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3 years ago

Please help me!!!30 pts

Chemistry

2 answers:

Tanya [424]3 years ago

5 0

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.

mars1129 [50]3 years ago

5 0

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

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2 years ago

Read 2 more answers

Consider the reaction: P4 + 6Cl2 = 4PCl3.

likoan [24]

Answer:

The answer to your question is below

Explanation:

Consider the reaction: P4 + 6Cl2 = 4PCl3.

a. How many grams of Cl2 are needed to react with 20.00 g of P4? ___68.7 g___________

P4 + 6Cl2 = 4PCl3

4(31) ---------- 12(35.5)

20 ---------- x

x = 20(12x35.5) / 4(31)

x = 8520 / 124

x = 68.7 g

b. You have 15.00 g. of P4 and 22.00 g. of Cl2, identify the limiting reactant and calculate the grams of PCl3 that can be produced as well as the grams of excess reactant remaining. LR____________ grams PCl3 _________ grams excess reactant ___________

P4 + 6Cl2 = 4PCl3

124g 426 g 4(31 + 3(35.5)) = 550g

15g 22g

I will use P4 to find the limiting reactant

x = (15 x 426) / 124 = 51.5 The limiting reactant is Chlorine

because we need 51.5 g and we only have 22g

Excess reactant

x = (22 x 124) / 426 = 6.4 g of P4

Excess P4 = 15 g - 6.4 = 8.6 g of P4 in excess

Grams of PCl3 produced

426 g of Cl2 ---------------- 550 g of PCl3

22g of Cl2 ------------- - x

x = (22 x 550) / 426 = 28.4 g of PCl3

c. If the actual amount of PCl3 recovered is 16.25 g., what is the percent yield? ______________

% yield = (16.25 - 28.4) / 28.4 x 100

% yield = 42.8

d. Given 28.00 g. of P4 and 106.30 g. of Cl2, identify the limiting reactant and calculate how many grams of the excess reactant will remain after the reaction. LR ______________ grams excess reactant

Limiting reactant

124g of P4 ------------- 426 g 6Cl2

28g --------------- x

x = (28 x 426) / 124

x = 96.2 g of Cl2 and we have 106.3 so Chlorine is the excess reactant and P4 is the limiting reactant.

Excess reactant = 106.3 - 96.2 = 10.1 g of Cl2 in excess

4 0

2 years ago

A chemistry student needs of acetic acid for an experiment. He has available of a w/w solution of acetic acid in acetone. Calcul

Volgvan

Answer:

49.4 g Solution

Explanation:

There is some info missing. I think this is the original question.

A chemistry student needs 20.0g of acetic acid for an experiment. He has 400.g available of a 40.5 % w/w solution of acetic acid in acetone.

Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button. Round your answer to 3 significant digits.

We have 400 g of solution and there are 40.5 g of solute (acetic acid) per 100 grams of solution. We can use this info to find the mass of acetic acid in the solution.

$400gSolution \times \frac{40.5gSolute}{100gSolution} = 162 g Solute$