1 / 53

BELLWORK – 1/29/19

BELLWORK – 1/29/19. What information does a chemical formula show?. Using Chemical Formulas. As you have seen, a chemical formula indicates the elements as well as the relative number of atoms or ions of each element present in a compound.

stimmons
Download Presentation

BELLWORK – 1/29/19

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. BELLWORK – 1/29/19 • What information does a chemical formula show?

  2. Using Chemical Formulas • As you have seen, a chemical formula indicates the elements as well as the relative number of atoms or ions of each element present in a compound. • Chemical formulas also allow chemists to calculate a number of characteristic values for a given compound.

  3. Calculating Formula Mass • The formula mass of any molecule, formula unit, or ion is sum of the average atomic masses of all atoms represented in its formula. • Ex: Find the formula mass of KClO3 1K atom = 39.1 amu 1 Cl atom = 35.5 amu3 3 O atoms = 48.0 atoms Formula Mass = 122.6 amu

  4. Molecular Mass • Molecular mass is a number equal to the sum of the atomic masses of the atoms in a molecule. • Ex: Find the molecular mass of CH4 • 1 atom of carbon = 12.0 amu • 4 atoms of hydrogen = 4.0 amu Molecular Mass = 16.0 amu

  5. The Mole • Before we can calculate molar mass, we must review what the mole is • A mole contains 6.02 x 1023 particles (atoms, ions, molecules, formula unit) • The number 6.02 x 1023 is known as Avogadro’s number. • One mole of any element contains Avogadro’s number of atoms. 1 mole Na = 6.02 x 1023 Na atoms 1 mole Au = 6.02 x 1023 Au atoms

  6. VIDEO – The Mole

  7. A Mole of Molecules • One mole of a covalent compound contains Avogadro’s number of molecules. 1 mole CO2 = 6.02 x 1023CO2 molecules 1 mole H2O = 6.02 x 1023H2O molecules • One mole of an ionic compound contains Avogadro’s number of formula units. 1 mole NaCl = 6.02 x 1023 NaCl formula units

  8. Samples of One Mole Quantities

  9. Avogadro’s Number • Avogadro’s number is written as conversion factors. 6.02 x 1023 particles and 1 mole 1 mole 6.02 x 1023 particles • The number of molecules in 0.50 mole of CO2 molecules is calculated as 0.50 mole CO2 molecules x 6.02 x 1023 CO2 molecules 1 mole CO2 molecules = 3.0 x 1023 CO2 molecules

  10. Learning Check A. Calculate the number of atoms in 2.0 moles of Al. B. Calculate the number of moles of S in 1.8 x 1024 S.

  11. Solution A. Calculate the number of atoms in 2.0 moles of Al.2.0 moles Al x 6.02 x 1023 Al atoms 1 mole Al =1.2 x 1024 Al atoms B. Calculate the number of moles of S in 1.8 x 1024 S. 1.8 x 1024 S atoms x 1 mole S 6.02 x 1023 S atoms = 3.0 mole S atoms

  12. Calculating Molar Mass- same method as Formula Mass, with different units on your answer. • The mass of one mole is called molar mass (g/mole). • The molar mass of an element is the atomic mass expressed in grams.

  13. One-Mole Quantities 32.1 g 55.9 g 58.5 g 294.2 g 342.3 g

  14. Molar Mass of K3PO4 Determine the molar mass of K3PO4 to 0.1 g. Molar Mass = [3(39.1) + 31.0 + 4(16)] = 212.3g/mol

  15. Learning Check Give the molar mass to the nearest 0.1 g. A. 1 mole of K atoms = ________ B. 1 mole of Sn atoms = ________

  16. Solution Give the molar mass to the nearest 0.1 g. A. 1 mole of K atoms = 39.1 g B. 1 mole of Sn atoms = 118.7 g

  17. Learning Check A. 1 mole of K2O = ______g B. 1 mole of antacid Al(OH)3 = ______g

  18. Solution A. 1 mole of K2O 2 (39.1) + 1 (16.0) = 94.2 g/mol 1mole K2O x 94.2g/mol = 94.2g B. 1 mole of antacid Al(OH)3 1 (27.0) + 3 (16.0) + 3 (1.0) = 78.0 g/mol 1mole Al(OH)3 x 78.0 g/mol = 78.0g

  19. Learning Check Prozac, C17H18F3NO, is an antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac?

  20. Solution Prozac, C17H18F3NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac? 17C (12.0) + 18H (1.0) + 3F (19.0) + 1N (14.0) + 1 O (16.0) = 204 + 18 + 57.0 + 14.0 + 16.0 = 309 g/mole

  21. SUMMARY • Formula Mass = atomic mass of ionic compounds in amu • Molecular Mass = atomic mass of molecular (covalent) compounds in amu. • Molar Mass = atomic mass of compounds in grams/mol

  22. Percent Composition • In a compound, the percent composition is the percent by mass of each element in the formula. • In one mole of CO2 there are 12.0 g of C and 32.0 g of O (molar mass 44.0 g/mol), 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 %

  23. 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

  24. 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

  25. CLASSWORK

  26. Calculations using Molar Mass • Mole factors are used to convert between the grams of a substance and the number of moles. Mole factor Grams Moles

  27. Molar Mass Factors Methane CH4 known as natural gas is used in gas cook tops and gas heaters. 1 mole CH4 = 16.0 g The molar mass of methane can be written as conversion factors. 16.0 g CH4 and 1 mole CH4 1 mole CH4 16.0 g CH4

  28. Learning Check Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass conversion factors for acetic acid.

  29. Solution Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass factors for acetic acid. 2(12.0) + 4(1.0) + 2(16) = 60.0g/mol 1 mole of acetic acid = 60.0 g acetic acid 1 mole acetic acid and 60.0 g acetic acid 60.0 g acetic acid 1 mole acetic acid

  30. Calculating Grams from Moles Aluminum is often used for the structure of lightweight bicycle frames. How many grams of Al are in 3.00 moles of Al? 3.00 moles Al x 27.0 g Al = 81.0 g Al 1 mole Al mole factor for Al

  31. Learning Check The artificial sweetener aspartame (Nutri-Sweet) C14H18N2O5 is used to sweeten diet foods, coffee and soft drinks. How many moles of aspartame are present in 225 g of aspartame?

  32. Solution Calculate the molar mass of C14H18N2O5. 14 (12.0) + 18 (1.0) + 2 (14.0) + 5(16.0) = 294 g/mole Set up the calculation using a mole factor. 225 g aspartame x 1 mole aspartame 294 g aspartame mole factor(inverted) = 0.765 mole aspartame

  33. 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

  34. Some Molecular and Empirical Formulas

  35. 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

  36. 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

  37. 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

  38. 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.

  39. 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.

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

  41. Learning Check A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula?

  42. Solution A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula? C3H4O3 = 3(12.0) + 4(1.0) + 3(16.0) = 88.0 g/mol 176.0 g/mol (molar mass) = 2.00 88.0 g/mol (empirical mass) Molecular formula = 2 (empirical formula) (C3H4O3 )2 = C6H8O6

  43. 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

  44. 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

  45. 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

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

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

  48. 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

  49. Solution (continued) Divide by the smallest number of moles. 5.00 moles C = 2.25 moles C 2.22 4.5 moles H = 2.0 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.

  50. 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

More Related