Chemistry · Ch 2 — Introduction to Analytical Chemistry

Stoichiometric problems

2.5.1

Stoichiometric problems

Problems based on stoichiometry generally fall into one of three categories, depending on what is given and what is asked for: problems based on a mass–mass relationship (a mass of one substance is given, and a mass of another substance in the same reaction is required); problems based on a mass–volume relationship (a mass is given and a gas volume is required, or vice versa); and problems based on a volume–volume relationship (a volume of one gas is given and a volume of another gas in the same reaction is required). Regardless of which category a problem falls into, the same four-step method solves it: first, write down the balanced chemical equation that represents the reaction; second, write the number of moles and the corresponding relative masses (or, for gases, volumes at STP) directly below the formula of each reactant and product involved; third, calculate those relative masses or volumes from each substance's formula mass, referring to STP conditions for any gas; and fourth, apply the unitary method — scaling the known relationship up or down in direct proportion — to calculate whatever unknown quantity the problem asks for. The four worked problems in this section ( …

Misc Problem 2.11Mass of carbon dioxide and water from the combustion of methane

Worked out. Worked example (mass-mass relationship): calculate the mass of carbon dioxide and water formed on complete combustion of 24 g of methane gas (atomic masses C = 12, H = 1, O = 16). Balanced equation: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)$ . Formula masses: $CH_4 = 16$ g, $CO_2 = 44$ g, $2H_2O = 36$ g. So 16 g of $CH_4$ on complete combustion produces 44 g of $CO_2$ ; therefore 24 g of $CH_4$ produces $(24/16) \times 44 = 66$ g of $CO_2$ . Similarly, 16 g of $CH_4$ produces 36 g of water, so 24 g of $CH_4$ produces $(24/16) \times 36 …

Misc Problem 2.12Mass of calcium oxide from decomposition of calcium carbonate

Worked out. Worked example (mass-mass relationship): how much CaO will be produced by decomposition of 5 g of $CaCO_3$ ? Balanced equation: $CaCO_3 \xrightarrow{\Delta} CaO + CO_2$ . Formula masses: $CaCO_3 = 40 + 12 + 3(16) = 100$ parts; $CaO = 40 + 16 = 56$ parts; $CO_2 = 12 + 2(16) = 44$ parts. So 100 g of $CaCO_3$ produces 56 g of CaO; therefore 5 g of $CaCO_3$ produces $(56/100) \times 5 = 2.8$ g of CaO. …

Misc Problem 2.13Volume of oxygen needed to burn propane completely

Worked out. Worked example (mass-volume relationship): how many litres of oxygen at STP are required to burn completely 2.2 g of propane, $C_3H_8$ ? Balanced equation: $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$ . Formula mass of $C_3H_8 = 3(12) + 8(1) = 44$ g; volume of $5O_2$ at STP $= 5 \times 22.4 = 112$ L (since 1 mole of an ideal gas occupies 22.4 L at STP). So 44 g of propane requires 112 L of oxygen; therefore 2.2 g of propane requires $(112/44) \times 2.2 = 5.6$ L …

Misc Problem 2.14Percentage purity of zinc from the volume of hydrogen liberated

Worked out. Worked example (mass-volume relationship): a 0.635 g piece of zinc treated with excess dilute $H_2SO_4$ liberates 200 cm3 of hydrogen at STP; find the percentage purity of the zinc sample (atomic mass Zn = 65). Balanced equation: $Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2$ , so 22.4 L of hydrogen at STP corresponds to 65 g of Zn. For $0.200$ L (200 cm3) of hydrogen at STP, the mass of Zn that reacted $= (65/22.4) \times 0.200 = 0.58$ g. Percentage purity of the zinc sample $= (0.58/0.635) \times 100 = …

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