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Reactivity 2. How much, how fast and how far? - How much? The amount of chemical change

Grade 12IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The mole (nn) is the SI unit for the amount of substance. One mole contains exactly 6.02×10236.02 \times 10^{23} elementary entities (atoms, molecules, or ions). This value is the Avogadro constant (LL or NAN_A).

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Relative atomic mass (ArA_r) is the weighted average mass of an atom of an element relative to 112\frac{1}{12} of the mass of an atom of carbon-12. Molar mass (MM) has the same numerical value but units of g mol−1\text{g mol}^{-1}.

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The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. The molecular formula shows the actual number of atoms and is a multiple of the empirical formula: molecular formula=(empirical formula)×n\text{molecular formula} = (\text{empirical formula}) \times n.

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Stoichiometry involves using the coefficients of a balanced chemical equation to determine the relative amounts of reactants and products. The molar ratio is the ratio of coefficients.

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The limiting reactant is the reactant that is completely consumed first in a chemical reaction, thereby determining the maximum amount of product that can be formed (the theoretical yield).

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Percentage yield measures the efficiency of a reaction: Percentage Yield=Experimental YieldTheoretical Yield×100%\text{Percentage Yield} = \frac{\text{Experimental Yield}}{\text{Theoretical Yield}} \times 100\%.

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Avogadro's Law states that equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules. At STP (273 K273 \text{ K} and 100 kPa100 \text{ kPa}), the molar volume of an ideal gas is 22.7 dm3 mol−122.7 \text{ dm}^3 \text{ mol}^{-1}.

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The Ideal Gas Equation PV=nRTPV = nRT relates pressure (PP in Pa\text{Pa}), volume (VV in m3\text{m}^3), temperature (TT in K\text{K}), and amount (nn in mol\text{mol}), where R=8.31 J K−1 mol−1R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}.

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Standard solutions have a known concentration (cc). Concentration is typically measured in mol dm−3\text{mol dm}^{-3} where 1 dm3=1000 cm31 \text{ dm}^3 = 1000 \text{ cm}^3.

📐Formulae

n=mMn = \frac{m}{M}

n=NNAn = \frac{N}{N_A}

n=c×Vn = c \times V

PV=nRTPV = nRT

P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

% Yield=Actual YieldTheoretical Yield×100\% \text{ Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100

% Atom Economy=Molar mass of desired productTotal molar mass of reactants×100\% \text{ Atom Economy} = \frac{\text{Molar mass of desired product}}{\text{Total molar mass of reactants}} \times 100

💡Examples

Problem 1:

Calculate the volume of carbon dioxide gas, in dm3\text{dm}^3, produced at STP when 5.00 g5.00 \text{ g} of calcium carbonate (CaCO3CaCO_3) is reacted with excess hydrochloric acid (HClHCl).

Solution:

M(CaCO3)=40.08+12.01+(3×16.00)=100.09 g mol−1M(CaCO_3) = 40.08 + 12.01 + (3 \times 16.00) = 100.09 \text{ g mol}^{-1} n(CaCO3)=5.00 g100.09 g mol−1=0.04995 moln(CaCO_3) = \frac{5.00 \text{ g}}{100.09 \text{ g mol}^{-1}} = 0.04995 \text{ mol} Equation: CaCO3(s)+2HCl(aq)→CaCl2(aq)+H2O(l)+CO2(g)CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g) From the stoichiometry, 1 mol CaCO31 \text{ mol } CaCO_3 produces 1 mol CO21 \text{ mol } CO_2. n(CO2)=0.04995 moln(CO_2) = 0.04995 \text{ mol} V(CO2)=n×Vm=0.04995 mol×22.7 dm3 mol−1=1.13 dm3V(CO_2) = n \times V_m = 0.04995 \text{ mol} \times 22.7 \text{ dm}^3 \text{ mol}^{-1} = 1.13 \text{ dm}^3

Explanation:

First, calculate the molar mass of the reactant. Then, find the moles of the reactant. Use the balanced equation to find the molar ratio between the known reactant and the required product. Finally, multiply the moles of the gas by the molar volume at STP (22.7 dm3 mol−122.7 \text{ dm}^3 \text{ mol}^{-1}).

Problem 2:

A gas occupies 250 cm3250 \text{ cm}^3 at a pressure of 1.00×105 Pa1.00 \times 10^5 \text{ Pa} and a temperature of 25 ∘C25 \text{ } ^\circ C. What will be its volume at 101.3 kPa101.3 \text{ kPa} and 0 ∘C0 \text{ } ^\circ C?

Solution:

Convert all units to SI: V1=250 cm3V_1 = 250 \text{ cm}^3, P1=100 kPaP_1 = 100 \text{ kPa}, T1=25+273=298 KT_1 = 25 + 273 = 298 \text{ K} P2=101.3 kPaP_2 = 101.3 \text{ kPa}, T2=0+273=273 KT_2 = 0 + 273 = 273 \text{ K} Use the combined gas law: P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} V2=P1V1T2T1P2V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} V2=100×250×273298×101.3V_2 = \frac{100 \times 250 \times 273}{298 \times 101.3} V2=226 cm3V_2 = 226 \text{ cm}^3

Explanation:

The combined gas law is used when conditions of pressure, volume, or temperature change for a fixed mass of gas. Ensure temperature is always in Kelvin (KK) and units for pressure/volume are consistent on both sides.