Question

Please help me!!!30 pts

Answer 1

Answer:

Question 7 to 12 are given in attached file because character limit is only 5000

Explanation:

1.  A 15.0 gram sample of a compound is found to contain 8.83 grams of sodium and 6.17 grams of  sulfur. Calculate the empirical formula of this compound.

Given data:

Mass of sample = 15 g

Mass of sodium = 8.83 g

Mass of sulfur = 6.17 g

Empirical formula = ?

Solution:

Number of gram atoms of Na = 8.83 / 23 = 0.4

Number of gram atoms of S = 6.17 / 32 = 0.2

Atomic ratio:

            Na               :               S          

           0.4/0.2         :            0.2/0.2  

            2                  :               1        

Na : S  = 2 :  1

Empirical formula is Na₂S.

2. Analysis of a 10.150 gram sample of a compound known to contain only phosphorus and oxygen  indicates a phosphorus content of 4.433 grams. What is the empirical formula of this compound?

Given data:

Mass of phosphorus = 4.433 g

Mass of oxygen = 10.150 g - 4.433 g = 5.717 g

Empirical formula = ?

Solution:

Number of gram atoms of P = 4.433 / 30.9738 = 0.1431

Number of gram atoms of O = 5.717/ 15.999 = 0.3573

Atomic ratio:

            P                        :               O          

        0.1431/0.1431         :            0.3573/0.1431

            1                         :                  2.5        

P : O  = 2(1 : 2.5)

Empirical formula is P₂O₅.

3. A compound is found to contain 36.48%Na, 25.41% S, and 38.11% O. Find its empirical formula.

Given data:

Percentage of sodium = 36.48%

Percentage of sulfur = 25.41%

Percentage of oxygen = 38.11%

Empirical formula = ?

Solution:

Number of gram atoms of Na = 36.48 / 23 = 1.6

Number of gram atoms of S = 25.41/ 32 = 0.8

Number of gram atoms of O = 38.11/ 16 = 2.4

Atomic ratio:

            Na              :               S              :      O

        1.6/0.8            :            0.8/0.8       :     2.4/0.8

            2                  :                1              :       3

Na: S : O  = 2 :  1 : 3

Empirical formula is Na₂SO₃.

4. A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula.

Given data:

Percentage of iron = 63.52%

Percentage of sulfur = 36.48%

Empirical formula = ?

Solution:

Number of gram atoms of Fe = 63.52 / 55.845 = 1.14

Number of gram atoms of S = 36.48 / 32 = 1.14

Atomic ratio:

            Fe                :                   S            

        1.14/1.14            :               1.14/1.14    

            1                  :                   1            

Fe : S  = 1 :  1

Empirical formula is FeS.

5. Qualitative analysis shows that a compound contains 32.38%sodium, 22.65% sulfur and 44.99 %  oxygen. Find the empirical formula of this compound.

Given data:

Percentage of sodium = 32.38%

Percentage of sulfur = 22.65%

Percentage of oxygen = 44.99%

Empirical formula = ?

Solution:

Number of gram atoms of Na = 32.38 / 23 = 1.4

Number of gram atoms of S = 22.65/ 32 = 0.7

Number of gram atoms of O = 44.99/ 16 = 2.8

Atomic ratio:

            Na              :               S               :            O

        1.4/0.7            :            0.7/0.7        :           2.8/0.7

            2                  :                1             :             4

Na: S : O  = 2 :  1 : 4

Empirical formula is Na₂SO₄.

6. Analysis of a 20.0 gram sample of a compound containing only calcium and bromine indicates that  4.00 grams of calcium are present. What is the empirical formula of the compound formed?

Given data:

Mass of sample = 20g

Mass of bromine = 20 g - 4 g = 16 g

Mass of calcium = 4 g

Empirical formula = ?

Solution:

Number of gram atoms of bromine = 16 / 80= 0.2

Number of gram atoms of calcium =  4/ 40= 0.1

Atomic ratio:

            Ca                :               Br      

        0.1/0.1              :            0.2/0.1

            1                   :                2      

Ca: Br  = 1 :  2

Empirical formula is CaBr₂.

Answer 2

Answer:

1)Na2S

2)P2O5

3)Na2SO3

4)FeS

5)Na2SO4

6)CaBr2

7)C8H20Pb

8)P4O10

9)C3Cl3N3O3

10)C2H4O2

11)C4H10O

12)C4H8O2

Explanation:

1. A 15.0 gram sample of a compound is found to contain 8.83 grams of sodium and 6.17 grams of  sulfur. Calculate the empirical formula of this compound.

Moles Na = 8.83 grams / 22.98 g/mol = 0.384 moles

Moles S = 6.17 grams / 32.065 g/mol = 0.192 moles

To find the mol ratio we divide by the smallest amount of moles

Na: 0.384  / 0.192  = 2

S: 0.192 /0.192 = 1

For each mol S we have 2 mol Na

The empirical formula is Na2S

2. Analysis of a 10.150 gram sample of a compound known to contain only phosphorus and oxygen  indicates a phosphorus content of 4.433 grams. What is the empirical formula of this compound?

Mass of oxygen = 10.150 - 4.433 = 5.717 grams

Moles P = 4.433 grams / 30.97 g/mol

Moles P = 0.143 moles

Moles O = 5.717 grams / 16.0 g/mol = 0.357 moles

To find the mol ratio we divide by the smallest amount of moles

P: 0.143 moles / 0.143 moles = 1

O = 0. 357 moles / 0.143 moles = 2.5

For each P atom we have 2.5 O atoms

The empirical formula is P2O5

3. A compound is found to contain 36.48%Na, 25.41% S, and 38.11% O. Find its empirical formula.

Suppose the mass of the compound = 100 grams

Moles Na = 36.48 grams / 22.98 g/mol = 1.587 moles

Moles S = 25.41 grams / 32.065 g/mol = 0.792 moles

Moles O = 38.11 grams / 16.0 g/mol = 2.382 moles

To find the mol ratio we divide by the smallest amount of moles

Na: 1.587 moles / 0.792 moles = 2

S: 0.792 moles / 0.792 moles = 1

O: 2.382 moles / 0.792 = 3

The empirical formula = Na2SO3

4. A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula.

Suppose the mass of the compound = 100 grams

Moles Fe = 63.52 grams / 55.845 g/mol = 1.137 moles

Moles S = 36.48 grams / 32.065 g/mol = 1.137 moles

The empirical formula is FeS