Ch.3 - Mass Relationships in Chemical Reactions

McMurry8th EditionChemistryISBN: 9781292336145Not the one you use?Change textbook

McMurry 8th Edition

Ch.3 - Mass Relationships in Chemical Reactions

Problem 69

Chapter 3, Problem 69

Aluminum reacts with oxygen to yield aluminum oxide. If 5.0 g of Al reacts with 4.45 g of O2, what is the empirical formula of aluminum oxide?

Verified step by step guidance

Step 1: Write the balanced chemical equation for the reaction: 4Al + 3O_2 → 2Al_2O_3.

Step 2: Calculate the moles of aluminum (Al) using its molar mass (26.98 g/mol): moles of Al = mass of Al / molar mass of Al.

Step 3: Calculate the moles of oxygen (O_2) using its molar mass (32.00 g/mol): moles of O_2 = mass of O_2 / molar mass of O_2.

Step 4: Determine the mole ratio of Al to O by dividing the moles of each by the smallest number of moles calculated in the previous steps.

Step 5: Use the mole ratio to determine the empirical formula of aluminum oxide by expressing the ratio in the smallest whole numbers.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It allows us to calculate the amounts of substances consumed and produced in a reaction based on balanced chemical equations. Understanding stoichiometry is essential for determining the empirical formula, as it helps in converting masses of reactants to moles and subsequently to the mole ratio of the elements.

Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of the elements present in that compound. It is derived from the mole ratio of the elements obtained from stoichiometric calculations. For aluminum oxide, determining the empirical formula involves finding the ratio of aluminum to oxygen after calculating the moles of each element based on the given masses.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is crucial for converting between grams and moles in stoichiometric calculations. For this question, knowing the molar masses of aluminum (approximately 27 g/mol) and oxygen (approximately 32 g/mol) allows for the accurate conversion of the given masses of Al and O2 into moles, which is necessary for determining the empirical formula.

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Related Practice

Textbook Question

Pure oxygen was first made by heating mercury(II) oxide: HgO --> (heat) Hg + O2 Unbalanced(a) Balance the equation. (b) How many grams of mercury and how many grams of oxygen are formed from 45.5 g of HgO?(c) How many grams of HgO would you need to obtain 33.3 g of O2

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Textbook Question

Titanium dioxide (TiO2), the substance used as the pigment in white paint, is prepared industrially by reaction of TiCl4 with O2 at high temperature. TiCl4 + O2 --> (heat) TiO2 + 2 Cl2 How many kilograms of TiO2 can be prepared from 5.60 kg of TiCl4?

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Textbook Question

Silver metal reacts with chlorine (Cl2) to yield silver chlo-ride. If 2.00 g of Ag reacts with 0.657 g of Cl2, what is the empirical formula of silver chloride?

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Textbook Question

The industrial production of hydriodic acid takes place by treatment of iodine with hydrazine (N2H4):2 I2 + N2H4 --> 4 HI + N2(b) How many grams of HI are produced from the reaction of 115.7 g of N2H4 with excess iodine

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Textbook Question

An alternative method for producing hydriodic acid is the reaction of iodine with hydrogen sulfide: H₂S + I₂ → 2 HI + S (a) How many grams of I₂ are needed to react with 49.2 g of H₂S?

501

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Textbook Question

An alternative method for producing hydriodic acid is the reaction of iodine with hydrogen sulfide:H₂S + I₂ → 2 HI + S (b) How many grams of HI are produced from the reaction of 95.4 g of H₂S with excess I₂?

461

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