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Chapter 3: Problem 54

The empirical formula of a compound is \(\mathrm{CH}\). If the molar mass of this compound is about \(78 \mathrm{~g},\) what is its molecular formula?

Short Answer

Expert verified
The molecular formula is \(\mathrm{C}_6\mathrm{H}_6\).

Step by step solution

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01

Determine the empirical formula mass

Calculate the molar mass of the empirical formula, CH. The atomic mass of carbon (C) is approximately 12 g/mol and hydrogen (H) is approximately 1 g/mol.\[\text{Empirical formula mass} = (12\, \text{g/mol}) + (1\, \text{g/mol}) = 13\, \text{g/mol}\]
02

Find the ratio of molar mass to empirical formula mass

Divide the given molar mass by the empirical formula mass to find the ratio, which tells you how many times larger the molecular formula is compared to the empirical formula.\[\text{Ratio} = \frac{78\, \text{g/mol}}{13\, \text{g/mol}} = 6\]
03

Determine the molecular formula

Multiply the subscripts in the empirical formula by the ratio obtained to get the molecular formula.Since the empirical formula is \(\text{CH}\) and the ratio is 6, the molecular formula is \(\mathrm{C}_6\mathrm{H}_6\).

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Empirical Formula
The empirical formula provides the simplest whole-number ratio of atoms within a chemical compound. For example, the empirical formula \(\mathrm{CH}\) indicates that carbon and hydrogen exist in a 1:1 ratio in the compound. Unlike the molecular formula, which tells us the exact number of each atom in a molecule, the empirical formula gives a simplified version.
Finding the empirical formula involves determining the relative number of each type of atom by analyzing experimental data, often from mass spectrometry or combustion analysis. It's a fundamental concept that helps chemists understand the composition of compounds, especially when molecular details are complex or unknown.
Molar Mass
The molar mass is the mass of one mole of a substance, and it's measured in grams per mole (g/mol). It's a key concept essential for converting between the amount of substance (in moles) and mass (in grams). Calculating the molar mass requires knowing the atomic masses of all elements in a compound and adding them together based on the compound's formula.
For instance, you would find the molar mass of \(\mathrm{CH}\) by adding the atomic masses of carbon (12 g/mol) and hydrogen (1 g/mol), resulting in a total of 13 g/mol. Understanding molar mass aids in predicting reactions and determining why some reactions might proceed faster than others based on mass interactions.
Chemical Compounds
Chemical compounds consist of two or more different elements joined together by chemical bonds. They can be classified into various types like ionic, covalent, and metallic compounds depending on the nature of the bond. For example, \(\mathrm{C}_6\mathrm{H}_6\), or benzene, is a covalent compound characterized by shared electrons among the atoms.
  • The properties of a compound are distinct from the properties of its constituent elements.
  • Compounds can be broken down through chemical reactions, unlike mixtures which can be separated by physical means.
  • Understanding the composition of compounds is crucial for industries like pharmaceuticals where precision is vital.
Compounds form the backbone of chemistry, acting as reactants, products, and intermediates in countless reactions.
Atomic Mass
Atomic mass is the mass of an individual atom, usually expressed in unified atomic mass units (u or amu). The atomic mass of an element reflects the total number of protons and neutrons in its nucleus. For example, carbon has an atomic mass of approximately 12 amu, while hydrogen is about 1 amu.
Knowing atomic mass is important because it allows chemists to calculate molar masses and form empirical and molecular formulas. This information helps not only in lab settings but also across fields like materials science and biochemistry, where the atomic composition of materials determines their properties and usefulness.
Step by Step Chemistry Solution
A step-by-step solution in chemistry breaks down complex problems into manageable parts. This approach is critical in solving empirical formula and molecular formula problems, among others.
For example, to find the molecular formula from an empirical formula, one first calculates the empirical formula mass. Next, the molar mass of the compound is measured. Then, by dividing the molar mass by the empirical formula mass, you find how many times the empirical formula fits into the molecular formula.
  • This method simplifies problem-solving by reducing potential errors common in holistic approaches.
  • Each step builds understanding, reinforcing the logic behind calculations.
  • This approach is versatile, applicable to other areas such as stoichiometry and reaction balancing.
In essence, step-by-step solutions transform daunting tasks into straightforward, solvable equations through clear, logical progression.

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Most popular questions from this chapter

The annual production of sulfur dioxide from burning coal and fossil fuels, auto exhaust, and other sources is about 26 million tons. The equation for the reaction is $$ \mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) $$ How much sulfur (in tons), present in the original materials. would result in that quantity of \(\mathrm{SO}_{2}\) ?

Heating \(2.40 \mathrm{~g}\) of the oxide of metal \(\mathrm{X}\) (molar mass of \(\mathrm{X}=55.9 \mathrm{~g} / \mathrm{mol}\) ) in carbon monoxide (CO) yields the pure metal and carbon dioxide. The mass of the metal product is \(1.68 \mathrm{~g}\). From the data given, show that the simplest formula of the oxide is \(\mathrm{X}_{2} \mathrm{O}_{3}\) and write a balanced equation for the reaction.

A mixture of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) is heated until all the water is lost. If \(5.020 \mathrm{~g}\) of the mixture gives \(2.988 \mathrm{~g}\) of the anhydrous salts, what is the percent by mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) in the mixture?

A sample of \(10.0 \mathrm{~g}\) of sodium reacts with oxygen to form \(13.83 \mathrm{~g}\) of sodium oxide \(\left(\mathrm{Na}_{2} \mathrm{O}\right)\) and sodium peroxide \(\left(\mathrm{Na}_{2} \mathrm{O}_{2}\right) .\) Calculate the percent composition of the product mixture.

Silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) can be prepared by heating Si in chlorine gas: $$ \mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(l) $$ In one reaction, \(0.507 \mathrm{~mol}\) of \(\mathrm{SiCl}_{4}\) is produced. How many moles of molecular chlorine were used in the reaction?
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