Formulae

State the meaning of the term empirical formula.

The term empirical formula is the simplest whole number ratio atoms of each element in a compound.

An unknown compound contains carbon, hydrogen and oxygen only. It was shown to contain 3.20 g carbon, 0.54 g hydrogen and 4.26 g oxygen.

Calculate the empirical formula of the unknown compound.

Divide each mass by relevant A_r

`C = 3.20 / 12.0 = 0.2667`

`H = 0.54 / 1.0 = 0.54`

`O = 4.26 / 16.0 = 0.26625`

Then, divide each number by the smallest number calculated to identify the ratio

`C = 0.2667 / 0.26625 = 1`

`H = 0.54 / 0.26625 = 2`

`O = 0.26625 / 0.26625 = 1`

Therefore, the empirical formula of this compound is `CH_2O`

i. 12.00 g of a metal oxide contains 10.66 g of copper and the remainder of the compound is oxygen. Calculate the empirical formula of the metal oxide.

ii. Magnesium oxide is formed by the combustion of magnesium.

Write the ionic equation for the formation of the magnesium oxide product.

i. The amount of oxygen in the compound `= 12.00 - 10.66 = 1.34 g`

Divide each mass by relevant A_r

`Cu = 10.66 / 63.54 = 0.1678`

`O = 1.34 / 16.0 = 0.08375`

Then, divide each number by the smallest number calculated to identify the ratio

`Cu = 0.1678 / 0.08375 = 2`

`O = 0.08375 / 0.08375 = 1`

Therefore, the empirical formula of this metal oxide is `Cu_2O`

ii. `Mg^(2+) + O^(2-) -> MgO`

Remarks:

• Any element in the Periodic Table by use of the group number, e.g. group 2 metals are `M^(2+`, group 7 nonmetals are `X^-`
• Specific anion, e.g. sulphate (`SO""_4^(2-`), nitrate (`NO""_3^-`), carbonate (`CO""_3^(2-`)
• Other ions should be memorized such as silicate (`SiO""_3^(2-`), phosphate (`PO""_4^(3-`)

‘Asbestos’ is a generic, commercial term used to describe a number of mineral fibres, which were once commonly found in most buildings. The most common type of asbestos, chrysotile, makes up the vast majority of all asbestos products. Its chemical composition is 28.03% magnesium, 21.60% silicon, 1.16% hydrogen and the rest is oxygen.

a. Calculate the empirical formula for chrysotile asbestos.

b. The M_r of chrysotile is 520.2. Calculate the actual molecular formula of chrysotile.

a. The amount of oxygen `= 100% - 28.03% - 21.60% - 1.16% = 49.21 %`

Divide each % composition by relevant A_r

`Mg = 28.03 / 24.3 = 1.1534`

`Si = 21.60 / 28.1 = 0.7686`

`H = 1.16 / 1.0 = 1.16`

`O = 49.21 / 16.0 = 3.075625`

Then, divide each number by the smallest number calculated to identify the ratio

`Mg = 1.15354 / 0.7686 = 1.5`

`Si = 0.7686 / 0.7686 = 1`

`H = 1.16 / 0.7686 = 1.5`

`O = 3.075625 / 0.7686 = 4`

All numbers must be multiplied by 2 to obtain a whole number ratio.

Therefore, the empirical formula of this compound is `Mg_3Si_2H_3O_8`

b. The M_r of the empirical formula `= 3 xx 24.3 + 2 xx 28.1 + 3 xx 1.0 + 8 xx 16.0 = 260.1 g`

The coefficient, n, between the actual chrysotile and the empirical formula `= 520.2 / 260.1 = 2`

Thus, the actual molecular formula of chrysotile is `Mg_6Si_4H_6O_16`

Strong heating of calcium nitrate will cause a thermal decomposition reaction to occur. During the reaction, calcium oxide, nitrogen dioxide and another gas are formed. 

Write a balanced chemical equation, with state symbols, for this reaction.

`2Ca(NO_3)_2 (s) -> 2CaO (s) + 4NO_2 (g) + O_2 (g)`

This is a thermal decomposition means that using heat to break down calcium nitrate => Add oxygen as a reactant would be wrong.

A hydrated salt, S, was analysed and found to have the following percentage composition by mass: 16.98% Ca; 11.86% N; 67.77% O; 3.39% H 

Determine the empirical formula of the compound, showing clearly that it is a hydrated salt.

Divide each % composition by relevant A_r

`Ca = 16.98 / 40.1 = 0.42344`

`N = 11.86 / 14.0 = 0.847142`

`H = 3.39 / 1.0 = 3.39`

`O = 67.77 / 16.0 = 4.235625`

Then, divide each number by the smallest number calculated to identify the ratio

`Ca = 0.42344 / 0.42344 = 1`

`N = 0.847142 / 0.42344 = 2`

`H = 3.39 / 0.42344 = 8`

`O = 4.235625 / 0.42344 = 10`

Thus the empirical formula of the hydrate salt is `Ca(NO_3)_(2).4H_2O`. 

Note that: `CaN_2O_10H_8` or `CaN_2O_(6).4H_2O` will be incorrect.

Write a net ionic equation for the reaction that occurs when excess nitric acid (aq) and sodium carbonate (aq) are combined.

The balanced molecular equation for the reaction must be taken into consideration before writing the net ionic equation for the reaction between excess nitric acid (`HNO_3`) and sodium carbonate (`Na_2CO_3`):

`2HNO_3​(aq)+Na_2​CO_3​(aq)→2NaNO_3​(aq)+CO_2​(g)+H_2​O(l)`

The whole ionic equation, which lists every ion present in the solution, can then be written. HNO3 and Na2CO3 will separate into their individual ions since they are soluble in water: 

`2H^+(aq)+2NO""_3^−​(aq)+2Na^+(aq)+CO""_3^(2−)​(aq)→2Na^(+)(aq)+2NO""_3^(−​)(aq)+CO2_​(g)+H_2​O(l)`

The spectator ions, or ions that show up on both sides of the equation but do not take part in the chemical reaction, can now be located and eliminated. The nitrate ions (`NO""_3^− (aq)`) and sodium ions (`Na^+`) are spectator ions in this instance. The net ionic equation is obtained by subtracting these from both sides:

`2H^+(aq)+CO""_3^(2−)​(aq)→CO_2(g)+H_2​O(l)`

According to this net ionic equation, carbonate ions from sodium carbonate and hydrogen ions from nitric acid react to form carbon dioxide gas and water.

Complete the table below with the oxidation states of the elements in the following species.

The oxidation state of a simple ion is the charge on the ion

• Group 1 metal ions, e.g Na⁺ have an oxidation state of +1
• Group 2 metal ions, e.g Mg²⁺ have an oxidation state of +2
• Halide ions , e.g. Cl^- have an oxidation state of -1
• Group 6 nonmetal ions, e.g. O^2- have an oxidation state of -2
• Transition metal ions have variable oxidation states, we need to take into account the charge e.g. vanadium has four common charge oxidation state: +2, +3, +4, +5

A solution of sulfuric acid contains? 

A. `H_2SO_4` molecules only 

B. `H^+` and `SO""_4^(2-)` ions only 

C. `H^+`, `SO""_4^(2-)` ions and `H_2SO_4` molecules 

D. `H^+` ions only

The answer is C.

Sulfuric acid (`H_2SO_4`) is a strong acid that dissociates completely in water. When sulfuric acid is dissolved in water, it ionizes to release hydrogen ions (`H^+`) and sulfate ions (`SO""_4^(2-`). This means that in a solution of sulfuric acid, you will find `H^+` ions, `SO""_4^(2-` ions, and undissociated `H_2SO_4` molecules. 

Which are the correct common oxidation states of fluorine, oxygen and hydrogen in compounds?

The answer is A.

Fluorine is the most electronegative elements so its oxidation state is always -1

The oxidation state of oxygen is always -2 except it is in peroxides which shows the oxidation number of oxygen is -1 and another situation in which oxygen reacts with fluorine to generate F₂O, with this case, oxygen is in +2 oxidation state

Hydrogen has always oxidation number +1 except when in metal hydrides where it is -1, for example, NaH, the oxidation number of hydrogen will be -1