Determining the Empirical and Molecular Formulas of Saccharin: A Step-by-Step Guide

Introduction

Saccharin is an artificial sweetener widely used in various food and beverage products. Its unique composition and molar mass provide an interesting case study for understanding empirical and molecular formulas. In this article, we will walk through the detailed steps to determine the empirical and molecular formulas of saccharin from its given percentage composition and molar mass.

Understanding Empirical and Molecular Formulas

The empirical formula of a compound shows the simplest whole number ratio of atoms of each element present in the compound. The molecular formula, on the other hand, provides the actual number of each type of atom in one molecule of the compound. These formulas are essential for understanding the structural and chemical properties of compounds like saccharin.

Gathering Data

The given percentage composition and molar mass of saccharin are as follows:

C: 45.89% H: 2.75% N: 7.65% O: 26.20% S: 17.50% Molar mass: 183.19 g/mol

Step 1: Converting Percentages to Grams

To determine the empirical and molecular formulas, we need to first convert the given percentages to grams, assuming a 100 g sample:

C: 45.89 g H: 2.75 g N: 7.65 g O: 26.20 g S: 17.50 g

Step 2: Converting Grams to Moles

Next, convert grams to moles using the molar masses of each element:

C: 12.01 g/mol H: 1.008 g/mol N: 14.01 g/mol O: 16.00 g/mol S: 32.07 g/mol

The calculations are as follows:

Moles of C: 45.89 g / 12.01 g/mol ≈ 3.82 mol Moles of H: 2.75 g / 1.008 g/mol ≈ 2.73 mol Moles of N: 7.65 g / 14.01 g/mol ≈ 0.55 mol Moles of O: 26.20 g / 16.00 g/mol ≈ 1.63 mol Moles of S: 17.50 g / 32.07 g/mol ≈ 0.55 mol

Step 3: Finding the Simplest Whole Number Ratio

To find the empirical formula, we divide each mole value by the smallest number of moles:

C: 3.82 mol / 0.55 mol ≈ 6.93 ≈ 7 H: 2.73 mol / 0.55 mol ≈ 4.96 ≈ 5 N: 0.55 mol / 0.55 mol 1 O: 1.63 mol / 0.55 mol ≈ 2.96 ≈ 3 S: 0.55 mol / 0.55 mol 1

Hence, the empirical formula for saccharin is:

C7H5NOS

The molar mass of the empirical formula (C7H5NOS) can be calculated as:

12*7 1*5 14 16*3 32 183.19 g/mol

Step 4: Determining the Molecular Formula

The molar mass of saccharin is given as 183.19 g/mol, which matches the molar mass of the empirical formula. Therefore, the molecular formula is the same as the empirical formula:

C7H5NOS

Conclusion

We have successfully determined the empirical and molecular formulas of saccharin using the given percentage composition and molar mass. This method can be applied to various other compounds to understand their structural and chemical properties.

Through this process, we obtained the empirical and molecular formulas and verified that they match the given molar mass, confirming the accuracy of our calculations.