Class 11 sbcc par VIII
•Download as PPTX, PDF•
1 like•27 views
This document discusses empirical formulas, molecular formulas, and how to calculate them given different types of chemical composition data. It provides step-by-step explanations and examples of calculating empirical formulas from percentage composition or mass data, and calculating molecular formulas given empirical formulas and molar or molecular masses. It concludes with some practice problems for the reader to try.
Report
Share
Report
Share
Recommended
Percent comp_empirical formula_molecular formula
This document provides information and examples for calculating percent composition, empirical formulas, and molecular formulas of compounds. It defines key terms like percent composition and empirical formula. It then works through examples of calculating the percent composition of magnesium and oxygen that form a compound, the percent composition and mass of carbon in propane, and determining empirical formulas from elemental percentages or mole ratios. The document explains how to calculate molecular formulas from empirical formulas and molar masses. Finally, it provides practice problems for the reader to work through.
Empirical and molecular formulas
A simplified approach to empirical and molecular formulas is presented here for high school chemistry and preliminary college level chemistry courses.
Chemistry Chapter 3
The document provides information about atoms and their structure. It defines key terms like protons, neutrons, electrons, nucleus and isotopes. It explains that the number of protons determines the element and distinguishes one atom from another. The mole is also defined as 6.02x10^23 particles and is used to measure amounts of substances on a macroscopic scale. Formulas are given to calculate molar mass and empirical formulas.
Empirical and molecula formula
This document provides steps for calculating empirical and molecular formulas from percent composition data and molar mass. It presents an example calculation showing that a compound that is 43.7% P and 56.3% O with a molar mass of 283.88 g/mol has an empirical formula of P2O5 and a molecular formula of P4O10. Several practice problems are provided for students to determine empirical and molecular formulas.
Chapter 7.4 : Determining Chemical Formulas
This document discusses determining chemical formulas, including:
1) Defining empirical and molecular formulas, and how to calculate empirical formulas from percentage or mass composition data.
2) Explaining how to determine a molecular formula from an empirical formula using the relationship between molecular formula mass and empirical formula mass.
3) Providing examples of calculating empirical formulas from composition data and determining molecular formulas from empirical formulas and molar masses.
Empirical and molecular formula class 11
The document discusses empirical and molecular formulas. An empirical formula shows the simplest whole number ratio of atoms in a compound, while a molecular formula shows the exact number of each atom. To calculate molecular formula from empirical formula: 1) make a table with elements, percentages, atomic masses, moles, and simplest ratios; 2) the empirical formula mass is the sum of the atoms' masses; 3) divide the molar mass by the empirical formula mass to get the common factor for the molecular formula. Two examples are given to calculate empirical and molecular formulas from percentage compositions and molar masses.
Chapter 7.3 : Using Chemical Formulas
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.
Empirical, molecular formulas & % Composition
The document discusses methods for calculating percent composition and molecular formulas from empirical data. It provides examples of calculating percent composition from mass percentages of elements in compounds and from experimental analysis of sample masses. It also demonstrates calculating empirical formulas from percent compositions or measured element masses and determining molecular formulas from empirical formulas and molar masses.
Recommended
Percent comp_empirical formula_molecular formula
This document provides information and examples for calculating percent composition, empirical formulas, and molecular formulas of compounds. It defines key terms like percent composition and empirical formula. It then works through examples of calculating the percent composition of magnesium and oxygen that form a compound, the percent composition and mass of carbon in propane, and determining empirical formulas from elemental percentages or mole ratios. The document explains how to calculate molecular formulas from empirical formulas and molar masses. Finally, it provides practice problems for the reader to work through.
Empirical and molecular formulas
A simplified approach to empirical and molecular formulas is presented here for high school chemistry and preliminary college level chemistry courses.
Chemistry Chapter 3
The document provides information about atoms and their structure. It defines key terms like protons, neutrons, electrons, nucleus and isotopes. It explains that the number of protons determines the element and distinguishes one atom from another. The mole is also defined as 6.02x10^23 particles and is used to measure amounts of substances on a macroscopic scale. Formulas are given to calculate molar mass and empirical formulas.
Empirical and molecula formula
This document provides steps for calculating empirical and molecular formulas from percent composition data and molar mass. It presents an example calculation showing that a compound that is 43.7% P and 56.3% O with a molar mass of 283.88 g/mol has an empirical formula of P2O5 and a molecular formula of P4O10. Several practice problems are provided for students to determine empirical and molecular formulas.
Chapter 7.4 : Determining Chemical Formulas
This document discusses determining chemical formulas, including:
1) Defining empirical and molecular formulas, and how to calculate empirical formulas from percentage or mass composition data.
2) Explaining how to determine a molecular formula from an empirical formula using the relationship between molecular formula mass and empirical formula mass.
3) Providing examples of calculating empirical formulas from composition data and determining molecular formulas from empirical formulas and molar masses.
Empirical and molecular formula class 11
The document discusses empirical and molecular formulas. An empirical formula shows the simplest whole number ratio of atoms in a compound, while a molecular formula shows the exact number of each atom. To calculate molecular formula from empirical formula: 1) make a table with elements, percentages, atomic masses, moles, and simplest ratios; 2) the empirical formula mass is the sum of the atoms' masses; 3) divide the molar mass by the empirical formula mass to get the common factor for the molecular formula. Two examples are given to calculate empirical and molecular formulas from percentage compositions and molar masses.
Chapter 7.3 : Using Chemical Formulas
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.
Empirical, molecular formulas & % Composition
The document discusses methods for calculating percent composition and molecular formulas from empirical data. It provides examples of calculating percent composition from mass percentages of elements in compounds and from experimental analysis of sample masses. It also demonstrates calculating empirical formulas from percent compositions or measured element masses and determining molecular formulas from empirical formulas and molar masses.
2011 topic 01 lecture 2 - empirical and molecular formulae
The document discusses empirical and molecular formulas. It provides examples to show how to determine the empirical formula from percent composition data and how to then determine the molecular formula from the empirical formula and molar mass. Specifically, it shows how to calculate the empirical formula of adipic acid is C3H5O2 from its percent composition by mass. It then shows that if the molar mass of adipic acid is 146 g/mol, and the molar mass of C3H5O2 is 73.08 g/mol, then the molecular formula of adipic acid must be C6H10O4.
Empirical formulas
This document discusses empirical formulas and how to determine them from percentage composition or experimental data. It provides examples of calculating empirical formulas from the mass percentages of elements in a compound or the grams of elements in a sample. It also explains how to determine the molecular formula of a compound from the empirical formula and experimental molar mass. Key steps include converting percentages to grams of elements, calculating moles of each element, and dividing by the smallest ratio of elements to give whole number ratios in the empirical formula. The molecular formula is found by dividing the experimental molar mass by the molar mass of the empirical formula.
Molecular formula
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.
Percent Composition, Empirical and Molecular Formula
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.
Percent composition
1. The document discusses calculating percent composition and determining empirical and molecular formulas.
2. To calculate percent composition, the formula mass is determined and the mass of each element is calculated as a percentage of the total mass.
3. An empirical formula shows the lowest whole number ratio of atoms in a compound, while a molecular formula shows the actual number of atoms. Molecular formulas for ionic compounds are always empirical, while molecular formulas for molecular compounds may or may not be empirical.
Honors1011 molar mass and percent composition
- The law of definite proportions states that a chemical compound always contains the same elements in the same proportions by mass, regardless of sample size. This allows determination of a compound's mass and elemental percentages.
- Molar mass is the mass of one mole of a substance and can be calculated by adding the masses of each element in a compound based on its chemical formula. Common units are grams per mole (g/mol).
- Percent composition by mass can be determined by calculating the mass of each element in a compound and expressing it as a percentage of the total mass of the compound.
Formula mass powerpoint
The document discusses concepts related to counting molecules and determining formula mass and molar mass. It provides examples of calculating formula mass for various compounds by identifying the elements in the chemical formula, looking up the atomic mass of each element, multiplying by the subscript if present, and summing the atomic masses. It also discusses hydrates, how to name them, and how to calculate their formula mass by including the mass of water molecules.
Percentage composition exercise
The document contains 6 chemistry problems asking for empirical and molecular formulas of compounds given their percentage compositions or products of combustion. Problem 1 asks for the empirical formula of a compound containing 65.5% carbon, 5.5% hydrogen, and 29.0% oxygen. Problem 2 asks for the molecular formula if the molar mass is 110 g/mol. Problem 3 asks for the empirical formula of a compound containing 18.7% lithium, 16.3% carbon, and 65.0% oxygen. Problem 4 asks for the molecular formula if the molar mass is 73.8 g/mol. Problems 5 and 6 similarly ask for the empirical and molecular formulas of acetic acid and aniline given combustion products and m
percent composition of compounds
Percent composition is calculated by finding the molar mass of a compound using the periodic table, determining the mass contributed by the component of interest, and dividing that mass by the total molar mass of the compound before multiplying by 100. This allows you to determine the percentage of a compound's total mass made up of a particular component by mass. The document provides an example calculating the percent composition of carbon in carbon dioxide (CO2).
Empirical and Molecular
The document contains examples and explanations of concepts related to chemistry calculations including:
- Percentage composition problems calculating the percent by mass of elements in compounds.
- Empirical formula problems finding the lowest whole number ratio of atoms in a compound from percentage or mole composition data.
- Molecular formula problems relating the empirical formula to molar mass to determine the actual formula.
- Hydrate problems using the formula for hydrated compounds to determine the amount of water bonded in the crystal structure from percentage or mass composition data.
11 24 What Is Molar Mass
This document contains a chemistry lesson on molar mass. It defines molar mass as the sum of the atomic masses in a molecule. It provides examples of calculating the molar mass of different compounds by adding up the atomic masses of each element. The lesson concludes by having students practice calculating molar masses in groups and answering an exit slip with questions.
3.3 chem myp mole formula
This document discusses the mole concept in chemistry. It defines key terms like mole, molar mass, relative atomic mass, and Avogadro's number. The main points are:
- A mole is the unit used to measure the number of elementary particles in a substance and represents 6.022x10^23 particles.
- The molar mass of a substance is the mass in grams of 1 mole of that substance. It can be used to convert between moles and mass.
- Formulas are given to calculate moles, mass, or volume of a substance using molar mass and moles as conversion factors between units.
- Examples show how to use these formulas and factor-
The mole concept
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.
Regular_AvogadroMole
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.
The Mole 9.5
The document provides examples for calculating percentage composition by mass of elements in compounds. It gives the formula for finding percentage composition and worked examples calculating the percentage of elements like nitrogen and hydrogen in ammonia, hydrogen and oxygen in hydrogen peroxide, and water in copper(II) sulfate and iron(III) chloride hexahydrate.
Chapter 10: The Mole
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.
Chapter 10 The Mole
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.
The Mole 9.6
The document discusses how to determine the formula of a compound through experimentation. It provides an example of finding the formula of magnesium oxide. Key steps include measuring the masses of reactants and products, calculating moles of each substance, and determining the ratio of elements in the compound based on the mole ratios. The formula of magnesium oxide in this example is MgO.
Percent composition
Percent composition is the percent by mass of each element in a compound, calculated by taking the mass of an element divided by the total mass of the compound and multiplying by 100. To calculate percent composition, you write the compound formula, do an atom inventory and find the molar mass, then use the formula: % composition of element = mass of element x 100% / mass of compound. For compounds with multiple elements, separate calculations must be done for each element. As an example, the percent composition of hydrogen in water is 11.1% and oxygen is 88.9%, which should total very closely to 100%.
Chapter 10 (2)
The document discusses key concepts related to moles, including:
1) A mole is defined as 6.02x1023 representative particles of a substance, which can be used to convert between the number of particles and moles.
2) The mass of one mole of a substance is its molar mass, which can be used to convert between mass and moles.
3) One mole of an ideal gas occupies a volume of 22.4L at STP, which can be used to convert between moles and volume.
4) Percent composition by mass and molar mass can be used to determine empirical and molecular formulas of compounds.
C05 the mole concept
This chapter discusses the mole concept, including defining the mole, deriving empirical and molecular formulas, stating Avogadro's Law, and applying the mole concept to ionic and molecular equations. It introduces the mole as the amount of substance containing 6x1023 particles. It provides examples of how to determine the empirical formula, molecular formula, and formula of a compound from composition data. It also discusses molar volume of gases and limiting reactants. Worked examples are included for many of these concepts.
Empirical_and_Molecular_Formulas.ppt
This document discusses empirical and molecular formulas. It provides examples of calculating molar mass and percentage composition of compounds. The key differences between empirical and molecular formulas are explained, with empirical formulas showing the lowest whole number ratio of atoms in a compound and molecular formulas showing the actual number of each atom. The document demonstrates how to determine the empirical formula of a compound given its elemental percentage composition. It also explains how the molecular formula can be derived from the empirical formula and molar mass.
More Related Content
What's hot
2011 topic 01 lecture 2 - empirical and molecular formulae
The document discusses empirical and molecular formulas. It provides examples to show how to determine the empirical formula from percent composition data and how to then determine the molecular formula from the empirical formula and molar mass. Specifically, it shows how to calculate the empirical formula of adipic acid is C3H5O2 from its percent composition by mass. It then shows that if the molar mass of adipic acid is 146 g/mol, and the molar mass of C3H5O2 is 73.08 g/mol, then the molecular formula of adipic acid must be C6H10O4.
Empirical formulas
This document discusses empirical formulas and how to determine them from percentage composition or experimental data. It provides examples of calculating empirical formulas from the mass percentages of elements in a compound or the grams of elements in a sample. It also explains how to determine the molecular formula of a compound from the empirical formula and experimental molar mass. Key steps include converting percentages to grams of elements, calculating moles of each element, and dividing by the smallest ratio of elements to give whole number ratios in the empirical formula. The molecular formula is found by dividing the experimental molar mass by the molar mass of the empirical formula.
Molecular formula
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.
Percent Composition, Empirical and Molecular Formula
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.
Percent composition
1. The document discusses calculating percent composition and determining empirical and molecular formulas.
2. To calculate percent composition, the formula mass is determined and the mass of each element is calculated as a percentage of the total mass.
3. An empirical formula shows the lowest whole number ratio of atoms in a compound, while a molecular formula shows the actual number of atoms. Molecular formulas for ionic compounds are always empirical, while molecular formulas for molecular compounds may or may not be empirical.
Honors1011 molar mass and percent composition
- The law of definite proportions states that a chemical compound always contains the same elements in the same proportions by mass, regardless of sample size. This allows determination of a compound's mass and elemental percentages.
- Molar mass is the mass of one mole of a substance and can be calculated by adding the masses of each element in a compound based on its chemical formula. Common units are grams per mole (g/mol).
- Percent composition by mass can be determined by calculating the mass of each element in a compound and expressing it as a percentage of the total mass of the compound.
Formula mass powerpoint
The document discusses concepts related to counting molecules and determining formula mass and molar mass. It provides examples of calculating formula mass for various compounds by identifying the elements in the chemical formula, looking up the atomic mass of each element, multiplying by the subscript if present, and summing the atomic masses. It also discusses hydrates, how to name them, and how to calculate their formula mass by including the mass of water molecules.
Percentage composition exercise
The document contains 6 chemistry problems asking for empirical and molecular formulas of compounds given their percentage compositions or products of combustion. Problem 1 asks for the empirical formula of a compound containing 65.5% carbon, 5.5% hydrogen, and 29.0% oxygen. Problem 2 asks for the molecular formula if the molar mass is 110 g/mol. Problem 3 asks for the empirical formula of a compound containing 18.7% lithium, 16.3% carbon, and 65.0% oxygen. Problem 4 asks for the molecular formula if the molar mass is 73.8 g/mol. Problems 5 and 6 similarly ask for the empirical and molecular formulas of acetic acid and aniline given combustion products and m
percent composition of compounds
Percent composition is calculated by finding the molar mass of a compound using the periodic table, determining the mass contributed by the component of interest, and dividing that mass by the total molar mass of the compound before multiplying by 100. This allows you to determine the percentage of a compound's total mass made up of a particular component by mass. The document provides an example calculating the percent composition of carbon in carbon dioxide (CO2).
Empirical and Molecular
The document contains examples and explanations of concepts related to chemistry calculations including:
- Percentage composition problems calculating the percent by mass of elements in compounds.
- Empirical formula problems finding the lowest whole number ratio of atoms in a compound from percentage or mole composition data.
- Molecular formula problems relating the empirical formula to molar mass to determine the actual formula.
- Hydrate problems using the formula for hydrated compounds to determine the amount of water bonded in the crystal structure from percentage or mass composition data.
11 24 What Is Molar Mass
This document contains a chemistry lesson on molar mass. It defines molar mass as the sum of the atomic masses in a molecule. It provides examples of calculating the molar mass of different compounds by adding up the atomic masses of each element. The lesson concludes by having students practice calculating molar masses in groups and answering an exit slip with questions.
3.3 chem myp mole formula
This document discusses the mole concept in chemistry. It defines key terms like mole, molar mass, relative atomic mass, and Avogadro's number. The main points are:
- A mole is the unit used to measure the number of elementary particles in a substance and represents 6.022x10^23 particles.
- The molar mass of a substance is the mass in grams of 1 mole of that substance. It can be used to convert between moles and mass.
- Formulas are given to calculate moles, mass, or volume of a substance using molar mass and moles as conversion factors between units.
- Examples show how to use these formulas and factor-
The mole concept
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.
Regular_AvogadroMole
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.
The Mole 9.5
The document provides examples for calculating percentage composition by mass of elements in compounds. It gives the formula for finding percentage composition and worked examples calculating the percentage of elements like nitrogen and hydrogen in ammonia, hydrogen and oxygen in hydrogen peroxide, and water in copper(II) sulfate and iron(III) chloride hexahydrate.
Chapter 10: The Mole
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.
Chapter 10 The Mole
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.
The Mole 9.6
The document discusses how to determine the formula of a compound through experimentation. It provides an example of finding the formula of magnesium oxide. Key steps include measuring the masses of reactants and products, calculating moles of each substance, and determining the ratio of elements in the compound based on the mole ratios. The formula of magnesium oxide in this example is MgO.
Percent composition
Percent composition is the percent by mass of each element in a compound, calculated by taking the mass of an element divided by the total mass of the compound and multiplying by 100. To calculate percent composition, you write the compound formula, do an atom inventory and find the molar mass, then use the formula: % composition of element = mass of element x 100% / mass of compound. For compounds with multiple elements, separate calculations must be done for each element. As an example, the percent composition of hydrogen in water is 11.1% and oxygen is 88.9%, which should total very closely to 100%.
Chapter 10 (2)
The document discusses key concepts related to moles, including:
1) A mole is defined as 6.02x1023 representative particles of a substance, which can be used to convert between the number of particles and moles.
2) The mass of one mole of a substance is its molar mass, which can be used to convert between mass and moles.
3) One mole of an ideal gas occupies a volume of 22.4L at STP, which can be used to convert between moles and volume.
4) Percent composition by mass and molar mass can be used to determine empirical and molecular formulas of compounds.
What's hot (20)
2011 topic 01 lecture 2 - empirical and molecular formulae
2011 topic 01 lecture 2 - empirical and molecular formulae
Percent Composition, Empirical and Molecular Formula
Percent Composition, Empirical and Molecular Formula
Similar to Class 11 sbcc par VIII
C05 the mole concept
This chapter discusses the mole concept, including defining the mole, deriving empirical and molecular formulas, stating Avogadro's Law, and applying the mole concept to ionic and molecular equations. It introduces the mole as the amount of substance containing 6x1023 particles. It provides examples of how to determine the empirical formula, molecular formula, and formula of a compound from composition data. It also discusses molar volume of gases and limiting reactants. Worked examples are included for many of these concepts.
Empirical_and_Molecular_Formulas.ppt
This document discusses empirical and molecular formulas. It provides examples of calculating molar mass and percentage composition of compounds. The key differences between empirical and molecular formulas are explained, with empirical formulas showing the lowest whole number ratio of atoms in a compound and molecular formulas showing the actual number of each atom. The document demonstrates how to determine the empirical formula of a compound given its elemental percentage composition. It also explains how the molecular formula can be derived from the empirical formula and molar mass.
PERCENTAGE COMPOSITION, EMPIRICAL AND MOLECULAR FORMULAS.ppt
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
Stoichiometry
The document discusses different types of stoichiometry calculations encountered in AS Organic Chemistry, including:
1) Percentage yield calculations from chemical reactions.
2) Determining empirical formulas from elemental analysis data involving masses, percentages, or combustion analysis.
3) Calculating molecular formulas using empirical formulas and experimentally determined molar masses.
14 the mole!!!
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.
C05 the mole concept
This document discusses the mole concept in chemistry. It defines the mole as the amount of substance containing 6.02x1023 particles. A mole of any substance has a mass in grams equal to its molar mass. The document explains how to determine empirical and molecular formulas from percentage composition data using mole calculations. It also discusses limiting reactants and using moles to calculate gas volumes based on Avogadro's Law. Several examples are provided to demonstrate determining formulas from mass or molar mass data.
Quantitative aspects of chemical change
The document discusses quantitative aspects of chemical changes including empirical and molecular formulas, molar mass calculations, percentage composition, and stoichiometry. It provides steps for calculating empirical formulas from elemental analysis data, determining molecular formulas from empirical formulas and molar masses, and using molar ratios to solve stoichiometric problems. An example problem demonstrates determining the empirical and molecular formulas for a chlorinated hydrocarbon from its elemental composition and molar mass.
Ch 10 PPT...The Mole.pptx
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 Mole Moncept
This document provides information about the mole concept in chemistry. It defines the mole as the amount of substance that contains as many elementary entities as there are atoms in 12 grams of carbon-12. It then discusses how to calculate the number of moles, mass of one mole of atoms or molecules, and molar mass. The document also explains how to derive empirical and molecular formulas from percentage composition data and relative molecular masses. It provides examples of calculating limiting reactants and the molar volume of gases.
Chapter 10 - Chemical Quantities
- The mole is a unit used to measure very large numbers of small particles like atoms or molecules. One mole equals 6.02 x 10^23 particles.
- The mass of one mole of an element is called its gram atomic mass (gam). The mass of one mole of a compound is called its gram molecular mass (gmm) or gram formula mass (gfm).
- At standard temperature and pressure, one mole of any gas occupies a volume of 22.4 L. This volume is known as the molar volume.
C05 the mole concept
This document provides an overview of key concepts related to the mole concept in chemistry. It defines the mole as the number of atoms or molecules in 1 gram of hydrogen or 12 grams of carbon. The mole concept allows chemists to relate mass, number of particles, and volume of gases. It discusses how to calculate empirical and molecular formulas, Avogadro's constant, molar mass, limiting reactants, and other mole-related calculations and applications. Worked examples are provided to demonstrate how to use the mole concept to find formulas of compounds from percentage composition data and other information.
Chapter 1-mole concept.ppt
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
Resource Empirical formula
1) The document outlines a 5-week unit on chemical bonds with a focus on molecular and empirical formulas over 3 sessions.
2) It provides learning objectives, references, and learning tasks including examples to determine percentage composition and calculate empirical and molecular formulas given percentage or mass data.
3) An example problem calculates the empirical formula NO2 and molecular formula N2O4 for a compound containing nitrogen and oxygen.
Percentcompempiricalformulamolecularformula 140213120449-phpapp01
An 8.20 g piece of magnesium combines completely with 5.40 g of oxygen to form a compound. The percentages of magnesium and oxygen in the compound are calculated to be 60.3% and 39.7% respectively, totaling 100%. Another example calculates the percentages of carbon (81.8%) and hydrogen (18.2%) in propane. The document then provides instructions and examples for calculating empirical formulas based on the percentages of elements in compounds.
14 the mole!!!
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.
Ch 11 notes complete
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.
Ch 11 notes complete
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.
Stoichiometry1.ppt
The document provides information about the mole concept in chemistry. Some key points:
- A mole is equal to Avogadro's number of particles of a substance, which is 6.022x1023 particles.
- One mole of any substance has a mass in grams equal to its formula mass or molar mass.
- Stoichiometry problems use the mole concept to calculate amounts of substances in chemical reactions based on molar ratios and molar masses.
- Examples are provided for calculating moles, masses, empirical formulas, and volumes of gases from given amounts of reactants.
Ch3 z53 stoich
This document discusses atomic mass and isotopes. It begins by explaining that an atomic mass unit (amu) is used to discuss the mass of atoms, where 1 amu is 1/12 the mass of a carbon-12 atom. Atomic masses listed in the periodic table are in amu. Isotopes have different atomic masses that result in the average atomic mass not being a whole number. Examples are provided to demonstrate calculating average atomic masses from the masses and abundances of isotopes.
8 More On The Mole!
The document discusses empirical formulas, stoichiometry, percentage yield calculations, and identifying limiting reagents. It provides examples of calculating empirical formulas from elemental composition percentages. It also gives examples of finding the limiting reagent, calculating theoretical yield, and determining percentage yield in chemical reactions.
Similar to Class 11 sbcc par VIII (20)
PERCENTAGE COMPOSITION, EMPIRICAL AND MOLECULAR FORMULAS.ppt
PERCENTAGE COMPOSITION, EMPIRICAL AND MOLECULAR FORMULAS.ppt
Percentcompempiricalformulamolecularformula 140213120449-phpapp01
Percentcompempiricalformulamolecularformula 140213120449-phpapp01
More from Aarti Soni
Crop production and management
This document provides information on various agricultural practices. It discusses the seven main practices of crop production: soil preparation through ploughing and levelling, sowing seeds, adding manure and fertilizers, irrigation, weed removal, harvesting, and storage. It also describes traditional and modern irrigation methods, the importance of weeding, and the differences between organic manure and chemical fertilizers. The overall document serves as a guide to the basic steps and considerations in crop cultivation.
Microorganisms Friends and Foe (Part 4)
The document discusses various methods for preserving food, including chemical, heat/cold treatments, and storage/packing. It notes that microorganisms can spoil food and make us ill if consumed, so preservation is important. Common preservation methods mentioned are canning, refrigeration, and drying. The document also contains information about the nitrogen cycle, noting that atmospheric nitrogen must be fixed before use by plants and animals through processes like lightning, rhizobium bacteria, and blue-green algae.
Microorganisms Friends and Foe Part 3
Viruses are microscopic organisms that can only reproduce inside host cells. They come in various shapes and sizes and can infect animals, plants, and bacteria. Useful microbes are used in food/beverage production, medicine, vaccine production, and environmental cleanup. Harmful microbes can cause communicable diseases in humans and plants via air, water, food, or contact. Antibiotics are drugs that kill germs and cure infections, while vaccines protect against diseases by triggering antibody production without causing illness. Proper vaccination prevents diseases like cholera, hepatitis, smallpox, and tuberculosis.
Microorganisms Friends And Foe Part 1
This document summarizes key information about microorganisms. It discusses the discovery of microorganisms in the 1600s by Hooke and van Leeuwenhoek. Microorganisms are classified into four main types: bacteria, fungi, algae, and protozoa. Bacteria are further described as unicellular microbes that can exist alone or in colonies, are found in diverse habitats, and have a variety of shapes. They lack a nucleus, have ribosomes, and some respire aerobically while others anaerobically. Bacteria play important roles in industries like dairy, wine, and medicine, and can break down waste, clean oil spills, and mine minerals.
CBSE-Class 11 - Chemistry -Chapter 14-Environmental Chemistry
This document discusses various causes and effects of soil pollution. It explains that pesticides can pollute soil through various processes like chemical degradation, microbial action, and photodegradation. Industrial waste, oil spills, acid rain, nuclear waste, and improper waste disposal can also degrade soil quality. Polluted soil can negatively impact human health, plant growth, soil structure, fertility, and biodiversity through biomagnification of toxic chemicals up the food chain. The document provides strategies like regulated hazardous waste disposal to minimize polluted areas and switching to non-polluting fuels to help control soil pollution.
CBSE-Class 11 - Chemistry -Chapter 14-Environmental Chemistry
The document discusses various types of environmental pollution including atmospheric and water pollution. It contains sections on particulate matter (PM) in the atmosphere, sources of water pollution, and biological oxygen demand (BOD) and chemical oxygen demand (COD) in water. The sections cover topics such as the size and composition of PM, common air and water pollutants, how BOD and COD are used to measure organic pollution in water, and standards for drinking water.
CBSE-Class 11 - Chemistry -Chapter 14-Environmental Chemistry
Environmental chemistry is the study of chemical phenomena that occur in natural environments and how human activity affects these phenomena. It involves understanding natural chemical concentrations and effects, in order to study human impacts. Major air pollutants include carbon monoxide, sulfur oxides, nitrogen oxides, and particulates produced from fossil fuel combustion. These primary pollutants can form secondary pollutants like ground-level ozone and acid rain through chemical reactions in the atmosphere. Air pollution is linked to various health issues and is a leading cause of noncommunicable diseases worldwide.
Ch 2 - Is matter around us Pure?
This chapter discusses solutions, their properties, and concentration. It describes various methods to separate mixtures, including evaporation, magnetic separation, centrifugation, separating funnel, sublimation, chromatography, distillation, and crystallization. The chapter also explains the steps to obtain components from blue/black ink using distillation and crystallization. Finally, it discusses physical and chemical changes, comparing the differences between physical and chemical changes.
Class 10 chemical reactions and equations
This document appears to be from a science textbook chapter that introduces several key chemistry concepts. It covers topics like thermal, electrolytic, and photolytic reactions; displacement reactions; combination and decomposition reactions; exothermic and endothermic reactions; redox reactions; and corrosion and rancidity. It also includes sample experiments, important questions, and contact information for the author.
Class 11 sbcc part IX
This document discusses stoichiometry and chemical reactions. It defines stoichiometry as the calculation of reactants and products in chemical reactions based on the law of conservation of mass. It explains how to identify the limiting reagent, which is the first reagent to be completely used up in a chemical reaction. Excess reagents remain after the limiting reagent is used up. The document provides examples of how to use balanced chemical equations and mole ratios to perform stoichiometric calculations converting between moles, mass, and particles of reactants and products.
Class 11 sbcc part XI
This document discusses stoichiometry calculations involving molarity and molality. It provides examples of calculating molarity, molality, mass percent, and mole fraction for various solutions. Molarity is moles per liter of solution, while molality is moles per kilogram of solvent. Examples include calculating the molality of NaCl in water, the molarity of HCl in a solution, and the molality, molarity, and mass of various solutes needed to make solutions of specific concentrations.
Class 11 sbcc part X
This document discusses various concepts related to stoichiometry including limiting reagents, concentration of solutions, mass percent, mole fraction, molarity, and molality. It provides examples of calculating the limiting reagent, mass percent, and mole fraction in chemical reactions and solutions. Key steps include calculating the moles of each reactant, determining which has the lower moles and is therefore limiting, and using mole ratios from a balanced equation to calculate grams of products and excess reactants.
Class 11 sbcc part VII
This document discusses empirical and molecular formulas. It provides examples to explain the process of determining empirical formulas from percentage composition data. The key steps are: 1) calculating grams of each element, 2) converting to moles, 3) dividing by the smallest number of moles to get ratios, 4) multiplying ratios to get near-whole numbers, 5) rounding to whole numbers, and 6) writing the empirical formula. A simple example finds the empirical formula of a compound containing 54.55% C, 9.09% H, and 36.36% O to be C2H4O. The document also discusses molecular formulas and how they differ from empirical formulas.
Class 11 sbcc part VI
The document discusses the mole concept in chemistry. It defines the mole as the amount of a substance that contains as many elementary entities (atoms, molecules, ions, etc.) as there are atoms in exactly 12 grams of carbon-12. This number is known as Avogadro's number and is equal to 6.022×1023. The mole allows chemists to convert between the mass of a sample and the number of particles it contains. It provides a link between macroscopic and microscopic scales in chemistry. The mole is a central unit for quantitative chemical calculations and experiments.
Chemistry Class 11 sbcc part V
Here are the answers:
1) Calculate the molecular mass of glucose, C6H12O6:
- 6 carbon atoms x 12.011 amu/atom = 72.066 amu
- 12 hydrogen atoms x 1.008 amu/atom = 12.096 amu
- 6 oxygen atoms x 16.00 amu/atom = 96.00 amu
- Total molecular mass = 72.066 amu + 12.096 amu + 96.00 amu = 180.162 amu
2) Find the relative formula mass of calcium phosphate, Ca3(PO4)2:
- 3 calcium atoms x 40.078 amu/atom = 120.234 am
Chemistry Class 11 sbcc part IV
John Dalton developed the first modern atomic theory in the early 1800s. He proposed that each chemical element is composed of unique atoms that cannot be created or destroyed through chemical reactions, but can combine to form compounds. Dalton hypothesized that differences in the masses of gas particles explained why water absorbed gases like carbon dioxide and nitrogen to different extents. His atomic theory marked the first truly scientific explanation of atoms. Dalton's concepts of atoms and molecules helped establish chemistry as a rigorous science.
Class 11 sbcc part III
This document compares Dalton's Atomic Theory from 1808 to the Modern Atomic Theory. Dalton's theory proposed that matter is made of indivisible atoms and that atoms of a given element are all the same, but different elements have different atom types. However, Dalton's theory could not explain gas volumes or atomic bonding. Later experiments and scientists expanded on atomic structure and properties, establishing the Modern Atomic Theory which accounts for atomic nuclei, isotopes, radioactive decay, and nuclear reactions.
Class 11 sbcc part II
This document discusses several key concepts in chemistry:
- The law of multiple proportions, which states that elements combine in fixed ratios of small whole numbers. For example, carbon and oxygen form two oxides with a mass ratio of 2:1.
- Gay-Lussac's law of definite proportions by volume, which states that compounds combine in fixed volume ratios. This law was explained by Avogadro's hypothesis relating equal volumes and numbers of molecules.
- Avogadro's law from 1811, which states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. This allowed gases to be used to determine relative molecular masses.
More from Aarti Soni (20)
CBSE-Class 11 - Chemistry -Chapter 14-Environmental Chemistry
CBSE-Class 11 - Chemistry -Chapter 14-Environmental Chemistry
CBSE-Class 11 - Chemistry -Chapter 14-Environmental Chemistry
CBSE-Class 11 - Chemistry -Chapter 14-Environmental Chemistry
CBSE-Class 11 - Chemistry -Chapter 14-Environmental Chemistry
CBSE-Class 11 - Chemistry -Chapter 14-Environmental Chemistry
Recently uploaded
The History of Stoke Newington Street Names
Presented at the Stoke Newington Literary Festival on 9th June 2024
www.StokeNewingtonHistory.com
How to Make a Field Mandatory in Odoo 17
In Odoo, making a field required can be done through both Python code and XML views. When you set the required attribute to True in Python code, it makes the field required across all views where it's used. Conversely, when you set the required attribute in XML views, it makes the field required only in the context of that particular view.
Your Skill Boost Masterclass: Strategies for Effective Upskilling
Your Skill Boost Masterclass: Strategies for Effective UpskillingExcellence Foundation for South Sudan
Strategies for Effective Upskilling is a presentation by Chinwendu Peace in a Your Skill Boost Masterclass organisation by the Excellence Foundation for South Sudan on 08th and 09th June 2024 from 1 PM to 3 PM on each day.বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
বাংলাদেশের অর্থনৈতিক সমীক্ষা ২০২৪ [Bangladesh Economic Review 2024 Bangla.pdf] কম্পিউটার , ট্যাব ও স্মার্ট ফোন ভার্সন সহ সম্পূর্ণ বাংলা ই-বুক বা pdf বই " সুচিপত্র ...বুকমার্ক মেনু 🔖 ও হাইপার লিংক মেনু 📝👆 যুক্ত ..
আমাদের সবার জন্য খুব খুব গুরুত্বপূর্ণ একটি বই ..বিসিএস, ব্যাংক, ইউনিভার্সিটি ভর্তি ও যে কোন প্রতিযোগিতা মূলক পরীক্ষার জন্য এর খুব ইম্পরট্যান্ট একটি বিষয় ...তাছাড়া বাংলাদেশের সাম্প্রতিক যে কোন ডাটা বা তথ্য এই বইতে পাবেন ...
তাই একজন নাগরিক হিসাবে এই তথ্য গুলো আপনার জানা প্রয়োজন ...।
বিসিএস ও ব্যাংক এর লিখিত পরীক্ষা ...+এছাড়া মাধ্যমিক ও উচ্চমাধ্যমিকের স্টুডেন্টদের জন্য অনেক কাজে আসবে ...
clinical examination of hip joint (1).pdf
described clinical examination all orthopeadic conditions .
Bed Making ( Introduction, Purpose, Types, Articles, Scientific principles, N...
Topic : Bed making
Subject : Nursing Foundation
Advanced Java[Extra Concepts, Not Difficult].docx
This is part 2 of my Java Learning Journey. This contains Hashing, ArrayList, LinkedList, Date and Time Classes, Calendar Class and more.
LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
This Dissertation explores the particular circumstances of Mirzapur, a region located in the
core of India. Mirzapur, with its varied terrains and abundant biodiversity, offers an optimal
environment for investigating the changes in vegetation cover dynamics. Our study utilizes
advanced technologies such as GIS (Geographic Information Systems) and Remote sensing to
analyze the transformations that have taken place over the course of a decade.
The complex relationship between human activities and the environment has been the focus
of extensive research and worry. As the global community grapples with swift urbanization,
population expansion, and economic progress, the effects on natural ecosystems are becoming
more evident. A crucial element of this impact is the alteration of vegetation cover, which plays a
significant role in maintaining the ecological equilibrium of our planet.Land serves as the foundation for all human activities and provides the necessary materials for
these activities. As the most crucial natural resource, its utilization by humans results in different
'Land uses,' which are determined by both human activities and the physical characteristics of the
land.
The utilization of land is impacted by human needs and environmental factors. In countries
like India, rapid population growth and the emphasis on extensive resource exploitation can lead
to significant land degradation, adversely affecting the region's land cover.
Therefore, human intervention has significantly influenced land use patterns over many
centuries, evolving its structure over time and space. In the present era, these changes have
accelerated due to factors such as agriculture and urbanization. Information regarding land use and
cover is essential for various planning and management tasks related to the Earth's surface,
providing crucial environmental data for scientific, resource management, policy purposes, and
diverse human activities.
Accurate understanding of land use and cover is imperative for the development planning
of any area. Consequently, a wide range of professionals, including earth system scientists, land
and water managers, and urban planners, are interested in obtaining data on land use and cover
changes, conversion trends, and other related patterns. The spatial dimensions of land use and
cover support policymakers and scientists in making well-informed decisions, as alterations in
these patterns indicate shifts in economic and social conditions. Monitoring such changes with the
help of Advanced technologies like Remote Sensing and Geographic Information Systems is
crucial for coordinated efforts across different administrative levels. Advanced technologies like
Remote Sensing and Geographic Information Systems
9
Changes in vegetation cover refer to variations in the distribution, composition, and overall
structure of plant communities across different temporal and spatial scales. These changes can
occur natural.
How to Setup Warehouse & Location in Odoo 17 Inventory
In this slide, we'll explore how to set up warehouses and locations in Odoo 17 Inventory. This will help us manage our stock effectively, track inventory levels, and streamline warehouse operations.
Solutons Maths Escape Room Spatial .pptx
Solutions of Puzzles of Mathematics Escape Room Game in Spatial.io
Philippine Edukasyong Pantahanan at Pangkabuhayan (EPP) Curriculum
(𝐓𝐋𝐄 𝟏𝟎𝟎) (𝐋𝐞𝐬𝐬𝐨𝐧 𝟏)-𝐏𝐫𝐞𝐥𝐢𝐦𝐬
𝐃𝐢𝐬𝐜𝐮𝐬𝐬 𝐭𝐡𝐞 𝐄𝐏𝐏 𝐂𝐮𝐫𝐫𝐢𝐜𝐮𝐥𝐮𝐦 𝐢𝐧 𝐭𝐡𝐞 𝐏𝐡𝐢𝐥𝐢𝐩𝐩𝐢𝐧𝐞𝐬:
- Understand the goals and objectives of the Edukasyong Pantahanan at Pangkabuhayan (EPP) curriculum, recognizing its importance in fostering practical life skills and values among students. Students will also be able to identify the key components and subjects covered, such as agriculture, home economics, industrial arts, and information and communication technology.
𝐄𝐱𝐩𝐥𝐚𝐢𝐧 𝐭𝐡𝐞 𝐍𝐚𝐭𝐮𝐫𝐞 𝐚𝐧𝐝 𝐒𝐜𝐨𝐩𝐞 𝐨𝐟 𝐚𝐧 𝐄𝐧𝐭𝐫𝐞𝐩𝐫𝐞𝐧𝐞𝐮𝐫:
-Define entrepreneurship, distinguishing it from general business activities by emphasizing its focus on innovation, risk-taking, and value creation. Students will describe the characteristics and traits of successful entrepreneurs, including their roles and responsibilities, and discuss the broader economic and social impacts of entrepreneurial activities on both local and global scales.
RHEOLOGY Physical pharmaceutics-II notes for B.pharm 4th sem students
Physical pharmaceutics notes for B.pharm students
Hindi varnamala | hindi alphabet PPT.pdf
हिंदी वर्णमाला पीपीटी, hindi alphabet PPT presentation, hindi varnamala PPT, Hindi Varnamala pdf, हिंदी स्वर, हिंदी व्यंजन, sikhiye hindi varnmala, dr. mulla adam ali, hindi language and literature, hindi alphabet with drawing, hindi alphabet pdf, hindi varnamala for childrens, hindi language, hindi varnamala practice for kids, https://www.drmullaadamali.com
Traditional Musical Instruments of Arunachal Pradesh and Uttar Pradesh - RAYH...
Traditional Musical Instruments of Arunachal Pradesh and Uttar Pradesh
Recently uploaded (20)
NEWSPAPERS - QUESTION 1 - REVISION POWERPOINT.pptx
NEWSPAPERS - QUESTION 1 - REVISION POWERPOINT.pptx
Your Skill Boost Masterclass: Strategies for Effective Upskilling
Your Skill Boost Masterclass: Strategies for Effective Upskilling
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
বাংলাদেশ অর্থনৈতিক সমীক্ষা (Economic Review) ২০২৪ UJS App.pdf
Bed Making ( Introduction, Purpose, Types, Articles, Scientific principles, N...
Bed Making ( Introduction, Purpose, Types, Articles, Scientific principles, N...
LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
How to Setup Warehouse & Location in Odoo 17 Inventory
How to Setup Warehouse & Location in Odoo 17 Inventory
Philippine Edukasyong Pantahanan at Pangkabuhayan (EPP) Curriculum
Philippine Edukasyong Pantahanan at Pangkabuhayan (EPP) Curriculum
RHEOLOGY Physical pharmaceutics-II notes for B.pharm 4th sem students
RHEOLOGY Physical pharmaceutics-II notes for B.pharm 4th sem students
Traditional Musical Instruments of Arunachal Pradesh and Uttar Pradesh - RAYH...
Traditional Musical Instruments of Arunachal Pradesh and Uttar Pradesh - RAYH...
Class 11 sbcc par VIII
- 3. Empirical Formula Molecular formula
- 4. In the previous part we have already seen about Percentage composition , empirical formula and molecular formula definition and steps involved in it. In this part 1) We will learn more simple and hard solved problems. 2) We will see how to use given data and solve the problem? 3) At the end some practice problems you need to try.
- 6. What is the molecular formula for a compound with the empirical formula HO and molar mass of 34 g ? Step 1 : Find the total mass of the empirical formula given to you 1g + 16 g = 17 g Step 2 : Divide the molar mass from the problem by the mass of the empirical formula 34 g / 17 g = 2 Step 3 : Multiply each subscript in the EF by the number we just found in step 2 Apply the formula MF = (EF)n = (OH)2 = H2O2
- 7. What is the empirical formula of a compound which is 25.90% nitrogen and 74.10% oxygen?
- 8. A compound was found to contain 13.5 g Ca, 10.8 g O, and 0.675 g H. What is the empirical formula of the compound?
- 10. The compound has EF weight as C+2H+O, vapor density is 30. Find the Molecular formula? Given = 12 + 2 X 1 + 16 = 30 g
- 11. Molar mass of the compound is 42.08g, EF is CH2 , What is molecular formula? Given: Molar mass = 42.08g EF is CH2 To find = MF 14.03 g
- 13. Molar mass of the compound is 120.12g, EF is CH2O , What is molecular formula? Given: Molar mass = 120.12g EF is CH2O To find = MF 30 g
- 14. The compound has 6.9899g C,1.1724g H and 9.309g O and Molecular mass is 90.079. Find the EF and MF.
- 15. What is the molecular formula of a compound which has empirical formula of CH4N and molecular mass of 60.12g? Step 1: Determine the formula of the empirical formula 1 C is 12.01 g/mol = 12.01 g 4 H is 1.01 g/mol = 4.04 g 1 N is 14.01 g/mol = 14.01 g Total = 30.06 g/mol Step 2: Determine how many times the molecular mass grater than the empirical formula mass 60.12 g/30.06 g = 2.000 Step 3: Multiply the empirical formula by the answer of step 2 to get the molecular formula (CH4N) x 2 = C2H8N2
- 16. The compound has 13.5g Ca, 10.8g O and 0.675g H. Find the empirical formula. Given: Ca = 13.5g, O = 10.8g, H = 0.675g To find: EF So the empirical formula of a compound is CaO2H2
- 18. The empirical formula for ethylene is CH2. Find the MF if the molecular mass is 28.1g/mol. Given: EF = CH2 Mass = 28.1g/mol To find: MF
- 19. A sample of a compound containing boron B and hydrogen H contains 6.444 g of B and 1.803 g of H. The molar mass of the compound is about 30 g. What is its molecular formula? What are the empirical formula and empirical mass for C3H6O3 ? Find the empirical formula of the following: 1) C6H6 2) C4H8 3) H2O 4) H4C4O8 5) P4O10 6) (NH4)2CO3 7) Mg(NO3)2