Question

Calculate the percent composition, determine the molecular formula and the empirical formula for the carbon-hydrogenoxygen compound that results when \(30.21 \mathrm{~g}\) of carbon, \(40.24 \mathrm{~g}\) of oxygen, and \(5.08 \mathrm{~g}\) of hydrogen are reacted to produce a product with a molar mass of \(180.18 \mathrm{~g}\).

Short answer

Empirical formula: CH2O. Molecular formula: C6H12O6. Percent composition: 40.00% C, 6.73% H, 53.27% O.

Step-by-step solution

- Calculate Moles of Each Element

First, determine the moles of each element by dividing the mass of each element by its atomic mass. For Carbon: o number of moles of carbon :\[ n_C = \frac{30.21 \mathrm{~g}}{12.01 \mathrm{~g/mol}} = 2.515 \text{ moles} \] For Oxygen: o number of moles of oxygen :\[ n_O = \frac{40.24 \mathrm{~g}}{16.00 \mathrm{~g/mol}} = 2.515 \text{ moles} \] For Hydrogen: o number of moles of hydrogen:\[ n_H = \frac{5.08 \mathrm{~g}}{1.01 \mathrm{~g/mol}} = 5.03 \text{ moles} \]

- Determine the Empirical Formula

Find the ratio of the moles of each element. 1. Divide all the moles by the smallest number of moles among the elements which is 2.515:\[\text{For Carbon:}\frac{2.515}{2.515} = 1. \] \[\text{For Oxygen:}\frac{2.515}{2.515} = 1. \] \[\text{For Hydrogen:}\frac{5.03}{2.515} \approx 2. \] Thus, the empirical formula is \( CH_2O \).

- Calculate the Empirical Formula Mass

Add the atomic masses of the elements in the empirical formula: For \( CH_2O \): \[\text{Molar mass} = 12.01 + 2(1.01) + 16.00 = 30.02 \text{ g/mol} \]

- Determine the Molecular Formula

Divide the molar mass of the compound by the empirical formula mass: \[\text{n} = \frac{180.18 \text{ g/mol}}{30.02 \text{ g/mol}} = 6 \] Therefore, multiply the empirical formula by 6 to get the molecular formula: \((CH_2O)_6 = C_6H_{12}O_6 \)

- Calculate Percent Composition

Calculate the percent composition of each element in the molecular formula: For Carbon:\[ \% C = \frac{6 \times 12.01}{180.18} \times 100 \approx 40.00 \% \] For Hydrogen: \[ \% H = \frac{12 \times 1.01}{180.18} \times 100 \approx 6.73 \% \] For Oxygen: \[ \% O = \frac{6 \times 16.00}{180.18} \times 100 \approx 53.27 \% \]

Key concepts

Percent Composition

Percent composition is the percentage by mass of each element in a compound. It is calculated by taking the mass of each element, dividing it by the total molar mass of the compound, and multiplying by 100. This helps understand the relative amount of each element present in the compound.

In our exercise, the percent compositions are:

• For Carbon: \[ \text{\text{% C}} = \frac{6 \times 12.01}{180.18} \times 100 \approx 40.00 \% \]
• For Hydrogen: \[ \text{\text{% H}} = \frac{12 \times 1.01}{180.18} \times 100 \approx 6.73 \% \]
• For Oxygen: \[ \text{\text{% O}} = \frac{6 \times 16.00}{180.18} \times 100 \approx 53.27 \% \]

Knowing percent composition can be useful when comparing compounds or determining the purity of a substance in various applications.

Empirical Formula

The empirical formula of a compound gives the simplest whole number ratio of atoms of each element. Determining the empirical formula starts with calculating the moles of each element. Divide each mole value by the smallest mole value. This gives the simplest ratio of atoms.

In our exercise:

• For Carbon: \[ \frac{2.515}{2.515} = 1 \]
• For Oxygen: \[ \frac{2.515}{2.515} = 1 \]
• For Hydrogen: \[ \frac{5.03}{2.515} \approx 2 \]

Hence, the empirical formula is \( \text{CH}_2 \text{O} \). This means in one molecule of the compound, the atoms of carbon, hydrogen, and oxygen are in the ratio 1:2:1.

Molecular Formula

The molecular formula indicates the actual number of atoms of each element in a molecule of the compound. To determine it, divide the compound's molar mass by the empirical formula mass, and then multiply the empirical formula by this value.

In our exercise:

• Empirical formula: \( \text{CH}_2 \text{O} \)
• Empirical formula mass: \[ 12.01 + 2(1.01) + 16.00 = 30.02 \text{ g/mol} \]
• Molecular formula factor: \[ \frac{180.18}{30.02} = 6 \]

Thus, multiply the empirical formula by 6: \[ (\text{CH}_2 \text{O})_6 = \text{C}_6 \text{H}_{12} \text{O}_6 \]. This molecular formula shows the actual number of atoms in each molecule.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It uses the relationships between the quantities of reactants and the products formed. In our exercise, stoichiometry helps us determine how the amounts of carbon, hydrogen, and oxygen combine to form the empirical and molecular formulas.

The balanced chemical equations and mole ratios play key roles here, ensuring that we start with the right amounts of reactants to predict the outcomes accurately. Understanding stoichiometry is crucial for predicting yields and scaling reactions up or down.

Mole Calculation

Mole calculation is central to chemical composition analysis. A mole represents \( 6.022 \times 10^{23} \) entities (atoms, molecules). It connects the macroscopic world (grams) to the atomic world (moles).

For our problem, we calculated moles as follows:

• Carbon: \[ \frac{30.21 \text{ g}}{12.01 \text{ g/mol}} = 2.515 \text{ moles} \]
• Oxygen: \[ \frac{40.24 \text{ g}}{16.00 \text{ g/mol}} = 2.515 \text{ moles} \]
• Hydrogen: \[ \frac{5.08 \text{ g}}{1.01 \text{ g/mol}} = 5.03 \text{ moles} \]

By converting the masses of each element to moles, we can better understand and calculate chemical formulas and reactions. This forms the backbone of empirical and molecular formula determination!