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Chemistry: Stoichiometry - Mass, Mole, and Gas-Volume Calculations

Grade 8IB

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

🔑Concepts

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The Mole (nn) is the SI unit for amount of substance. One mole contains exactly 6.022×10236.022 \times 10^{23} elementary entities (Avogadro's number, NAN_A).

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Molar Mass (MM) is the mass of one mole of a substance, expressed in g⋅mol−1g \cdot mol^{-1}. It is numerically equal to the relative atomic/molecular mass (ArA_r or MrM_r).

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Stoichiometry involves using the balanced chemical equation to calculate the relative quantities of reactants and products. The coefficients in a balanced equation represent the mole ratio.

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Molar Volume (VmV_m): At Standard Temperature and Pressure (STP), one mole of any ideal gas occupies a fixed volume. In the IB curriculum, STP is defined as 0∘C0^{\circ}C (273 K273 \text{ K}) and 100 kPa100 \text{ kPa}, where Vm=22.7 dm3⋅mol−1V_m = 22.7 \text{ dm}^3 \cdot mol^{-1}.

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Law of Conservation of Mass: The total mass of reactants must equal the total mass of products in a chemical reaction.

📐Formulae

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

N=n×NAN = n \times N_A

V=n×VmV = n \times V_m

Percentage Yield=(Actual YieldTheoretical Yield)×100%\text{Percentage Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%

💡Examples

Problem 1:

Calculate the mass of 2.5 moles2.5 \text{ moles} of Magnesium Oxide (MgOMgO). Given: Ar(Mg)=24.31,Ar(O)=16.00A_r(Mg) = 24.31, A_r(O) = 16.00.

Solution:

M(MgO)=24.31+16.00=40.31 g/molM(MgO) = 24.31 + 16.00 = 40.31 \text{ g/mol} m=n×M=2.5 mol×40.31 g/mol=100.775 gm = n \times M = 2.5 \text{ mol} \times 40.31 \text{ g/mol} = 100.775 \text{ g}

Explanation:

First, find the molar mass of the compound by summing the atomic masses. Then, multiply the number of moles by the molar mass to find the total mass.

Problem 2:

What volume would 11.0 g11.0 \text{ g} of Carbon Dioxide (CO2CO_2) gas occupy at STP? (Use Vm=22.7 dm3⋅mol−1V_m = 22.7 \text{ dm}^3 \cdot mol^{-1} and M(CO2)=44.01 g/molM(CO_2) = 44.01 \text{ g/mol})

Solution:

n=mM=11.0 g44.01 g/mol≈0.25 moln = \frac{m}{M} = \frac{11.0 \text{ g}}{44.01 \text{ g/mol}} \approx 0.25 \text{ mol} V=n×Vm=0.25 mol×22.7 dm3/mol=5.675 dm3V = n \times V_m = 0.25 \text{ mol} \times 22.7 \text{ dm}^3/mol = 5.675 \text{ dm}^3

Explanation:

Convert the mass of the gas to moles first using the molar mass. Once you have the moles, multiply by the molar volume constant for STP to find the volume.

Problem 3:

In the reaction 2H2(g)+O2(g)→2H2O(l)2H_2(g) + O_2(g) \rightarrow 2H_2O(l), how many moles of O2O_2 are required to react completely with 4.0 g4.0 \text{ g} of H2H_2 gas?

Solution:

n(H2)=4.0 g2.02 g/mol≈1.98 moln(H_2) = \frac{4.0 \text{ g}}{2.02 \text{ g/mol}} \approx 1.98 \text{ mol} From the equation, the ratio of H2:O2H_2 : O_2 is 2:12:1. n(O2)=12×n(H2)=1.982=0.99 moln(O_2) = \frac{1}{2} \times n(H_2) = \frac{1.98}{2} = 0.99 \text{ mol}

Explanation:

Convert the given mass of hydrogen to moles. Use the stoichiometric ratio from the balanced equation (2:1) to determine the required moles of oxygen.