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What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{CH}_{2}\), molar mass \(=84 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{NH}_{2} \mathrm{Cl}\), molar mass \(=51.5 \mathrm{~g} / \mathrm{mol}\)

Short Answer

Expert verified
The molecular formula of the compounds are: (a) \(\mathrm{C}_6\mathrm{H}_{12}\) and (b) \(\mathrm{NH}_2\mathrm{Cl}\).

Step by step solution

01

Determine the molar mass of the empirical formula

First, find the molar mass of the empirical formula, CH₂. The molar mass of carbon (C) is 12 g/mol, and the molar mass of hydrogen (H) is 1 g/mol. So, the total molar mass of CH₂ is: \[12 + 2(1) = 14 \mathrm{~g/mol}\]
02

Determine the whole-number multiple

Next, divide the given molar mass of the compound by the molar mass of the empirical formula: \[\frac{84 \mathrm{~g/mol}}{14 \mathrm{~g/mol}} = 6\] So, the molecular formula is 6 times larger than the empirical formula.
03

Calculate the molecular formula

Multiply each element in the empirical formula, CH₂, by the whole-number multiple, 6: \[6(CH_2) \Rightarrow C_6H_{12}\] So, the molecular formula of this compound is \(\mathrm{C}_6\mathrm{H}_{12}\). #Part (b):#
04

Determine the molar mass of the empirical formula

First, find the molar mass of the empirical formula, NH₂Cl. The molar mass of nitrogen (N) is 14 g/mol, the molar mass of hydrogen (H) is 1 g/mol, and the molar mass of chlorine (Cl) is 35.5 g/mol. So, the total molar mass of NH₂Cl is: \[14 + 2(1) + 35.5 = 51.5 \mathrm{~g/mol}\]
05

Determine the whole-number multiple

Next, divide the given molar mass of the compound by the molar mass of the empirical formula: \[\frac{51.5 \mathrm{~g/mol}}{51.5 \mathrm{~g/mol}} = 1\] So, the molecular formula is 1 time larger than the empirical formula (meaning they are the same).
06

Calculate the molecular formula

Multiply each element in the empirical formula, NH₂Cl, by the whole-number multiple, 1: \[1(NH_2Cl) \Rightarrow NH_2Cl\] So, the molecular formula of this compound is \(\mathrm{NH}_2\mathrm{Cl}\).

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Key Concepts

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

Empirical Formula
The empirical formula represents the simplest whole-number ratio of atoms in a compound. It doesn't provide information about the actual number of atoms, just the ratio between them. For instance, in the empirical formula \(CH_2\), there are always twice as many hydrogen atoms as carbon atoms. This doesn't mean there are only two hydrogen atoms; instead, it reflects the relative proportions.

To derive an empirical formula, you start by determining the number of moles of each element in a sample. Then, you divide each element's mole value by the smallest mole value among them. Often, this results in whole numbers, but if it doesn’t, you can multiply each number by the same factor to get whole numbers.

The empirical formula is crucial as it serves as a stepping stone toward finding the molecular formula, providing insight into the simplest composition of a molecule.
Molar Mass Calculation
Molar mass is the mass of one mole of a substance, expressed in grams per mole \((\mathrm{g/mol})\). It's calculated as the sum of the atomic masses of all atoms in a molecule. For example, for the compound \(CH_2\), the molar mass is calculated by adding the atomic masses of one carbon atom \((12 \mathrm{~g/mol})\) and two hydrogen atoms \((2 \times 1 \mathrm{~g/mol} = 2 \mathrm{~g/mol})\), resulting in a molar mass of \(14 \mathrm{~g/mol}\).

Calculating the molar mass is vital for several purposes:
  • Quantifying the amount of a substance.
  • Converting between grams and moles when working with chemical equations.
  • Determining the molecular formula of a compound when combined with its empirical formula.
Knowing the molar mass allows chemists to scale up empirical formulas to the actual molecular formula using the compound's known molar mass.
Chemical Compounds
Chemical compounds are substances composed of two or more different types of atoms bonded together. These bonds can be ionic, covalent, or metallic, and they define the structure and properties of the compound.

To describe a chemical compound's composition, scientists use formulas. There are two main types: *empirical formulas* and *molecular formulas*. The empirical formula shows the simplest ratio of elements, while the molecular formula reveals the actual number of each type of atom in a molecule. For example, the molecular formula \(C_6H_{12}\) indicates that the molecule contains six carbon and twelve hydrogen atoms, which could be simplified to the empirical formula \(CH_2\).

Understanding these formulas is essential for studying chemical reactions and properties, predicting compound behavior, and designing new materials.

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Most popular questions from this chapter

(a) What is the mass, in grams, of a mole of \({ }^{12} \mathrm{C} ?\) (b) How many carbon atoms are present in a mole of \({ }^{12} \mathrm{C} ?\)

What is the mass, in kilograms, of an Avogadro's number of people, if the average mass of a person is \(160 \mathrm{lb}\) ? How does this compare with the mass of Earth, \(5.98 \times 10^{24} \mathrm{~kg}\) ?

When ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) reacts with chlorine \(\left(\mathrm{Cl}_{2}\right)\), the main product is \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\), but other products containing \(\mathrm{Cl}\), such as \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2},\) are also obtained in small quantities. The formation of these other products reduces the yield of \(\mathrm{C}_{2} \mathrm{H}_{5}\) Cl. (a) Calculate the theoretical yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) when \(125 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6}\) reacts with \(255 \mathrm{~g}\) of \(\mathrm{Cl}_{2}\), assuming that \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{Cl}_{2}\) react only to form \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{Cl}\) and HCl. (b) Calculate the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) if the reaction produces \(206 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\).

Without doing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing number of atoms: \(0.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\), \(23 \mathrm{~g} \mathrm{Na}, 6.0 \times 10^{23} \mathrm{~N}_{2}\) molecules.

Copper is an excellent electrical conductor widely used in making electric circuits. In producing a printed circuit board for the electronics industry, a layer of copper is laminated on a plastic board. A circuit pattern is then printed on the board using a chemically resistant polymer. The board is then exposed to a chemical bath that reacts with the exposed copper, leaving the desired copper circuit, which has been protected by the overlaying polymer. Finally, a solvent removes the polymer. One reaction used to remove the exposed copper from the circuit board is $$ \begin{aligned} \mathrm{Cu}(s)+\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}(a q)+4 \mathrm{NH}_{3}(a q) & \longrightarrow \\ & 2 \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}(a q) \end{aligned} $$ A plant needs to produce 5000 circuit boards, each with a surface area measuring 2.0 in. \(\times 3.0\) in. The boards are covered with a \(0.65-\mathrm{mm}\) layer of copper. In subsequent processing, \(85 \%\) of the copper is removed. Copper has a density of \(8.96 \mathrm{~g} / \mathrm{cm}^{3} .\) Calculate the masses of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) needed to produce the circuit boards, assuming that the reaction used gives a \(97 \%\) yield.

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