clarity.

Understanding moles helps in deducing the simplest whole-number ratio of elements, crucial for determining the empirical formula.