Chemistry : Chemical Laws And Calculations

Chemical laws in chemistry

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Law Of Conservation Of Mass

The law of conservation of mass states that matter can neither be created nor destroyed in course of chemical reactions. This law is also referred to as law of indestructibility of matter.

Experiment To Verify Law Of Conservation Of Mass

Reagents : Hydrogen tetraoxosulphate (iv) (H2SO4) and Potassium chloride (KCl)

Equation : H2SO4(aq) + 2KCl(aq) = 2HCl + K2SO4

K2SO4 and HCl are the new products, despite the fact that reaction has taken place, the weight of the reactant before reaction and the weight of the product after reaction are found to be the same. This simply means the mass of reactant before reaction is equal to the mass of product after reaction.top↑

Conclusion

Since there is no change in mass after the formation of the products, hence it confirms that matter can neither created nor destroyed in the course of chemical reaction.

Note: not all chemical reactions obey law of conservation of matter, in some chemical reactions, there are slight change in the mass of the product and reactant before and after reaction, that is why Dalton’s atomic theory about indestructibility of matter was prove wrong.
To read more on why Dalton’s atomic theory was prove wrong, click the this link Read more

Law Of Definite Proportion

The law of definite proportion states that all pure samples of the chemical compound contain the same elements combined in the same proportion by mass.

Experiment To Verify Law Of Definite Proportion

To find out if different samples of a compound will have the same proportion by mass of constituent elements.top↑

Methods : prepare two samples of copper (ii) oxide by different method.

1st Method

put a small amount of copper metal in a beaker.
Carefully add some concentrated trioxonitrate (v) acid until all the copper dissolves.
The equation for the reaction is;
Cu(s) + HNO3(aq) = Cu(NO3)2 + H2O(l)  +  NO2(g)
The product will be evaporated to dryness in order to remove elements of water.
The residue copper (ii) trioxonitrate (v) is heated strongly in order to remove the brown fumes of nitrogen (iv) oxide.
The equation for the reaction is;
2Cu(NO3)2 = 2CuO + 4NO2 + O2(g)
The black residue copper (ii) oxide is obtained which can be store in a dessicator to in order to keep it dry.

Note : dessicator are use to store gases in order to keep them dry.top↑

2nd Method

Put a little green copper (ii) trioxocarbonate (iv) (CuCO3) salt in a dry crucible and apply heat. The Copper trioxocarbonate (iv) decomposes on heating, turning from green to black copper (ii) oxide and Carbon (iv) oxide which is given off.
The equation for the reaction is;
CuCO3 = CuO + CO2(g)
The black residue copper (ii) oxide (CuO) is then put in a dessicator in order to keep it dry.

Analyse the samples of copper (ii) oxide, getting from the two methods,
Weigh two dry porcelain boats and record their masses, let record the mass of the two boats as Mass A
Place 1½g of each of the prepared samples of copper (ii) oxide inside a porcelain boat.
Weigh the two porcelain boats again and record the new masses also. Let record this mass, as Mass B

Pass hydrogen gas through the reaction and heat the two samples strongly, a reddish brown copper will be formed.top↑
The equation for the reaction is;
CuO(s) + H2 = Cu(s) + H2O(l)
Let record this mass as Mass C
Allow the two samples to cool down and weigh the two samples again and record their masses. Now calculate the amount of copper obtained from each sample of oxide.
In order to calculate the amount of copper obtained from each sample,
You subtract Mass A from Mass B

And divide the mass of the copper obtained in the last process by the result of Mass A – Mass B and then multiply by 100
Doing that for both samples,

Conclusion

it will be observe that the percentage of copper in both sample of the copper is approximately the same.top↑

Law Of Multiple Proportion

The law of multiple proportion states that if two elements A and B combine together to form more than one compound, then the several masses of A combined separately with a fixed mass of B are in a simple ratio to one another.

Experiment To Verify The Law Of Multiple Proportion

In order to demonstrate the law of multiple proportion, we make use of copper and oxygen forming two oxides, copper (ii) oxide (CuO) and copper (i) oxide (Cu2O)

Method

weigh two porcelain boats, record their masses of the porcelain boats as MassA, put some sample of copper (ii) oxide (CuO) in one porcelain boat and copper (i) oxide in the other boat and weigh again and record the masses of both oxides, record this one as Mass B

Place the two boats containing the oxides in a hard combustion tube and carry out their reduction reaction by passing hydrogen over them as was done in the first experiment. Remove the source of heat and allow the samples to cool down and then reweigh the samples and record their new masses of the metallic copper as Mass C, now calculate how much copper is obtained from each sample of oxide.top↑

Note : In the reaction hydrogen acts as a reducing agent. Hydrogen was used in the reaction to reduce copper (ii) oxide and copper (i) oxide to copper.

In order to calculate the amount of copper obtained from each sample,
You subtract Mass A from Mass B

And divide the mass of the copper obtained in the last process by the result of Mass A – Mass B and then multiply by 100,
It will be observe that, the two samples are in simple ratio to one another.top↑

Chemical Equation And Balancing

It is certain that before any equation is written one must check that the formula of each compound reacting is written correctly. No equation can be balanced ones the formula of any of the reacting substance or products is wrong.

Hence, the first thing to do when writing a chemical equation is to write the correct formula of the reacting compound.

After ensuring that the formulae of compound or substances are correct, then write the equation as shown below;

Example 1

A + B = C + D

A + B = reactant side while
C + D = product side
When writing a chemical equation, the substances reacting ( the reactant) are written on the left hand side while the products of the reaction are written on the right hand side.top↑
E.g write the chemical equation of the reaction between hydrogen and oxygen to produce water
First, write the correct formula of the reacting element,
The reacting elements are Hydrogen (H) and Oxygen (O)

H2 + O2 = H2O

On checking the equation, it is found that the formulae of hydrogen, oxygen and water are correctly written, but is the equation balanced? NO it is not.

In order to balance the equation, these are the procedure;
In balancing any chemical equation, the total number of atoms of each element present on the reactant side, that is, the left hand side must be equal to the total number of atoms of each element on the product side ( right hand side) in accordance with the law of conservation of matter.top↑

So back to our unbalanced equation

H2 + O2 = H2O

The reason why the equation is not balance is because, there are two atoms of oxygen on the left hand side (reactant side) while we have only one atom of oxygen on the right hand side ( product side),
In order to balance the equation we have to add to the atom of oxygen at the product side as written below;

H2 + O2 = 2H2O
In attempting to satisfy the number of oxygen atoms on both sides, we now have two atoms of hydrogen on the reactant side and four atoms on the product side.
So therefore, we will amend the reactant side and write the equation as written below;top↑

2H2 + O2 = 2H2O
Now, the equation is thus balanced.

Note : hydrogen and oxygen are diatomic gas, that is, they occur as H2 and O2, not as H or O in a chemical reaction.

Example 2

Write the chemical equation for Carbon (C) and Oxygen (O) reacting together to form Carbon (iv) oxide (CO2)

Solution

First write down the correct formula of the reacting elements
Carbon = C, and oxygen = O

C + O2 = CO2
You can see that in this type of equation, there is no need to balance the equation because it already balanced,
One atom of Carbon and one atom of oxygen at reactant side is equal to the same thing as in the product side. Hence the equation as been balanced already.
That example is to cheap right? Yea, try this onetop↑

Example 3

Write the chemical equation for the reaction between Sodium hydroxide (NaOH) and tetraoxosulphate (iv) acid (H2SO4) to give sodium tetraoxosulphate (iv) salt and water.

Solution

First write down the correct formulae of the reacting elements
Sodium hydroxide = NaOH
Tetraoxosulphate (iv) acid = H2SO4
Sodium tetraoxosulphate (iv) salt = Na2SO4
Water = H2O

H2SO4 + NaOH = Na2SO4 + H2O
Looking at the above equation, you will see that;
1) There are 1 atoms of sodium at the reactant side and 2 atoms of sodium at the product side

2) There are 5 atoms of oxygen at the product side and 5 atoms of oxygen at the reactant side

3) There are 1 atom of sulphur at the product side and 1 atom at the reactant side.top↑

4) There are 3 atoms of hydrogen at the reactant side and 2 atoms of hydrogen at the product side.
Hence, the reaction is not balanced.

In order to balance the equation, we have to add;

1) 1 atom of sodium to the reactant side

2) 1 atom of hydrogen to the reactant side
balancing the equation we will have;

H2SO4 + 2NaOH = Na2SO4 + 2H20
Therefore the equation is balanced.
This means that 2 moles of sodium hydroxide (NaOH) react with 1 mole of hydrogen tetraoxosulphate (vi) (H2SO4) to give 1 mole of sodium tetraoxosulphate (vi) (Na2SO4) salt and 2 moles of water (H2O) top↑

Calculation Of Mass

Balancing equation enables us to calculate the masses of substances used and formed in a chemical reactions.

Example 1

What mass of copper is produced when zinc is added to a solution containing 3.19g of copper (ii) tetraoxosulphate (vi) (CuSO4) solution.
[Cu = 63.5, S =32, O = 16 Zn = 65)

Solution

First write down the correct formula for the reaction
Zn(s) + CuSO4(aq) = ZnSO4(aq) + Cu
1 mole of Zn + 1 mole of CuSO4 = 1 mole of ZnSO4 + 1 mole of Cu
As we are only asked to find the mass of copper produced from copper (ii) tetraoxosulphate (vi)
So the only compound that we will need in our calculation is where copper came from, which is from copper (ii) tetraoxosulphate (vi) top↑
1 mole of CuSO4 gives 1 mole of Cu
Molar mass of 1 mole of CuSO4
= 63.5 + 32 + (4 × 16)
= 95.5 + 64
= 159.5g

Molar mass of 1 mole of Cu = 63.5g

Hence,
159.5g of CuSO4 produces 63.5g of Cu
So here is the question;
If 159.5g of CuSO4 produces 63.5g of Cu, then what will 31.9g of CuSO4 produce.
That will be;

31.9 × 63.5
159.5
= 12.7g of Cu

Hence 12.7g of copper is deposited when zinc is added to a solution containing 31.9g of copper (ii) tetraoxosulphate (vi).

Example 2

What mass of Calcium oxide (CaO)is formed when 10g of Calcium trioxocarbonate (iv) (CaCO3) decompose on heat.top↑

Solution

First write down the correct balanced equation for the reaction.

CaCO3(s) = CaO(s) + CO2(g)

In the reaction, 1 mole of CaCO3 decomposes to produce 1 mole of CaO and 1 mole of CO2 gas.
Molar mass of 1 mole of CaCO3
= 40 + 12 + (3 × 16)
= 52 + 48
= 100g

Molar mass of 1 mole of CaO = 40 + 16 = 56g

Molar mass of 1 mole of CO2
= 12 + (16 × 2)
= 12 + 32
= 44g

If 1 mole of CaCO3 produce 1 mole of CaO, then
How many mole will 10g of CaCO3 produce
But in terms of mass top↑

If 100g of CaCO3 produces 56g of CaO
What will 10g of CaCO3 produce?
That will be;

10 × 56
100
= 5.6g of CaO
Hence, 5.6g of CaO is produced when 10g of CaCO3 decomposes to form CaO and CO2.

Note : K2CO3, Na2CO3 and (NH4)2CO3 do not decompose on heating. Please note this.

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