Question

A tomato weighing \(60.0 \mathrm{~g}\) is analyzed for sulfur content by digestion in concentrated \(\mathrm{HNO}_{3}\) followed by precipitation with Ba to form \(\mathrm{BaSO}_{4} .0 .466 \mathrm{~g}\) of \(\mathrm{BaSO}_{4}\) is obtained. If all of the sulfur is converted to \(\mathrm{BaSO}_{4}\), what is the percentage of sulfur in the tomato?

Short answer

The percentage of sulfur in the tomato is approximately \(2.66\%\).

Step-by-step solution

Calculate the amount of sulfur in BaSO4

Using the periodic table, we find the molar masses of the elements in BaSO4. The molar mass of Barium (Ba) is \(137.33~\mathrm{g/mol}\), Sulfur (S) is \(32.07~\mathrm{g/mol}\), and Oxygen (O) is \(16.00~\mathrm{g/mol}\). The molar mass of BaSO4 is the sum of the molar masses of each element: \(m_{\mathrm{BaSO}_{4}} = 137.33 + (32.07 + 4 \times 16.00) = 233.43~\mathrm{g/mol}\) Now we can find the fraction of sulfur in BaSO4 by dividing the molar mass of sulfur by the molar mass of BaSO4: \[\frac{m_\mathrm{S}}{m_{\mathrm{BaSO}_{4}}} = \frac{32.07}{233.43}\]

Find the mass of sulfur in the tomato

We are given that \(0.466~\mathrm{g}\) of BaSO4 is obtained. Since we know the fraction of sulfur in BaSO4, we can find the mass of sulfur in the tomato: \[m_\mathrm{S_{Tomato}} = 0.466 \cdot \frac{m_\mathrm{S}}{m_{\mathrm{BaSO}_{4}}} = 0.466 \cdot \frac{32.07}{233.43}\]

Calculate the percentage of sulfur in the tomato

Now that we have the mass of sulfur in the tomato, we can calculate the percentage of sulfur with respect to the total mass of the tomato (\(60.0~\mathrm{g}\)): \[\mathrm{Percentage}~\mathrm{of}~\mathrm{Sulfur} = \frac{m_\mathrm{S_{Tomato}}}{m_\mathrm{Tomato}} \times 100 \%\] \[= \frac{0.466 \cdot \frac{32.07}{233.43}}{60.0} \times 100\%\] Calculating the percentage, we find that: \[ \mathrm{Percentage}~\mathrm{of}~\mathrm{Sulfur} = 2.66\% \] Therefore, the percentage of sulfur in the tomato is approximately \(2.66\%\).

Key concepts

Molar Mass

Understanding the concept of molar mass is crucial for many calculations in chemistry. It is defined as the mass of one mole of a substance, typically expressed in grams per mole (\texttt{g/mol}). The molar mass of a compound is calculated by summing up the molar masses of all the atoms in a single molecule of that compound.

For instance, in our exercise, we calculated the molar mass of barium sulfate (\texttt{BaSO}\(_4\)) by adding up the molar masses of barium (Ba), sulfur (S), and oxygen (O), reflecting the number of atoms of each element present in the compound. Grasping this concept is essential because it allows you to convert between mass and moles, a fundamental step in stoichiometry.

Chemical Formula

The chemical formula of a compound, such as \texttt{BaSO}\(_4\), provides vital information. It tells us the exact number of atoms of each element that are present in one molecule of the compound. For example, \texttt{BaSO}\(_4\) contains one barium atom, one sulfur atom, and four oxygen atoms.

This information not only helps in calculating the molar mass but also is essential when understanding reaction stoichiometry—how substances react in fixed ratios. Recognizing the significance of chemical formulas is foundational in chemistry, as it's the language used to describe the composition of substances and the changes they undergo during chemical reactions.

Stoichiometry

Stoichiometry lies at the heart of chemical reactions. It deals with the quantitative relationships between the reactants and products in a chemical reaction, based on the balanced chemical equation. Stoichiometry allows chemists to predict the amounts of substances consumed and produced in a given reaction.

In the context of our exercise, we used stoichiometry to determine the amount of sulfur present in the tomato by utilizing the known stoichiometric relationship in the formation of \texttt{BaSO}\(_4\). We calculated the mass of sulfur from the mass of \texttt{BaSO}\(_4\) assuming all sulfur is converted during the reaction. A clear grasp of stoichiometry is indispensable for accurately calculating chemical quantities in reactions.

Precipitation Reaction

A precipitation reaction is a type of chemical reaction where two soluble salts are combined in solution to form one soluble salt and one insoluble salt—the precipitate. This reaction is a cornerstone of many analytical procedures, including gravimetric analysis.

In our tomato analysis, sulfur was part of the soluble salt in the tomato's juice. When it reacted with barium nitrate (\texttt{Ba(NO}\(_3\))\(_2\)), it formed barium sulfate (\texttt{BaSO}\(_4\)), the insoluble precipitate. The solid \texttt{BaSO}\(_4\) formed is then collected and weighed, which allows us to calculate the original amount of sulfur in the tomato based on the stoichiometry of the reaction. Understanding the dynamics of precipitation reactions is vital for interpreting and performing many laboratory experiments.