Question

Determine the molecular formulas to which the following empirical formulas and molar masses pertain. a. SNH \((188.35 \mathrm{~g} / \mathrm{mol})\) b. \(\mathrm{NPCl}_{2}(347.64 \mathrm{~g} / \mathrm{mol})\) c. \(\mathrm{CoC}_{4} \mathrm{O}_{4}(341.94 \mathrm{~g} / \mathrm{mol})\) d. SN \((184.32 \mathrm{~g} / \mathrm{mol})\)

Short answer

The molecular formulas for the given empirical formulas and molar masses are: a. S₄N₄H₄ b. N₃P₃Cl₆ c. Co₂C₈O₈ d. S₄N₄

Step-by-step solution

Calculate the molar mass of the empirical formula

First, find the molar mass of the empirical formula, SNH. Use the periodic table to find the atomic masses of each element: S (32.07 g/mol), N (14.01 g/mol), H (1.01 g/mol). Now, add the atomic masses together to determine the molar mass of the empirical formula: \(32.07 + 14.01 + 1.01 = 47.09 \mathrm{~g/mol}\)

Find the ratio

Divide the given molar mass by the molar mass of the empirical formula to calculate the ratio: \(188.35 \mathrm{~g/mol} \div 47.09 \mathrm{~g/mol} = 4\)

Determine the molecular formula

Multiply the empirical formula by the ratio obtained in step 2 to obtain the molecular formula: SNH × 4 = S₄N₄H₄. So the molecular formula is S₄N₄H₄. #b. NPCl₂(347.64 g/mol)#

Calculate the molar mass of the empirical formula

Calculate the molar mass of the empirical formula, NPCl₂: N (14.01 g/mol), P (30.97 g/mol), Cl (35.45 g/mol). Add the atomic masses together: \(14.01 + 30.97 + 2(35.45) = 116.33 \mathrm{~g/mol}\)

Find the ratio

Divide the given molar mass by the molar mass of the empirical formula: \(347.64 \mathrm{~g/mol} \div 116.33 \mathrm{~g/mol} = 3\)

Determine the molecular formula

Multiply the empirical formula by the ratio: NPCl₂ × 3 = N₃P₃Cl₆. The molecular formula is N₃P₃Cl₆. #c. CoC₄O₄(341.94 g/mol)#

Calculate the molar mass of the empirical formula

Calculate the molar mass of the empirical formula, CoC₄O₄: Co (58.93 g/mol), C (12.01 g/mol), O (16.00 g/mol). Add the atomic masses together: \(58.93 + 4(12.01) + 4(16.00) = 170.97 \mathrm{~g/mol}\)

Find the ratio

Divide the given molar mass by the molar mass of the empirical formula: \(341.94 \mathrm{~g/mol} \div 170.97 \mathrm{~g/mol} = 2\)

Determine the molecular formula

Multiply the empirical formula by the ratio: CoC₄O₄ × 2 = Co₂C₈O₈. The molecular formula is Co₂C₈O₈. #d. SN (184.32 g/mol)#

Calculate the molar mass of the empirical formula

Calculate the molar mass of the empirical formula, SN: S (32.07 g/mol), N (14.01 g/mol). Add the atomic masses together: \(32.07 + 14.01 = 46.08 \mathrm{~g/mol}\)

Find the ratio

Divide the given molar mass by the molar mass of the empirical formula: \(184.32 \mathrm{~g/mol} \div 46.08 \mathrm{~g/mol} = 4\)

Determine the molecular formula

Multiply the empirical formula by the ratio: SN × 4 = S₄N₄. The molecular formula is S₄N₄.

Key concepts

Empirical Formula

An empirical formula represents the simplest whole-number ratio of atoms of each element in a compound. It doesn't necessarily reflect the exact number of atoms found in one molecule, but it provides a basic representation of the compound’s composition.
The empirical formula is derived based on the proportions of each element within a compound. This information can be determined from experimental data, such as the percentage composition by mass of each element in a sample.

• **Example**: The empirical formula for glucose is CH₂O, which shows the ratio of carbon, hydrogen, and oxygen in the compound.
• It is important to remember that **the empirical formula does not indicate the actual number of atoms**, only the ratio in the compound.

Using empirical formulas helps in identifying unknown substances and predicting their chemical behavior.

Molar Mass

Molar mass is a key concept in chemistry that is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It allows chemists to convert between the mass of a substance and the amount in moles, a basic unit in chemistry.
Molar mass can be calculated by adding together the atomic masses of all the elements expressed in the empirical or molecular formula, using the periodic table as a reference.

• **Calculation step**: For water (H₂O), the molar mass would be calculated as follows: H (1.01 g/mol) × 2 + O (16.00 g/mol) = 18.02 g/mol.
• **Importance**: It helps in determining how much of a substance participates in a chemical reaction.

Recognizing the molar mass is crucial when evaluating substances, as it relates to both the empirical and molecular formula calculations.

Atomic Mass

Atomic mass is the mass of an individual atom of a chemical element, often measured in atomic mass units (amu) or daltons (Da). It reflects the number of protons and neutrons in the nucleus of an atom.
While the periodic table lists the average atomic mass for each element, it is important to remember this number can vary slightly due to the presence of isotopes.

• **Average mass**: The atomic mass is calculated based on the weighted average of an element's isotopes found naturally.
• **Significance in calculations**: Using atomic masses, chemists can determine the **molar mass** of compounds and the proportions of different elements in any substance.

Having a solid understanding of atomic mass is fundamental in performing accurate chemical calculations and predicting reaction outcomes.

Periodic Table

The periodic table is a fundamental tool in chemistry that organizes all known chemical elements in a tabular format according to their atomic number, electron configuration, and recurring chemical properties.

• Each element is identified by its chemical symbol and atomic number.
• The table is structured in such a way that elements with similar properties fall into the same columns, known as groups or families.

The position of an element in the periodic table can provide insight into its properties and behaviors, enabling scientists to predict how it might interact with other elements.
By using the periodic table, chemists can easily find essential data such as atomic masses, electronegativity, and valence electrons, which are useful for **molecular formula** calculations and understanding chemical reactivity.

Chemical Calculations

Chemical calculations involve the use of mathematical concepts to solve problems related to chemical reactions and formulas. These calculations help to predict the quantity of products or reactants in a chemical reaction.
There are several types of calculations crucial for chemistry studies:

• **Molarity and Concentration**: For finding the concentration of a solution.
• **Stoichiometry**: Used to calculate the amounts of reactants and products in chemical equations using balanced chemical equations and their molar masses.
• **Limiting Reactant and Theoretical Yield**: These calculations determine which reactant will run out first and the maximum amount of product that can be produced.

Being proficient with chemical calculations not only aids in laboratory work but also gives a clearer understanding of the principles governing the transformations that occur at the molecular level.