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#64 Using your answers from Problem 63, calculate the number of grams of these elements.

#64 Using your answers from Problem 63, calculate the number of grams of these elements. a. sulfur in 3.54 g H 2 S. % comp. S is: 3 2.1 x 100 = 94.1% 34.1. Therefore the number of grams of sulfur is: .941 x 3.54 = 3.33 g Sulfur.

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#64 Using your answers from Problem 63, calculate the number of grams of these elements.

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  1. #64 Using your answers from Problem 63, calculate the number of grams of these elements. a. sulfur in 3.54 g H2S % comp. S is: 32.1 x 100 = 94.1% 34.1 Therefore the number of grams of sulfur is: .941 x 3.54 = 3.33 g Sulfur B. nitrogen in 25.0 g (NH4)2C2O4 % comp. N is: 28.0 x 100 = 22.6% 124.0 Therefore the number of grams of nitrogen is: .226 x 25.0 = 5.65 g N

  2. #64 Using your answers from Problem 63, calculate the number of grams of these elements. c. magnesium in 97.4 g Mg(OH)2 % comp. Mg is: 24.3 x 100 = 41.7% 82.6 Therefore the number of grams of magnesium is: .417 x 97.4 = 40.6 g Magnesium d. phosphorus in 804 g Na3PO4 % comp. P is: 31.0 x 100 = 18.9% 164.0 Therefore the number of grams of nitrogen is: .189 x 804.0 = 152 g P

  3. #65 Which of the following compounds has the highest iron content? a. FeCl2 % composition Fe is: 55.8 x 100 = 44.0% 126.8 % composition Fe is: 55.8 x 100 = 32.3% 55.8+(12x6)+9+96 b. Fe(C2H3O2)3 % composition Fe is: 55.8 x 100 = 62.1% 89.8 c. Fe(OH)2 % composition Fe is: 55.8 x 100 = 77.7% 71.8 d. FeO

  4. #66 You find that 7.36 g of a compound has decomposed to give 6.93 g of oxygen. The only other element in the compound is hydrogen. If the molar mass of the compound is 34.0 g/mole, what is its molecular formula? 1. Go from grams to an % composition first: 6.93 x 100 = 94.2 % Oxygen 7.36 7.36-6.93 x 100 = 5.8 % Hydrogen 7.36 continued next slide

  5. 94.2 % Oxygen 5.8 % Hydrogen 2. Assume a 100g sample gives 94.2 g O 5.8 g H 3. Convert to moles: x 1 mole = 5.88 moles 16.0 g O x 1 mole = 5.88 moles 1.0 g H continued next slide

  6. 4. divide by the smaller of the two numbers: (in this case the numbers are the same: 5.88 moles O/5.88 = 1 5.88 moles H/ 5.88 = 1 The empirical formula (lowest whole number ratio) is HO or a 1 to 1 ratio 5. Since the problem states the molar mass is 34.0 you have to multiply the ratios until you get 34.1 grams for a molecular formula: continued next slide

  7. HO is 17.0 g/mole H2O2 is 34.0 g/mole which matches our problem Therefore the molecular formula is H2O2

  8. #67 Which of the following can be classified as an empirical formula? a. S2Cl2 NO because you can reduce this ratio to S1Cl1 Therefore it is not the smallest whole number ratio. b. C6H10O4 NO because you can reduce this ratio to C3H5O2 Therefore it is not the smallest whole number ratio. continued next slide

  9. #67 Which of the following can be classified as an empirical formula? c. Na2SO4 YES because you cannot reduce down any further Ionic compounds are already given in their empirical formula.

  10. #68. What is the molecular formula for each compound? Each compound’s empirical formula and molar mass is given. a. CH2O 90 g/mole to get this you just multiply the molar mass of the given formula by small whole numbers until you get the answer: CH2O is 30g/mole. Therefore this x 3 gives the 90 g/ mole. The molecular formula is: C3H6O3

  11. #68. What is the molecular formula for each compound? Each compound’s empirical formula and molar mass is given. b. HgCl 472.2 g/mole this is tricky because it appears to be an ionic compound but it is molecular therefore the molar mass of the given empirical formula is: 236.1g/mole multiplying by 2 given 472.2 therefore the molecular formula is Hg2Cl2

  12. #69 Determine the molecular formula for each compound. a. 94.1 % O 5.9 % H molar mass: 34g 1. Assume a 100 g sample: 94.1g O 5.9g H continued next slide

  13. 2. convert to moles: 94.1g O x 1 mole = 5.88 16.0 g O 5.9g H x 1 mole = 5.9 1.0 g H 3. divide by the smaller number to get a ratio: 5.88/5.88 = 1 5.9/5.88 = 1 So we see a ratio of 1 to 1 thus the empirical formula is HO continued next slide

  14. The molecular formula is a multiple of HO which has a molar mass of 17.0 The original problem stated that the molecule has a molar mass of 34 g Therefore we can see that the empir- ical formula times 2 gives: 17.0 x 2 + 34.0 g The formula is: H2O2

  15. #69b Determine the molecular formula for each compound. b. 50.7 % C 4.2 % H 45.1 % O molar mass: 142g The book is wrong 1. Assume a 100 g sample: 50.7g C 4.2g H 45.1 g O continued next slide

  16. 2. convert to moles: 50.7g C x 1 mole = 4.22 12.0 g O 4.2g H x 1 mole = 4.2 1.0 g H 45.1g O x 1 mole = 2.84 16.0 g H continued next slide

  17. 3. divide by the smaller number to get a ratio: 4.22/2.84 = 1.48 4.2/2.84 = 1.47 2.84/2.84 = 1 x 2 = 3 x 2 = 3 x 2 = 2 These numbers are almost right, but they need to be whole numbers so multipy all by 2. Rounding you get C3H3O2 continued next slide

  18. 4. Finally, taking the empirical formula and multiplying the subscipts until I get the molar mass of the original problem (142g/mole) you get: C3H3O2 is molar mass of: 71g/mole times 2 is C6H6O4 which has a molar mass of: 142 which is the answer!

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