The Mole, Chemical Formulas, and Stoichiometry

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High School
Course:
High School Chemistry
Academic year:

2025
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The Mole, Chemical Formulas, and Stoichiometry The Mole ● Definition: The mole (mol) is the SI unit for the amount of a substance. ● Avogadro's Constant NA ● N ● ● A ● ● : 1 mol = 6.022×1023 ● 6.022×10 ● ● 23 ● particles (atoms, molecules, ions, formula units). It is used to convert between moles and the number of particles. ● Molar Mass (M): The mass of one mole of a substance (g/mol). It is numerically equal to the atomic/molecular/formula mass in atomic mass units (u). Relationships ● Moles to Particles: N=n×NA ● N=n×N ● ● A ● ● (where N ● N = # of particles, ● n ● n = # of moles) ● ● Particles to Moles: n=NNA ● n= ● ● N ● A ● ● N ● ● ● Moles to Mass: m=n×M ● m=n×M (where ● m ● m = mass in grams) ● ● Mass to Moles: n=mM ● n= ● ● M ● m ● ● Calculating Molar Mass 1. Find the atomic masses of each element from the periodic table. 2. Multiply each atomic mass by the number of atoms of that element in the chemical formula. 3. Add the results to get the molar mass in g/mol. Example: Find the molar mass of water ( H2O H 2 O). 1. Atomic mass of H = 1.008 g/mol. Atomic mass of O = 16.00 g/mol. 2. (2 * 1.008 g/mol) + (1 * 16.00 g/mol) 2. 016 g/mol + 16.00 g/mol = 18.016 g/mol Therefore, the molar mass of water is 18.016 g/mol. Chemical Proportions and Formulas ● Law of Definite Proportions: A chemical compound always contains the same elements in the same proportions by mass. ● Mass Percent: ● Mass of elementMass of compound×100% ● Mass of compound ● Mass of element ● ● ×100% ● Percentage Composition: The percent by mass of each element in a compound. ● Empirical Formula: The simplest whole-number ratio of atoms in a compound. ● Molecular Formula: The actual number of atoms of each element in a molecule. ● Molecular Formula=(Empirical Formula)×n ● Molecular Formula=(Empirical Formula)×n ● where n ● n is a whole number. ● ● Hydrate: A compound with a specific number of water molecules bound to its structure. ● Anhydrous: A substance without water molecules. Calculations Mass Percent ● Given mass data: ● Mass of elementMass of compound×100% ● Mass of compound ● Mass of element ● ● ×100% ● From chemical formula: ● Mass of element in 1 molMolar mass of compound×100% ● Molar mass of compound ● Mass of element in 1 mol ● ● ×100% Empirical Formula 1. Assume a 100g sample if given percentages. Convert % to grams. 2. Convert grams to moles for each element ( 3. massmolar mass 4. molar mass 5. mass 6. 7. ). 8. Divide each mole value by the smallest mole value to get the mole ratio. 9. If necessary, multiply by a whole number to get whole number subscripts. Example: A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen. Find the empirical formula. 1. Assume 100g: 40g C, 6.7g H, 53.3g O. nC=40 g12.01 g/mol=3.33 mol 3. n 2. 4. C 5. 6. = 7. 12.01 g/mol 8. 40 g 9. 10.=3.33 mol 11.nH=6.7 g1.008 g/mol=6.65 mol 12.n 13.H 14. 15.= 16.1.008 g/mol 17.6.7 g 18. 19.=6.65 mol 20.nO=53.3 g16.00 g/mol=3.33 mol 21.n 22.O 23. 24.= 25.16.00 g/mol 26.53.3 g 27. 28.=3.33 mol 29.Divide by the smallest (3.33): C = 1, H = 2, O = 1. 30.The empirical formula is 31.CH2O 32.CH 33.2 34. 35.O. Molecular Formula 1. Calculate the molar mass of the empirical formula. n=Molar mass of molecular formulaMolar mass of empirical formula 3. n= 2. 4. Molar mass of empirical formula 5. Molar mass of molecular formula 6. 7. 8. Multiply the subscripts in the empirical formula by n 10.n. 9. Example: A compound has an empirical formula of CH2O CH 2 O and a molar mass of 180 g/mol. Find the molecular formula. 1. Molar mass of CH2O 3. CH 2. 4. 2 5. O = 12.01 + (2 * 1.008) + 16.00 = 30.026 g/mol. 7. n=180 g/mol30.026 g/mol=6 8. n= 6. 9. 30.026 g/mol 10.180 g/mol 11. 12.=6 13.C1×6H2×6O1×6 14.C 15.1×6 16. 17.H 18.2×6 19. 20.O 21.1×6 22. 23. The molecular formula is 24.C6H12O6 25.C 26.6 27. 28.H 29.12 30. 31.O 32.6 33. 34.. Hydrate Formula 1. Find the mass of water lost (mass of hydrate - mass of anhydrous compound). 2. Convert the mass of water and anhydrous compound to moles. 3. Divide moles of water by moles of anhydrous compound to find x 5. x in the formula: Salt · 6. xH2O 7. xH 4. 8. 2 9. 10.O. Example: A hydrate of CuSO4 CuSO 4 weighs 2.50 g. After heating, the anhydrous salt weighs 1.59 g. Find the formula of the hydrate. 1. Mass of water lost = 2.50 g - 1.59 g = 0.91 g. nCuSO4=1.59 g159.61 g/mol=0.00996 mol 3. n 2. 4. CuSO 5. 4 6. 7. 8. = 9. 159.61 g/mol 10.1.59 g 11. 12.=0.00996 mol 13.nH2O=0.91 g18.016 g/mol=0.0505 mol 14.n 15.H 16.2 17. 18.O 19. 20.= 21.18.016 g/mol 22.0.91 g 23. 24.=0.0505 mol 25.0.0505 mol0.00996 mol=5.07≈5 26.0.00996 mol 27.0.0505 mol 28. 29.=5.07≈5 The formula of the hydrate is 30.CuSO4⋅5H2O 31.CuSO 32.4 33. 34.⋅5H 35.2 36. 37.O. Stoichiometry ● Stoichiometry: Study of quantitative relationships in chemical reactions. ● Mole Ratio: Ratio of moles of any two substances in a balanced chemical equation. ● Limiting Reactant: Reactant completely consumed in a reaction, limiting the amount of product formed. ● Excess Reactant: Reactant present in a greater amount than necessary to react completely with the limiting reactant. ● Theoretical Yield: The maximum amount of product that can be formed from a given amount of reactants, assuming complete conversion of the limiting reactant. ● Actual Yield: The amount of product actually obtained from a chemical reaction. ● Percentage Yield: ● Actual YieldTheoretical Yield×100% ● Theoretical Yield ● Actual Yield ● ● ×100% Calculations 1. Balance the Chemical Equation: This is crucial for determining correct mole ratios. 2. Convert to Moles: Convert the given masses of reactants to moles using their molar masses. 3. Determine the Limiting Reactant: ● Calculate the moles of product that each reactant could produce, assuming the other reactant is in excess. ● The reactant that produces the least amount of product is the limiting reactant. 4. Calculate the Theoretical Yield: Use the moles of the limiting reactant and the mole ratio from the balanced equation to calculate the theoretical yield (in moles) of the desired product. Convert this to mass using the molar mass of the product. 5. Calculate the Percentage Yield: 6. Actual YieldTheoretical Yield×100% 7. Theoretical Yield 8. Actual Yield 9. 10.×100% Example: Consider the reaction: C6H12O6→2C2H5OH+2CO2 C 6 H 12 O 6 →2C 2 H 5 OH+2CO 2 . If 20.0 g of glucose ( C6H12O6 C 6 H 12 O 6 ) produces 1.40 g of ethanol ( C2H5OH C 2 H 5 OH), what is the percentage yield? 1. The equation is balanced. nC6H12O6=20.0 g180.16 g/mol=0.111 mol 3. n 2. 4. C 5. 6 6. 7. H 8. 12 9. 10.O 11.6 12. 13. 14.= 15.180.16 g/mol 16.20.0 g 17. 18.=0.111 mol 19.Since there's only one reactant, glucose is the limiting reactant. 20.From the balanced equation, 1 mole of glucose produces 2 moles of ethanol. Therefore, 0.111 mol of glucose can produce 21.2×0.111 mol=0.222 mol 22.2×0.111 mol=0.222 mol of ethanol. Theoretical yield of ethanol: 23.0.222 mol×46.07 g/mol=10.23 g 24.0.222 mol×46.07 g/mol=10.23 g 25.Percentage yield: 26.1.40 g10.23 g×100%=13.7% 27.10.23 g 28.1.40 g 29. 30.×100%=13.7% Factors Affecting Yield ● ● ● ● ● Impure Reactants: Contaminants lower actual yield. Competing Reactions: Side reactions reduce the desired product. Reaction Conditions: Vessel surface, shape affect collisions. Lab Techniques: Loss during transfer, filtration, etc. Solubility: Product loss if slightly soluble.