Chemistry Exam Notes: Formulae, Conservation of Mass, and Ionic Equations

Empirical, Molecular & Structural Formulae: Determining Empirical Formula

Example: Determining Empirical Formula from Mass Data

Problem: When 2.4 ext{g} of an element X (relative atomic mass 24 ) reacts with excess chlorine, 9.5 ext{g} of the anhydrous chloride is formed. What is the empirical formula for the chloride? (Given: ext{A}r( ext{Cl}) = 35.5 )

Solution Approach:

Determine the mass of chlorine that reacted:
- Apply the Law of Conservation of Mass: \text{Mass of X} + \text{Mass of Cl} = \text{Mass of chloride}
- \text{Mass of Cl} = \text{Mass of chloride} - \text{Mass of X}
- \text{Mass of Cl} = 9.5 ext{g} - 2.4 ext{g} = 7.1 ext{g}
Calculate the moles of element X:
- \text{Moles of X} = \frac{\text{Mass of X}}{\text{Relative atomic mass of X}}
- \text{Moles of X} = \frac{2.4 ext{g}}{24 ext{ g/mol}} = 0.1 \text{ mol}
Calculate the moles of chlorine:
- \text{Moles of Cl} = \frac{\text{Mass of Cl}}{\text{Relative atomic mass of Cl}}
- \text{Moles of Cl} = \frac{7.1 ext{g}}{35.5 ext{ g/mol}} = 0.2 \text{ mol}
Determine the simplest whole-number ratio of moles:
- Divide both mole values by the smallest number of moles ( 0.1 ext{ mol} ):
  - For X: \frac{0.1}{0.1} = 1
  - For Cl: \frac{0.2}{0.1} = 2
Write the empirical formula:
- The ratio of X to Cl is 1:2 . Therefore, the empirical formula is \text{XCl}_2 .

Law of Conservation of Mass

Example: Combustion Analysis and Mass Conservation

Problem: 1.20 ext{g} of carbon were burnt in excess oxygen gas. The gaseous product formed was absorbed in potassium hydroxide solution. The increase in mass of the solution was 4.40 ext{g} . (Given: ext{A}r( ext{C})=12 ; ext{A}r( ext{O})=16 )

a. Name the gaseous product formed.

When carbon is burnt in excess oxygen, the complete combustion product is carbon(IV) oxide (carbon dioxide, \text{CO}_2 ).

b. Write balanced chemical equations to illustrate the reaction between:

i. Carbon and excess oxygen gas:
- \text{C(s)} + \text{O}2\text{(g)} \rightarrow \text{CO}2\text{(g)}
ii. The gaseous product formed (carbon dioxide) and potassium hydroxide:
- \text{CO}2\text{(g)} + 2\text{KOH(aq)} \rightarrow \text{K}2\text{CO}3\text{(aq)} + \text{H}2\text{O(l)}

c. Name the products in b(ii) above.

The products are potassium carbonate and water.

d. Calculate the mass of oxygen gas reacted.

Concept: The increase in mass of the potassium hydroxide solution is due to the absorption of the gaseous product, which is \text{CO}2 . Therefore, the mass of \text{CO}2 formed is 4.40 ext{g} .
Applying the Law of Conservation of Mass to the combustion reaction \text{C(s)} + \text{O}2\text{(g)} \rightarrow \text{CO}2\text{(g)} , we know:
- \text{Mass of C} + \text{Mass of O}2 = \text{Mass of CO}2
Rearranging to find the mass of oxygen:
- \text{Mass of O}2 = \text{Mass of CO}2 - \text{Mass of C}
- \text{Mass of O}_2 = 4.40 ext{g} - 1.20 ext{g} = 3.20 ext{g}
Thus, the mass of oxygen gas reacted is 3.20 ext{g} .

Ionic Equations

Writing Balanced Ionic Equations

General Steps:

Write the balanced molecular equation.
Dissociate all strong electrolytes (soluble ionic compounds, strong acids, strong bases) into their ions. Keep solids, liquids, gases, and weak electrolytes in their molecular forms.
Identify and cancel spectator ions (ions that appear on both sides of the equation without changing).
Write the net ionic equation with remaining species.

Problems: Write down balanced ionic equations to represent the following reactions:

\text{Fe(s)} + \text{H}2\text{SO}4\text{(aq)} \rightarrow \text{FeSO}4\text{(aq)} + \text{H}2\text{(g)}
- Molecular Equation (already balanced): \text{Fe(s)} + \text{H}2\text{SO}4\text{(aq)} \rightarrow \text{FeSO}4\text{(aq)} + \text{H}2\text{(g)}
- Complete Ionic Equation: \text{Fe(s)} + 2\text{H}^+\text{(aq)} + \text{SO}4^{2-}\text{(aq)} \rightarrow \text{Fe}^{2+}\text{(aq)} + \text{SO}4^{2-}\text{(aq)} + \text{H}_2\text{(g)}
- Spectator Ions: \text{SO}_4^{2-}\text{(aq)}
- Net Ionic Equation: \text{Fe(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Fe}^{2+}\text{(aq)} + \text{H}_2\text{(g)}
\text{MnO}2\text{(s)} + \text{HCl(aq)} \rightarrow \text{MnCl}2\text{(aq)} + \text{H}2\text{O(l)} + \text{Cl}2\text{(g)}
- Balanced Molecular Equation: \text{MnO}2\text{(s)} + 4\text{HCl(aq)} \rightarrow \text{MnCl}2\text{(aq)} + 2\text{H}2\text{O(l)} + \text{Cl}2\text{(g)}
- Complete Ionic Equation: \text{MnO}2\text{(s)} + 4\text{H}^+\text{(aq)} + 4\text{Cl}^-\text{(aq)} \rightarrow \text{Mn}^{2+}\text{(aq)} + 2\text{Cl}^-\text{(aq)} + 2\text{H}2\text{O(l)} + \text{Cl}_2\text{(g)}
- Spectator Ions: None, as \text{Cl}^- is both a spectator ion and a reactant (oxidized to \text{Cl}_2 ). In this redox reaction, not all \text{Cl}^- ions are spectators.
- Net Ionic Equation: \text{MnO}2\text{(s)} + 4\text{H}^+\text{(aq)} + 2\text{Cl}^-\text{(aq)} \rightarrow \text{Mn}^{2+}\text{(aq)} + 2\text{H}2\text{O(l)} + \text{Cl}_2\text{(g)}
\text{K}2\text{CrO}4\text{(aq)} + \text{Pb(NO}3\text{)}2\text{(aq)} \rightarrow \text{PbCrO}4\text{(s)} + \text{KNO}3\text{(aq)}
- Balanced Molecular Equation: \text{K}2\text{CrO}4\text{(aq)} + \text{Pb(NO}3\text{)}2\text{(aq)} \rightarrow \text{PbCrO}4\text{(s)} + 2\text{KNO}3\text{(aq)}
- Complete Ionic Equation: 2\text{K}^+\text{(aq)} + \text{CrO}4^{2-}\text{(aq)} + \text{Pb}^{2+}\text{(aq)} + 2\text{NO}3^-\text{(aq)} \rightarrow \text{PbCrO}4\text{(s)} + 2\text{K}^+\text{(aq)} + 2\text{NO}3^-\text{(aq)}
- Spectator Ions: 2\text{K}^+\text{(aq)} , 2\text{NO}_3^-\text{(aq)}
- Net Ionic Equation: \text{CrO}4^{2-}\text{(aq)} + \text{Pb}^{2+}\text{(aq)} \rightarrow \text{PbCrO}4\text{(s)}
\text{NaOH(aq)} + \text{HClO}4\text{(aq)} \rightarrow \text{NaClO}4\text{(aq)} + \text{H}_2\text{O(l)}
- Molecular Equation (already balanced): \text{NaOH(aq)} + \text{HClO}4\text{(aq)} \rightarrow \text{NaClO}4\text{(aq)} + \text{H}_2\text{O(l)}
- Complete Ionic Equation: \text{Na}^+\text{(aq)} + \text{OH}^-\text{(aq)} + \text{H}^+\text{(aq)} + \text{ClO}4^-\text{(aq)} \rightarrow \text{Na}^+\text{(aq)} + \text{ClO}4^-\text{(aq)} + \text{H}_2\text{O(l)}
- Spectator Ions: \text{Na}^+\text{(aq)} , \text{ClO}_4^-\text{(aq)}
- Net Ionic Equation: \text{OH}^-\text{(aq)} + \text{H}^+\text{(aq)} \rightarrow \text{H}_2\text{O(l)}

Empirical, Molecular & Structural Formulae: Determining Molecular Formula

Example: Determining Empirical and Molecular Formula from Percentage Composition and Vapour Density

Problem: A hydrocarbon with vapour density of 29 contains 82.76\% carbon and 17.24\% hydrogen. Determine its:
a. empirical formula
b. molecular formula
(Given: ext{A}r( ext{H})=1 ; ext{A}r( ext{C})=12 )

a. Determining the Empirical Formula:

Assume a 100 ext{g} sample:
- Mass of Carbon (C) = 82.76 ext{g}
- Mass of Hydrogen (H) = 17.24 ext{g}
Convert mass to moles:
- Moles of C = \frac{82.76 ext{g}}{12 ext{ g/mol}} \approx 6.897 \text{ mol}
- Moles of H = \frac{17.24 ext{g}}{1 ext{ g/mol}} = 17.24 \text{ mol}
Find the simplest mole ratio:
- Divide moles by the smallest number of moles ( 6.897 ):
  - For C: \frac{6.897}{6.897} \approx 1.00
  - For H: \frac{17.24}{6.897} \approx 2.50
Obtain whole numbers:
- Since 2.50 is not a whole number, multiply both ratios by 2 :
  - For C: 1.00 \times 2 = 2
  - For H: 2.50 \times 2 = 5

Empirical Formula: \text{C}2\text{H}5

b. Determining the Molecular Formula:

Calculate the Empirical Formula Mass (EFM):
- EFM of \text{C}2\text{H}5 = (2 \times 12) + (5 \times 1) = 24 + 5 = 29 ext{ g/mol}
Calculate the Molecular Molar Mass (MM) from Vapour Density:
- Assuming vapour density is relative to hydrogen ( \text{H}_2 ):
  - \text{MM} = 2 \times \text{Vapour Density}
  - \text{MM} = 2 \times 29 = 58 ext{ g/mol}
Determine the integer factor (n):
- n = \frac{\text{MM}}{\text{EFM}}
- n = \frac{58 ext{ g/mol}}{29 ext{ g/mol}} = 2
Calculate the Molecular Formula:
- Multiply the subscripts of the empirical formula by n .
- Molecular Formula = (\text{C}2\text{H}5)2 = \text{C}4\text{H}_{10}

Molecular Formula: \text{C}4\text{H}{10}

Empirical, Molecular & Structural Formulae: Determining Empirical Formula from Combustion Data

Example: Determining Empirical Formula from Combustion Products

Problem: On complete combustion in oxygen, 0.42 ext{g} of a gaseous hydrocarbon, Z, gave 1.32 ext{g} of carbon(IV) oxide ( \text{CO}2 ) and 0.54 ext{g} of steam ( \text{H}2\text{O} ). Calculate the empirical formula of Z. (Given: ext{A}r( ext{H})=1 ; ext{A}r( ext{C})=12 ; ext{A}r( ext{O})=16 )

Solution Approach:

Calculate the mass of Carbon (C) in the products:
- All carbon in the hydrocarbon (Z) is converted to \text{CO}_2 .
- Molar mass of \text{CO}_2 = 12 + (2 \times 16) = 44 ext{ g/mol}
- Mass of C = \text{Mass of CO}2 \times \frac{\text{A}r(\text{C})}{\text{M}r(\text{CO}2)}
- Mass of C = 1.32 ext{g} \times \frac{12}{44} = 0.36 ext{g}
Calculate the mass of Hydrogen (H) in the products:
- All hydrogen in the hydrocarbon (Z) is converted to \text{H}_2\text{O} .
- Molar mass of \text{H}_2\text{O} = (2 \times 1) + 16 = 18 ext{ g/mol}
- Mass of H = \text{Mass of H}2\text{O} \times \frac{2 \times \text{A}r(\text{H})}{\text{M}r(\text{H}2\text{O)}}
- Mass of H = 0.54 ext{g} \times \frac{2}{18} = 0.06 ext{g}
Verify the masses (optional but good practice for hydrocarbons):
- Total mass of C and H from Z = 0.36 ext{g} + 0.06 ext{g} = 0.42 ext{g}
- This matches the initial mass of hydrocarbon Z ( 0.42 ext{g} ), confirming that Z is indeed a hydrocarbon (composed only of C and H).
Convert masses of C and H to moles:
- Moles of C = \frac{0.36 ext{g}}{12 ext{ g/mol}} = 0.03 \text{ mol}
- Moles of H = \frac{0.06 ext{g}}{1 ext{ g/mol}} = 0.06 \text{ mol}
Determine the simplest whole-number ratio of moles:
- Divide both mole values by the smallest number of moles ( 0.03 ext{ mol} ):
  - For C: \frac{0.03}{0.03} = 1
  - For H: \frac{0.06}{0.03} = 2
Write the empirical formula:
- The ratio of C to H is 1:2 . Therefore, the empirical formula of Z is \text{CH}_2 .