Question

What is the percentage of 'free \(\mathrm{SO}_{3}^{\prime}\) in a sample of oleum labelled as '104.5\%'? (a) \(20 \%\) (b) \(40 \%\) (c) \(60 \%\) (d) \(80 \%\)

Short answer

The percentage of free \(\mathrm{SO}_3\) in the oleum sample labelled as '104.5\%' is 40\%.

Step-by-step solution

Understanding Oleum Percentage

Oleum is labelled with a percentage that indicates the amount of free \(\mathrm{SO}_3\) it contains on top of the \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7\) present. A label of 104.5% means that there is 104.5 grams of free \(\mathrm{SO}_3\) in 100 grams of \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7\).

Calculate the Mass of free \(\mathrm{SO}_3\)

To find the mass of free \(\mathrm{SO}_3\) in 100 grams of \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7\), calculate the mass in excess of 100%. The mass of \(\mathrm{SO}_3\) is 104.5% - 100% = 4.5% of the oleum sample. Therefore, the mass of free \(\mathrm{SO}_3\) is 4.5 grams since 100% corresponds to 100 grams.

Convert Mass to Moles

To convert the mass of free \(\mathrm{SO}_3\) to moles, use its molar mass. The molar mass of \(\mathrm{SO}_3\) is approximately 80.066 g/mol. Moles of \(\mathrm{SO}_3\) = \(\frac{4.5 \text{ g}}{80.066 \text{ g/mol}}\).

Calculate Total Moles in Oleum Sample

The Oleum sample is essentially 100% \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7\) plus the additional \(\mathrm{SO}_3\). The molar mass of \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7\) is approximately 82.07 x 2 + 16.00 x 7 = 178.14 g/mol. Moles of \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7\) in 100 g = \(\frac{100 \text{ g}}{178.14 \text{ g/mol}}\).

Calculate Percentage of free \(\mathrm{SO}_3\)

Percent of free \(\mathrm{SO}_3\) is calculated by the ratio of moles of \(\mathrm{SO}_3\) to the total moles in the sample times 100%. Total moles = Moles of \(\mathrm{SO}_3\) + Moles of \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7\). Percent = \(\left(\frac{\text{Moles of } \mathrm{SO}_3}{\text{Total Moles}}\right) \times 100\%\).

Apply the Calculations to Find the Answer

First calculate the moles of \(\mathrm{SO}_3\) and \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7\), then calculate the percentage of free \(\mathrm{SO}_3\). Moles of \(\mathrm{SO}_3\): \(\frac{4.5}{80.066}\). Moles of \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7\): \(\frac{100}{178.14}\). Then calculate the percentage using the formula provided in the previous step.

Key concepts

Percent Composition

Percent composition is a way to express the relative amounts of each element within a compound. It answers the question, 'What percentage of the total mass of a compound is due to each constituent element?' To find the percent composition, you divide the mass of a specific element by the total mass of the compound and then multiply by 100%. This is particularly useful in determining the purity of substances and in calculating the empirical formula of a compound.

For instance, in the exercise of finding the percentage of free \(\mathrm{SO}_3\) in oleum, the percent composition reflects how much of the oleum's total weight is made up of free \(\mathrm{SO}_3\) compared to the \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7\) present. Understanding percent composition allows students to appreciate the real-world applications of concepts like the purity of chemicals and the stoichiometry of reactions.

Mole Concept

The mole concept is a fundamental principle in chemistry that links the micro-world of atoms and molecules to the macro-world of grams and liters. One mole is essentially a specific number (Avogadro's number: approximately \(6.022 \times 10^{23}\)) of particles. It acts as a bridge between an element's atomic mass in amu (atomic mass units) and its molar mass in grams.

When we say that one mole of \(\mathrm{SO}_3\) has a mass of approximately 80.066 grams, we are using the mole concept to equate the molar mass of \(\mathrm{SO}_3\) with the number of molecules present in that mass. It's like counting oranges by the dozen, but instead, we count atoms and molecules by the mole. This concept allows students to measure and calculate with quantities of substances even at the microscopic level.

Molar Mass

Molar mass is the weight of one mole of a substance and is expressed in grams per mole (g/mol). It is numerically equivalent to the atomic or molecular mass of a substance but scaled from atomic mass units to grams. The molar mass allows scientists to weigh out amounts of a substance in a lab in such a way that a known number of molecules are obtained.

For example, \(\mathrm{SO}_3\) has a molar mass of 80.066 g/mol, meaning that 80.066 grams of \(\mathrm{SO}_3\) contain one mole (or \(6.022 \times 10^{23}\) molecules) of \(\mathrm{SO}_3\). Learning to convert between mass and moles using the molar mass is essential for many stoichiometric calculations in chemistry.

Chemical Stoichiometry

Chemical stoichiometry is the area of chemistry that pertains to the quantitative relationships between the reactants and products in a chemical reaction. It's based on the conservation of mass and the concept of the mole. Stoichiometry allows chemists to predict the outcomes of reactions, determine what amounts of substances are needed, and calculate yields.

In this oleum example, stoichiometry involves using molar masses of \(\mathrm{SO}_3\) and \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7\) to convert mass of these compounds into moles to find the percent of free \(\mathrm{SO}_3\) present. Understanding stoichiometry is crucial as it’s not just about knowing the steps in a process; it’s about understanding the interplay between the quantities of substances involved.