1 / 14

Ch. 7 Review

Ch. 7 Review. Molar Mass Calculations. Question #1 (moles mass). What is the mass of 7.50 moles of sulfur dioxide, SO 2 ? S = 32.07 g/mol O = 16.00 g/mol. g SO 2. 7.50 mol SO 2. = g SO 2. X. 1 mol SO 2. SO 2. 32.07 g. 16.00 g. x 1. x 2. 32.07 g.

vivek
Download Presentation

Ch. 7 Review

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. Ch. 7 Review Molar Mass Calculations

  2. Question #1 (molesmass) What is the mass of 7.50 moles of sulfur dioxide, SO2? S = 32.07 g/mol O = 16.00 g/mol g SO2 7.50 mol SO2 = g SO2 X 1 mol SO2

  3. SO2 32.07 g 16.00 g x 1 x 2 32.07 g + 32.00 g = 64.07 g SO2 g SO2 7.50 mol SO2 = g SO2 X 1 mol SO2

  4. SO2 32.07 g 16.00 g x 1 x 2 32.07 g + 32.00 g = 64.07 g SO2 64.07 g SO2 7.50 mol SO2 = g SO2 480.525 481 X 1 mol SO2

  5. Question 2: (MassMoles) How many moles are there in 993.6 g of potassium sulfate K2SO4? K=39.10 g/mol; S=32.07 g/mol; O=16.00 g/mol 1 mol K2SO4 993.6 g K2SO4 = mol K2SO4 X g K2SO4

  6. K2SO4 39.10 g 32.07 g 16.00 g x 2 x 1 x 4 78.20 g + 32.07 g + 64.00 g = 174.27 g 1 mol K2SO4 993.6 g K2SO4 = mol K2SO4 5.701 X g K2SO4 174.27

  7. Question 3: (MolesMolecules) How many molecules are there in 4.55 moles of nitrogen (N2)? 1 mole = 6.02 x 1023molecules 6.02 x 1023 molecules N2 4.55 mol N2 =molecules N2 2.74 x 1024 X 1 mol N2

  8. Question 4: (Molesmoleculesatoms) How many atoms of nitrogen, N, are there in 2.18 moles of nitrogen (N2)? 1 mole = 6.02 x 1023molecules 6.02 x 1023 molecules N2 atoms N 2 2.18 mol N2 X X 1 mol N2 1 molecules N2 = atoms N 2.62 x 1024

  9. Question 5: (moleculesmoles) How many moles are there in 1.326 x 1012 molecules of carbon tetrachloride (CCl4)? 1 mole = 6.02 x 1023molecules 1 mol CCl4 1.326 x 1012molecules CCl4 X molecules CCl4 6.02 x 1023 = mol CCl4 2.202 x 10-12

  10. Question 6: (molesvolume) What is the volume occupied by 4.20 moles of oxygen gas (O2) at STP? 1 mole = 22.4 L for gas @STP L O2 22.4 4.20 mol O2 = L O2 94.08 X 1 mol O2

  11. Question 7: (volumemoles) How many moles are there in 0.335 dm3 of argon gas (Ar) at STP? Note: 1 L = 1 dm3 & 1 mole = 22.4 dm3 for gas @STP 1 mol Ar 0.335 dm3 Ar = mol Ar 0.0150 X 22.4 dm3 Ar

  12. Question 8: (Percent composition) N2O4 14.01 g 16.00 g x 2 x 4 28.02 g + 64.00 g = 92.02 g N2O4 28.02 g = %N X 100 = 30.4 % N 92.02 g 64.00 g = %O X 100 = 69.6 % O 92.02 g

  13. Question 9: (Percent composition) A sample of a compound that has a mass of 0.432 g is analyzed. The sample is found to be made up of oxygen and fluorine only. Given that the sample contains 0.128 g of oxygen, calculate the percentage composition of the compound. 0.128 g = %O X 100 = 29.6 % O 0.432 g = %F 100% - 29.6% = 70.4 % F Alternative method: 0.432 g – 0.128 g = 0.304 g F 0.304 g = %F X 100 = 70.4 % F 0.432 g

  14. Question 10: (Empirical formula) Find the empirical formula of a compound, given that the compound is found to be: 47.9% zinc (Zn) and 52.1% chlorine (Cl) by mass. 17 Cl 35.45 30 Zn 65.39 Assume 100g sample. 1 mol Zn x 47.9 g Zn = 0.733 mol Zn 65.39 g Zn = 1 0.733 1 mol Cl x 52.1 g Cl = 1.47 mol Cl 35.45 g Cl = 2 0.733 ZnCl2

More Related