An organic compound was found to contain only \(\mathrm{C}, \mathrm{H},\) and Cl. When a 1.50 -g sample of the compound was completely combusted in air, \(3.52 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) was formed. In a separate experiment the chlorine in a 1.00 -g sample of the compound was converted to \(1.27 \mathrm{~g}\) of AgCl. Determine the empirical formula of the compound.

Short Answer

Expert verified
The empirical formula of the compound is \(C_{2}H_{3}Cl\).

Step by step solution

01

Finding moles of carbon

Determine the amount of carbon in the sample by using information about the combustion of the compound, which forms CO₂. Given CO₂ mass: 3.52 g. Obtain the CO₂ molar mass: \[44.01 \frac{g}{mol}\ (12.01 \,g \,of \,C + 2 \times 16.0 \,g \,of \,O).\] To find the moles of \(C\), divide the mass of CO₂ produced by the molar mass of CO₂: \[moles\,of\,C = \frac{mass\,of\,CO_2}{molar\,mass\,of\,CO_2} = \frac{3.52 \,g}{44.01 \frac{g}{mol}}.\]
02

Finding moles of chlorine

Determine the amount of chlorine in the sample by using information about the formation of AgCl. Given AgCl mass: 1.27 g. Obtain the AgCl molar mass: \[143.32 \frac{g}{mol}\ (35.45 \,g \,of \,Cl + 107.87 \,g \,of \,Ag).\] To find moles of \(Cl\), divide the mass of AgCl produced by the molar mass of AgCl: \[moles\,of\,Cl = \frac{mass\,of\,AgCl}{molar\,mass\,of\,AgCl} = \frac{1.27 \,g}{143.32 \frac{g}{mol}}.\]
03

Finding the mass of hydrogen

Presuming that the substance contains only C, H, and Cl, we can find the mass of hydrogen using the law of conservation of mass. \[mass\,of\,H = mass\,of\,sample - (mass\,of\,C + mass\,of\,Cl).\] Given the mass of sample: 1.50 g. Calculate the mass of C and Cl: - Mass of C: moles of C × molar mass of C - Mass of Cl: moles of Cl × molar mass of Cl
04

Finding moles of hydrogen

Find moles of H by dividing its mass by the molar mass of hydrogen. \[moles\,of\,H = \frac{mass\,of\,H}{molar\,mass\,of\,H} = \frac{mass\,of\,H}{1.01 \frac{g}{mol}}.\]
05

Finding the empirical formula

Determine the empirical formula by finding the lowest whole-number ratio between the moles of C, H, and Cl. Divide the moles of C, H, and Cl by the smallest mole value of the three. Round the resulting numbers to the nearest whole numbers, which will give the empirical formula of the compound.

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Key Concepts

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

Stoichiometry
The concept of stoichiometry is central to the field of chemistry and involves the calculation of the quantities of reactants and products involved in chemical reactions. It is based on the balanced chemical equation for the reaction and the mole concept, which allows chemists to relate the mass of a substance to the number of particles it contains.

Stoichiometry enables us to predict how much product will form from a given amount of reactants, or how much reactant is needed to produce a certain amount of product. In the case of the empirical formula determination exercise, stoichiometry allows us to calculate the moles of carbon, hydrogen, and chlorine from their masses and molar masses, giving us the vital ratios needed to deduce the empirical formula of the compound.

  • Understanding the mole ratios in a balanced chemical equation.
  • Using the mole concept to convert between mass and number of moles.
  • Applying stoichiometry to infer the amounts of reactants and products.
Molar Mass
Molar mass is defined as the mass of one mole of a given substance and is expressed in grams per mole (g/mol). It is equivalent to the atomic or molecular weight of a substance, taken from the periodic table, and normalized to grams.

The molar mass serves as a conversion factor between the mass of a substance and the amount in moles. In the textbook exercise, the molar mass of CO₂ and AgCl is used to determine the number of moles of carbon and chlorine, respectively. The mass of CO₂ and AgCl formed during the experiments is divided by their respective molar masses to find the moles of each element present.

Key points:

  • The molar mass is used as a conversion factor in stoichiometric calculations.
  • Obtaining the molar mass from the sum of atomic weights of elements in a compound.
  • The molar mass helps us establish the relationship between mass and mole in chemical equations.
Combustion Analysis
Combustion analysis is a laboratory method used to determine the elemental composition of a compound, especially those containing carbon and hydrogen. By igniting the compound in the presence of excess oxygen, it will combust fully to produce CO₂ and water. The masses of these combustion products can be measured to deduce the amounts of the constituent elements of the original compound.

In the empirical formula determination problem, combustion analysis allows us to find out how much carbon is in the sample as it is converted to CO₂. From the mass of the CO₂ produced, we can calculate the moles of carbon. This technique also plays a crucial role in ensuring the complete conversion of carbon, which aligns with the principle of the law of conservation of mass.
  • Essential in identifying the proportion of carbon in organic compounds.
  • Provides an accurate method to measure the content of combustible elements.
  • Combustion products are analyzed to calculate the amount of the original elements.
Law of Conservation of Mass
The law of conservation of mass states that matter cannot be created or destroyed in an isolated system. In the context of a chemical reaction, this law implies that the total mass of reactants must equal the total mass of products.

When carrying out stoichiometric calculations, as shown in the empirical formula derivation, this law helps us determine masses of elements within a compound that were not directly measured. For example, in the problem at hand, once we know the mass of carbon and chlorine, we use the law of conservation of mass to infer the mass of hydrogen by subtracting the mass of carbon and chlorine from the total mass of the compound. Ensuring mass balance is integral to accurately calculating the empirical formula.

  • Foundational principle in all chemical equations and reactions.
  • Ensures accurate stoichiometric computations based on mass balance.
  • Guides us in finding unmeasured masses within a compound.

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Most popular questions from this chapter

At least \(25 \mu \mathrm{g}\) of tetrahydrocannabinol (THC), the active ingredient in marijuana, is required to produce intoxication. The molecular formula of THC is \(\mathrm{C}_{21} \mathrm{H}_{30} \mathrm{O}_{2}\). How many moles of THC does this \(25 \mu \mathrm{g}\) represent? How many molecules?

What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{HCO}_{2},\) molar mass \(=90.0 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O},\) molar mass \(=88 \mathrm{~g} / \mathrm{mol}\).

The source of oxygen that drives the internal combustion engine in an automobile is air. Air is a mixture of gases, principally \(\mathrm{N}_{2}(\sim 79 \%)\) and \(\mathrm{O}_{2}(\sim 20 \%) .\) In the cylinder of an automobile engine, nitrogen can react with oxygen to produce nitric oxide gas, NO. As \(\mathrm{NO}\) is emitted from the tailpipe of the car, it can react with more oxygen to produce nitrogen dioxide gas. (a) Write balanced chemical equations for both reactions. (b) Both nitric oxide and nitrogen dioxide are pollutants that can lead to acid rain and global warming; collectively, they are called " \(\mathrm{NO}_{\mathrm{x}}^{\prime \prime}\) gases. In \(2007,\) the United States emitted an estimated 22 million tons of nitrogen dioxide into the atmosphere. How many grams of nitrogen dioxide is this? (c) The production of \(\mathrm{NO}_{\mathrm{x}}\) gases is an unwanted side reaction of the main engine combustion process that turns octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), into \(\mathrm{CO}_{2}\) and water. If \(85 \%\) of the oxygen in an engine is used to combust octane and the remainder used to produce nitrogen dioxide, calculate how many grams of nitrogen dioxide would be produced during the combustion of 500 grams of octane.

At least \(25 \mu \mathrm{g}\) of tetrahydrocannabinol \((\mathrm{THC}),\) the active ingredient in marijuana, is required to produce intoxication. The molecular formula of \(\mathrm{THC}\) is \(\mathrm{C}_{21} \mathrm{H}_{30} \mathrm{O}_{2}\). How many moles of THC does this 25 \mug represent? How many molecules? (b) Caffeine, a stimulant found in coffee, contains \(49.5 \% \mathrm{C}\), \(5.15 \% \mathrm{H}, 28.9 \% \mathrm{~N},\) and \(16.5 \% \mathrm{O}\) by mass and has a molar mass of \(195 \mathrm{~g} / \mathrm{mol}\). (c) Monosodium glutamate (MSG), a flavor enhancer in certain foods, contains \(35.51 \% \mathrm{C}, 4.77 \% \mathrm{H}, 37.85 \% \mathrm{O},\) \(8.29 \% \mathrm{~N},\) and \(13.60 \% \mathrm{Na},\) and has a molar mass of \(169 \mathrm{~g} / \mathrm{mol}\)

(a) You are given a cube of silver metal that measures \(1.000 \mathrm{~cm}\) on each edge. The density of silver is \(10.5 \mathrm{~g} / \mathrm{cm}^{3}\). How many atoms are in this cube? (b) Because atoms are spherical, they cannot occupy all of the space of the cube. The silver atoms pack in the solid in such a way that \(74 \%\) of the volume of the solid is actually filled with the silver atoms. Calculate the volume of a single silver atom. (c) Using the volume of a silver atom and the formula for the volume of a sphere, calculate the radius in angstroms of a silver atom.

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