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

Short Answer

Expert verified
The compound's molecular formula is \(C_6H_6\).

Step by step solution

01

Determine the molar mass of the empirical formula

You calculate the molar mass of the empirical formula \(CH\). Carbon (C) has an atomic mass of approximately \(12 \mathrm{~g/mol}\) and Hydrogen (H) has an atomic mass of approximately \(1 \mathrm{~g/mol}\). Add these together to find the molar mass of the empirical formula, which equates to \(13 \mathrm{~g/mol}\).
02

Determine the ratio of the actual molar mass to the empirical formula mass

You divide the given molar mass of the compound \(78 \mathrm{~g/mol}\) by the molar mass of the empirical formula \(13 \mathrm{~g/mol}\). The result is 6. Therefore, the molecular formula contains 6 times the atoms as the empirical formula.
03

Write out the Molecular Formula

Since we found out that the molecular formula contains six times the atoms as the empirical formula, we can write out the molecular formula of the compound as \(C_6H_6\).

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Key Concepts

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

Empirical Formula
An empirical formula represents the simplest whole-number ratio of the atoms in a compound. It shows which elements are present and in what ratio. For instance, in the compound with an empirical formula of CH, the ratio of carbon to hydrogen is 1:1.
  • Provides basic information about the composition of the compound.
  • Helps chemists to understand the basic structure of unknown compounds.
However, the empirical formula does not tell us how many atoms are actually present in the molecule. It must sometimes be used as a stepping stone to discover the molecular formula. Understanding the empirical formula is crucial to unraveling the full dimensions of a compound's composition.
Molar Mass Calculation
Calculating molar mass is essential in converting between moles and grams, allowing us to measure the substance's quantity. To find the molar mass of a compound given its empirical formula, you add together the atomic masses of each element according to the formula. Using our example, the empirical formula CH consists of:
  • Carbon (C) with an atomic mass of approximately 12 g/mol
  • Hydrogen (H) with an atomic mass of approximately 1 g/mol
Adding these together results in a molar mass of 13 g/mol for the empirical formula.
It's important to double-check the atomic masses from a reliable periodic table. Molar mass calculation is a pivotal step in determining the molecular formula, especially when given an actual molar mass of the compound, as it helps establish the "multiplying factor" needed to derive the molecular formula.
Atomic Mass
Atomic mass is the mass of an atom, typically expressed in atomic mass units (amu). It is approximately equivalent to the number of protons and neutrons in the atom. For chemical calculations, it's expressed in grams per mole (g/mol), which aids in conversions in chemical equations and formulas. For example, the atomic mass of:
  • Carbon (C) is about 12 g/mol.
  • Hydrogen (H) is about 1 g/mol.
This knowledge allows chemists to calculate the molar mass of compounds and understand their composition on a much deeper level. Having a handle on atomic mass is vital to apply this understanding practically in balancing equations and stoichiometric calculations.
Chemical Composition
Chemical composition refers to the identity and proportion of the elements within a compound. It's a detailed fingerprint of a compound that highlights how each element contributes to the overall structure and function. To find a compound's full chemical composition:
  • Start with its empirical formula.
  • Use the empirical formula to compute the molar mass.
  • Determine the molecular formula using the provided molar mass.
In our example exercise, starting with the empirical formula (CH), and given the molar mass of roughly 78 g/mol, we can decode the compound's specific make-up as benzene, with the molecular formula C₆H₆. Understanding chemical composition is not just about memorizing formulas. It's about unlocking insights into how compounds interact, behave, and react under different conditions. This knowledge is indispensable for innovating and solving complex chemical problems.

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Most popular questions from this chapter

A compound X contains 63.3 percent manganese (Mn) and 36.7 percent O by mass. When \(X\) is heated, oxygen gas is evolved and a new compound Y containing 72.0 percent \(\mathrm{Mn}\) and 28.0 percent \(\mathrm{O}\) is formed. (a) Determine the empirical formulas of X and Y. (b) Write a balanced equation for the conversion of \(\mathrm{X}\) to \(\mathrm{Y}\)

Platinum forms two different compounds with chlorine. One contains 26.7 percent \(\mathrm{Cl}\) by mass, and the other contains 42.1 percent Cl by mass. Determine the empirical formulas of the two compounds.

Disulfide dichloride \(\left(\mathrm{S}_{2} \mathrm{Cl}_{2}\right)\) is used in the vulcanization of rubber, a process that prevents the slippage of rubber molecules past one another when stretched. It is prepared by heating sulfur in an atmosphere of chlorine:$$\mathrm{S}_{8}(l)+4 \mathrm{Cl}_{2}(g) \longrightarrow 4 \mathrm{~S}_{2} \mathrm{Cl}_{2}(l)$$What is the theoretical yield of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\) in grams when \(4.06 \mathrm{~g}\) of \(\mathrm{S}_{8}\) are heated with \(6.24 \mathrm{~g}\) of \(\mathrm{Cl}_{2} ?\) If the actual yield of \(\mathrm{S}_{2} \mathrm{Cl}_{2}\) is \(6.55 \mathrm{~g}\), what is the percent yield?

In a natural product synthesis, a chemist prepares a complex biological molecule entirely from nonbiological starting materials. The target molecules are often known to have some promise as therapeutic agents, and the organic reactions that are developed along the way benefit all chemists. The overall synthesis, however, requires many steps, so it is important to have the best possible percent yields at each step. What is the overall percent yield for such a synthesis that has 24 steps with an 80 percent yield at each step?

Because of its detrimental effect on the environment, the lead compound described in Problem 3.148 has been replaced by methyl tert-butyl ether (a compound of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) ) to enhance the performance of gasoline. (This compound is also being phased out because of its contamination of drinking water.) When \(12.1 \mathrm{~g}\) of the compound are burned in an apparatus like the one shown in Figure \(3.6,30.2 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(14.8 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are formed. What is the empirical formula of the compound?

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