The document provides information about moles, molar mass, empirical and molecular formulas, percentage composition of compounds, and analytical techniques like mass spectrometry and combustion analysis. It includes examples of calculations involving moles, molar mass, empirical formulas from percentage composition data, determining molecular formulas from empirical formulas and molar masses, and calculating hydration in salts.
This document introduces stoichiometry, which is the quantitative study of amounts of reactants and products in a chemical reaction. It defines stoichiometry, coefficients, mole ratios, and how to use mole ratios to solve stoichiometry problems. Specifically, it explains that mole ratios allow you to relate the amounts of different substances in a chemical equation. It provides examples demonstrating how to write equations, identify coefficients and mole ratios, and use mole ratios to calculate the amount of products formed from a given amount of reactants.
This document provides an overview of stoichiometry concepts in chemistry. It defines stoichiometry as calculations with chemical formulas and equations. Key concepts covered include the law of conservation of mass, types of chemical reactions, anatomy of chemical equations, formula weights and molecular weights, moles, mole ratios, empirical formulas, combustion analysis, stoichiometric calculations, limiting reactants, theoretical yield, and percent yield.
This document discusses the key gas laws and their relationships. It begins by introducing the four main properties that determine gas behavior: pressure, volume, amount of gas, and temperature. It then explains each gas law in more detail: Boyle's Law describes the inverse relationship between pressure and volume; Charles' Law specifies the direct relationship between volume and temperature; Gay-Lussac's Law shows the direct link between pressure and temperature; and the Combined Gas Law incorporates all three. The document also presents sample problems demonstrating applications of the gas laws.
The document discusses the concept of a dozen and introduces Avogadro's number as a more appropriate unit than a dozen for counting very small particles like atoms and molecules. It notes that a dozen equals 12 items or objects, while Avogadro's number (6.022 x 1023) is more suitable for microscopic entities since it represents the number of constituent particles in one mole of a substance. The document suggests replacing the term "dozen" with "mole" and 12 with Avogadro's number when calculating quantities of very small particles.
The document describes the main types of chemical reactions: synthesis reactions where two or more reactants form one product, decomposition reactions where one reactant breaks down into multiple products, replacement reactions including single replacement where one element replaces another in a compound and double replacement where ions switch between compounds, and combustion reactions where a substance reacts with oxygen releasing energy.
The document discusses stoichiometry, which is the calculation of quantities in chemical reactions based on a balanced equation. It explains how to interpret balanced equations in terms of particles, moles, mass, and gas volume. It also covers how to perform stoichiometric calculations using mole-mole, mole-mass, and volume-volume conversions based on molar ratios from balanced equations.
The document discusses measured numbers and significant figures. It provides examples of measuring lengths using a meter stick and estimating digits. Measured numbers are obtained through measurement, while exact numbers come from counting or definitions. Numbers are classified as either exact or measured, with measured requiring a measuring tool and exact coming from counts or definitions.
This document discusses moles, molar mass, and Avogadro's number. It explains that a mole is the amount of a substance that contains 6.022x1023 particles, known as Avogadro's number. It also defines molar mass as the mass in grams of one mole of a substance. The document provides examples of calculating molar mass from atomic masses and using molar mass to determine the number of moles or particles in a given mass of a substance.
This document introduces stoichiometry, which is the quantitative study of amounts of reactants and products in a chemical reaction. It defines stoichiometry, coefficients, mole ratios, and how to use mole ratios to solve stoichiometry problems. Specifically, it explains that mole ratios allow you to relate the amounts of different substances in a chemical equation. It provides examples demonstrating how to write equations, identify coefficients and mole ratios, and use mole ratios to calculate the amount of products formed from a given amount of reactants.
This document provides an overview of stoichiometry concepts in chemistry. It defines stoichiometry as calculations with chemical formulas and equations. Key concepts covered include the law of conservation of mass, types of chemical reactions, anatomy of chemical equations, formula weights and molecular weights, moles, mole ratios, empirical formulas, combustion analysis, stoichiometric calculations, limiting reactants, theoretical yield, and percent yield.
This document discusses the key gas laws and their relationships. It begins by introducing the four main properties that determine gas behavior: pressure, volume, amount of gas, and temperature. It then explains each gas law in more detail: Boyle's Law describes the inverse relationship between pressure and volume; Charles' Law specifies the direct relationship between volume and temperature; Gay-Lussac's Law shows the direct link between pressure and temperature; and the Combined Gas Law incorporates all three. The document also presents sample problems demonstrating applications of the gas laws.
The document discusses the concept of a dozen and introduces Avogadro's number as a more appropriate unit than a dozen for counting very small particles like atoms and molecules. It notes that a dozen equals 12 items or objects, while Avogadro's number (6.022 x 1023) is more suitable for microscopic entities since it represents the number of constituent particles in one mole of a substance. The document suggests replacing the term "dozen" with "mole" and 12 with Avogadro's number when calculating quantities of very small particles.
The document describes the main types of chemical reactions: synthesis reactions where two or more reactants form one product, decomposition reactions where one reactant breaks down into multiple products, replacement reactions including single replacement where one element replaces another in a compound and double replacement where ions switch between compounds, and combustion reactions where a substance reacts with oxygen releasing energy.
The document discusses stoichiometry, which is the calculation of quantities in chemical reactions based on a balanced equation. It explains how to interpret balanced equations in terms of particles, moles, mass, and gas volume. It also covers how to perform stoichiometric calculations using mole-mole, mole-mass, and volume-volume conversions based on molar ratios from balanced equations.
The document discusses measured numbers and significant figures. It provides examples of measuring lengths using a meter stick and estimating digits. Measured numbers are obtained through measurement, while exact numbers come from counting or definitions. Numbers are classified as either exact or measured, with measured requiring a measuring tool and exact coming from counts or definitions.
This document discusses moles, molar mass, and Avogadro's number. It explains that a mole is the amount of a substance that contains 6.022x1023 particles, known as Avogadro's number. It also defines molar mass as the mass in grams of one mole of a substance. The document provides examples of calculating molar mass from atomic masses and using molar mass to determine the number of moles or particles in a given mass of a substance.
This document provides examples of how to calculate the limiting reactant and maximum product formed in chemical reactions. It demonstrates calculating moles of each reactant, comparing reactant ratios to determine the limiting reactant, and using mole ratios to calculate maximum product. An alternative method is proposed to systematically solve these types of problems.
The mole is a unit used in chemistry to express amounts of substances. It represents 6.022x10^23 elementary entities, such as atoms, molecules, ions or other particles of a substance. This number is known as Avogadro's constant, after scientist Amedeo Avogadro who proposed that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. The mass of one mole of a substance, known as its molar mass, can be used to determine the number of moles in a given mass of that substance and vice versa through calculation.
Percent Composition, Empirical and Molecular FormulaEllebasy Tranna
This document provides examples and explanations for calculating percent composition, empirical formulas, and molecular formulas. It begins with examples of calculating percent composition of elements in compounds and mixtures. It then defines empirical and molecular formulas, and provides steps for calculating empirical formulas from mass percentages of elements or experimental data. Several examples are worked through. The document emphasizes that empirical formulas show the simplest whole number ratio of elements in a compound, while molecular formulas indicate the actual number of each type of atom in a molecule.
Law of multiple proportions and law of definite proportionsNikki Wilkinson
The document discusses several laws related to chemical reactions and composition:
The Law of Definite Proportions states that a chemical compound always has the same proportions by mass of elements. The Law of Multiple Proportions states that when two elements form multiple compounds, the ratios of one element's masses are small whole numbers. The Law of Conservation of Mass states the total mass remains the same in a chemical reaction. Examples and practice problems demonstrate applying these laws.
The document provides an introduction to stoichiometry including the four types of stoichiometry problems: (1) given and unknown amounts in moles, (2) given amount in moles and unknown in grams, (3) given amount in grams and unknown in moles, and (4) given and unknown amounts in grams. It then works through example problems of each type, calculating amounts of products or reactants using molar ratios derived from balanced chemical equations.
Worksheet for working out the Percentage by Mass of various compounds. Pupils will need a data sheet or a list of relative atomic masses to be able to complete the questions.
Here are brief explanations of the key concepts:
- The composition of an element is fixed because elements are pure substances made of only one type of atom.
- The composition of a compound is also fixed, but compounds contain two or more elements combined in a fixed ratio.
- Properties of mixtures can vary because mixtures are combinations of two or more substances that are not chemically combined. Their compositions are not fixed.
- Mixtures can be classified as solutions, suspensions, or colloids based on whether the mixed substances are uniformly dispersed (solutions), settle over time (suspensions), or are dispersed with particles too small to settle but large enough to scatter light (colloids).
- Every sample of
The document provides instructions for an experiment demonstrating the Law of Conservation of Mass. Students measure the mass of a system containing baking soda in a flask and vinegar in a balloon before and after the vinegar is added, triggering a reaction, and again after popping the balloon. Any changes in mass must be explained by the redistribution of atoms between reactants and products, not by the creation or destruction of atoms.
This document discusses the rules for determining significant figures in measurements:
1. All non-zero digits are significant, as are zeros between non-zero digits. Leading and trailing zeros are only significant if a decimal is present.
2. When adding or subtracting measurements, the answer should have the same number of decimal places as the least precise measurement.
3. When multiplying or dividing measurements, the answer should have the same number of significant figures as the factor with the fewest significant figures.
Discusses the chemical of slightly soluble compounds. Ksp and factors affecting solubility are included as well as solved problems.
**More good stuff available at:
www.wsautter.com
and
http://www.youtube.com/results?search_query=wnsautter&aq=f
This document discusses stoichiometry, which uses balanced chemical equations to determine amounts of reactants and products in chemical reactions. It provides examples of using mole ratios from chemical equations to solve stoichiometry problems involving moles of substances or conversions between moles and grams. The key aspects are that chemical equations provide mole ratios that can be used as conversion factors, and problems must be worked in moles since equations relate substances in moles.
This document provides an overview of basic chemistry concepts including:
- Dalton's atomic theory and modern atomic theory which established atoms as the fundamental units of matter and that they can exist as isotopes.
- Berzelius hypothesis and Avogadro's law which established that equal volumes of gases under similar conditions contain equal numbers of molecules.
- Definitions of atoms, molecules, atomic mass, molecular mass, gram atomic mass, gram molecular mass, formula mass, and gram formula mass.
- Introduction of the mole concept based on Avogadro's number, which established a mole as a specific number of particles (atoms or molecules).
A chemical reaction involves the transformation of reactants into different products through rearrangement of atoms. Chemical reactions conserve mass as atoms are not destroyed or created, but instead are reorganized into new substances. Balancing chemical equations ensures the same number and type of atoms are on both sides of the reaction.
This document provides an overview of moles, molar mass, and stoichiometry. It defines a mole as 6.02x1023 particles of an element or compound. Avogadro's number is equal to this quantity of particles. Molecular mass is the mass of one molecule in atomic mass units (amu) while molar mass is the mass of one mole of a substance in grams. Examples are given for calculating molecular and molar mass. The document then discusses balancing chemical equations and using molar ratios from balanced equations to solve stoichiometry problems involving moles, grams, and particles of reactants and products.
This document provides an introduction to chemistry, including defining matter and its three states (solid, liquid, gas). It discusses the particles that compose matter (atoms, molecules, ions) and their properties. Methods for separating mixtures are also outlined, such as filtration, distillation, evaporation, and chromatography. Consumer products often contain mixtures to appeal to customers, with examples given of household cleaning supplies and personal care items.
The document discusses different types of chemical reactions including combination, decomposition, single replacement, double replacement, and combustion reactions. It provides examples of each type of reaction by showing the starting reactants and products. Combination reactions involve two or more reactants directly combining to form a single product. Decomposition reactions involve a single reactant breaking into two or more products.
The document discusses chemical reactions and stoichiometry. It defines stoichiometry as using ratios to determine quantities of reactants and products in a chemical reaction. It explains that coefficients in a balanced chemical equation represent molar ratios and can be used to determine moles, mass, and volume of substances in a reaction. It provides examples of solving stoichiometry problems, including determining limiting reactants.
Stoichiometry allows us to use balanced chemical equations to determine the amounts of reactants and products involved in chemical reactions. It treats the chemical equation like a recipe, using mole ratios derived from the coefficients to solve mole-mole, mole-mass, and mass-mass problems. For example, if 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water, how many moles of oxygen are needed to produce 4 moles of water? By using the 1:1 mole ratio of oxygen to water given in the balanced equation, we can determine that 4 moles of oxygen are needed.
The document defines a solution as a homogeneous mixture composed of a solvent and one or more solutes. It provides examples of natural solutions like air, rain, and seawater as well as manufactured solutions like vinegar and gasoline. The remainder of the document quizzes the reader to identify the solute and solvent in various solutions like blue dye in water, orange juice in water, oxygen and nitrogen gas, coffee powder in hot water, sugar in water, and sodium chloride in water. It concludes that a mixture of corn oil and water is not a solution because the mixture is not homogeneous.
This document discusses concepts related to stoichiometry including empirical formulas, molecular formulas, percentage composition, and hydrates. It provides examples of calculating empirical formulas from mass percentages of elements in compounds and using mole ratios. It also distinguishes between empirical formulas that give the lowest whole number ratio of atoms in a compound and molecular formulas that give the actual ratio in compounds.
The document discusses the mole concept in chemistry. It defines the mole as 6.022x10^23 particles, which is known as Avogadro's number. The mole can refer to individual atoms, molecules, ions or other particles. The document provides examples of how to calculate the number of particles or moles of a substance using the formula N=n×NA. It also discusses molar mass and how to calculate the mass of a substance using the formula m=n×M.
The document discusses the mole concept in chemistry. Some key points:
- A mole is the amount of substance containing Avogadro's number (6.022x1023) of elementary entities like atoms, molecules, formula units.
- One mole of any substance has a mass in grams equal to its formula/molar mass. For example, 1 mole of iron (Fe) has a mass of 55.85 g.
- The mole can be used to convert between the number of particles/formula units and the mass of a substance using molar mass and the definition of 1 mole.
- Common calculations include determining moles from mass or vice versa using molar mass, as well as
This document provides examples of how to calculate the limiting reactant and maximum product formed in chemical reactions. It demonstrates calculating moles of each reactant, comparing reactant ratios to determine the limiting reactant, and using mole ratios to calculate maximum product. An alternative method is proposed to systematically solve these types of problems.
The mole is a unit used in chemistry to express amounts of substances. It represents 6.022x10^23 elementary entities, such as atoms, molecules, ions or other particles of a substance. This number is known as Avogadro's constant, after scientist Amedeo Avogadro who proposed that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. The mass of one mole of a substance, known as its molar mass, can be used to determine the number of moles in a given mass of that substance and vice versa through calculation.
Percent Composition, Empirical and Molecular FormulaEllebasy Tranna
This document provides examples and explanations for calculating percent composition, empirical formulas, and molecular formulas. It begins with examples of calculating percent composition of elements in compounds and mixtures. It then defines empirical and molecular formulas, and provides steps for calculating empirical formulas from mass percentages of elements or experimental data. Several examples are worked through. The document emphasizes that empirical formulas show the simplest whole number ratio of elements in a compound, while molecular formulas indicate the actual number of each type of atom in a molecule.
Law of multiple proportions and law of definite proportionsNikki Wilkinson
The document discusses several laws related to chemical reactions and composition:
The Law of Definite Proportions states that a chemical compound always has the same proportions by mass of elements. The Law of Multiple Proportions states that when two elements form multiple compounds, the ratios of one element's masses are small whole numbers. The Law of Conservation of Mass states the total mass remains the same in a chemical reaction. Examples and practice problems demonstrate applying these laws.
The document provides an introduction to stoichiometry including the four types of stoichiometry problems: (1) given and unknown amounts in moles, (2) given amount in moles and unknown in grams, (3) given amount in grams and unknown in moles, and (4) given and unknown amounts in grams. It then works through example problems of each type, calculating amounts of products or reactants using molar ratios derived from balanced chemical equations.
Worksheet for working out the Percentage by Mass of various compounds. Pupils will need a data sheet or a list of relative atomic masses to be able to complete the questions.
Here are brief explanations of the key concepts:
- The composition of an element is fixed because elements are pure substances made of only one type of atom.
- The composition of a compound is also fixed, but compounds contain two or more elements combined in a fixed ratio.
- Properties of mixtures can vary because mixtures are combinations of two or more substances that are not chemically combined. Their compositions are not fixed.
- Mixtures can be classified as solutions, suspensions, or colloids based on whether the mixed substances are uniformly dispersed (solutions), settle over time (suspensions), or are dispersed with particles too small to settle but large enough to scatter light (colloids).
- Every sample of
The document provides instructions for an experiment demonstrating the Law of Conservation of Mass. Students measure the mass of a system containing baking soda in a flask and vinegar in a balloon before and after the vinegar is added, triggering a reaction, and again after popping the balloon. Any changes in mass must be explained by the redistribution of atoms between reactants and products, not by the creation or destruction of atoms.
This document discusses the rules for determining significant figures in measurements:
1. All non-zero digits are significant, as are zeros between non-zero digits. Leading and trailing zeros are only significant if a decimal is present.
2. When adding or subtracting measurements, the answer should have the same number of decimal places as the least precise measurement.
3. When multiplying or dividing measurements, the answer should have the same number of significant figures as the factor with the fewest significant figures.
Discusses the chemical of slightly soluble compounds. Ksp and factors affecting solubility are included as well as solved problems.
**More good stuff available at:
www.wsautter.com
and
http://www.youtube.com/results?search_query=wnsautter&aq=f
This document discusses stoichiometry, which uses balanced chemical equations to determine amounts of reactants and products in chemical reactions. It provides examples of using mole ratios from chemical equations to solve stoichiometry problems involving moles of substances or conversions between moles and grams. The key aspects are that chemical equations provide mole ratios that can be used as conversion factors, and problems must be worked in moles since equations relate substances in moles.
This document provides an overview of basic chemistry concepts including:
- Dalton's atomic theory and modern atomic theory which established atoms as the fundamental units of matter and that they can exist as isotopes.
- Berzelius hypothesis and Avogadro's law which established that equal volumes of gases under similar conditions contain equal numbers of molecules.
- Definitions of atoms, molecules, atomic mass, molecular mass, gram atomic mass, gram molecular mass, formula mass, and gram formula mass.
- Introduction of the mole concept based on Avogadro's number, which established a mole as a specific number of particles (atoms or molecules).
A chemical reaction involves the transformation of reactants into different products through rearrangement of atoms. Chemical reactions conserve mass as atoms are not destroyed or created, but instead are reorganized into new substances. Balancing chemical equations ensures the same number and type of atoms are on both sides of the reaction.
This document provides an overview of moles, molar mass, and stoichiometry. It defines a mole as 6.02x1023 particles of an element or compound. Avogadro's number is equal to this quantity of particles. Molecular mass is the mass of one molecule in atomic mass units (amu) while molar mass is the mass of one mole of a substance in grams. Examples are given for calculating molecular and molar mass. The document then discusses balancing chemical equations and using molar ratios from balanced equations to solve stoichiometry problems involving moles, grams, and particles of reactants and products.
This document provides an introduction to chemistry, including defining matter and its three states (solid, liquid, gas). It discusses the particles that compose matter (atoms, molecules, ions) and their properties. Methods for separating mixtures are also outlined, such as filtration, distillation, evaporation, and chromatography. Consumer products often contain mixtures to appeal to customers, with examples given of household cleaning supplies and personal care items.
The document discusses different types of chemical reactions including combination, decomposition, single replacement, double replacement, and combustion reactions. It provides examples of each type of reaction by showing the starting reactants and products. Combination reactions involve two or more reactants directly combining to form a single product. Decomposition reactions involve a single reactant breaking into two or more products.
The document discusses chemical reactions and stoichiometry. It defines stoichiometry as using ratios to determine quantities of reactants and products in a chemical reaction. It explains that coefficients in a balanced chemical equation represent molar ratios and can be used to determine moles, mass, and volume of substances in a reaction. It provides examples of solving stoichiometry problems, including determining limiting reactants.
Stoichiometry allows us to use balanced chemical equations to determine the amounts of reactants and products involved in chemical reactions. It treats the chemical equation like a recipe, using mole ratios derived from the coefficients to solve mole-mole, mole-mass, and mass-mass problems. For example, if 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water, how many moles of oxygen are needed to produce 4 moles of water? By using the 1:1 mole ratio of oxygen to water given in the balanced equation, we can determine that 4 moles of oxygen are needed.
The document defines a solution as a homogeneous mixture composed of a solvent and one or more solutes. It provides examples of natural solutions like air, rain, and seawater as well as manufactured solutions like vinegar and gasoline. The remainder of the document quizzes the reader to identify the solute and solvent in various solutions like blue dye in water, orange juice in water, oxygen and nitrogen gas, coffee powder in hot water, sugar in water, and sodium chloride in water. It concludes that a mixture of corn oil and water is not a solution because the mixture is not homogeneous.
This document discusses concepts related to stoichiometry including empirical formulas, molecular formulas, percentage composition, and hydrates. It provides examples of calculating empirical formulas from mass percentages of elements in compounds and using mole ratios. It also distinguishes between empirical formulas that give the lowest whole number ratio of atoms in a compound and molecular formulas that give the actual ratio in compounds.
The document discusses the mole concept in chemistry. It defines the mole as 6.022x10^23 particles, which is known as Avogadro's number. The mole can refer to individual atoms, molecules, ions or other particles. The document provides examples of how to calculate the number of particles or moles of a substance using the formula N=n×NA. It also discusses molar mass and how to calculate the mass of a substance using the formula m=n×M.
The document discusses the mole concept in chemistry. Some key points:
- A mole is the amount of substance containing Avogadro's number (6.022x1023) of elementary entities like atoms, molecules, formula units.
- One mole of any substance has a mass in grams equal to its formula/molar mass. For example, 1 mole of iron (Fe) has a mass of 55.85 g.
- The mole can be used to convert between the number of particles/formula units and the mass of a substance using molar mass and the definition of 1 mole.
- Common calculations include determining moles from mass or vice versa using molar mass, as well as
This document provides an overview of key concepts in chemical quantities including:
1) The mole is a unit used to measure very large amounts of substances and is defined as 6.02x10^23 items. It can represent atoms, molecules, ions, or formula units.
2) Molar mass is the mass of one mole of a substance in grams and can be used to determine the mass of any amount of a substance.
3) Molar conversions allow calculations between the number of moles, mass in grams, and number of particles using molar mass and Avogadro's number.
The document discusses moles, molar mass, and empirical and molecular formulas.
It defines key terms like mole, Avogadro's number, and molar mass. A mole represents 6.02x1023 particles of a substance. Molar mass is the mass in grams of one mole of a substance.
Examples are provided for calculating moles from mass and vice versa using molar mass. Empirical formulas represent the lowest whole number ratio of elements in a compound, while molecular formulas specify the actual number of each atom in a molecule or formula unit.
The document discusses moles, molar mass, and empirical and molecular formulas.
It defines key terms like mole, Avogadro's number, and molar mass. A mole represents 6.02x1023 particles of a substance. Molar mass is the mass in grams of one mole of a substance.
Examples are provided for calculating moles from mass and vice versa using molar mass. Empirical formulas give the lowest whole number ratio of elements in a compound, while molecular formulas specify the actual number of each atom in a molecule or formula unit.
This document discusses calculating formula masses, molar masses, and percentage compositions of chemical compounds. It provides examples of calculating the formula mass of compounds by adding the atomic masses of each element. Molar mass is defined as the mass of one mole of a compound and is numerically equal to formula mass. The document demonstrates using molar mass to convert between mass and moles of a compound. It also gives examples of calculating the percentage by mass of each element in compounds like copper(I) sulfide and sodium carbonate decahydrate.
PERCENTAGE COMPOSITION, EMPIRICAL AND MOLECULAR FORMULAS.pptELMERBELZA
Here are the steps to solve these sample problems:
1. Empirical Formula: 0.44 g H = 0.044 mol H = 2 mol H
6.92 g O = 0.433 mol O = 1 mol O
Empirical Formula = H2O
Molecular Formula = H2O (same as empirical)
2. Empirical Formula: 36.48% Na = 36.48 g Na = 1.5 mol Na
25.41% S = 25.41 g S = 1 mol S
38.11% O = 38.11 g O = 2.375 mol O
Empirical Formula = NaSO2
Molar mass of NaSO2 = 105 g/mol
- The mole concept allows chemists to conveniently keep track of large numbers of particles. A mole is defined as 6.02 x 1023 particles, whether atoms, molecules, ions, etc.
- The formula mass (or molar mass) of a compound is the sum of the atomic masses of each element in its chemical formula. It has units of grams per mole (g/mol).
- Calculations involving moles, mass, particles and formula mass allow conversions between the microscopic and macroscopic scales in chemistry.
Okay, here are the steps:
1) Convert the mass percentages to grams of each element in 100 g of the compound:
K: 24.75% = 24.75 g
Mn: 34.77% = 34.77 g
O: 40.51% = 40.51 g
2) Calculate the moles of each element:
K: 24.75 g / 39.10 g/mol (molar mass of K) = 0.634 mol
Mn: 34.77 g / 54.94 g/mol (molar mass of Mn) = 0.634 mol
O: 40.51 g / 16.00 g/mol (molar mass of O)
The document discusses moles, molar mass, molar concentration, and stoichiometry. It defines a mole as 6.02 x 1023 particles, whether atoms, molecules, ions, or electrons. Molar mass is the mass of one mole of a substance in grams. Molar concentration, or molarity, expresses the moles of solute per liter of solution. Stoichiometry uses mole ratios from balanced chemical equations to calculate amounts of reactants and products.
This document discusses different methods of measuring matter, including counting, mass, and volume. It then explains how to convert between units using the factor-label method. A mole is defined as 6.02x1023 representative particles of an element or compound. The mass of one mole of a substance is called its molar mass. Molar mass can be used to convert between moles and grams. The volume of one mole of gas at standard temperature and pressure is 22.4 L.
The document discusses the concept of the mole in chemistry. It defines a mole as 6.02 x 1023 representative particles, which can be atoms, molecules, or formula units. It provides examples of calculating the number of moles and mass of different substances. It also explains that 1 mole of any gas occupies 22.4 liters at standard temperature and pressure. Key terms discussed include molar mass, percent composition, empirical formula, and molecular formula.
1. The document discusses mass relationships in chemical reactions including atomic mass, molar mass, molecular mass, formula mass, and percent composition.
2. It defines the mole as the amount of a substance containing 6.022x1023 elementary entities, such as atoms or molecules. Molar mass is the mass in grams of one mole of a substance.
3. Examples are provided for calculating the number of atoms, moles of substances, and masses involved in chemical reactions using molar mass and mole ratios from balanced chemical equations.
The document discusses the mole concept in chemistry. Some key points:
- 1 mole is equal to 6.02 x 1023 particles, whether atoms, molecules, ions, etc. This number is known as Avogadro's number.
- The mass of 1 mole of a substance in grams is equal to the substance's molar mass shown on the periodic table. For example, 1 mole of carbon is 12.01 g.
- Calculations can be done to convert between moles, mass, particles, and volume (at STP) using molar mass and Avogadro's number.
- Percent composition problems calculate the percentage of each element in a compound based on the
Stoichiometry deals with the numerical relationships of elements and compounds and the mathematical proportions of reactants and products in chemical transformations
This document provides instruction on key chemistry concepts related to Avogadro's constant, the mole, chemical formulas, and stoichiometry calculations. It begins with a review of scientific notation and unit conversions. It then defines Avogadro's number as 6.022x1023, explains that it represents the number of particles in 1 mole, and provides examples of mole-particle conversions. Subsequently, it introduces molar mass and shows examples of mole-mass conversions. The document concludes by explaining empirical and molecular formulas, providing examples of determining formulas from percentage composition and molar mass data. Worked practice problems are included throughout to illustrate the application of these concepts.
Stoichiometry deals with the numerical relationships between elements and compounds in chemical reactions. It involves calculating the masses of reactants and products using moles, molar mass, molecular mass, and Avogadro's number. Percentage composition is the percentage by mass of each element in a compound. Understanding moles, molar mass, and balancing chemical equations allows solving stoichiometry problems involving mass relationships in chemical reactions.
The document discusses atomic and molecular masses, moles, molar masses, and calculating empirical and molecular formulas. It provides examples of calculating moles from masses and vice versa using molar masses. It also discusses calculating percentage compositions and determining molecular formulas from empirical formulas using molar masses.
This document discusses stoichiometry, which is the quantitative relationships between reactants and products in chemical reactions. Some key points covered include:
- Stoichiometric amounts refer to the exact molar amounts of reactants and products in a balanced chemical equation.
- Atomic masses are given in atomic mass units (amu) which is based on carbon-12 having a mass of exactly 12 amu.
- Average atomic masses take into account the natural abundances of isotopes.
- The mole is the unit used to express amounts of substances in chemistry and 1 mole contains 6.022x1023 elementary entities.
- Molar mass is the mass in grams of 1 mole of a substance and is equal to the sum of the
1) The document discusses concepts related to chemistry including relative atomic mass, relative molecular mass, moles, and empirical and molecular formulae.
2) It provides examples of calculating relative atomic mass, relative molecular mass, moles, percentage composition by mass, and determining empirical and molecular formulae.
3) Key formulas and concepts covered include using the atomic masses of elements to determine relative atomic mass and molecular mass, the definition of a mole in relation to Avogadro's number, and methods for calculating the number of moles and atoms in a sample.
1) Atomic radius generally increases down a group and decreases across a period due to shielding effects of additional electron shells and increasing nuclear charge, respectively.
2) Ionization energy generally increases down a group and across a period as it becomes increasingly difficult to remove an electron due to greater nuclear attraction from less shielding and more protons. Exceptions occur when removing an electron would destabilize a full shell.
3) Electronegativity generally increases across a period as the effective nuclear charge felt by valence electrons increases with fewer shielding shells and more protons.
This document provides information on wave quantum mechanics and electron configurations. It discusses:
- Erwin Schrodinger's contributions to developing quantum mechanics and proposing the wave-like nature of electrons.
- How electrons occupy distinct energy levels and orbitals around the nucleus, rather than defined circular orbits. Electrons have wave-like properties.
- The shapes of s, p, d and f orbitals and how electrons fill these orbitals according to various principles like Aufbau and Hund's rule.
- Exceptions to the Aufbau principle seen in some elements.
- How to represent electron configurations using both energy level diagrams and shorthand notation.
- Gases have negligible volume and molecules are far apart with mostly empty space between them. Gas molecules move randomly in straight lines and collide elastically.
- The kinetic molecular theory describes gas properties including high translational energy of gas molecules that move randomly in all directions.
- Pressure, volume, temperature of gases are directly or inversely related as described by Boyle's, Charles', and Gay-Lussac's laws and the combined gas law.
- Elements in the same group have similar properties due to having the same number of valence electrons. Atoms get smaller from left to right in a period as protons are added, and larger from top to bottom in a group as energy levels are added.
- Electronegativity, ionization energy, and electron affinity all generally increase from left to right across a period as the effective nuclear charge increases. Electronegativity also increases up a group as the distance from valence electrons to the nucleus decreases.
- Ionization energy increases from top to bottom within a group as it is easier to remove an electron from an atom with fewer energy levels. Electron affinity generally increases from left to right across a period and decreases
This document discusses acids and bases. It defines their key properties including reacting with metals, carbonates, conducting electricity, turning litmus paper colors, and neutralizing each other. It explains the theories of Arrhenius, Brønsted-Lowry, and Lewis on acids and bases. It also covers acid-base reactions, indicators, pH, titrations, strong/weak acids and bases, and acid-base stoichiometry.
Gases are composed of molecules that are far apart from each other and occupy the entire volume of their container. Gas molecules move randomly in straight lines and collide elastically with each other and the container walls. The kinetic molecular theory describes gases as having negligible intermolecular forces and volume. Pressure, volume, temperature of a gas are related by the combined gas law and ideal gas law. Gas calculations involve using stoichiometry, the ideal gas law, and gas laws like Boyle's, Charles', and Gay-Lussac's to determine volume, pressure, amount, or temperature changes.
The document discusses the key properties and reactions of acids and bases. It defines acids as substances that produce hydrogen ions (H+) in water and bases as substances that produce hydroxide ions (OH-). Acids react with metals, carbonates, conduct electricity, turn litmus paper red, and neutralize bases. Bases conduct electricity, turn litmus paper blue, and neutralize acids. Theories of acids and bases including Arrhenius, Brønsted-Lowry, and Lewis are explained. Strong and weak acids/bases, monoprotic/diprotic/triprotic acids, pH, titrations, and acid-base indicators are also covered.
This document provides an overview of solution stoichiometry and includes 3 sample problems demonstrating how to use a balanced chemical equation to solve for unknown quantities. The general steps are: 1) Write the balanced equation, 2) Convert given amounts to moles, 3) Use the mole ratio to calculate moles of the unknown, 4) Convert moles to the required unit. Sample Problem 1 calculates molar concentration of sulfuric acid. Sample Problem 2 determines the minimum volume of sodium carbonate needed. Sample Problem 3 calculates the mass of silver chromate precipitate formed.
This document discusses water treatment processes including water softening. Water softening is a treatment process that uses chemicals to remove minerals like calcium and magnesium from hard water to make it softer. Acceptable concentrations of chemicals used in water treatment processes are also mentioned.
- Concentration can be expressed in several ways including mass/volume percent, mass/mass percent, volume/volume percent, parts per million (ppm), and molar concentration (mol/L).
- Mass/volume percent is the mass of solute divided by the volume of solution. Volume/volume percent is the volume of solute divided by the volume of solution.
- Mass/mass percent is the mass of solute divided by the mass of solution. Parts per million (ppm) and parts per billion (ppb) express very small concentrations.
- Molar concentration expresses the number of moles of solute per liter of solution and is the standard unit for expressing concentration in chemistry
This document discusses acids and bases according to several theories. It begins by describing the properties of acids, including reacting with metals and carbonates, conducting electricity, turning litmus paper colors, and neutralizing bases. It then discusses the properties of bases. The Arrhenius theory defines acids as substances that produce H+ ions in water and bases as those that produce OH- ions. However, this theory has limitations and does not account for all acids and bases. The Brønsted-Lowry theory broadens the definition to any substance that can donate or accept protons. Strong acids fully dissociate in water while weak acids only partially dissociate. The pH scale measures the concentration of H+ ions on a
The document discusses key concepts about solutions including:
- The components of a solution are a solvent and one or more solutes. Common solvents include water and gases.
- Factors that affect solubility and the rate of dissolving include temperature, agitation, and surface area of the solute. Increasing these factors generally increases solubility.
- Solutions can be classified as saturated, unsaturated, or supersaturated depending on the amount of solute dissolved compared to the maximum amount possible at a given temperature.
- Solubility is also affected by properties of the solute molecules like polarity, charge, and size - more polar or charged molecules or smaller ions tend to be more soluble.
- Solub
The document discusses percentage yield and percentage purity calculations for chemical reactions. It provides an example calculation for percentage yield where 1.72g of ammonia is obtained from a reaction using 7.5g of nitrogen. The percentage yield is calculated to be 18.9%. An example is also given for calculating percentage purity of an impure iron pyrite sample where the purity is found to be 86.3%. Formulae for percentage yield and percentage purity are defined. Key steps like using stoichiometry and unit conversions are highlighted.
This document discusses the rules for determining significant digits in measurements and calculations. It defines four main rules:
1. All non-zero digits are significant, as are zeros between non-zero digits.
2. Leading zeros are not significant.
3. Trailing zeros may or may not be significant, depending on whether a decimal is present.
4. In calculations, the answer should match the least number of significant digits in the input values.
It provides examples like 0.08006 having 4 significant digits, and 1000mL having 1 significant digit. The document also covers rounding rules and calculating volumes and perimeters based on measurements to the correct number of significant digits.
This document discusses isotopes and nuclear reactions. It defines isotopes as atoms of the same element with different numbers of neutrons. It also describes how atomic mass is calculated based on isotope abundances. The document then discusses four types of nuclear reactions: fusion, fission, alpha decay, and beta decay. It provides examples of writing balanced nuclear equations and calculating half-life. Artificial transmutation and uses of nuclear technology like reactors and weapons are also summarized.
Isotopes are atoms of the same element that have different numbers of neutrons. Atoms of the same element can have different mass numbers depending on their number of neutrons. The atomic mass listed on the periodic table is an average that takes into account the relative abundances of each isotope of that element. Atomic mass calculations involve multiplying the mass and abundance percentage of each isotope and adding them together.
This document defines and provides examples of different types of chemical reactions including synthesis, decomposition, single displacement, double displacement, neutralization, and combustion reactions. Key aspects that determine the reaction type are whether reactants are elements, compounds, or if oxygen is involved. The position of elements in the periodic table also provides clues about reactivity in displacement reactions.
The document provides step-by-step instructions for balancing chemical equations. It demonstrates how to balance equations with monatomic and polyatomic ions, including making coefficients even numbers when needed. Examples show balancing equations such as H2O2 → H2O + O2, CaSO4 + KOH → Ca(OH)2 + K2SO4, and ZnS + O2 → ZnO + SO2. Key steps include writing element symbols, choosing coefficients, and balancing the numbers of each element on both sides of the equation.
2. The MOLE!!!
People use words to represent specific
quantities all the time
Dozen eggs Pair of gloves Six-Pack
3. The MOLE!!!
1 mole = 6.02214199 x 1023 particles
1 mole = 6.02 x 1023
(the short form for mole is “mol”)
4. The MOLE!!!
1 mol = 6.02 x 1023
Called the Avogadro constant or
Avogadro’s number
(devised through experiments that
determined how many carbon atoms were
present in exactly 12 grams of carbon)
5. The MOLE!!!
1 mole = 602214199000000000000000 molecules
A very big
number!
6. The MOLE!!!
Converting moles to number of particles:
Number of
moles
N = n X NA
Number of Avogadro’s
particles number
7. The MOLE!!!
N = n X NA
A sample contains 1.25 mol of NO2.
a) How many molecules are in the
sample?
b) How many atoms are in the
sample?
8. The MOLE!!!
N = n X NA
a) A sample contains 1.25 mol of NO2.
How many molecules are in the
sample?
N = n X NA
N = (1.25mol) X (6.02 x 1023 molecules/mol)
N = 7.52 x 1023 molecules
.: there are 7.52 x 1023 molecules in 1.25 mol of NO2
9. The MOLE!!!
N = n X NA
A sample contains 1.25 mol of NO2.
b) How many atoms are in the sample?
(7.52 x 1023 molecules) x (3 atoms/molecule)
= 2.26 x 1024 atoms
.: there are 2.26 x 1024 atoms in 1.25 mol of NO2
10. The MOLE!!!
Rearranging the formula…
n=N
NA
How many moles are present in a sample of CO2
made up of 5.83 x 1024 molecules?
11. The MOLE!!!
How many moles are present in a sample of CO2
made up of 5.83 x 1024 molecules?
n= N
NA
= (5.83 x 10 24
molecules CO2)
(6.02 x 1023 molecules/mol)
= 9.68 mol CO 2
.: there are 9.68 mol of CO2 in the sample
12. MOLAR MASS
M = molar mass
Molar mass of H =
1.0079 grams per mole
Molar mass of Li =
6.941 grams per mole
MNa = 22.990g/mol
13. MOLAR MASS
Molar mass of compounds
MBeO = 9.01g/mol + 16.00g/mol
= 25.01g/mol
MCO =
2
12.01g/mol + 2x16.00g/mol
= 44.01g/mol
14. MOLAR MASS
Number of
moles
m=nxM
mass Molar mass
m
n M
15. MOLAR MASS
A flask contains 0.750 mol of CO2. What mass of
CO2 is in this sample?
GIVEN: n = 0.750mol
M = 12.01g/mol + 2 x 16.00g/mol
= 44.01g/mol
m=?
m = nxM
= (0.750mol) x (44.01g/mol)
= 33.0g
.: the mass of CO2 is 33.0g
17. MOLAR MASS
How many moles of CH3COOH are in a 23.6g sample?
GIVEN: m = 23.6g
M = (2 x 12.01g/mol C) + (4 x 1.008g/mol H) + (2 x 16.00g/mol O)
= 60.06g
n=?
n= m
M
= (23.6g)/(60.06g/mol CH3COOH)
= 0.393mol CH3COOH
.: there are 0.393mol of CH3COOH in 23.6g of CH3COOH
19. PERCENTAGE COMPOSITION
Law of Definite Proportions:
The elements in a compound are always present
in the same proportions by mass
Example:
Water = 11.2% hydrogen, 88.8% oxygen
MH O = 18.016g/mol MH = 1.008g/mol
2
% of H in H2O = mass of H/mass of water
= (1.008g/mol x 2)/(18.016g/mol)
= 0.112 11.2%
20. PERCENTAGE COMPOSITION
A compound with a mass of 48.72g contains
32.69g of Zn & 16.03g of S. What is the percent
composition of the compound?
%Zn = 32.69g/48.72g
= 0.6710 67.10% .: the percentage
composition for Zn is
67.10% and the
%S = 16.03g/48.72g percentage composition
= 0.3290 32.90% for S is 32.90%
21. EMPIRICAL FORMULA
A compound is 81.9% carbon, 6.12% hydrogen, and
12.1% oxygen by mass. What is the empirical
formula of the compound?
Molecular formula = Actual formula of compound
Empirical formula = simplest formula (shows the
lowest number ratio of the elements)
Example: Benzene
Molecular formula = C6H6
Empirical formula = CH
22. EMPIRICAL FORMULA
A compound is 81.9% carbon, 6.12% hydrogen, and
12.1% oxygen by mass. What is the empirical
formula of the compound?
STEP1: Assume the sample is 100g
STEP2: Find the number of moles of each element
STEP3: Divide all answers from STEP2 by the LOWEST
answer from STEP2
23. EMPIRICAL FORMULA
A compound is 81.9% carbon, 6.12% hydrogen, and
12.1% oxygen by mass. What is the empirical
formula of the compound?
STEP1: Assume the sample is 100g
So...
C = 81.9g H = 6.12g O = 12.1g
24. EMPIRICAL FORMULA
A compound is 81.9% carbon, 6.12% hydrogen, and
12.1% oxygen by mass. What is the empirical
formula of the compound?
STEP2: Find the number of moles of each element
C = 81.9g/12.01g/mol H = 6.12g/1.008g/mol
= 6.819mol = 6.0714mol
O = 12.1g/16.00g/mol
= 0.75625mol SMALLEST ANSWER
25. EMPIRICAL FORMULA
A compound is 81.9% carbon, 6.12% hydrogen, and
12.1% oxygen by mass. What is the empirical
formula of the compound?
STEP3: Divide all answers by the smallest answer
C = 6.819/0.75625 H = 6.0714/0.75625
= 9.0168 = 8.028
O = 0.75625/0.75625
=1
26. EMPIRICAL FORMULA
A compound is 81.9% carbon, 6.12% hydrogen, and
12.1% oxygen by mass. What is the empirical
formula of the compound?
C = 6.819/0.75625 H = 6.0714/0.75625
= 9.0168 = 8.028
O = 0.75625/0.75625
=1
= C9H8O
27. MOLECULAR FORMULA
The empirical formula of ribose is CH2O. The molar
mass of this compound was determined to be
150g/mol. What is the molecular formula of ribose?
GIVEN: Empirical formula (1 C, 2 H, 1 O)
M = 150g/mol
STEP1: Determine the molar mass of the empirical
formula
STEP2: Divide the given molar mass by your answer
from STEP1
STEP3: Multiply your empirical formula by your
answer from STEP2
28. MOLECULAR FORMULA
The empirical formula of ribose is CH2O. The molar
mass of this compound was determined to be
150g/mol. What is the molecular formula of ribose?
STEP1: Determine the molar mass of the empirical
formula
12g/mol + 2 x 1.008g/mol + 16g/mol = 30g/mol
STEP2: Divide the given molar mass by your answer
from STEP1
150g/mol / 30g/mol = 5
29. MOLECULAR FORMULA
The empirical formula of ribose is CH2O. The molar
mass of this compound was determined to be
150g/mol. What is the molecular formula of ribose?
STEP3: Multiply your empirical formula by your answer
from STEP2
C1x5H2x5O1x5 = C5H10O5
31. MASS SPECTROMETER
1) Upload sample
2) Same is vapourized
3) Sample is ionized
4) Ions accelerated by electric field
5) Detection to mass-to-charge ratio based on details of motion
6) Ions assorted according to mass-to-charge ratio
32. CARBON-HYDROGEN
COMBUSTION ANALYZER
1)Weigh H2O absorber and CO2 absorber before
experiment
2)Sample is burned
3)Absorbers are weighed again
33. CARBON-HYDROGEN COMBUSTION ANALYZER
A 1.000g sample of pure compound, containing
only carbon and hydrogen, was combusted in a
carbon-hydrogen combustion analyzer. The
combustion produced 0.6919g of water and 3.338g
of carbon dioxide.
a)Calculate the masses of the hydrogen and the
carbon
b)Find the empirical formula of the compound.
34. CARBON-HYDROGEN COMBUSTION ANALYZER
A 1.000g sample…The combustion produced 0.6919g of
water and 3.338g of carbon dioxide.
a) Calculate the masses of the hydrogen and the carbon
Mass of H = 2.02g/mol H2 x 0.6919g H2O
always 18.02g/mol H2O
0.112 This gives you the percent
= 0.07756g H2 composition of hydrogen in
water
Mass of C = 12.01g/mol C x 3.338g CO2
always 44.01g/mol CO2
0.27289
= 0.9109g C
Therefore there was 0.0775g of H and 0.911g of C
35. CARBON-HYDROGEN COMBUSTION ANALYZER
A 1.000g sample…The combustion produced 0.6919g of
water and 3.338g of carbon dioxide.
b) Find the empirical formula of the compound
Moles of H = 0.07756g Moles of C = 0.9109g
1.008g/mol 12.01g/mol
= 0.07694mol = 0.07584mol
Empirical formula = C0.07584/0.07584H0.07694/0.07584
SMALLEST ANSWER
= C1.0H1.0
= CH
Therefore the empirical formula is CH
36. HYDRATED SALTS
A 50.0g sample of Ba(OH)2·XH2O contains
27.2g of Ba(OH)2.
a)Calculate the percent, by mass, of water in
the sample
b)Find the value of X
37. HYDRATED SALTS
A 50.0g sample of Ba(OH)2·XH2O contains 27.2g of Ba(OH)2.
a) Calculate the percent, by mass, of water in the sample
= (total mass of sample) – (mass of Ba(OH)2 in sample) x 100%
(total mass of sample)
= 50.0g – 27.2g x 100%
50.0g
= 45.6%
Therefore the percent by mass of water is 45.6%
38. HYDRATED SALTS
A 50.0g sample of Ba(OH)2·XH2O contains 27.2g of Ba(OH)2.
b) Find the value of X
nBa(OH)2 = m/M
= (27.2g)
(171.3g/mol)
= 0.159 mol Ba(OH)2 SMALLEST ANSWER
nH2O = m/M
= (50.0g – 27.2g)
(18.02g/mol) this is the molar mass of H2O
= 1.27mol H2O
0.159/0.159 mol Ba(OH)2·1.27/0.159 mol H2O
Ba(OH)2·8H2O
Therefore X is 8