SlideServe
Télécharger Obtenir Pro
Menu
Accueil Pro
Préférences
ApparenceLight Aide et commentaires
© SlideServe · v2026

Chapter 7 Chemical Quantities

elke avatar
elke
25 diapositives · 640 Views · Aug 09, 2014

Chapter 7 Chemical Quantities. 7.4 Percent Composition and Empirical Formulas. Percent Composition. In a compound, the percent composition is the percent by mass of each element in the formula. In one mole of CO 2 (molar mass 44.0 g), there are 12.0 g of C and 32.0 g of O.

#chapter 7 chemical quantities #chapter 7 chemical quantities ppt #powerpoint #presentation
Chapter 7 Chemical Quantities - slide 1
01 / 25Chapter 7 Chemical Quantities
1 / 17

Télécharger la présentation

Chapter 7 Chemical Quantities
Une image / un lien ci-dessous est fourni (tel quel) pour télécharger la présentation

Politique de téléchargement : le contenu du site vous est fourni TEL QUEL pour votre information et votre usage personnel ; il ne peut être vendu, concédé sous licence ni partagé sur d’autres sites sans l’accord de son auteur.Le contenu vous est fourni TEL QUEL, pour votre information et votre usage personnel uniquement. Téléchargez la présentation en cliquant sur ce lien. Lors du téléchargement, si vous ne parvenez pas à télécharger une présentation, l’éditeur a peut-être supprimé le fichier de son serveur.Pendant le téléchargement, si vous ne parvenez pas à obtenir une présentation, le fichier a peut-être été supprimé par l’éditeur.

F I N

Plus de contenu associé

Transcription de la présentation

  1. Chapter 7 Chemical Quantities 7.4 Percent Composition and Empirical Formulas

  2. Percent Composition • In a compound, the percent composition is the percent by mass of each element in the formula. • In one mole of CO2 (molar mass 44.0 g), there are 12.0 g of C and 32.0 g of O. 12.0 g C x 100 = 27.3 % C 44.0 g CO2 32.0 g O x 100 = 72.7 % O 44.0 g CO2100.0 %

  3. Learning Check What is the percent carbon in C5H8NNaO4 (MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? 1) 7.10 %C 2) 35.5 %C 3) 60.0 %C

  4. Solution 2) 35.5 %C Molar mass = 169.1 g % = total g C x 100 total g MSG = 60.0 g C x 100 = 35.5 % C 169.1 g MSG

  5. Types of Formulas • The molecular formula is the true or actual number of the atoms in a molecule. • The empirical formula is the simplest whole number ratio of the atoms. • The empirical formula is calculated by dividing the subscripts in the molecular formula by a whole number to give the lowest ratio.C5H10O5  5 = C1H2O1 = CH2Omolecular empirical formulaformula

  6. Some Molecular and Empirical Formulas

  7. Learning Check A. What is the empirical formula for C4H8? 1) C2H4 2) CH2 3) CH B. What is the empirical formula for C8H14? 1) C4H7 2) C6H12 3) C8H14 C. Which is a possible molecular formula for CH2O? 1) C4H4O4 2) C2H4O2 3) C3H6O3

  8. Solution A. What is the empirical formula for C4H8? 2) CH2C4H8  4 B. What is the empirical formula for C8H14? 1) C4H7C8H14 2 C. Which is a possible molecular formula for CH2O? 2) C2H4O2 3) C3H6O3

  9. Learning Check If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain. 1) SN 2) SN4 3) S4N4

  10. Solution If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain. 3) S4N4 If the molecular formula has 4 atoms of N, and S and N are related 1:1, then there must also be 4 atoms of S.

  11. Relating Empirical and Molecular Formulas • A molecular formula is equal to or a multiple of the empirical formula. • Thus, the molar mass is equal to or a multiple of the empirical mass. molar mass = a whole number empirical mass • Multiply the empirical formula by the whole numberto determine the molecular formula.

  12. Finding the Molecular Formula Determine the molecular formula of compound that has a molar mass of 78.0 and an empirical formula of CH. 1. Empirical mass of CH = 13.0 g 2. Divide the molar mass by the empirical mass. 3. 78.0 g = 6.00 13.0 g 4. Multiply the subscripts in CH by 6. 5. Molecular formula = (CH)6 = C6H6

  13. Learning Check A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula? 1) C3H4O3 2) C6H8O6 3) C9H12O9

  14. Solution A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula? 2) C6H8O6 C3H4O3 = 88.0 g/EF 176.0 g (molar mass) = 2.00 88.0 g (empirical mass) Molecular formula = 2 (empirical formula) (C3H4O3 )2 = C6H8O6

  15. Finding the Molecular Formula A compound is C 24.27%, H 4.07%, and Cl 71.65%. The molar mass is known to be 99.0 g. What are the empirical and molecular formulas? 1. Write the mass percents as the grams in a 100.00-g sample of the compound. C 24.27 g H 4.07 g Cl 71.65 g

  16. Finding the Molecular FormulaContinued 2. Calculate the number of moles of each element. 24.27 g C x 1 mole C = 2.02 moles C 12.0 g C 4.07 g H x 1 mole H = 4.03 moles H 1.01 g H 71.65 g Cl x 1 mole Cl = 2.02 moles Cl 35.5 g Cl

  17. Finding the Molecular Formula(continued) 3. Divide each by the smallest 2.02 moles C = 1 mole C 2.024.03 moles H = 2 moles H 2.022.02 moles Cl = 1 mole Cl 2.02 Empirical formula = C1H2Cl1 = CH2Cl

  18. Finding the Molecular Formula(continued) 4. Calculate empirical mass (EM)empirical mass CH2Cl = 49.5 g 5. Divide molar mass by empirical mass Molar mass = 99.0 g = 2Empirical mass 49.5 g 6. Determine Molecular formula(CH2Cl)2 = C2H4Cl2

  19. Learning Check Aspirin is 60.0% C, 4.5 % H and 35.5 % O. Calculate its empirical (simplest) formula.

  20. Solution (continued) In 100 g of aspirin, there are 60.0 g C, 4.5 g H, and 35.5 g O. 60.0 g C x 1 mole C = 5.00 moles C 12.0 g C 4.5 g H x 1 mole H= 4.5 moles H 1.01 g H 35.5 g O x 1mole O= 2.22 moles O 16.0 g O

  21. Solution (continued) Divide by the smallest number of moles. 5.00 moles C = 2.25 moles C 2.22 4.5 moles H = 2.00 moles H 2.22 2.22 moles O = 1.00 mole O 2.22 Note that the results are not all whole numbers. To obtain whole numbers, multiply by a factor to give whole numbers.

  22. Solution (continued) Multiply each number of moles by 4 C: 2.25 moles C x 4 = 9 moles C H: 2.0 moles H x 4 = 8 moles H O: 1.00 mole O x 4 = 4 moles O Use the whole numbers as subscripts to obtain the simplest formula C9H8O4

  23. Learning Check A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g, what is the molecular formula?

  24. Solution In 100.0 g, there are 27.4 g S, 12.0 g N and 60.6 g Cl. 27.4 g S x 1 mole S =0.854 mole S 32.1 g S 12.0 g N x 1 mole N= 0.857moles N 14.0 g N 60.6 g Cl x 1mole Cl = 1.71 moles Cl 35.5 g Cl

  25. Solution (continued) Dividing by the smallest number of moles 0.854 mole S /0.854 = 1.00 mole S 0.857 mole N/0.854 = 1.00 mole N 1.71 moles Cl/0.854 = 2.00 moles Cl Empirical formula = SNCl2 = 117.1 g Molar Mass/ Empirical mass 351 = 3117.1 Molecular formula = (SNCl2)3 = S3N3Cl6