Question

"Smelling salts," which are used to revive someone who has fainted, typically contain ammonium carbonate, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\). Ammonium carbonate decomposes readily to form ammonia, carbon dioxide, and water. The strong odor of the ammonia usually restores consciousness in the person who has fainted. The unbalanced equation is $$ \left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Calculate the mass of ammonia gas that is produced if \(1.25 \mathrm{~g}\) of ammonium carbonate decomposes completcly.

Short answer

The mass of ammonia gas (\(\mathrm{NH}_3\)) produced when 1.25 grams of ammonium carbonate (\(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\)) decompose completely is 0.221 grams.

Step-by-step solution

Balance the chemical equation

Balance the given equation by adjusting the coefficients: $$ \mathrm{NH}_{4} \mathrm{CO}_{3}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Here, we can see that the number of atoms of each element is already balanced, so the balanced equation becomes: $$ \mathrm{NH}_{4} \mathrm{CO}_{3}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$

Convert mass of ammonium carbonate to moles

To convert the mass of ammonium carbonate to moles, we will use its molar mass. The molar mass of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) is: \(M = 2(14.01) + 6(1.008) + 12.01 + 3(16.00) = 96.12 \thinspace \mathrm{g/mol}\) Now, convert the mass of ammonium carbonate (1.25 g) to moles: $$ \frac{1.25 \thinspace \mathrm{g}}{96.12 \thinspace \mathrm{g/mol}} = 0.0130 \thinspace \mathrm{mol} $$

Use stoichiometry to find moles of ammonia

From the balanced equation, we can see that 1 mole of ammonium carbonate decomposes to produce 1 mole of ammonia. So, the moles of ammonia produced will be the same as the moles of ammonium carbonate: $$ 0.0130 \thinspace \mathrm{mol} \thinspace \mathrm{NH}_{4}\mathrm{CO}_{3} \rightarrow 0.0130 \thinspace \mathrm{mol} \thinspace \mathrm{NH}_{3} $$

Convert moles of ammonia to mass

Now, we will convert the moles of ammonia to mass using its molar mass. The molar mass of \(\mathrm{NH}_{3}\) is: \(M = 14.01 + 3(1.008) = 17.034 \thinspace \mathrm{g/mol}\) Now, convert the moles of ammonia to mass: $$ (0.0130 \thinspace \mathrm{mol}) \times (17.034 \mathrm{\thinspace g/mol}) = 0.221 \thinspace \mathrm{g} $$ Hence, the mass of ammonia gas produced when 1.25 grams of ammonium carbonate decompose completely is 0.221 grams.

Key concepts

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It is a quantitative aspect of chemistry that involves using balanced chemical equations to determine the relative amounts of reactants needed or products formed.

Consider the equation from the exercise which represents the decomposition of ammonium carbonate to form ammonia, carbon dioxide, and water. Stoichiometry allows us to use this equation to predict that one mole of ammonium carbonate decomposes to yield one mole of ammonia, one mole of carbon dioxide, and one mole of water. The stoichiometric coefficients (the numbers placed before the molecules in the equation) are all implied to be one, because the equation is balanced as is.

Using the principles of stoichiometry, we can relate the mass of substances consumed or produced in a reaction to the balanced chemical equation. By converting the mass of a reactant (in this case, ammonium carbonate) to moles using its molar mass, we can determine the number of moles of ammonia that will form, because in a stoichiometric relationship, the amount of reactant and product is fixed by their mole ratio in the balanced equation.

Chemical Equations

A chemical equation is a symbolic representation of a chemical reaction where the reactants are written on the left side and the products on the right, typically separated by an arrow indicating the direction of the reaction. In our exercise, the unbalanced chemical equation is initially given as \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\). To make it balanced, we ensure that the number of atoms of each element is the same on both sides of the equation.

In reality, the given equation is already balanced; each side has the same number of nitrogen, hydrogen, carbon, and oxygen atoms. When working with chemical equations, it is essential to remember that mass and atoms are conserved in the reaction, meaning nothing is lost or gained, only transformed. A correctly balanced equation is fundamental for applying stoichiometry in order to calculate quantities in a reaction, which is exactly what is demonstrated in this exercise.

Molar Mass

The molar mass of a substance is the mass of one mole of that substance, typically expressed in grams per mole (g/mol). It is calculated by summing the average atomic masses of all the atoms represented in the molecular formula of the substance.

In our exercise, we calculate the molar mass of ammonium carbonate by adding together the atomic masses of the constituent atoms: nitrogen, hydrogen, carbon, and oxygen. For instance, ammonium carbonate \(\left(\mathrm{NH}_{4}\right)_{2}\mathrm{CO}_{3}\) has a molar mass of \(96.12\thinspace \mathrm{g/mol}\), computed by adding twice the molar mass of ammonium \(\left(\mathrm{NH}_{4}\right)\) and one molar mass of carbonate \(\mathrm{CO}_{3}\).

Understanding molar mass is essential because it serves as a conversion factor between the mass of a substance and the number of moles. This allows us to interconvert between mass and moles of a reactant or product — a key step in stoichiometry that we carry out in the exercise to find the amount of ammonia produced from the decomposition of ammonium carbonate.