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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.

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Ch. 7 Review

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  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

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