Question

A compound contains only \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{N}\). Combustion of \(35.0 \mathrm{mg}\) of the compound produces \(33.5 \mathrm{mg} \mathrm{CO}_{2}\) and \(41.1 \mathrm{mg}\) \(\mathrm{H}_{2} \mathrm{O}\). What is the empirical formula of the compound?

Short answer

The empirical formula of the compound containing C, H, and N is \(C_3H_7N\).

Step-by-step solution

Determine moles of carbon and hydrogen

We know that one mole of CO₂ contains one mole of carbon, and one mole of H₂O contains two moles of hydrogen. Given the mass of CO₂ and H₂O, we can find the moles of carbon and hydrogen produced after combustion. Moles of Carbon = \( \frac{mass~of~CO_2}{molar~mass~of~CO_2} \times moles~of~C~in~CO_2 \) Moles of Carbon = \( \frac{33.5~mg}{44.01~g/mol} \times 1 \) (convert mg to g: 33.5mg = 0.0335g) Moles of Hydrogen = \( \frac{mass~of~H_2O}{molar~mass~of~H_2O} \times moles~of~H~in~H_2O \) Moles of Hydrogen = \( \frac{41.1~mg}{18.02~g/mol} \times 2 \) (convert mg to g: 41.1mg = 0.0411g)

Determine mass of nitrogen

The compound contains only C, H, and N. We can find the mass of nitrogen by subtracting the masses of carbon and hydrogen from the total mass of the compound. Mass of Nitrogen = Total mass of compound - (Mass of Carbon + Mass of Hydrogen) Initially, find the masses of Carbon and Hydrogen: Mass of Carbon = Moles of Carbon * Molar mass of C Mass of Hydrogen = Moles of Hydrogen * Molar mass of H

Calculate moles of nitrogen

Now that we have the mass of nitrogen, we need to find the moles of nitrogen in the compound. Moles of Nitrogen = \( \frac{mass~of~N}{molar~mass~of~N} \)

Find the whole number ratio

Divide the moles of each element (C, H, and N) by the smallest value among the three and round them off to the nearest whole number.

Write the empirical formula

Write the empirical formula of the compound using the whole numbered ratio of the elements (C, H, and N) obtained in step 4. This will give us the empirical formula of the compound.

Key concepts

Combustion Analysis

Combustion analysis is a technique used to determine the elemental composition of a compound. When a substance combusts, it reacts with oxygen to produce carbon dioxide and water if it contains carbon and hydrogen, respectively. In this process, knowing the amounts of \(\text{CO}_2\)and \(\text{H}_2\text{O}\)produced allows us to deduce the moles of carbon and hydrogen in the original compound.
This method is particularly useful for organic compounds, which often consist of carbon, hydrogen, and sometimes other elements like nitrogen.
To perform a combustion analysis, you begin by precisely measuring the masses of the products after combustion.

• From the \(\text{CO}_2\) mass, the amount of carbon in the original compound can be deduced, since each \(\text{CO}_2\) molecule contains one carbon atom.
• Similarly, the amount of hydrogen can be determined from the \(\text{H}_2\text{O}\) produced, as each molecule of water contains two hydrogen atoms.

This provides a clear picture of how carbon and hydrogen are transformed during combustion.

Mole Calculations

Mole calculations are foundational in chemistry, allowing us to translate between mass and the number of atoms or molecules. The mole is a unit that represents \(6.022 \times 10^{23}\) entities (Avogadro's number) and is key when dealing with molecular quantities.
This concept is essential when interpreting the products of combustion analysis.

• To find moles of carbon in a combustion problem, divide the mass of \(\text{CO}_2\) produced by its molar mass (44.01 g/mol). This gives the number of moles of carbon atoms, since each \(\text{CO}_2\) contains one carbon atom.
• For hydrogen, divide the mass of \(\text{H}_2\text{O}\) by its molar mass (18.02 g/mol), then multiply by two, as each water molecule contains two hydrogen atoms.

These calculations transform experimental data into quantitative chemical information, crucial for further analysis.

Elemental Composition

Elemental composition involves determining the amounts of various elements within a compound. In the context of combustion analysis, this typically focuses on quantifying carbon, hydrogen, and any other elements present, such as nitrogen.
After finding the moles of carbon and hydrogen, the composition dictates the leftover portion must consist of other elements in the compound.

• The initial step involves calculating the mass of each element from the moles derived earlier.
• Next, subtract the mass of carbon and hydrogen from the total compound mass to find the mass of other elements like nitrogen.

By accurately determining the mass and moles of each element, one can construct a detailed breakdown of the compound's components, which is critical for deriving empirical formulas.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions, often starting with balanced equations. However, when determining empirical formulas, stoichiometry primarily focuses on the mole ratios of elements.
To establish these ratios, divide the moles of each element in the compound by the smallest amount found. Doing this aligns these values into the simplest whole-number ratio, which directly informs the empirical formula.

• This step standardizes the relative amounts of each element.
• A clear whole-number ratio solidifies the proportion in which elements combine within a compound.

This stoichiometric ratio is the backbone of empirical formulas, offering insight into a compound’s foundational structure. With these ratios, one can translate quantitative details from combustion analyses into comprehensible chemical formulas.