Problem 160 - Introductory Chemistry Atoms First

Question

Saccharin, which is used as an artificial sweetener, is made up of \(45.90 \%\) by mass \(C, 2.75 \%\) by mass \(\mathrm{H}, 26.20 \%\) by mass \(\mathrm{O}, 7.65 \%\) by mass \(\mathrm{N}\), and \(17.50 \%\) by mass \(\mathrm{S}\). If the molar mass of saccharin is \(183.19 \mathrm{~g} / \mathrm{mol}\), what is its molecular formula?

Step-by-Step Solution

Answer

The molecular formula of saccharin is C7H5O3NS.

Step 1: Convert mass percentages to moles

To convert the mass percentages into moles, we will assume that we have 100g of saccharin. This means that we have 45.9g of C, 2.75g of H, 26.2g of O, 7.65g of N, and 17.5g of S present in 100g of saccharin. Now, we will calculate the number of moles for each element. For carbon (C): \(\dfrac{45.90 \text{g}}{12.01 \text{g/mol}} = 3.82 \text{mol} \) For hydrogen (H): \(\dfrac{2.75 \text{g}}{1.008 \text{g/mol}} = 2.73 \text{mol} \) For oxygen (O): \(\dfrac{26.20 \text{g}}{16.00 \text{g/mol}} = 1.64 \text{mol}\) For nitrogen (N): \(\dfrac{7.65 \text{g}}{14.01 \text{g/mol}} = 0.546 \text{mol}\) For sulfur (S): \(\dfrac{17.50 \text{g}}{32.07 \text{g/mol}} = 0.546 \text{mol}\)

Step 2: Find empirical formula

To find the empirical formula, divide the number of moles for each element by the smallest number of moles, in this case, 0.546. For carbon (C): \(\dfrac{3.82 \text{mol}}{0.546 \text{mol}} = 7 \) (rounded) For hydrogen (H): \(\dfrac{2.73 \text{mol}}{0.546 \text{mol}} = 5 \) (rounded) For oxygen (O): \(\dfrac{1.64 \text{mol}}{0.546 \text{mol}} = 3 \) (rounded) For nitrogen (N): \( \dfrac{0.546 \text{mol}}{0.546 \text{mol}} = 1 \) (rounded) For sulfur (S): \( \dfrac{0.546 \text{mol}}{0.546 \text{mol}} = 1 \) (rounded) So, the empirical formula of saccharin is C7H5O3N1S1 or C7H5O3NS.

Step 3: Calculate empirical molar mass

Calculate the molar mass of the empirical formula we just found: C7H5O3NS = 7(12.01) + 5(1.008) + 3(16) + 14.01 + 32.07 = 163.142 g/mol

Step 4: Determine molecular formula

Now, we can use the molar mass of saccharin provided and divide it by the empirical molar mass to find the ratio between both: Molecular formula ratio: \(\dfrac{183.19 \text{g/mol}}{163.142 \text{g/mol}} = 1.12\) Since the molecular formula ratio is close to 1, we can assume that the empirical and molecular formulas are the same due to rounding errors. Therefore, the molecular formula of saccharin is C7H5O3NS.

Key Concepts

Empirical Formula

An empirical formula reveals the simplest whole-number ratio of atoms present in a compound. To determine the empirical formula, we begin by converting the percentage composition of each element into moles, assuming we have 100 grams of the compound. This allows us to directly use the percentages as grams. Once we have the mass of each element, we convert it to moles using the respective atomic masses. For instance, in saccharin, we calculated the moles of carbon (C) as follows: \[ \text{Moles of } C = \frac{45.90 \text{ g}}{12.01 \text{ g/mol}} = 3.82 \text{ mol} \]After calculating moles for all elements, we identify and divide each by the smallest mole value, ensuring we get a ratio that represents the simplest formula possible. Sometimes it's necessary to round these values to the nearest whole number, especially if the resulting fractions are close enough, indicating a small measurement variance. Thus, for saccharin, the simplest ratio was found to be C\(_7\)H\(_5\)O\(_3\)N\(_1\)S\(_1\), or C\(_7\)H\(_5\)O\(_3\)NS. This formula represents the relative number of each type of atom, providing a fundamental idea of the compound's basic structure.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is crucial for converting between the mass of a compound and the amount of substance in moles. The molar mass of an elemental atom equals the atomic weight for that element, found on the periodic table, and for a compound, it is the sum of the molar masses of its component atoms.For saccharin, once we established its empirical formula as C\(_7\)H\(_5\)O\(_3\)NS, we calculate the molar mass by summing the products of the number of each type of atom and their atomic mass:\[ \text{Molar Mass of } C_7H_5O_3NS = 7(12.01) + 5(1.008) + 3(16.00) + 14.01 + 32.07 = 163.142 \text{ g/mol} \]Understanding molar mass helps in elucidating crucial details about a compound, enabling the quantitative analysis necessary to progress from the empirical to the molecular formula.

Mass Percentage

Mass percentage indicates how much of each element is present in a compound, expressed as a percentage of the total mass. It's calculated using the formula:\[ \text{Mass Percentage} = \left( \frac{\text{Mass of the Element}}{\text{Total Mass of Compound}} \right) \times 100 \%\]In the context of determining the empirical formula, knowing the mass percentage is helpful because it allows direct conversion to grams, assuming you have a 100 g sample size. This simplifies calculations, allowing you to easily use the percentage values for computing moles without any further adjustments.For saccharin, the values given were \(45.90\%\) carbon, \(2.75\%\) hydrogen, \(26.20\%\) oxygen, \(7.65\%\) nitrogen, and \(17.50\%\) sulfur. Using these values help us convert directly to moles and lay the groundwork for determining more detailed molecular information.

Mole Ratio

A mole ratio is a proportion that expresses the relative number of moles of any two substances involved in a chemical reaction or within a compound. In relation to empirical formula calculation, the mole ratio helps determine the simplest formula based on how atoms combine.After calculating the moles of each element from their mass percentages, the next step is refining these values to a simplest ratio. This necessitates dividing all moles by the smallest value among them. The resulting numbers often need rounding to the nearest integer to represent the smallest whole number ratio in the empirical formula. In the saccharin problem, we calculated the following mole ratios for our elements: C, H, O, N, and S. These ratios allowed us to define the empirical formula as C\(_7\)H\(_5\)O\(_3\)NS by setting up a simple integer proportion across the elements. Such ratios provide the basic insight needed to transition from raw data to meaningful chemical representation, guiding further molecular formula calculations.