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Can you determine the empirical formula for the following compounds and show work. 6....

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lkehoe | Valedictorian

Posted November 1, 2013 at 11:03 PM via web

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Can you determine the empirical formula for the following compounds and show work.

6.  A glaze used to produce iridescent effects on ceramics contains 48.8% cadmium, 20.8% carbon, 2.62% hydrogen, and 27.8% oxygen.  What is the formula of this compound?  What is this compound's name?

7.  Can you determine the molecular formula for the following compounds by showing your work:  It contains 92.3% carbon and 7.7% hydrogen with a molar mass of about 78 grams per mole.  a. Determine the empirical formula, b. Calculate the Molar Mass of the empirical formula, c. Divide the known (given) Molar Mass by the calculated empirical formula Molar Mass, d.  Multiply that amount through the subscripts of the empirical formula to obtain the molecular formula.

8.  Do the same as you showed for #7.  It contains 33.3% calcium, 40.0% oxygen, and 26.7% sulfur with a molar mass of about 120g/mol.

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llltkl | College Teacher | Valedictorian

Posted November 2, 2013 at 2:02 AM (Answer #1)

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I am to answer your first couple of questions and leave the last one for your practice.

6.

The compound contains 48.8% cadmium, 20.8% carbon, 2.62% hydrogen, and 27.8% oxygen. Overall, sum of the given data exceeds 100. Rounding off the experimental data to one digit after decimal, eliminates that element of complicacy of the given data.

Mass ratio of Cd:C:H:O = 48.8:20.8:2.6:27.8

Atom ratio of Cd:C:H:O = (48.8/112.41):(20.8/12):(2.6/1.008):(27.8/16) =0.434:1.733:2.579:1.737 `~~1:4:6:4` Empirical formula of the compound constituting the glaze is: `CdC_4H_6O_4` Assuming this to be the molecular formula too (ideally, molar mass determined through another independent experiment is required), the compound is, most probably, `Cd(CH_3CO_2)_2` i.e. Cadmium acetate.

7.

The compound contains 92.3% carbon and 7.7% hydrogen.

Mass ratio of C:H = 92.3:7.7

Atom ratio of C:H = (92.3/12):(7.7/1.008) = 7.69:7.64 `~~1:1` Therefore the empirical formula of the compound is CH. Mass of the empirical formula = (12+1.008) = 13.008 g/unit Molar mass of the compound is 78. `78/13.008~~6` Hence the molecular formulaof the given compound is `C_(6*1)H_(6*1)=C_6H_6`

Sources:

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Yojana_Thapa | Student , Grade 10 | eNoter

Posted January 25, 2014 at 6:42 PM (Answer #2)

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6.  A glaze used to produce iridescent effects on ceramics contains 48.8% cadmium, 20.8% carbon, 2.62% hydrogen, and 27.8% oxygen.  What is the formula of this compound?  What is this compound's name?
  • Cadmium: 48.8% 
        Atomic mass : 112

       # of moles: (48.8% / 112)= 0.435

        Simplest ratio of moles: (0.435/0.435) = 1

  • Carbon: 20.8%

       Atomic mass : 12

       # of moles: (20.8% / 12)= 1.733

       Simplest ratio of moles: (1.733/0.435) = 4

  • Hydrogen: 2.62%

      Atomic mass : 1

      # of moles: (2.62%/ 1) = 2.62

      Simplest ratio of moles: (2.62/0.435)= 6

  • Oxygen: 27.8%

     Atomic Mass: 16

     # of Moles: (27.8%/16)= 1.737

     Simplest ration of moles : (1.737/0.435)= 4

Empirical Formula = CdC4H6O4

7.  It contains 92.3% carbon and 7.7% hydrogen with a molar mass of about 78 grams per mole.  

     a. Determine the empirical formula

     b. Calculate the Molar Mass of the empirical formula

     c. Divide the known (given) Molar Mass by the calculated empirical formula Molar Mass,

     d.  Multiply that amount through the subscripts of the empirical formula to obtain the molecular formula.

  • Carbon: 92.3%

     Atomic mass : 12

     # of moles: (92.3%/ 12) = 8

     Simplest ratio of moles: (8/8)= 1

  • Hydrogen: 7.7%

    Atomic mass : 1

    # of moles: (7.7%/ 1) = 8

    Simplest ratio of moles: (8/8) = 1

Empirical Formula = CH

- Molecular formula = Molar mass/ EF Mass= Factor

- Molar mass= 78g/moL

- EF mass= (12 + 1)= 13

- 78/13= 6

- 6 is the factor

(CH)6 = C6H6

Molecular Formula = C6H6

8.  Do the same as you showed for #7.  It contains 33.3% calcium, 40.0% oxygen, and 26.7% sulfur with a molar mass of about 120g/mol.

  • Calcium: 33.3%

     Atomic mass : 40.08

     # of moles: (33.3/40.08) = 0.828

     Simplest ratio of moles: (0.828/0.828) = 1

  • Oxygen: 40.0%

     Atomic mass : 16

     # of moles: (40%/ 16)= 2.5

     Simplest ratio of moles: (2.5/0.828)= 3

  • Sulfur: 26.7%

    Atomic mass : 32

    # of moles: (26.7/32)= 0.834

    Simplest ratio of moles: (0.834/0.828)= 1

Empirical Formula= CO3S

Molecular formula = Molar mass/ EF Mass= Factor

- Molar mass= 120g/moL

- EF mass= (12)+ (3 x 16) + (32)= 92

- 120/92= 1

- 1 is the factor

(CO3S)1 = CO3S

Molecular Formula = CO3S

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parama9000 | Student , Grade 11 | Valedictorian

Posted January 30, 2014 at 7:30 AM (Answer #3)

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6.CdC4H6O4

7.C6H6

8.CO3S

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