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scZoUnD [109]
2 years ago
9

Determine the empirical formula for a compound that contains 48.6%C, 8.2% H, and 43.2% S by mass.

Chemistry
1 answer:
vagabundo [1.1K]2 years ago
4 0

The empirical formula for a compound that contains 48.6%C, 8.2% H, and 43.2% S by mass is C_3H_6S.

<h3>What is the empirical formula?</h3>

An empirical formula tells us the relative ratios of different atoms in a compound.

We need to calculate the number of moles

Number of a mole of carbon =

48.6 g X (\frac{1 mole }{ 12.0107 g}) =4.05 mole

Number of a mole of hydrogen =

8.2g X (\frac{1 mole}{.00784g}) =8.14 mole

Number of moles of sulphur =

43.2g X (\frac{ mole}{32.065g}) = 1.35 mole

Dividing each mole using the smallest number that is divided by 1.35 moles.

Carbon= \frac{4.05 mole }{1.35 mole} =3

Oxygen= \frac{8.14 mole}{1.35 mole} =6

Sulfur= \frac{1.35 mole}{1.35 mole} =1

Empirical formula is C_3H_6S

Learn more about empirical formula here:

brainly.com/question/14044066

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. Sulfur dioxide can be produced in the laboratory by the reaction of hydrochloric acid and a sulfite salt such as sodium sulfit
shutvik [7]

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Mass of SO₂ can be made from 25.0 g of Na₂SO₃ and 22 g of HCl = 12.672 g

Explanation:

SO₂( sulfur dioxide) can be produced in the lab. by the reaction of hydrochloric acid & sulphite salt such as sodium.

        the balanced chemical equation is as follows

                       Na₂SO₃ + 2 HCl → 2 NaCl + SO₂ + H₂O

Moles of Na₂SO₃ = \frac{Mass}{Molecular mass} =\frac{25}{126} = 0.198

Moles of HCl = \frac{mass}{molecular mass}=\frac{22}{36.5}= 0.6

using mole ratio method to find limiting reagent

      For sodium sulfite \frac{mole}{stoichiometry}  = \frac{0.198}{1}= 0.198

 for HCl \frac{mole}{stoichiometry}  = \frac{0.6}{2}= 0.3

since <u>sodium sulfite</u> is <u>limiting reactant</u> for above chemical reaction

1 mole of Na₂SO₃ produce 1 mole of SO₂

0.198 mole of Na₂SO₃ produce 0.198 mole of SO₂

∴ Mass of SO₂ produce = mole x molar mass of SO₂

                                       = 0.198 x 64

                                       = 12.672 g

8 0
3 years ago
Solid solutions that are mixtures of elements with metallic properties are known as_____
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4 0
3 years ago
A solution contains 0.036 M Cu2+ and 0.044 M Fe2+. A solution containing sulfide ions is added to selectively precipitate one of
Ratling [72]

Answer:

The precipitate is CuS.

Sulfide will precipitate at  [S2-]= 3.61*10^-35 M

Explanation:

<u>Step 1: </u>Data given

The solution contains 0.036 M Cu2+ and 0.044 M Fe2+

Ksp (CuS) = 1.3 × 10-36

Ksp (FeS) = 6.3 × 10-18

Step 2:  Calculate precipitate

CuS → Cu^2+ + S^2-         Ksp= 1.3*10^-36

FeS → Fe^2+ + S^2-      Ksp= 6.3*10^-18

Calculate the minimum of amount needed to form precipitates:

Q=Ksp

<u>For copper</u>  we have:  Ksp=[Cu2+]*[S2-]

Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]

[S2-]= 3.61*10^-35 M

<u>For Iron</u>  we have: Ksp=[Fe2+]*[S2-]

Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]

[S2-]= 1.43*10^-16 M

CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.

The precipitate is CuS.

Sulfide will precipitate at  [S2-]= 3.61*10^-35 M

3 0
3 years ago
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