Problem 130 - Introductory Chemistry Atoms First

Question

Vitamin \(\mathrm{C}\), also known as ascorbic acid, contains carbon, hydrogen, and possibly oxygen. A \(0.160 \mathrm{~g}\) sample of ascorbic acid is subjected to combustion analysis, yielding \(40.93 \%\) C and \(4.58 \%\) H. If the molar mass of ascorbic acid is approximately \(176 \mathrm{~g} / \mathrm{mol}\), what is its molecular formula?

Step-by-Step Solution

Answer

The molecular formula of ascorbic acid is C6H6O6.

Step 1: Find the mass of carbon, hydrogen, and oxygen in the sample

From the given information, we can find the mass of carbon and hydrogen in the sample. Then, we will find the mass of oxygen by subtracting the obtained values from the total mass of the sample. - Mass of carbon = (40.93 / 100) * 0.160 g = 0.0655 g - Mass of hydrogen = (4.58 / 100) * 0.160 g = 0.0073 g - Mass of oxygen = 0.160 g - (0.0655 g + 0.0073 g) = 0.0872 g

Step 2: Convert the mass of each element to moles

Next, we will convert the mass of carbon, hydrogen, and oxygen to moles using their molar masses: 1. Molar mass of carbon (C) = 12.01 g/mol 2. Molar mass of hydrogen (H) = 1.008 g/mol 3. Molar mass of oxygen (O) = 16.00 g/mol - Moles of carbon = \( \frac{0.0655 \text{ g}}{12.01 \text{ g/mol}} = 0.00545 \text{ mol} \approx 5.45 \times 10^{-3} \text{ mol} \) - Moles of hydrogen = \( \frac{0.0073 \text{ g}}{1.008 \text{ g/mol}} = 0.00724 \text{ mol} \approx 7.24 \times 10^{-3} \text{ mol} \) - Moles of oxygen = \( \frac{0.0872 \text{ g}}{16.00 \text{ g/mol}} = 0.00545 \text{ mol} \approx 5.45 \times 10^{-3} \text{ mol} \)

Step 3: Determine the mole ratio to find the empirical formula

Now we will find the mole ratios of each element by dividing the moles by the smallest number of moles. - Mole ratio of carbon (C) = \( \frac{5.45 \times 10^{-3} \text{ mol}}{5.45 \times 10^{-3} \text{ mol}} = 1.00 \) - Mole ratio of hydrogen (H) = \( \frac{7.24 \times 10^{-3} \text{ mol}}{5.45 \times 10^{-3} \text{ mol}} = 1.33 \) - Mole ratio of oxygen (O) = \( \frac{5.45 \times 10^{-3} \text{ mol}}{5.45 \times 10^{-3} \text{ mol}} = 1.00 \) Now round them to the nearest whole number: - Carbon: 1 - Hydrogen: 1 - Oxygen: 1 The empirical formula is CH1O1, or simply CH0.

Step 4: Use the molar mass to find the molecular formula

To find the molecular formula, we first find the molar mass of the empirical formula and then compare it to the given molar mass of ascorbic acid. - Molar mass of CH0 = (12.01 + 1.008 + 16.00) g/mol = 29.018 g/mol Now we find the ratio between the molar mass of the molecular formula and the empirical formula: - \( \frac{176 \text{ g/mol}}{29.018 \text{ g/mol}} = 6.066 \approx 6 \) Since the ratio is approximately 6, we can multiply the empirical formula by 6 to find the molecular formula: - C1 x 6 = C6 - H1 x 6 = H6 - O1 x 6 = O6 The molecular formula is C6H6O6, also known as ascorbic acid.

Key Concepts

Molar Mass

Understanding molar mass is fundamental when diving into the heart of chemistry and specifically, in performing combustion analysis chemistry exercises. Simply put, the molar mass is the weight of one mole (Avogadro's number, \( 6.022 \times 10^{23} \) particles) of any chemical substances. This mass is usually expressed in grams per mole (g/mol) and is unique for each element as listed on the periodic table.

For instance, carbon has a molar mass of \( 12.01 g/mol \), which means one mole of carbon weighs \( 12.01 grams \). To convert the mass of an element to moles in a combustion analysis, like in the provided example with vitamin C, you would divide the given mass by the molar mass. This conversion is the cornerstone of stoichiometry, allowing chemists to relate quantities of different substances in a reaction.

Empirical Formula

Now, let's focus on the empirical formula. This formula represents the simplest whole-number ratio of elements in a compound. It's the reduced form of the molecular formula, which may have multiple instances of the empirical formula. For example, in the given exercise, we witness the process of deriving an empirical formula from combustion analysis data.

We start by determining the moles of each element present in a compound and then find the simplest ratio by dividing by the smallest number of moles. The approximations are necessary to round off to the nearest whole number, giving us the simplest ratio of atoms present in a molecule. Empirical formulas are particularly useful when the chemical compound's actual molecular formula is complex and where simplification aids in understanding the composition of the molecule.

Molecular Formula

In comparison, the molecular formula gives the exact number of atoms of each element in a molecule, making it a multiple of the empirical formula. After you have the empirical formula, as in the vitamin C example, you can determine the molecular formula by comparing the empirical formula's molar mass to the actual molar mass of the compound.

If the molar mass is known, like the approximate molar mass provided for ascorbic acid (\( 176 g/mol \)), you would divide this by the molar mass of the empirical formula to get a ratio. The empirical formula is then multiplied by this ratio, resulting in the molecular formula, which is a true representation of the actual numbers and kinds of atoms in a molecule. Through this process, students grasp the connection between the empirical and molecular formulas, allowing them to understand complex molecular structures and their corresponding weights.