Question

What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{CH}_{3} \mathrm{O}\), molar mass \(=62.0 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{NH}_{2}\), molar mass \(=32.0 \mathrm{~g} / \mathrm{mol}\)

Short answer

The molecular formulas of the given compounds are: (a) C2H6O2 (b) N2H4

Step-by-step solution

(a) Finding the molar mass of the empirical formula C3H3O

To find the molar mass of the empirical formula, add the atomic masses of the elements present in the formula. The atomic masses are: Carbon (C) = 12.01 g/mol Hydrogen (H) = 1.008 g/mol Oxygen (O) = 16.00 g/mol Molar mass of the empirical formula (CH3O) = 1 × 12.01 + 3 × 1.008 + 1 × 16.00 = \(12.01 + 3.024 + 16.00 = 31.034 \mathrm{~g/mol}\)

(a) Finding the molecular formula

Now, we will find the multiple needed to obtain the given molar mass of the compound. Divide the given molar mass by the molar mass of the empirical formula: Multiple = \(\frac{62.0\mathrm{~g/mol}}{31.034 \mathrm{~g/mol}} = 2 \) Now, multiply the empirical formula by this multiple to find the molecular formula: Molecular formula = 2 × (CH3O) = C2H6O2 So, the molecular formula for compound (a) is C2H6O2.

(b) Finding the molar mass of the empirical formula NH2

To find the molar mass of the empirical formula, add the atomic masses of the elements present in the formula. The atomic masses are: Nitrogen (N) = 14.01 g/mol Hydrogen (H) = 1.008 g/mol Molar mass of the empirical formula (NH2) = 1 × 14.01 + 2 × 1.008 = \(14.01 + 2.016 = 16.026 \mathrm{~g/mol}\)

(b) Finding the molecular formula

Now, we will find the multiple needed to obtain the given molar mass of the compound. Divide the given molar mass by the molar mass of the empirical formula: Multiple = \(\frac{32.0\mathrm{~g/mol}}{16.026\mathrm{~g/mol}} = 2 \) Now, multiply the empirical formula by this multiple to find the molecular formula: Molecular formula = 2 × (NH2) = N2H4 So, the molecular formula for compound (b) is N2H4.

Key concepts

Empirical Formula

The empirical formula of a compound is the simplest whole-number ratio of atoms of each element present in a compound. It doesn't necessarily reflect the exact number of atoms in a molecule, but rather the simplest ratio. For example, a compound with the empirical formula \(\text{CH}_3\text{O}\) means that for every carbon (C) atom, there are three hydrogen (H) atoms, and one oxygen (O) atom, making its simplest ratio 1:3:1.

This form is used because it provides a basic understanding of a compound's makeup, which is essential when deriving the molecular formula. Identifying the empirical formula involves analyzing the percentage composition of each element in the compound and converting these percentages to moles to determine the simplest ratio.

• It is useful in chemical analysis to give insight into the basic composition.
• The empirical formula is not always the molecular formula.

While the empirical formula gives the smallest integer ratio, the molecular formula provides the actual number of atoms in each molecule of the substance. Understanding the relationship between these two is key in deducing the chemistry behind many compounds.

Molar Mass

Molar mass is a crucial concept in chemistry that refers to the mass of one mole of a given substance or element. It is expressed in grams per mole (g/mol) and helps to convert between the number of atoms, molecules, or formula units of a substance and its mass. To determine the molar mass, you add up the atomic masses of all atoms in the molecule's formula.

For example, in the case of \(\text{CH}_3\text{O}\), the molar mass is calculated by adding the mass of 1 carbon atom (12.01 g/mol), 3 hydrogen atoms (3 x 1.008 g/mol), and 1 oxygen atom (16.00 g/mol). This gives a total molar mass of 31.034 g/mol.

• Molar mass allows chemists to determine how much of a substance is present in a sample.
• This makes it easier to conduct stoichiometric calculations in reactions.

Understanding and calculating molar mass accurately is essential for practical applications in laboratory settings, making it fundamental in the study of chemical reactions and compounds.

Atomic Masses

Atomic masses, also known as atomic weights, are the masses of individual atoms, typically expressed in atomic mass units (amu). They are crucial for calculating molar mass and, consequently, molecular formulas of compounds. Each element on the periodic table has a specific atomic mass, which represents an average mass of one atom of the element, factoring in all its isotopes.

Consider carbon with an atomic mass of 12.01 amu. This value considers the naturally occurring isotopes of carbon and their abundance. Hydrogen, with an atomic mass of 1.008 amu, and oxygen, at 16.00 amu, play similar roles in weighted averages of their isotopes.

• Atomic masses are fundamental for balancing chemical equations.
• They help convert between moles and grams for elements in a compound.

Having a solid grasp of atomic masses empowers students to navigate through chemistry problems that require meticulous calculations regarding mass relationships and chemical compositions.