A balanced chemical equation shows the relative number of moles of each reactant and product involved in a chemical reaction. Stoichiometry is the study of these quantitative relationships in chemical reactions.
State Symbols
Every balanced chemical equation in A-Level Chemistry should include state symbols to specify the physical state of each species:
- (s) = solid (e.g. \( \text{Mg(s)} \))
- (l) = liquid (e.g. \( \text{H}_2\text{O(l)} \))
- (g) = gas (e.g. \( \text{CO}_2\text{(g)} \))
- (aq) = aqueous solution, dissolved in water (e.g. \( \text{HCl(aq)} \))
Reacting Mass Calculation Pathway
To calculate the mass of a product or reactant from a known mass of another species, use this standard three-step pathway:
Step 1: Find the moles of the reactant with known mass, \( \text{Fe}_2\text{O}_3 \).
\( M_r(\text{Fe}_2\text{O}_3) = (2 \times 55.8) + (3 \times 16.0) = 159.6 \text{ g mol}^{-1} \)
\[ n(\text{Fe}_2\text{O}_3) = \frac{16.0}{159.6} = 0.100 \text{ mol} \]
Step 2: Use the stoichiometric ratio from the balanced equation to find moles of \( \text{Fe} \).
The equation shows 1 mole of \( \text{Fe}_2\text{O}_3 \) produces 2 moles of \( \text{Fe} \). Therefore, the ratio is \( 1 : 2 \).
\[ n(\text{Fe}) = 0.100 \times 2 = 0.200 \text{ mol} \]
Step 3: Convert moles of \( \text{Fe} \) to mass.
\( A_r(\text{Fe}) = 55.8 \text{ g mol}^{-1} \)
\[ m(\text{Fe}) = 0.200 \times 55.8 = 11.2 \text{ g} \]
Percentage Yield
In practice, the actual mass of product obtained in a reaction is almost always less than the theoretical maximum mass calculated. The percentage yield tells us how efficient the process is:
Reasons for a yield of less than 100% include:
- The reaction may be reversible and reach equilibrium.
- Side reactions may occur, producing unexpected products.
- Some product may be lost during filtration, transfer, or purification.
Plug the values into the yield equation:
\[ \text{Percentage Yield} = \frac{9.0}{11.2} \times 100 = 80.4\% \]
Atom Economy
Atom economy is a measure of the proportion of starting materials that end up in the desired product rather than as waste products.
When calculating atom economy, you must multiply the relative formula mass of each species by its balancing coefficient from the balanced equation. For example, if a reactant is \( 3\text{H}_2 \), you must use \( 3 \times 2.0 = 6.0 \) in the total mass calculation.
Step 1: Find the total formula mass of the desired product.
Desired product = \( 3\text{H}_2 \). Mass = \( 3 \times (2 \times 1.0) = 6.0 \text{ g mol}^{-1} \)
Step 2: Find the sum of the formula masses of all reactants.
Reactants = \( \text{CH}_4 + \text{H}_2\text{O} \)
Sum = \( [12.0 + (4 \times 1.0)] + [(2 \times 1.0) + 16.0] = 16.0 + 18.0 = 34.0 \text{ g mol}^{-1} \)
Step 3: Calculate atom economy.
\[ \text{Atom Economy} = \frac{6.0}{34.0} \times 100 = 17.6\% \]
Economic and Environmental Benefits
Developing reactions with high atom economies and yields is a central goal of industrial green chemistry:
- Economic benefits: High atom economy reduces the money spent on raw materials that just end up as useless waste. It also reduces waste treatment and disposal costs.
- Environmental benefits: Reducing waste limits environmental pollution. It preserves raw resource reserves, making the process more sustainable.
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