What is the molar mass of NaN3

Molar Mass of Sodium Azide

To find the molar mass of sodium azide ($\text{NaN}_3$), we need to determine the sum of the atomic masses of each element in the compound.

Atomic Masses:

• Sodium (Na): 22.99 g/mol
• Nitrogen (N): 14.01 g/mol

Calculations:

Sodium azide consists of one sodium atom and three nitrogen atoms. Therefore, the total molar mass is calculated as follows:

Inserting the atomic masses:

Performing the multiplication and summation:

Thus, the molar mass of $\text{NaN}_3$ is:

Hence, the molar mass of sodium azide ($\text{NaN}_3$) is 65.02 g/mol.

The Significance of Atomic Masses

Atomic masses play a crucial role in the field of chemistry and physics. Understanding atomic masses is essential for a variety of reasons:

1. Mass of Atoms: The atomic mass (or atomic weight) represents the average mass of an atom of an element, measured in atomic mass units (amu). This average accounts for the different isotopes of the element and their relative abundances.

2. Isotopic Composition: Many elements exist in multiple forms known as isotopes, each with a different number of neutrons. The atomic mass reflects the weighted average of all these isotopes.

3. Molecular Weight Calculation: The atomic masses of elements are instrumental in calculating the molecular weights of compounds. For instance, the molecular weight of water (H$_2$O) is determined by summing the atomic masses of its constituent atoms:

4. Given that the atomic mass of hydrogen (H) is approximately 1.01 amu and oxygen (O) is about 16.00 amu, we get:

5. Chemical Reactions and Stoichiometry: Atomic masses allow chemists to balance chemical equations and determine the proportions of reactants and products. This is fundamental in stoichiometry, where molar ratios are used to predict the outcomes of chemical reactions.

6. Avogadro's Number: The concept of atomic mass is closely related to Avogadro's number (6.022 \times 10^{23} , \text{atoms/mole}), which is used to convert between the mass of a substance and the number of atoms or molecules it contains.

7. Important Formula:

Understanding atomic masses thus provides a foundation for more advanced topics, including chemical bonding, reaction kinetics, and thermodynamics. It is a fundamental concept that underscores many aspects of both theoretical and applied chemistry.

Explanation of Molecular Formula Calculation

Molecular formula calculation is a fundamental process in chemistry that involves determining the exact number of atoms of each element in a molecule. This enables chemists to understand the composition and structure of different compounds.

Steps for Calculation

1. Determine the Empirical Formula: The empirical formula represents the simplest whole-number ratio of atoms in a compound. To find this, follow these steps:

   • Convert the mass of each element to moles by dividing by their atomic masses.
   • Divide each element's mole quantity by the smallest mole quantity among them.
   • If necessary, multiply these ratios by a common factor to get whole numbers.

2. Calculate the Molar Mass of the Empirical Formula: Sum the products of the number of atoms of each element (in the empirical formula) and their atomic masses.

   $$M_{emp} = \sum (n_i \times M_i)$$

   Where:

   • $M_{emp}$ is the molar mass of the empirical formula.
   • $n_i$ is the number of atoms of element $i$ in the empirical formula.
   • $M_i$ is the atomic mass of element $i$.

3. Determine the Molecular Formula: To find the actual number of atoms in the molecule, you need the molar mass of the compound ($M_{mol}$) and the molar mass of the empirical formula ($M_{emp}$).

   Calculate the ratio:

   $$n = \frac{M_{mol}}{M_{emp}}$$

   Multiply the subscripts in the empirical formula by this ratio $n$ to get the molecular formula.

Example

Suppose you have a compound whose empirical formula is CH2 and its molar mass is 56 g/mol.

1. Empirical formula mass:

   $$M_{emp} = 12.01\, \text{g/mol} + 2 \times 1.008\, \text{g/mol} = 14.026\, \text{g/mol}$$

2. Calculate the ratio $n$:

   $$n = \frac{56\, \text{g/mol}}{14.026\, \text{g/mol}} \approx 4$$

3. Determine the molecular formula:

   By multiplying the subscripts in the empirical formula CH2 by 4,

   $$C_{4}H_{8}$$

Thus, the molecular formula is $C_{4}H_{8}$.

Importance

Understanding molecular formulas is crucial because they provide precise information about the molecular structure and the chemical properties of compounds. This knowledge is essential for various applications, including chemical synthesis, pharmaceuticals, and material science.