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 is ${\mathrm{CH}}_6{\mathrm{N}}_2$.

Step-by-step solution

Calculate the amount of carbon in the compound

To find the amount of carbon in the compound, first, let's determine the amount of carbon present in ${\mathrm{CO}}_2$. Since the molar mass of carbon is 12.01g/mol and the molar mass of ${\mathrm{CO}}_2$ is 44.01 g/mol, we get the proportion of carbon in ${\mathrm{CO}}_2$ as: $$\frac{12.01}{44.01}\approx 0.273$$ Now, multiply the mass of ${\mathrm{CO}}_2$ produced (33.5 mg) by the proportion of carbon in ${\mathrm{CO}}_2$ to find the mass of carbon in the compound: $$33.5\mathrm{mg}\cdot 0.273\approx 9.14\mathrm{mg}$$

Calculate the amount of hydrogen in the compound

Similarly, we'll find the amount of hydrogen in the compound. Since the molar mass of hydrogen is 1.01 g/mol and the molar mass of ${\mathrm{H}}_2\mathrm{O}$ is 18.02 g/mol, we get the proportion of hydrogen in ${\mathrm{H}}_2\mathrm{O}$ as: $$\frac{2\cdot 1.01}{18.02}\approx 0.112$$ Now, multiply the mass of ${\mathrm{H}}_2\mathrm{O}$ produced (41.1 mg) by the proportion of hydrogen in ${\mathrm{H}}_2\mathrm{O}$ to find the mass of hydrogen in the compound: $$41.1\mathrm{mg}\cdot 0.112\approx 4.60\mathrm{mg}$$

Calculate the amount of nitrogen in the compound

The compound contains only C, H, and N. We know the total mass of the compound is 35.0 mg. So, to find the mass of nitrogen, we subtract the masses of carbon and hydrogen we calculated in steps 1 and 2: $$35.0\mathrm{mg}-9.14\mathrm{mg}-4.60\mathrm{mg}\approx 21.26\mathrm{mg}$$

Find moles of each element

To determine the empirical formula, we need to find the simplest whole number ratio of each element in the compound. For this, divide the mass of each element by its respective molar mass to find the moles: Moles of carbon: $$\frac{9.14\mathrm{mg}}{12.01\mathrm{g}/\mathrm{mol}\cdot 1000\mathrm{mg}/\mathrm{g}}\approx 0.760\mathrm{mol}$$ Moles of hydrogen: $$\frac{4.60\mathrm{mg}}{1.01\mathrm{g}/\mathrm{mol}\cdot 1000\mathrm{mg}/\mathrm{g}}\approx 4.56\mathrm{mol}$$ Moles of nitrogen: $$\frac{21.26\mathrm{mg}}{14.01\mathrm{g}/\mathrm{mol}\cdot 1000\mathrm{mg}/\mathrm{g}}\approx 1.52\mathrm{mol}$$

Find the simplest whole number ratio of elements

Divide the moles of each element by the smallest number of moles (0.760 in this case) to find the simplest whole number ratio: $$\frac{0.760}{0.760}:\frac{4.56}{0.760}:\frac{1.52}{0.760}\approx 1:6:2$$

Write the empirical formula

Based on the simplest whole number ratio of elements, the empirical formula of the compound is: $${\mathrm{CH}}_6{\mathrm{N}}_2$$

Key concepts

Combustion Analysis

Combustion analysis is a powerful method used to determine the empirical formula of organic compounds. By burning the compound in oxygen, we measure the amounts of carbon dioxide (${\mathrm{CO}}_2$) and water (${\mathrm{H}}_2\mathrm{O}$) produced. Since the initial compound is combusted, and knowing that ${\mathrm{CO}}_2$ contains all the carbon and ${\mathrm{H}}_2\mathrm{O}$ contains all the hydrogen originally present, we can backtrack to deduce how much of each element was in the original compound.
This process helps us figure out the chemical composition of the compound, as with the given problem where we had to find the empirical formula by analyzing the products of combustion.

Chemical Composition

The chemical composition refers to the different elements that make up a compound and their relative amounts. In many chemical analyses, including combustion analysis, finding the chemical composition is a key step to understanding the nature and structure of a compound.
In our example, we started with a compound consisting of carbon ($\mathrm{C}$), hydrogen ($\mathrm{H}$), and nitrogen ($\mathrm{N}$). By analyzing the combustion products (${\mathrm{CO}}_2$ and ${\mathrm{H}}_2\mathrm{O}$), we were able to determine the masses of the individual elements within the original compound. This is crucial because it forms the basis for further calculations to derive the compound’s empirical formula.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It plays a critical role in calculations like those needed for combustion analysis.
Using stoichiometry, we calculated the masses of carbon and hydrogen by first understanding their fractional contributions in ${\mathrm{CO}}_2$ and ${\mathrm{H}}_2\mathrm{O}$, respectively. These ratios allowed us to determine the exact amounts of carbon and hydrogen in the original substance, which were subsequently used in finding the moles needed to solve for the empirical formula.
Stoichiometry ensures that our calculations maintain balance according to the law of conservation of mass, which states that mass is neither created nor destroyed.

Mass Percentage

The mass percentage of an element in a compound tells us how much of that element is present in comparison to the whole compound. This is usually represented as a percentage.
While from a quantitative standpoint we deal with masses in milligrams or grams, understanding the mass percentage can sometimes provide quicker insights into the composition. In the context of the initial problem, while we directly used mass, knowing the mass percentages could immediately hint at the dominance or minor presence of certain elements. However, the step-by-step calculation is more precise for deriving the empirical formula.

Molar Mass

Molar mass is the mass in grams of one mole of a substance. It is expressed in units of g/mol. This concept is pivotal when moving from determining mass to moles of a substance, an essential step in empirical formula calculations.
In the given exercise, the molar masses of carbon (12.01 g/mol), hydrogen (1.01 g/mol), and nitrogen (14.01 g/mol) were crucial in converting the masses of $\mathrm{C}$, $\mathrm{H}$, and $\mathrm{N}$ into moles. Knowing the molar mass allows us to transform the mass of an element into a more universally comparable unit ($\mathrm{mol}$), thereby allowing for the calculation of the stoichiometric ratios needed for solving the empirical formula.
Understanding molar mass is thus fundamental for quantitative chemical analyses.