[{"data":1,"prerenderedAt":229},["ShallowReactive",2],{"topic-the-mole-and-chemical-formulas":3},{"topic":4,"prev":214,"next":222},{"id":5,"slug":6,"title":7,"shortTitle":8,"lesson":5,"lastAlignmentAudit":9,"category":10,"description":11,"metaDescription":12,"objectives":13,"conceptSections":35,"workedExample":125,"oerSources":136,"relatedTopicSlugs":139,"relatedElements":143,"relatedReferences":149,"keyEquations":150,"howTo":164,"conceptProbe":173,"practiceTeaser":177,"seoKeywords":185,"selfStudyQuestions":191,"objectiveCount":213},8,"the-mole-and-chemical-formulas","The Mole and Chemical Formulas","Moles & Composition","2026-06-22","quantitative","Understand the mole concept, molar mass, percent composition, empirical formulas, and molecular formulas.","Master the mole concept: molar mass calculations, percent composition, empirical and molecular formula determination. Avogadro's number explained clearly.",[14,17,20,23,26,29,32],{"id":15,"text":16},"8.1","Define the mole and use Avogadro's number to convert between moles and number of particles",{"id":18,"text":19},"8.2","Calculate molar mass of elements and compounds",{"id":21,"text":22},"8.3","Interconvert among moles, mass, and number of particles using molar mass and Avogadro's number",{"id":24,"text":25},"8.4","Calculate percent composition by mass from a chemical formula",{"id":27,"text":28},"8.5","Determine an empirical formula from percent composition or mass data",{"id":30,"text":31},"8.6","Determine a molecular formula from the empirical formula and molar mass",{"id":33,"text":34},"8.7","Determine an empirical formula from combustion analysis data",[36,48,57,66,75,84,93,98,107,116,121],{"heading":37,"content":38,"relatedObjectives":39,"deepDive":41},"The Mole and Avogadro's Number","\u003Cp>Atoms and molecules are far too small to count individually, so chemists use a counting unit called the \u003Cstrong>mole\u003C/strong> (mol). One mole of anything contains exactly 6.022 × 10²³ items — this is \u003Cstrong>Avogadro's number\u003C/strong> (N\u003Csub>A\u003C/sub>).\u003C/p>\u003Cp>To put that in perspective, one mole of pennies stacked up would reach the sun and back more than a billion times. Yet one mole of water molecules is just 18 grams — roughly a tablespoon.\u003C/p>\u003Cp>The mole bridges the gap between the atomic scale (where mass is measured in amu) and the laboratory scale (grams). This makes it the chemist's most essential conversion unit: every quantitative calculation in chemistry passes through moles at some point.\u003C/p>",[40],54,[42,45],{"label":43,"body":44},"How anyone counted to Avogadro’s number","\u003Cp>Nobody tallied a mole of atoms one by one; the number was pinned down by measuring in bulk and dividing. One modern route: grow a nearly perfect silicon sphere, measure its volume, then use X-ray diffraction to measure the exact spacing of atoms in the crystal, which tells you how many atoms fit inside. The counting became so precise that in 2019 the logic was inverted: the mole is now \u003Cem>defined\u003C/em> as exactly 6.02214076 × 10²³ entities, a fixed constant of the SI system rather than a measured property of carbon-12.\u003C/p>",{"label":46,"body":47},"Why this matters: a sip of water outnumbers the ocean","\u003Cp>A single glass of water holds more molecules than the entire ocean holds glasses of water. The arithmetic is quick: 250 mL of water is about 14 moles, roughly 8 × 10²⁴ molecules, while all of Earth’s oceans (about 1.3 × 10²¹ L) contain only around 5 × 10²¹ glass-sized portions. That gap of more than a thousandfold is why chemistry needs the mole: the objects are so absurdly numerous that only a bulk counting unit makes the numbers speakable.\u003C/p>",{"heading":49,"content":50,"relatedObjectives":51,"deepDive":53},"Calculating Molar Mass","\u003Cp>The \u003Cstrong>molar mass\u003C/strong> of a substance is the mass of one mole of that substance, expressed in grams per mole (g/mol). Numerically, it equals the formula mass in amu.\u003C/p>\u003Cp>To calculate the molar mass of a compound, sum the atomic masses of every atom in the formula:\u003C/p>\u003Col>\u003Cli>Write the chemical formula (e.g., H₂O)\u003C/li>\u003Cli>Multiply each element's atomic mass by its subscript\u003C/li>\u003Cli>Add the contributions: H₂O = 2(1.008) + 1(16.00) = 18.02 g/mol\u003C/li>\u003C/ol>\u003Cp>For ionic compounds, the same procedure applies using the formula unit. For example, NaCl has a molar mass of 22.99 + 35.45 = 58.44 g/mol.\u003C/p>\u003Cp>\u003Cstrong>Compounds with parenthesized polyatomic ions.\u003C/strong> When a polyatomic ion has an outer subscript, multiply the masses of every atom inside the parentheses by that subscript. For Fe(NO\u003Csub>3\u003C/sub>)\u003Csub>3\u003C/sub>: 1 Fe + 3 N + 9 O = 55.85 + 3(14.01) + 9(16.00) = 241.88 g/mol.\u003C/p>\u003Cp>\u003Cstrong>Hydrates\u003C/strong> include water of crystallization; the centered dot indicates that many H\u003Csub>2\u003C/sub>O per formula unit. For CuSO\u003Csub>4\u003C/sub>·5H\u003Csub>2\u003C/sub>O: 63.55 + 32.07 + 4(16.00) + 5(18.02) = 249.72 g/mol.\u003C/p>\u003Cp>\u003Cstrong>Doubly-parenthesized double salts\u003C/strong> work the same way: each parenthesis distributes its subscript over its contents. For Fe(NH\u003Csub>4\u003C/sub>)\u003Csub>2\u003C/sub>(SO\u003Csub>4\u003C/sub>)\u003Csub>2\u003C/sub>: 1 Fe + 2 N + 8 H + 2 S + 8 O = 284.07 g/mol.\u003C/p>",[52],55,[54],{"label":55,"body":56},"Common mistake: forgetting the parentheses multiply everything","\u003Cp>In Ca(NO₃)₂, the subscript 2 multiplies the entire nitrate group: two N atoms and \u003Cem>six\u003C/em> O atoms, not two N and three O. The molar mass is 40.08 + 2(14.01) + 6(16.00) = \u003Cspan class=\"nowrap\">164.10 g/mol.\u003C/span> Dropping the distribution (counting only three oxygens) gives 116.10, an error of nearly 30% that then poisons every downstream calculation.\u003C/p>\u003Cp>Reliable habit: before summing, expand the formula into a full atom inventory (Ca: 1, N: 2, O: 6) and only then reach for the periodic table.\u003C/p>",{"heading":58,"content":59,"relatedObjectives":60,"deepDive":62},"Converting Between Moles, Mass, and Particles","\u003Cp>Three quantities sit at the heart of chemical calculations: \u003Cstrong>mass\u003C/strong> (grams), \u003Cstrong>moles\u003C/strong>, and \u003Cstrong>number of particles\u003C/strong>. Two conversion factors connect them:\u003C/p>\u003Cul>\u003Cli>\u003Cstrong>Molar mass\u003C/strong> (g/mol) converts between grams and moles\u003C/li>\u003Cli>\u003Cstrong>Avogadro's number\u003C/strong> (6.022 × 10²³/mol) converts between moles and particles\u003C/li>\u003C/ul>\u003Cp>The conversion map looks like this:\u003C/p>\u003Cp style='text-align:center;'>\u003Cstrong>Mass (g)\u003C/strong> ↔ \u003Cem>÷ or × molar mass\u003C/em> ↔ \u003Cstrong>Moles\u003C/strong> ↔ \u003Cem>÷ or × N\u003Csub>A\u003C/sub>\u003C/em> ↔ \u003Cstrong>Particles\u003C/strong>\u003C/p>\u003Cp>For example, to find how many molecules are in 36.0 g of water: 36.0 g ÷ 18.02 g/mol = 2.00 mol, then 2.00 mol × 6.022 × 10²³ = 1.20 × 10²⁴ molecules. Always pass through moles as the central hub when converting between mass and particle counts.\u003C/p>",[61],56,[63],{"label":64,"body":65},"Common mistake: multiplying when you should divide","\u003Cp>The fastest error-catcher in mole work is a sense of scale. Moles of a lab-sized sample should be a modest number (a few grams of most substances is well under one mole). Particle counts should be astronomical: anything visible contains upwards of 10²⁰ particles. So if your “number of molecules” comes out as 10⁻²³, you divided by Avogadro’s number where you should have multiplied; if your “moles” from a 5-gram sample come out in the hundreds, the molar mass went in upside down.\u003C/p>\u003Cp>Sanity-check every answer against those two magnitudes before moving on; the check takes two seconds and catches the two most common inversions.\u003C/p>",{"heading":67,"content":68,"relatedObjectives":69,"deepDive":71},"Percent Composition","\u003Cp>\u003Cstrong>Percent composition\u003C/strong> tells you the mass percentage of each element in a compound. It answers the question: what fraction of the compound's total mass comes from each element?\u003C/p>\u003Cp>To calculate from a formula:\u003C/p>\u003Cdiv class='chem-equation'>% element = (atoms of element × atomic mass) / molar mass of compound × 100%\u003C/div>\u003Cp>For example, in water (H₂O, molar mass 18.02 g/mol):\u003C/p>\u003Cul>\u003Cli>% H = (2 × 1.008) / 18.02 × 100% = 11.19%\u003C/li>\u003Cli>% O = 16.00 / 18.02 × 100% = 88.81%\u003C/li>\u003C/ul>\u003Cp>Percent composition is useful for comparing how much of an active element different compounds provide. Farmers choose nitrogen fertilizers partly based on the mass percent of nitrogen: NH₃ (82.2% N) is much richer in nitrogen than NH₄NO₃ (35.0% N).\u003C/p>",[70],57,[72],{"label":73,"body":74},"Why this matters: the numbers on a fertilizer bag","\u003Cp>Farmers buy nitrogen by percent composition. Urea, CO(NH₂)₂, delivers \u003Cspan class=\"nowrap\">28.02 / 60.06 = 46.7%\u003C/span> nitrogen by mass, the highest of any common solid fertilizer, while ammonium nitrate carries 35.0%. Those percentages, computed exactly the way this section teaches, determine shipping costs, application rates, and prices per acre.\u003C/p>\u003Cp>The same calculation runs through ore refining (which iron ore is richest?) and nutrition labels. Percent composition is the answer to a very commercial question: how much of what I paid for is the element I actually wanted?\u003C/p>",{"heading":76,"content":77,"relatedObjectives":78,"deepDive":80},"Empirical Formulas from Percent Composition","\u003Cp>The \u003Cstrong>empirical formula\u003C/strong> gives the simplest whole-number ratio of atoms in a compound. Determining it from experimental data follows a consistent procedure:\u003C/p>\u003Col>\u003Cli>Start with percent composition (or mass data). If given percentages, assume a 100-g sample so that percentages become grams directly.\u003C/li>\u003Cli>Convert grams of each element to moles by dividing by atomic mass.\u003C/li>\u003Cli>Divide every mole value by the \u003Cem>smallest\u003C/em> mole value to get mole ratios.\u003C/li>\u003Cli>If the ratios are not whole numbers, multiply all by the smallest integer that makes them whole (e.g., multiply by 2 if you see a 0.5, by 3 if you see a 0.33).\u003C/li>\u003C/ol>\u003Cp>For example, a compound that is 40.0% C, 6.7% H, and 53.3% O yields a mole ratio of C : H : O = 1 : 2 : 1, giving the empirical formula CH₂O.\u003C/p>",[79],58,[81],{"label":82,"body":83},"Common mistake: rounding 1.33 to 1","\u003Cp>After dividing by the smallest mole value, ratios like 1.33, 1.5, or 1.25 are not rounding errors to squash into whole numbers; they are exact small fractions announcing a multiplier. A ratio of 1.33 is 4/3 (multiply everything by 3); 1.5 is 3/2 (multiply by 2); 1.25 is 5/4 (multiply by 4). Rounding 1.33 down to 1 quietly changes the compound.\u003C/p>\u003Cp>Rule of thumb: treat a value within about 0.05 of a whole number as that whole number (experimental data are never perfect), but treat anything near .25, .33, .5, .67, or .75 as a fraction demanding its multiplier.\u003C/p>",{"heading":85,"content":86,"relatedObjectives":87,"deepDive":89},"Molecular Formulas from Empirical Formulas","\u003Cp>The \u003Cstrong>molecular formula\u003C/strong> shows the actual number of atoms per molecule. It is always a whole-number multiple of the empirical formula.\u003C/p>\u003Cp>To determine the molecular formula, you need two things:\u003C/p>\u003Col>\u003Cli>The empirical formula (and its mass)\u003C/li>\u003Cli>The compound's molar mass (from experiment — often measured via gas density, mass spectrometry, or other techniques)\u003C/li>\u003C/ol>\u003Cp>Calculate the multiplier:\u003C/p>\u003Cdiv class='chem-equation'>n = molar mass / empirical formula mass\u003C/div>\u003Cp>Then multiply every subscript in the empirical formula by n. For example, if the empirical formula is CH₂O (mass 30.03 g/mol) and the molar mass is 180.18 g/mol, then n = 180.18 / 30.03 = 6, and the molecular formula is C₆H₁₂O₆ (glucose).\u003C/p>",[88],59,[90],{"label":91,"body":92},"Why this matters: three compounds, one empirical formula","\u003Cp>Formaldehyde (CH₂O), a preservative you would never eat; acetic acid (C₂H₄O₂), the tang in vinegar; and glucose (C₆H₁₂O₆), the sugar your cells burn: all three reduce to the same empirical formula, CH₂O. Composition analysis alone literally cannot tell a poison from a sugar.\u003C/p>\u003Cp>That is why the molar mass is the essential second measurement: 30 g/mol picks out formaldehyde, 60 picks acetic acid, 180 picks glucose. The empirical formula narrows the suspects; the molar mass makes the identification.\u003C/p>",{"heading":94,"content":95,"relatedObjectives":96},"Using Avogadro's Number in Calculations","\u003Cp>Avogadro's number (6.022 × 10²³ mol\u003Csup>−1\u003C/sup>) converts between moles and individual particles — atoms, molecules, ions, or formula units, depending on the substance.\u003C/p>\u003Cp>Common calculation patterns:\u003C/p>\u003Cul>\u003Cli>\u003Cstrong>Moles → particles:\u003C/strong> Multiply by N\u003Csub>A\u003C/sub>. Example: 0.50 mol O₂ × 6.022 × 10²³ = 3.01 × 10²³ molecules of O₂.\u003C/li>\u003Cli>\u003Cstrong>Particles → moles:\u003C/strong> Divide by N\u003Csub>A\u003C/sub>. Example: 1.81 × 10²⁴ atoms Fe ÷ 6.022 × 10²³ = 3.00 mol Fe.\u003C/li>\u003C/ul>\u003Cp>Remember that Avogadro's number always relates moles to whatever particle is specified by the formula. One mole of H₂O gives 6.022 × 10²³ \u003Cem>molecules\u003C/em>, but since each molecule contains 3 atoms, it contains 3 × 6.022 × 10²³ = 1.81 × 10²⁴ total atoms.\u003C/p>",[97],62,{"heading":99,"content":100,"relatedObjectives":101,"deepDive":103},"Empirical Formulas from Combustion Analysis","\u003Cp>\u003Cstrong>Combustion analysis\u003C/strong> is a common experimental technique for determining the empirical formula of organic compounds containing C, H, and sometimes O, N, or S.\u003C/p>\u003Cp>In a combustion analysis:\u003C/p>\u003Col>\u003Cli>A known mass of the compound is burned completely in excess O₂\u003C/li>\u003Cli>All carbon is converted to CO₂ (collected and weighed)\u003C/li>\u003Cli>All hydrogen is converted to H₂O (collected and weighed)\u003C/li>\u003Cli>Oxygen (if present) is found by difference: mass O = sample mass − mass C − mass H\u003C/li>\u003C/ol>\u003Cp>From the masses of CO₂ and H₂O produced, you can calculate grams of C and H, convert to moles, and determine the empirical formula using the standard procedure. For example, if burning 1.000 g of a compound yields 1.500 g CO₂ and 0.409 g H₂O, then the mass of C = 0.4093 g, mass of H = 0.0458 g, and mass of O (by difference) = 0.545 g, leading to the empirical formula.\u003C/p>",[102],63,[104],{"label":105,"body":106},"Why this matters: the instrument that identified organic chemistry","\u003Cp>For over 150 years, burning a sample and weighing the CO₂ and H₂O was \u003Cem>the\u003C/em> way to determine what an organic compound was made of; Justus von Liebig industrialized the technique in the 1830s with a five-bulb glass apparatus so effective it became the emblem of chemistry itself. Modern CHN analyzers automate the same combustion into a benchtop instrument, and pharmaceutical labs still run elemental analysis to confirm that a newly synthesized drug has exactly the composition its formula claims. The problems you solve here are that instrument’s arithmetic.\u003C/p>",{"heading":108,"content":109,"relatedObjectives":110,"deepDive":112},"Counting Atoms in a Given Mass","\u003Cp>To determine how many atoms are in a sample of a substance, chain together two conversions: mass → moles → atoms.\u003C/p>\u003Col>\u003Cli>Convert mass to moles: divide by the molar mass\u003C/li>\u003Cli>Convert moles to atoms: multiply by Avogadro's number\u003C/li>\u003Cli>If the substance is molecular, account for how many atoms of the target element exist per molecule\u003C/li>\u003C/ol>\u003Cp>For example, how many oxygen atoms are in 50.0 g of CaCO₃ (molar mass 100.09 g/mol)?\u003C/p>\u003Cul>\u003Cli>50.0 g ÷ 100.09 g/mol = 0.4996 mol CaCO₃\u003C/li>\u003Cli>Each formula unit has 3 oxygen atoms\u003C/li>\u003Cli>0.4996 mol × 3 × 6.022 × 10²³ = 9.02 × 10²³ oxygen atoms\u003C/li>\u003C/ul>\u003Cp>The key is to always identify how many atoms of the element of interest appear in each formula unit or molecule.\u003C/p>",[111],64,[113],{"label":114,"body":115},"Try it: count the atoms in a flake of gold","\u003Cp>A 1.0 g flake of gold: how many atoms is that? Work it, then check below.\u003C/p>\u003Cp>\u003Cstrong>Answer:\u003C/strong> \u003Cspan class=\"nowrap\">1.0 g ÷ 196.97 g/mol\u003C/span> = 5.1 × 10⁻³ mol, then \u003Cspan class=\"nowrap\">5.1 × 10⁻³ mol\u003C/span> × 6.022 × 10²³ atoms/mol = \u003Cstrong>3.1 × 10²¹ atoms\u003C/strong> (2 sig figs, from the 1.0 g). Both scale checks pass: the mole count is small, the atom count is astronomical. Three thousand billion billion atoms in a flake lighter than a paperclip: the two-step chain (mass → moles → atoms) is short, but it crosses twenty orders of magnitude.\u003C/p>",{"heading":117,"content":118,"relatedObjectives":119},"Formula Subscripts as Mole Ratios","\u003Cp>The subscripts in a chemical formula do double duty: they tell you both the number of atoms per molecule \u003Cem>and\u003C/em> the number of moles of each element per mole of compound.\u003C/p>\u003Cp>For example, in glucose (C₆H₁₂O₆):\u003C/p>\u003Cul>\u003Cli>One molecule contains 6 C atoms, 12 H atoms, and 6 O atoms\u003C/li>\u003Cli>One \u003Cem>mole\u003C/em> contains 6 mol C, 12 mol H, and 6 mol O\u003C/li>\u003C/ul>\u003Cp>This equivalence makes molar-level calculations straightforward. Need the moles of hydrogen in 3.0 mol of glucose? Simply multiply: 3.0 mol C₆H₁₂O₆ × 12 mol H / 1 mol C₆H₁₂O₆ = 36 mol H.\u003C/p>\u003Cp>Subscript ratios also directly define the empirical formula. If analysis shows a compound contains elements in a 1:2:1 mole ratio, the empirical formula is XY₂Z (where X, Y, Z are the elements). The subscripts \u003Cem>are\u003C/em> the mole ratio.\u003C/p>",[120],65,{"heading":122,"content":123,"relatedObjectives":124},"A Practical Problem Map for Mole and Formula Calculations","\u003Cp>Most quantitative composition problems become simple once you identify the target and route:\u003C/p>\u003Col>\u003Cli>\u003Cstrong>Target = particles?\u003C/strong> Use moles → particles with Avogadro’s number.\u003C/li>\u003Cli>\u003Cstrong>Target = mass?\u003C/strong> Use moles → mass with molar mass.\u003C/li>\u003Cli>\u003Cstrong>Target = empirical formula?\u003C/strong> Convert each element to moles, divide by the smallest, then scale to whole numbers.\u003C/li>\u003Cli>\u003Cstrong>Target = molecular formula?\u003C/strong> Find empirical-formula mass, compute multiplier = (molar mass)/(empirical mass), then multiply subscripts.\u003C/li>\u003C/ol>\u003Cp>Common pitfalls: rounding subscripts too early, skipping unit tracking, and forgetting that subscripts represent mole ratios. Keep at least one guardrail in every solution: each conversion factor must cancel units cleanly before moving to the next step.\u003C/p>",[79,88,102,120],{"title":126,"problem":127,"steps":128,"answer":135},"Determining Molecular Formula from Percent Composition and Molar Mass","A compound is 40.00% carbon, 6.71% hydrogen, and 53.29% oxygen by mass. Its molar mass is 180.16 g/mol. Determine the empirical and molecular formulas.",[129,130,131,132,133,134],"Assume a 100.00 g sample: 40.00 g C, 6.71 g H, 53.29 g O","Convert to moles: C = 40.00 / 12.01 = 3.331 mol; H = 6.71 / 1.008 = 6.657 mol; O = 53.29 / 16.00 = 3.331 mol","Divide by the smallest (3.331): C = 1.000, H = 1.998 ≈ 2, O = 1.000","Empirical formula = CH₂O with empirical formula mass = 12.01 + 2(1.008) + 16.00 = 30.03 g/mol","Find the multiplier: n = 180.16 / 30.03 = 6.00","Multiply all subscripts by 6: molecular formula = C₆H₁₂O₆","Empirical formula: CH₂O. Molecular formula: C₆H₁₂O₆ (glucose).",[137,138],"OpenStax Chemistry 2e, Ch 3.1: Formula Mass and the Mole Concept (CC BY 4.0)","OpenStax Chemistry 2e, Ch 3.2: Determining Empirical and Molecular Formulas (CC BY 4.0)",[140,141,142],"stoichiometry","solutions-and-concentration","measurement-and-significant-figures",[144,145,146,147,148],"C","H","O","N","S",[],[151,154,158,160],{"label":152,"equation":153},"Moles from Mass","moles = mass (g) / molar mass (g/mol)",{"label":155,"equation":156,"note":157},"Number of Particles","particles = moles x 6.022 x 10^23","Avogadro's number (NA)",{"label":67,"equation":159},"% element = (atoms of element x atomic mass) / molar mass of compound x 100%",{"label":161,"equation":162,"note":163},"Molecular Formula Multiplier","n = molar mass / empirical formula mass","Multiply all empirical subscripts by n",{"title":165,"steps":166},"How to Determine an Empirical Formula from Percent Composition",[167,168,169,170,171,172],"Assume a 100.0 g sample so that each percentage converts directly to grams.","Convert grams of each element to moles by dividing by that element's atomic mass from the periodic table.","Divide every mole value by the smallest mole value to obtain mole ratios.","If any ratio is not close to a whole number, multiply all ratios by the smallest integer that converts them to whole numbers (multiply by 2 for halves, by 3 for thirds, etc.).","Write the empirical formula using the whole-number ratios as subscripts.","If the molar mass is known, divide it by the empirical formula mass to find the multiplier n, then multiply all subscripts by n to get the molecular formula.",{"question":174,"answer":175,"type":176},"Two compounds have the same empirical formula, CH2O. Compound A has a molar mass of 30 g/mol and Compound B has a molar mass of 180 g/mol. Are these the same substance? What are their molecular formulas?","They are not the same substance. Compound A has n = 30/30 = 1, so its molecular formula is CH2O (formaldehyde). Compound B has n = 180/30 = 6, so its molecular formula is C6H12O6 (glucose). The empirical formula only gives the simplest ratio of atoms, so different compounds can share the same empirical formula while having very different molecular formulas and properties.","conceptual",[178,182],{"id":179,"problem":180,"type":181},"pt-8-1","A compound contains 52.14% carbon, 13.13% hydrogen, and 34.73% oxygen by mass. Determine its empirical formula.","calculation",{"id":183,"problem":184,"type":181},"pt-8-2","How many individual oxygen atoms are present in 25.0 g of calcium carbonate (CaCO3, molar mass 100.09 g/mol)?",[186,187,188,189,190],"mole concept","molar mass","percent composition","empirical formula","Avogadro's number",[192,194,196,199,201,203,205,207,209,211],{"question":193},"What is the mole and what is Avogadro’s number?",{"question":195},"What is molar mass and how does it relate to the periodic table?",{"question":197,"hint":198},"How do you convert between grams, moles, and number of particles?","Two conversion factors are involved — one uses molar mass, the other uses Avogadro’s number.",{"question":200},"What is percent composition by mass and how is it calculated from a formula?",{"question":202},"What is the difference between an empirical formula and a molecular formula?",{"question":204},"How do you determine an empirical formula from percent composition data?",{"question":206},"How do you find a molecular formula from an empirical formula?",{"question":208},"What is combustion analysis and what information does it provide?",{"question":210},"How are mole ratios used as conversion factors?",{"question":212},"Why is the mole concept central to all quantitative chemistry?",7,{"id":213,"slug":215,"lesson":213,"title":216,"shortTitle":217,"description":218,"category":219,"objectiveCount":220,"problemCount":221},"redox-reactions-and-redox-balancing","Redox Reactions and Redox Balancing","Redox & Balancing","Assign oxidation numbers, recognize redox reactions, identify oxidizing and reducing agents, write half-reactions, and balance redox equations by both the oxidation-number and half-reaction methods.","reactions",6,60,{"id":223,"slug":141,"lesson":223,"title":224,"shortTitle":225,"description":226,"category":10,"objectiveCount":227,"problemCount":228},9,"Solutions and Concentration","Solutions & Molarity","Learn about solution preparation, molarity, dilution, and concentration units used in chemistry.",11,75,1785108608071]