Problem 92 - Chemistry The Molecular Science

Question

About a century ago, Paul Ehrlich discovered Salvarsan, the first arsenical antibiotic that cured a targeted disease, syphilis. The compound was the six hundred and sixth compound he tried against the disease, but the first one found to be effective against it. The compound has molar mass \(549.102 \mathrm{~g} / \mathrm{mol}\) and this mass percent composition: \(39.37 \% \mathrm{C} ; 3.304 \% \mathrm{H} ; 7.653 \% \mathrm{~N} ;\) \(8.741 \% \mathrm{O} ; 40.93 \% \mathrm{As}\) (a) Calculate the empirical formula of Salvarsan. (b) Determine the molecular formula of this compound.

Step-by-Step Solution

Answer

Empirical formula: \(C_6H_6NOAs\). Molecular formula: \(C_{18}H_{18}N_3O_3As_3\).

Step 1: Convert Percentages to Masses

Assume you have 100 g of the compound. The mass of each element is then equal to its percentage: 39.37 g of C, 3.304 g of H, 7.653 g of N, 8.741 g of O, and 40.93 g of As.

Step 2: Convert Masses to Moles

Divide the mass of each element by its atomic mass to find the number of moles. Atomic masses: C = 12.01 g/mol, H = 1.008 g/mol, N = 14.01 g/mol, O = 16.00 g/mol, As = 74.92 g/mol. - For C: \( \frac{39.37}{12.01} = 3.278 \text{ moles} \)- For H: \( \frac{3.304}{1.008} = 3.278 \text{ moles} \) - For N: \( \frac{7.653}{14.01} = 0.546 \text{ moles} \) - For O: \( \frac{8.741}{16.00} = 0.546 \text{ moles} \) - For As: \( \frac{40.93}{74.92} = 0.546 \text{ moles} \)

Step 3: Find Empirical Formula Ratios

Divide the moles of each element by the smallest number of moles calculated in the previous step. The smallest number is 0.546 moles.- For C: \( \frac{3.278}{0.546} = 6 \)- For H: \( \frac{3.278}{0.546} = 6 \)- For N: \( \frac{0.546}{0.546} = 1 \)- For O: \( \frac{0.546}{0.546} = 1 \)- For As: \( \frac{0.546}{0.546} = 1 \)Thus, the empirical formula is \( C_6H_6NOAs \).

Step 4: Calculate Empirical Formula Molar Mass

Calculate the molar mass of the empirical formula \( C_6H_6NOAs \). - C: \(6 \times 12.01 = 72.06\) g/mol- H: \(6 \times 1.008 = 6.048\) g/mol- N: \(1 \times 14.01 = 14.01\) g/mol- O: \(1 \times 16.00 = 16.00\) g/mol- As: \(1 \times 74.92 = 74.92\) g/molTotal empirical mass = 183.038 g/mol.

Step 5: Determine Molecular Formula

Find the ratio of the compound's molar mass (549.102 g/mol) to the empirical formula mass (183.038 g/mol).\( \text{Ratio} = \frac{549.102}{183.038} \approx 3 \)Multiply the subscripts in the empirical formula \(C_6H_6NOAs\) by 3 to get the molecular formula:\(C_{18}H_{18}N_3O_3As_3\).

Key Concepts

Understanding Chemistry in Empirical and Molecular Formulas

Chemistry often deals with the study of substances and their interactions. Empirical and molecular formulas are fundamental concepts within this field, describing the composition of chemical compounds. An empirical formula represents the simplest whole-number ratio of elements in a compound. For example, in Salvarsan, the empirical formula is determined as \(C_6H_6NOAs\). This shows us that in the simplest form, for every 6 carbon and hydrogen atoms, there is 1 nitrogen, 1 oxygen, and 1 arsenic atom.
On the other hand, a molecular formula provides the actual number of each type of atom in a molecule. In Salvarsan's case, the molecular formula is \(C_{18}H_{18}N_3O_3As_3\), illustrating the precise count of all atoms in one molecule of the compound. This difference underscores a key chemistry principle: two compounds can share the same empirical formula but have different molecular ones, reflecting unique properties.

Role of Molar Mass in Determining Chemical Formulas

Molar mass is a concept in chemistry that allows you to connect mass percentages to the actual amounts of a substance within a compound. It is the mass of one mole of a substance, usually reported in grams per mole \( ext{g/mol}\). Knowing the molar mass is essential for calculating both empirical and molecular formulas.
The molar mass of Salvarsan is given as \(549.102 \text{ g/mol}\), but to find the empirical formula, this value is not immediately needed. Instead, we first calculate the molar masses for each element within the compound and adjust their ratios to the simplest form.

• This helps to derive the empirical formula molar mass, found by summing up the atomic masses based on the empirical formula (i.e., \(183.038 \text{ g/mol}\) for \(C_6H_6NOAs\)).
• Finally, by comparing the actual molar mass to this derived empirical molar mass, we find a multiplying factor (which is 3 here) to determine the molecular formula.

This usage of molar mass ensures accuracy in understanding the actual molecular arrangement of the compound.

Elemental Analysis: A Tool for Discovering Chemical Composition

Elemental analysis is a crucial method in chemistry used to figure out the elemental composition of a compound. This technique helps chemists determine the percentage by mass of each element present, as seen in Salvarsan, where elements like carbon, hydrogen, nitrogen, oxygen, and arsenic show specific mass percentages.
Understanding these percentages allows scientists to deduce how these elements combine to form the compound's structure. For instance, by assuming a 100 g sample, the mass of each element is directly equal to the percentages given, simplifying calculations.

• Next, each element's mass is converted to moles using its respective atomic mass. This step is crucial as chemical reactions and molecule formulations are based on moles, not mass alone.
• Once moles are calculated, determining the smallest mole proportion through division leads to the empirical formula.

Elemental analysis thus provides a systematic approach to breaking down complex compounds into understandable components, aiding in precise formula identification and further analysis of chemical substances.