Simplest004
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1) The document describes the simplest formula calculation for combustion of hydrocarbons in excess oxygen. It involves calculating the grams of carbon, hydrogen, and oxygen from the combustion products CO2 and H2O, then converting to moles and determining the ratios of each element. 2) The calculation shows combustion of 12.795g of hydrocarbon produces 32.689g of CO2 and 13.377g of H2O. From this the grams of C, H, and O are calculated. 3) Converting the grams of each element to moles and dividing the mole amounts by the smallest value gives molar ratios of 5:10:1 for C:H:O, yielding the simplest formula
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Worked-Problems-on-Stoichiometry.pptx
This document provides worked examples and explanations of key concepts in stoichiometry, including balancing chemical equations, calculating molar mass, determining percentage composition of compounds, interconverting between masses and moles of substances, and using balanced chemical equations to determine quantitative relationships between reactants and products. It discusses using the mole concept and Avogadro's number to relate numbers of particles and masses. Examples are provided to illustrate limiting reactants in chemical reactions.
Stoichiometry
This document discusses stoichiometry and the mole concept. It explains how to use molar masses and balanced chemical equations to calculate reactant and product masses. It provides examples of calculating masses, volumes, and percentages for various chemical reactions. Exercises at the end provide practice problems involving calculations using mole ratios and balanced equations.
Chem Unit5
The document discusses various chemistry concepts including:
1. Avogadro's number defines the number of particles in one mole of a substance as 6.02x1023.
2. Molar mass is the mass of one mole of a pure substance and is used to convert between grams and moles.
3. Empirical and molecular formulas can be determined from percent composition data using mole ratios and molar mass.
4. Chemical equations represent chemical reactions and must be balanced to satisfy the law of conservation of mass.
Ch3 stoichiometry
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
Chapter_3_Mass_Relationships.ppt
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.
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
molecular-formula.ppt
This document provides instructions for calculating empirical and molecular formulas from elemental composition data. It includes examples of determining the percent composition, calculating moles of elements, obtaining the simplest ratio of elements, and deriving the empirical formula. For molecular formulas, it describes dividing the molar mass by the mass of the empirical formula to obtain a factor to multiply the subscripts. Several practice problems are worked through with the full steps shown.
Ch 10 PPT...The Mole.pptx
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- 1. Simplest Formula Prepared by: Stephen Boylan October 20, 2009
- 2. Simplest Formula . . .
- 3. Simplest Formula - Simple001 A. Combustion in Excess oxygen 12.795 g. HC ----> 32.689 g. CO2 + 13.377 g. H2O B. 5 Step Calculation C. The Simplest Formula . Cx Hy Oz
- 4. Molar Mass CO2 Element Number Atomic Mass Formula n Units, amu Mass, C 1 12.01 12.01 O 2 16.00 32.00 44.01grams per mole CO2
- 7. Element Number Atomic Mass Formula
- 15. 2.0 Calculate the grams of carbon, hydrogen and oxygen.
- 17. 12.795g. HC --> 32.689 g CO2 + 13.377 g. H2O
- 18. C:
- 19. 32.689 grams CO2 12.01 grams C = 8.921 g. C
- 21. H:
- 22. 13.377 grams H2O 2x1.008 grams H = 1.497 g H
- 24. O:
- 25. 12.795 g HC - 8.921g C - 1.497 g H= 2.377 g O
- 28. C: 8.921 g. C 1 mole C = 0.7429 moles C
- 31. H: 1.497 g. H 1 mole H = 1.485 mole H
- 35. O: 2.377 g. O 1 mole O = 0.1485 mole O
- 38. C: 0.7429 moles C 1 = 5.003 ~ 5 moles C
- 40. H: 1.485 mole H 1 = 10 moles H
- 42. O: 0.1485 mole O 1 = 1 moles O