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Química Geral · Atoms, moles and stoichiometry

Mols e fórmulas químicas

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O que você vai aprender

  • Use Avogadro's number to convert between moles and numbers of molecules, formula units, ions and atoms.
  • Calculate molar masses of molecules, ionic compounds and hydrates, and convert between mass, moles and particles.
  • Calculate the mass percentage of each element in a compound and use percent composition to compare or identify compounds.
  • Find an empirical formula from mass percentages, element masses or combustion-analysis data.
  • Find a molecular formula from an empirical formula and a molar mass, and recognise when a molar mass is needed to decide it.

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The mole and Avogadro's number

A mole is a counting unit, like a dozen. One mole is exactly 6.02214076 × 10²³ specified entities (atoms, molecules, ions or formula units); for calculations use NA = 6.022 × 10²³ mol⁻¹. Always say which entity you are counting. One mole of water (drawn) is 6.022 × 10²³ H₂O molecules, but each molecule has three atoms, so it holds 3 × 6.022 × 10²³ = 1.807 × 10²⁴ atoms: 2 mol of H atoms and 1 mol of O atoms. The subscripts in a formula are mole ratios: 1 mol of Al₂(SO₄)₃ contains 2 mol Al³⁺ and 3 mol SO₄²⁻, 5 mol of ions in all, and 12 mol of O atoms. Moles × NA = particles; particles ÷ NA = moles.

Molar mass

The molar mass M is the mass of one mole, in g/mol. Add the atomic masses of every atom in the formula, each multiplied by its subscript; a subscript outside parentheses multiplies everything inside. Mg(OH)₂: 24.305 + 2 × (15.999 + 1.008) = 58.319 g/mol. For a hydrate, add the water molecules shown after the dot: BaCl₂·2H₂O is 208.23 + 2 × 18.015 = 244.26 g/mol. The number is the same as the formula mass in atomic mass units because NA is, to better than one part in a billion, the number of atomic mass units in one gram: one water molecule has a mass of 18.015 u, and 6.022 × 10²³ of them have a mass of 18.015 g. The mass of a single particle is therefore M ÷ NA; one H₂O molecule weighs 18.015 ÷ 6.022 × 10²³ = 2.991 × 10⁻²³ g.

Converting mass, moles and particles

Moles sit at the centre of every conversion: mass ÷ M = moles, moles × M = mass, moles × NA = particles. Never jump straight from grams to particles. Example: 10.0 g of NaHCO₃ (M = 84.006 g/mol) is 10.0 ÷ 84.006 = 0.119 mol, which is 0.119 × 6.022 × 10²³ = 7.17 × 10²² formula units and, with three O per formula unit, 2.15 × 10²³ O atoms. For a liquid measured by volume, first use the density to get its mass. Equal masses of different substances contain different numbers of moles: the smaller the molar mass, the more moles per gram. To compare numbers of atoms, multiply the moles of substance by the atoms in each formula unit.

Percent composition

The mass percentage of an element = (number of its atoms in the formula × its atomic mass ÷ molar mass) × 100%. It is a ratio of masses, not of atom counts. In ammonium sulfate, (NH₄)₂SO₄ (drawn), M = 132.134 g/mol and % N = 2 × 14.007 ÷ 132.134 × 100% = 21.20%. Multiply the mass fraction by a sample mass to find how much of the element it contains: 10.0 kg of ammonium sulfate holds 2.12 kg of nitrogen. The water in a hydrate is treated the same way: BaCl₂·2H₂O is 2 × 18.015 ÷ 244.26 = 14.75% water. A fixed compound always has the same composition, whatever the sample size. Percent composition depends only on the atom ratio, so compounds with the same empirical formula (C₂H₂ and C₆H₆, both 92.26% C) cannot be told apart by it.

Empirical formulas from mass percentages

The empirical formula is the smallest whole-number ratio of atoms. Assume 100 g of compound so that each percentage becomes grams, convert each to moles with its atomic mass, divide every value by the smallest, then clear any fraction by multiplying all values by the same small integer: x.50 → × 2, x.33 or x.67 → × 3, x.25 or x.75 → × 4. Round only values within about 0.1 of a whole number; 1.33 is not 1. Ascorbic acid (vitamin C, drawn) is 40.92% C, 4.58% H and 54.50% O: 40.92 ÷ 12.011 = 3.407 mol C, 4.58 ÷ 1.008 = 4.544 mol H, 54.50 ÷ 15.999 = 3.406 mol O. Dividing by 3.406 gives 1 : 1.334 : 1, and multiplying by 3 gives C₃H₄O₃.

Combustion analysis and molecular formulas

In combustion analysis a weighed sample burns in excess O₂; all its carbon ends up in CO₂ and all its hydrogen in H₂O, which are trapped and weighed. Mass of C = mass of CO₂ × 12.011 ÷ 44.009 (27.29%); mass of H = mass of H₂O × 2 × 1.008 ÷ 18.015 (11.19%). Oxygen cannot be found from the products, because part of the oxygen in them came from the O₂ supply, so oxygen is found by difference: sample mass − C − H. Example: 0.3000 g of a compound of C, H and O gives 0.4397 g CO₂ and 0.1800 g H₂O, so C = 0.1200 g (0.00999 mol), H = 0.02014 g (0.01998 mol) and O = 0.3000 − 0.1200 − 0.0201 = 0.1599 g (0.00999 mol): a 1 : 2 : 1 ratio, empirical formula CH₂O. The molecular formula is a whole-number multiple n of the empirical formula, with n = molar mass ÷ empirical formula mass. If the molar mass is 60.05 g/mol, n = 60.05 ÷ 30.026 = 2 and the compound is C₂H₄O₂; for ascorbic acid, 176.12 ÷ 88.06 = 2 gives C₆H₈O₆. Composition data alone can never fix n, because CH₂O, C₂H₄O₂ and C₆H₁₂O₆ have identical mass percentages; a separate molar-mass measurement, such as mass spectrometry or vapour density, is always needed.

What to carry forward

Route every calculation through moles: grams ÷ molar mass = moles, moles × 6.022 × 10²³ = particles, and the formula's subscripts are mole ratios. Mass percent is (atoms × atomic mass ÷ molar mass) × 100. For an empirical formula, turn masses into moles, divide by the smallest, and multiply to clear fractions instead of rounding them away; get C from CO₂, H from H₂O and O by difference. The molecular formula is the empirical formula times molar mass ÷ empirical formula mass.

Problemas de prática (40)

Praticar é grátis com uma conta.

Preveja o produto 1

  • Products from burning propan-1-olPredict the amounts of CO₂ and H₂O formed when a known amount of a C, H, O compound burns completely.

Classificar 2

Moles and Avogadro's number

  • What one mole of water containsDistinguish the number of molecules from the number of atoms in one mole of a molecular compound.

Empirical and molecular formulas

  • Spot the formula that is already empiricalClassify molecular formulas by whether they are already in lowest whole-number ratio.

Comparar 5

Molar mass and conversions

  • Which gram contains the most atoms?Compare the total number of atoms in equal masses of substances that differ in both molar mass and atoms per molecule.
  • Rank samples given in different units by massRank samples specified as moles or as numbers of molecules by converting each to a mass.

Percent composition

  • Rank iron minerals by iron contentRank several compounds of one element by its mass percentage, computed from their formulas.

Moles and Avogadro's number

  • Rank equal masses by number of moleculesRank equal masses of different molecular substances by number of molecules using their molar masses.
  • Rank samples by total number of atomsRank samples given in moles of different substances by their total number of atoms.

Calcular 19

Molar mass and conversions

  • Molar mass of caffeineCalculate the molar mass of a molecular compound from its formula and atomic masses.
  • Moles in 50.0 g of calcium carbonateConvert a mass of an ionic compound into moles with its molar mass.
  • Molecules in a measured volume of ethanolConvert a volume of a pure liquid into a number of molecules using its density and molar mass.
  • Molar mass of a hydrateCalculate the molar mass of a hydrated salt, including its water of crystallization.
  • Mass of a number of nitrogen moleculesConvert a number of molecules into a mass by going through moles.
  • Mass of 0.150 mol of sulfuric acidConvert an amount in moles into a mass with the molar mass.

Empirical and molecular formulas

  • Carbon in a sample from the CO₂ it formsCalculate the mass of carbon in a burned sample from the mass of carbon dioxide collected.
  • Empirical formula with a fractional ratioConvert a non-integer mole ratio into a whole-number empirical formula by multiplying rather than rounding.
  • Empirical formula from mass percentagesDetermine an empirical formula from mass percentages by converting to moles and dividing by the smallest.

Moles and Avogadro's number

  • Moles from a number of moleculesConvert a number of molecules into moles by dividing by Avogadro's number.
  • Hydrogen atoms in a sample of methaneCount the atoms of one element in a sample by combining moles, formula subscripts and Avogadro's number.
  • Moles of ions from a mass of saltConvert the mass of an ionic compound into the total amount of ions it releases using its formula.
  • Mass of a single gold atomFind the mass of one atom by dividing the molar mass by Avogadro's number.
  • Oxygen atoms in calcium nitrateCount the atoms of one element in a weighed sample of an ionic compound whose formula contains a parenthesised polyatomic ion.

Percent composition

  • Nitrogen delivered by a bag of ureaUse a mass percentage to find the mass of an element contained in a large sample.
  • Percent nitrogen in ammonium nitrateCalculate the mass percentage of an element that appears in two different ions of a salt.
  • Percent oxygen in glucoseCalculate the mass percentage of oxygen in a molecular compound from its formula.
  • Composition of a NaCl–KCl mixtureFind the composition of a two-salt mixture from the total mass of a shared element using the mass fraction of that element in each salt.
  • Water driven off a hydrateUse the mass fraction of water in a hydrate to predict the mass lost on complete dehydration.

Encontre o erro 4

Molar mass and conversions

  • Find the error in a molar-mass calculationDiagnose a molar mass that ignores a subscript applied to a parenthesised group.

Empirical and molecular formulas

  • Find the error in an empirical formulaDiagnose an empirical-formula calculation that rounds a one-third fraction instead of clearing it.

Moles and Avogadro's number

  • Find the error in a molecule countDiagnose a mass-to-molecules calculation that uses the atomic mass of an element instead of the molar mass of its diatomic molecule.

Percent composition

  • Find the error in a percent compositionDiagnose a percent-composition calculation that leaves out the subscript of an element.

Interprete as evidências 1

  • Comparing the analyses of two samplesInterpret elemental analyses of two samples of different mass by comparing their mass percentages.

Trabalhe de trás para frente 4

Empirical and molecular formulas

  • Molecular formula from combustion analysisCombine combustion-analysis data with a molar mass to choose between molecular formulas of the same molar mass.
  • Molecular formula from empirical formula and molar massDetermine a molecular formula by multiplying every subscript of the empirical formula by the ratio of molar mass to empirical formula mass.
  • Empirical formula from a synthesisFind an empirical formula from the masses of an element and of the compound it forms, clearing a half-integer mole ratio.

Percent composition

  • Identify a metal from its chloride's compositionWork backwards from a mass percentage and a formula type to the molar mass and identity of an unknown element.

Planeje uma síntese 1

  • What to measure after combustion analysisChoose the measurement needed to turn an empirical formula into a molecular formula.

Explique por quê 3

Moles and Avogadro's number

  • Why u and g/mol share the same numberExplain why a formula mass in atomic mass units equals the molar mass in grams per mole.

Empirical and molecular formulas

  • Why oxygen is found by differenceExplain why the oxygen content of a compound cannot be read from the masses of its combustion products.

Percent composition

  • Two hydrocarbons with identical compositionExplain why compounds that share an empirical formula have identical percent compositions.

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