Chemistry plz HELP!!!!!?

Question

Problems.

1. Determine the empirical formula and the molecular formula of a compound containing 40.0%C,6.7%H and 53.3% O. The molecular weight of the compounds is 60.02 g.

2. Calculate the percentage composition of a compound containing 27.93g Fe,24.05gS,and 48.00 g O.

3.How many grams of silver is present in 3.45g Ag2S?<numbers are subscripts>

4.The reaction 2Al+3MnO---->Al2O2+3Mn proceeds until the limiting reagent is all consumed. A mixture containing 110g Al and 200g MnO was heated to initiate the reaction. Which substance is the excess reagent and by how much?

5. Calculate the theoretical yield, potassium chloride produced from the reaction of 2.56g of K and 3.85 g Cl2 if an actual experimental produces 3.81g KCL. Calculate the percentage yield.

thanks for helping ^^!!

Answer 1

1. Empirical formula of a compound is the lowest whole ratio of each atom in a molecule. The molecular formula is the exact number of each atom that is present in a molecule.

You can work out an empirical formula from the mass of each atom in a sample.

Firstly, assume that you have 100 g of sample. So in 100 g

The mass ratio of Carbon : Hydrogen : Oxygen

is 40.0 g : 6.7 g : 53.3 g

You need to convert this to moles in order to get the moles ratio of elemnts in the sample. To convert to moles divide each mass by the atomic mass of the atom because

moles = mass / molecular mass

So the mole ratio of C : H : O

is (40.0/12.0) : (6.7 / 1.008) : (53.3/16.0)

= 3.33 : 6.64 : 3.33

Now, to get the lowest whole ratio divide each number by the lowest number

C : H : O

= (3.33 / 3.33) : (6.64/3.33) : (3.33/3.33)

= 1 : 2 : 1

So the empirical formula is CH2O

The molecular weight of CH2O = 12.01 + (1.008 x 2) + 16

= 30.026 g/mol

You are told that the molecular formula of the actual molecule is 60.02 g/mol. So dividing the molecular formular weight weight by the empirical formula weight we get 60.02 / 30.026

= 2

So molecular formula is 2x empirical formula

therfore molecular formula = C2H4O2

2. The percentage composition ins the percentage mass of each atom in the total sample.

Total mass of your sample = 27.93 g (Fe) + 24.05 g (S) + 48.00 g (O) =  99.98 g.

Now % of Fe in total mass = (mass Fe / total mass) x 100/1

= (27.93 / 99.98) x 100

= 27.94 %

Do the same for the other two.

3. For this question you need the equation.

moles = mass / molecular weight.

First, work out the number of moles of Ag2S that you have.

The molecular weight Ag2S = (2 x 107.9) + 32.07 = 247.87g/mol

So now if moles = mass / molecular weight

then moles Ag2S = 3.45 g / 247.87 g/mol

= 0.0139 moles

Now, 1 molecule of Ag2S contains 2 atoms of Ag and 1 atom of S.

Therfore in 0.0139 moles of Ag2S there will be

(0.0139 x 2) moles of Ag

and 0.0139 moles of S

Moles Ag = 2 x 0.0139 = 0.0278 moles Ag

mass of Ag = (moles of Ag) x (molecular weight of Ag)

= (0.0278 mol) x (107.9 g/mol)

= 3.00 g of Ag

4. 2Al + 3MnO -------&gt; Al2O3 + 3Mn

nb, note I have changed Al2O2 to Al2O3. I believe this is correct, as oxidation state of Al is always 3+ and otherwise the equation is not balanced (Not enogh O on the right hand side). If I have it wrong you can still apply the method I use.

You need to work out the number of moles of your two starting reagents first using the equation

moles = mass / molecular mass

molecular mass Al = 26.98 g/mol

molecular mass MnO = 54.94 + 16 = 70.94 g/mol

moles Al = (mass Al) / (molecular mass Al)

= (110 g) / (26.98 g/mol)

= 4.077 moles of Al at start of reaction

moles MnO = (200 g)/ (70.94 g/mol)

=2.82 moles MnO at start of reaction.

From the balanced equation we know that

2 moles of Al react with 3 moles of MnO

or 1 moles Al needs 3/2 moles MnO

therfore if we start with 4.077 moles of Al we will need 6.12 moles of MnO.

We have worked out that we only have 2.82 moles of MnO, which is not enough to react with our 4.077 moles of Al. This means that our limiting reagent is MnO and the Al is said to be in excess.

Now, we said 1 mole Al needs 3/2 moles MnO

So 1 mole of MnO will require 2/3 mole of Al

So 2.82 moles nees (2.82 x 2/3) moles of Al

= 1.88 moles.

You only need 1.88 moles of Al

So unused (Excess) Al = 4.077mol  -1.88 mol

= 2.197 moles

mass unused = moles x molecular weight = 2.197 x 26.98 g/mol = 59.28 g

5. this is similar to 4.

1st do a balanced equation

2K + Cl2 -----&gt; 2KCl

molecular mass K = 39.10 g/mol

molecular mass Cl2 = (2x35.45) = 70.90 g/mol

molecular mass KCl = 39.10 + 35.45 = 74.55 g/mol

To work out theoretical yield you 1st need to know the limiting reagent. So work out the number of moles of each reactant in the reaction mixture.

moles K = mass / molecular mass = 2.56 g / 39.1 g/mol

= 0.0655 moles K

moles Cl2 = 3.85 / 70.90

= 0.0543 moles Cl2

From equation 1 moles of Cl2 react with 2 mole of K

So 0.0543 moles of Cl2 require (2x 0.0543) = 0.1086 moles K

We only have 0.0655 moles of K, therefore K is the limiting reagent and Cl2 is in excess.

The theoretical yield in the maximum yield we can possibly attain from teh quantities started with. In this case, since K is the limiting reagent, out theoretical yield will be the amount of KCl that can be produced from 0.0655 mol of K.

Now 1 mole Cl2 + 2 moles K ---------&gt; 2 moles of KCl

Therefore 1 mole of K ---------- 1 mole KCl

Therfore 0.0655 mol K ---------&gt; 0.0655 mol KCl

So maximum weight KCl possible is

mass KCl = (moles KCl) x (molecular weight KCl)

= (0.0655 mol) x (74.55 g/mol)

= 4.883 g maximum theoretical yield of KCl.

The percentage yield is the percentage of the theoretical yield you actually achieved.

% yield = (actual yield)/ (theoretical yield) x (100/1)

= (3.81 g / 4.883 g) x 100/1

= 78.0 %

Answer 2

Too many questions that need soace &amp; calc. Break them apart