Seconde \u2014 La quantit\u00e9 de mati\u00e8re<\/title>\n<style>\n:root{\n --deep:#101827; --blue:#2563eb; --cyan:#0891b2; --teal:#14b8a6;\n --purple:#7c3aed; --pink:#db2777; --orange:#f97316; --red:#b83227;\n --green:#0f766e; --amber:#f59e0b; --ink:#1f2937; --paper:#fff; --line:#d9e2ec;\n}\n*{box-sizing:border-box} html{scroll-behavior:smooth}\nbody{\n margin:0;font-family:Arial,Helvetica,sans-serif;color:var(--ink);line-height:1.65;\n background:\n radial-gradient(circle at 10% 8%,rgba(20,184,166,.13),transparent 24%),\n radial-gradient(circle at 88% 12%,rgba(245,158,11,.13),transparent 24%),\n linear-gradient(180deg,#f8fbff,#eef4f8);\n}\nheader{\n background:\n radial-gradient(circle at 78% 22%,rgba(245,158,11,.34),transparent 19%),\n linear-gradient(135deg,#0f172a,#0f766e,#2563eb);\n color:white;text-align:center;padding:56px 22px\n}\nheader h1{margin:0;font-size:clamp(2.1rem,4vw,3.7rem)}\nheader p{font-size:1.15rem;margin:10px 0 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.step{color:var(--deep);font-weight:800}\n.grid2{display:grid;grid-template-columns:repeat(auto-fit,minmax(320px,1fr));gap:22px;align-items:start}\n.grid3{display:grid;grid-template-columns:repeat(auto-fit,minmax(250px,1fr));gap:18px;align-items:start}\n.schema{background:#fbfdff;border:1px solid var(--line);border-radius:20px;padding:18px;text-align:center;overflow-x:auto;margin:14px 0}\nsvg{max-width:100%;height:auto}\n.btns{text-align:center;margin:18px 0 4px}\nbutton,.linkbtn{\n border:0;background:var(--blue);color:white;text-decoration:none;padding:14px 22px;border-radius:14px;\n font-size:17px;cursor:pointer;margin:6px;box-shadow:0 7px 18px rgba(37,99,235,.20);display:inline-block\n}\nbutton.stop{background:#333}.cyanbtn{background:var(--cyan)}.pinkbtn{background:var(--pink)}.greenbtn{background:#0f766e}.orangebtn{background:var(--orange)}\ntable{border-collapse:collapse;width:100%;margin:14px 0;background:white}\nth,td{border:1px solid #9aa7bd;padding:10px;text-align:center}th{background:#f0f4fb}\n.card{border:2px solid #e1e8f0;border-top:6px solid var(--teal);border-radius:18px;padding:18px;background:#fff}\n.mm-center{grid-column:1\/-1;text-align:center;background:linear-gradient(135deg,var(--deep),var(--teal));color:white;border-radius:18px;padding:18px;font-size:1.3rem;font-weight:900}\n.small{font-size:.92rem;color:#5b6770}\n@media print{button,.btns,.linkbtn{display:none}body{background:white}.section{box-shadow:none;break-inside:avoid}}\n<\/style>\n<\/head>\n<body>\n<header>\n<h1>La quantit\u00e9 de mati\u00e8re<\/h1>\n<p>Seconde physique-chimie \u2022 Mole \u2022 Constante d\u2019Avogadro \u2022 Masse molaire \u2022 Calculs avec m, M, V, \u03c1 et d<\/p>\n<\/header>\n\n<div class=\"container\">\n\n<div class=\"btns\">\n<button onclick=\"playAudioSummary()\">\ud83d\udd0a \u00c9couter le r\u00e9sum\u00e9 audio clair<\/button>\n<button class=\"stop\" onclick=\"stopAudioSummary()\">\u23f9 Arr\u00eater<\/button><br>\n<a class=\"linkbtn cyanbtn\" href=\"#fiche-bilan\">\ud83d\udccc Aller \u00e0 la fiche bilan<\/a>\n<a class=\"linkbtn pinkbtn\" href=\"#exercices\">\ud83e\uddea Exercices contextualis\u00e9s<\/a>\n<a class=\"linkbtn greenbtn\" href=\"#type-bac\">\ud83c\udf93 Partie type bac \/ \u00e9valuation<\/a>\n<a class=\"linkbtn orangebtn\" href=\"#liquides\">\ud83d\udca7 Cas des liquides<\/a>\n<\/div>\n\n<div id=\"audioText\" style=\"display:none;\">\nBienvenue dans le chapitre la quantit\u00e9 de mati\u00e8re.\nSi on veut compter les mol\u00e9cules ou les atomes dans un \u00e9chantillon, on r\u00e9alise que ces nombres sont immenses.\nOn d\u00e9finit donc un paquet de r\u00e9f\u00e9rence : la mole.\nUne mole est un paquet d\u2019atomes ou de mol\u00e9cules qui contient 6,02 fois 10 puissance 23 entit\u00e9s.\nCe nombre est la constante d\u2019Avogadro.\nLa quantit\u00e9 de mati\u00e8re se note n et s\u2019exprime en mole.\nLe nombre d\u2019entit\u00e9s N est reli\u00e9 \u00e0 la quantit\u00e9 de mati\u00e8re par la relation N \u00e9gale n fois NA.\nLa masse molaire atomique est la masse d\u2019une mole d\u2019atomes. On la note M et elle s\u2019exprime en gramme par mole.\nPour une mol\u00e9cule, on calcule la masse molaire mol\u00e9culaire en additionnant les masses molaires des atomes qui la constituent.\nPour un ion, on fait comme pour une mol\u00e9cule, car on n\u00e9glige la masse des charges \u00e9lectriques.\nPour calculer une quantit\u00e9 de mati\u00e8re \u00e0 partir d\u2019une masse, on utilise n \u00e9gale m divis\u00e9 par M.\nPour un liquide pur, si on conna\u00eet son volume et sa masse volumique, on calcule d\u2019abord la masse avec m \u00e9gale rho fois V, puis on calcule n avec n \u00e9gale m divis\u00e9 par M.\nSi on donne la densit\u00e9 d\u2019un liquide, on calcule sa masse volumique avec rho liquide \u00e9gale d fois rho eau, et rho eau vaut 1000 grammes par litre.\nFin du r\u00e9sum\u00e9.\n<\/div>\n\n<section class=\"section\">\n<h2>Objectifs du chapitre<\/h2>\n<p>\nCe cours reprend les phrases du cahier comme base : la mole, la constante d\u2019Avogadro, les masses molaires,\nle calcul d\u2019une quantit\u00e9 de mati\u00e8re \u00e0 partir d\u2019une masse, puis \u00e0 partir du volume d\u2019un liquide pur avec la masse volumique ou la densit\u00e9.\n<\/p>\n<div class=\"grid3\">\n<div class=\"card\"><h3>Comprendre<\/h3><p>Pourquoi on utilise la mole pour compter des entit\u00e9s microscopiques.<\/p><\/div>\n<div class=\"card\"><h3>Calculer<\/h3><p>Une masse molaire, une quantit\u00e9 de mati\u00e8re, une masse ou un volume.<\/p><\/div>\n<div class=\"card\"><h3>R\u00e9diger<\/h3><p>Formule du cours en rouge, transformation litt\u00e9rale, application num\u00e9rique, r\u00e9sultat soulign\u00e9.<\/p><\/div>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>\ud83d\udd34 M\u00e9thode obligatoire pour les exercices<\/h2>\n<p>\nDans chaque correction, on part de la <span class=\"red\">formule du cours encadr\u00e9e<\/span>, puis on fait le\n<span class=\"blue\">travail litt\u00e9ral<\/span>, et enfin l\u2019application num\u00e9rique avec les unit\u00e9s.\n<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Exemple : isoler la masse dans n = m \/ M<\/p>\n<div class=\"formule\">n = m \/ M<\/div>\n<p class=\"step\">On multiplie par M :<\/p>\n<div class=\"formule bluebox\">n \u00d7 M = m<\/div>\n<p class=\"step\">Donc :<\/p>\n<div class=\"formule bluebox\">m = n \u00d7 M<\/div>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>I. La mole<\/h2>\n\n<p>\nSi on veut compter les mol\u00e9cules qu\u2019il y a dans un \u00e9chantillon, on r\u00e9alise que ces nombres sont immenses.\n<\/p>\n\n<div class=\"note\">\n<p>Exemple du cahier : dans 1 g de carbone \u00b9\u00b2C, on trouve environ N = 4,98 \u00d7 10\u00b2\u00b2 atomes. Ce nombre est gigantesque.<\/p>\n<\/div>\n\n<p class=\"red\">\nOn va donc d\u00e9finir un paquet de r\u00e9f\u00e9rence : la mole.\n<\/p>\n<p>\nOn a choisi Avogadro : un paquet tr\u00e8s facile \u00e0 peser.\n<\/p>\n<p class=\"red\">\nUne mole est un paquet d\u2019atomes ou de mol\u00e9cules qui contient N<sub>A<\/sub> = 6,02 \u00d7 10\u00b2\u00b3 entit\u00e9s.\n<\/p>\n\n<div class=\"formule\">N<sub>A<\/sub> = 6,02 \u00d7 10\u00b2\u00b3 mol\u207b\u00b9<\/div>\n\n<p><span class=\"red\">N<sub>A<\/sub><\/span> : constante d\u2019Avogadro.<\/p>\n<p>\nC\u2019est le nombre d\u2019atomes qu\u2019il y a dans A = 12 g de carbone \u00b9\u00b2C.\n<\/p>\n\n<p class=\"red\">\nOn va d\u00e9finir n : quantit\u00e9 de mati\u00e8re, qui s\u2019exprimera en mole, mol.\n<\/p>\n<p>\nOn peut retrouver N, le nombre d\u2019atomes ou de mol\u00e9cules, en utilisant la relation \u00e9vidente :\n<\/p>\n\n<div class=\"formule\">N = n \u00d7 N<sub>A<\/sub><\/div>\n\n<ul>\n<li><span class=\"red\">N<\/span> : nombre d\u2019entit\u00e9s, sans unit\u00e9 ;<\/li>\n<li><span class=\"red\">n<\/span> : quantit\u00e9 de mati\u00e8re en mol ;<\/li>\n<li><span class=\"red\">N<sub>A<\/sub><\/span> : constante d\u2019Avogadro en mol\u207b\u00b9.<\/li>\n<\/ul>\n\n<div class=\"schema\">\n<svg width=\"850\" height=\"360\" viewBox=\"0 0 850 360\">\n<rect x=\"55\" y=\"45\" width=\"740\" height=\"240\" rx=\"22\" fill=\"#fbfdff\" stroke=\"#D9E2EC\"\/>\n<text x=\"90\" y=\"85\" font-size=\"20\" font-weight=\"900\" fill=\"#0f172a\">La mole : un paquet de r\u00e9f\u00e9rence<\/text>\n<circle cx=\"205\" cy=\"175\" r=\"78\" fill=\"#e0f2fe\" stroke=\"#2563EB\" stroke-width=\"4\"\/>\n<text x=\"155\" y=\"165\" fill=\"#2563EB\" font-size=\"20\" font-weight=\"900\">1 mole<\/text>\n<text x=\"130\" y=\"200\" fill=\"#2563EB\" font-size=\"17\" font-weight=\"900\">6,02 \u00d7 10\u00b2\u00b3 entit\u00e9s<\/text>\n<line x1=\"310\" y1=\"175\" x2=\"505\" y2=\"175\" stroke=\"#0f172a\" stroke-width=\"5\"\/>\n<polygon points=\"505,175 480,161 480,189\" fill=\"#0f172a\"\/>\n<text x=\"540\" y=\"150\" font-size=\"20\" font-weight=\"900\">N = n \u00d7 N\u2090<\/text>\n<text x=\"540\" y=\"190\" fill=\"#B83227\" font-size=\"17\" font-weight=\"900\">La mole sert \u00e0 compter sans \u00e9crire des nombres immenses.<\/text>\n<\/svg>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Exercice type \u2014 Calculer le nombre d\u2019entit\u00e9s<\/h3>\n<p>On pr\u00e9l\u00e8ve n = 1,0 \u00d7 10\u207b\u00b2 mol de mol\u00e9cules. Calculer le nombre de mol\u00e9cules.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">N = n \u00d7 N<sub>A<\/sub><\/div>\n<p class=\"step\">La formule est d\u00e9j\u00e0 sous la bonne forme.<\/p>\n<\/div>\n<p>N = 1,0 \u00d7 10\u207b\u00b2 \u00d7 6,02 \u00d7 10\u00b2\u00b3<\/p>\n<p class=\"resultat\">N = 6,02 \u00d7 10\u00b2\u00b9 mol\u00e9cules<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>II. Masses molaires<\/h2>\n\n<h3>1. Masse molaire atomique<\/h3>\n<p>\nOn a choisi N<sub>A<\/sub> pour que 1 mole d\u2019atome <sup>A<\/sup><sub>Z<\/sub>X p\u00e8se A grammes.\n<\/p>\n\n<p class=\"red\">\nMasse molaire atomique : c\u2019est la masse d\u2019une mole d\u2019atomes.\n<\/p>\n<div class=\"formule\">M<sub>X<\/sub> = A g\u00b7mol\u207b\u00b9<\/div>\n\n<p>Il suffit de regarder dans le tableau p\u00e9riodique la valeur de A.<\/p>\n\n<table>\n<tr><th>Atome<\/th><th>Masse molaire atomique<\/th><\/tr>\n<tr><td>\u00b9\u00b2C<\/td><td>M<sub>C<\/sub> = 12 g\u00b7mol\u207b\u00b9<\/td><\/tr>\n<tr><td>\u00b9H<\/td><td>M<sub>H<\/sub> = 1 g\u00b7mol\u207b\u00b9<\/td><\/tr>\n<tr><td>\u00b9\u2076O<\/td><td>M<sub>O<\/sub> = 16 g\u00b7mol\u207b\u00b9<\/td><\/tr>\n<tr><td>Na<\/td><td>M<sub>Na<\/sub> = 23 g\u00b7mol\u207b\u00b9<\/td><\/tr>\n<tr><td>Cl<\/td><td>M<sub>Cl<\/sub> = 35,5 g\u00b7mol\u207b\u00b9<\/td><\/tr>\n<\/table>\n\n<h3>2. Masse molaire mol\u00e9culaire<\/h3>\n<p class=\"red\">\nOn ajoute les nombres molaires des atomes qui constituent la mol\u00e9cule.\n<\/p>\n\n<div class=\"formule\">M<sub>CxHyOz<\/sub> = x \u00d7 M<sub>C<\/sub> + y \u00d7 M<sub>H<\/sub> + z \u00d7 M<sub>O<\/sub><\/div>\n\n<div class=\"exercice\">\n<h3>Exemples du cahier<\/h3>\n<p><span class=\"red\">Glucose C\u2086H\u2081\u2082O\u2086 :<\/span><\/p>\n<p>M<sub>C\u2086H\u2081\u2082O\u2086<\/sub> = 6 \u00d7 12 + 12 \u00d7 1 + 6 \u00d7 16<\/p>\n<p class=\"resultat\">M<sub>C\u2086H\u2081\u2082O\u2086<\/sub> = 180 g\u00b7mol\u207b\u00b9<\/p>\n\n<p><span class=\"red\">Chlorure de sodium NaCl :<\/span><\/p>\n<p>M<sub>NaCl<\/sub> = 1 \u00d7 23 + 1 \u00d7 35,5<\/p>\n<p class=\"resultat\">M<sub>NaCl<\/sub> = 58,5 g\u00b7mol\u207b\u00b9<\/p>\n\n<p><span class=\"red\">\u00c9thanol C\u2082H\u2086O :<\/span><\/p>\n<p>M<sub>C\u2082H\u2086O<\/sub> = 2 \u00d7 12 + 6 \u00d7 1 + 1 \u00d7 16<\/p>\n<p class=\"resultat\">M<sub>C\u2082H\u2086O<\/sub> = 46 g\u00b7mol\u207b\u00b9<\/p>\n\n<p><span class=\"red\">Eau H\u2082O :<\/span><\/p>\n<p>M<sub>H\u2082O<\/sub> = 2 \u00d7 1 + 1 \u00d7 16<\/p>\n<p class=\"resultat\">M<sub>H\u2082O<\/sub> = 18 g\u00b7mol\u207b\u00b9<\/p>\n<\/div>\n\n<h3>3. Masse molaire ionique<\/h3>\n<p class=\"red\">\nC\u2019est la masse d\u2019une mole d\u2019ion.\n<\/p>\n<p class=\"red\">\nOn n\u00e9glige la masse des charges \u00e9lectriques + ou \u2212.\n<\/p>\n<p>Donc on fait comme pour une mol\u00e9cule.<\/p>\n\n<div class=\"exercice\">\n<h3>Exemple \u2014 Ion hypochlorite ClO\u207b<\/h3>\n<p>M<sub>ClO\u207b<\/sub> = M<sub>Cl<\/sub> + M<sub>O<\/sub><\/p>\n<p>M<sub>ClO\u207b<\/sub> = 35,5 + 16<\/p>\n<p class=\"resultat\">M<sub>ClO\u207b<\/sub> = 51,5 g\u00b7mol\u207b\u00b9<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>III. Calcul d\u2019une quantit\u00e9 de mati\u00e8re<\/h2>\n\n<h3>1. \u00c0 partir de la masse m d\u2019un \u00e9chantillon<\/h3>\n<div class=\"formule\">n = m \/ M<\/div>\n<ul>\n<li><span class=\"red\">n<\/span> : quantit\u00e9 de mati\u00e8re en mol ;<\/li>\n<li><span class=\"red\">m<\/span> : masse de l\u2019\u00e9chantillon en g ;<\/li>\n<li><span class=\"red\">M<\/span> : masse molaire en g\u00b7mol\u207b\u00b9.<\/li>\n<\/ul>\n\n<div class=\"exercice\">\n<h3>Exercice type 1 \u2014 Glucose<\/h3>\n<p>Calculer la quantit\u00e9 de mati\u00e8re qu\u2019il y a dans 2,0 g de glucose C\u2086H\u2081\u2082O\u2086.<\/p>\n<p>Donn\u00e9e : M<sub>C\u2086H\u2081\u2082O\u2086<\/sub> = 180 g\u00b7mol\u207b\u00b9.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n = m \/ M<\/div>\n<p class=\"step\">La formule est d\u00e9j\u00e0 sous la bonne forme.<\/p>\n<\/div>\n<p>n = 2,0 \/ 180<\/p>\n<p class=\"resultat\">n = 0,011 mol = 1,1 \u00d7 10\u207b\u00b2 mol<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Exercice type 2 \u2014 Sel NaCl<\/h3>\n<p>Calculer la quantit\u00e9 de mati\u00e8re dans 3,0 g de NaCl.<\/p>\n<p>Donn\u00e9e : M<sub>NaCl<\/sub> = 58,5 g\u00b7mol\u207b\u00b9.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n = m \/ M<\/div>\n<\/div>\n<p>n = 3,0 \/ 58,5<\/p>\n<p class=\"resultat\">n = 0,051 mol = 5,1 \u00d7 10\u207b\u00b2 mol<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Exercice type 3 \u2014 Masse \u00e0 peser<\/h3>\n<p>Calculer la masse \u00e0 peser pour pr\u00e9lever n = 2,5 \u00d7 10\u207b\u00b2 mol d\u2019\u00e9thanol C\u2082H\u2086O.<\/p>\n<p>Donn\u00e9e : M<sub>C\u2082H\u2086O<\/sub> = 46 g\u00b7mol\u207b\u00b9.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n = m \/ M<\/div>\n<p class=\"step\">On multiplie par M :<\/p>\n<div class=\"formule bluebox\">n \u00d7 M = m<\/div>\n<p class=\"step\">Donc :<\/p>\n<div class=\"formule bluebox\">m = n \u00d7 M<\/div>\n<\/div>\n<p>m = 2,5 \u00d7 10\u207b\u00b2 \u00d7 46<\/p>\n<p class=\"resultat\">m = 1,15 g<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\" id=\"liquides\">\n<h2>IV. \u00c0 partir du volume d\u2019un liquide pur<\/h2>\n\n<h3>1. En utilisant la masse volumique \u03c1<\/h3>\n<div class=\"formule\">\u03c1 = m \/ V<\/div>\n<ul>\n<li><span class=\"red\">\u03c1<\/span> : masse volumique du liquide en g\u00b7L\u207b\u00b9 ;<\/li>\n<li><span class=\"red\">m<\/span> : masse de l\u2019\u00e9chantillon en g ;<\/li>\n<li><span class=\"red\">V<\/span> : volume de l\u2019\u00e9chantillon en L.<\/li>\n<\/ul>\n\n<p class=\"red\">\nOn va utiliser \u03c1 et V pour trouver la masse de l\u2019\u00e9chantillon m.\n<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">\u03c1 = m \/ V<\/div>\n<p class=\"step\">On multiplie par V :<\/p>\n<div class=\"formule bluebox\">\u03c1 \u00d7 V = m<\/div>\n<p class=\"step\">Donc :<\/p>\n<div class=\"formule bluebox\">m = \u03c1 \u00d7 V<\/div>\n<p class=\"step\">Ensuite on utilise :<\/p>\n<div class=\"formule\">n = m \/ M<\/div>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Exercice type \u2014 Eau<\/h3>\n<p>Calculer la quantit\u00e9 de mati\u00e8re qu\u2019il y a dans 10 mL d\u2019eau.<\/p>\n<p>Donn\u00e9es : \u03c1<sub>eau<\/sub> = 1000 g\u00b7L\u207b\u00b9 ; M<sub>H\u2082O<\/sub> = 18 g\u00b7mol\u207b\u00b9.<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule 1 :<\/p>\n<div class=\"formule\">\u03c1 = m \/ V<\/div>\n<p class=\"step\">On isole m :<\/p>\n<div class=\"formule bluebox\">m = \u03c1 \u00d7 V<\/div>\n<p class=\"start\">Formule 2 :<\/p>\n<div class=\"formule\">n = m \/ M<\/div>\n<p class=\"step\">On injecte la formule 1 dans la formule 2 :<\/p>\n<div class=\"formule bluebox\">n = \u03c1 \u00d7 V \/ M<\/div>\n<\/div>\n\n<p class=\"red\">Conversion :<\/p>\n<p>V = 10 mL = 10 \u00d7 10\u207b\u00b3 L = 1,0 \u00d7 10\u207b\u00b2 L<\/p>\n<p>Application num\u00e9rique :<\/p>\n<p>n = (1000 \u00d7 10 \u00d7 10\u207b\u00b3) \/ 18<\/p>\n<p class=\"resultat\">n = 0,55 mol<\/p>\n<\/div>\n\n<h3>2. En utilisant la densit\u00e9 d\u2019un liquide<\/h3>\n\n<div class=\"formule\">d = \u03c1<sub>liquide<\/sub> \/ \u03c1<sub>eau<\/sub><\/div>\n<ul>\n<li><span class=\"red\">d<\/span> : densit\u00e9 d\u2019un liquide, sans unit\u00e9 ;<\/li>\n<li><span class=\"red\">\u03c1<sub>liquide<\/sub><\/span> : masse volumique du liquide ;<\/li>\n<li><span class=\"red\">\u03c1<sub>eau<\/sub><\/span> : masse volumique de l\u2019eau = 1000 g\u00b7L\u207b\u00b9.<\/li>\n<\/ul>\n\n<p class=\"red\">On va utiliser d pour trouver \u03c1<sub>liquide<\/sub>.<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">d = \u03c1<sub>liquide<\/sub> \/ \u03c1<sub>eau<\/sub><\/div>\n<p class=\"step\">On multiplie par \u03c1<sub>eau<\/sub> :<\/p>\n<div class=\"formule bluebox\">d \u00d7 \u03c1<sub>eau<\/sub> = \u03c1<sub>liquide<\/sub><\/div>\n<p class=\"step\">Donc :<\/p>\n<div class=\"formule bluebox\">\u03c1<sub>liquide<\/sub> = d \u00d7 \u03c1<sub>eau<\/sub><\/div>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Exercice type \u2014 Essence<\/h3>\n<p>Calculer la quantit\u00e9 de mati\u00e8re qu\u2019il y a dans 15 mL d\u2019essence C\u2088H\u2081\u2088 de densit\u00e9 d = 0,75.<\/p>\n<p>Donn\u00e9es : M<sub>C<\/sub> = 12 g\u00b7mol\u207b\u00b9 ; M<sub>H<\/sub> = 1 g\u00b7mol\u207b\u00b9 ; \u03c1<sub>eau<\/sub> = 1000 g\u00b7L\u207b\u00b9.<\/p>\n\n<p class=\"red\">\u00c9tape 1 : masse molaire<\/p>\n<p>M<sub>C\u2088H\u2081\u2088<\/sub> = 8 \u00d7 M<sub>C<\/sub> + 18 \u00d7 M<sub>H<\/sub><\/p>\n<p>M<sub>C\u2088H\u2081\u2088<\/sub> = 8 \u00d7 12 + 18 \u00d7 1<\/p>\n<p>M<sub>C\u2088H\u2081\u2088<\/sub> = 114 g\u00b7mol\u207b\u00b9<\/p>\n\n<p class=\"red\">\u00c9tape 2 : masse volumique du liquide<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule :<\/p>\n<div class=\"formule\">d = \u03c1<sub>liq<\/sub> \/ \u03c1<sub>eau<\/sub><\/div>\n<p class=\"step\">On isole \u03c1<sub>liq<\/sub> :<\/p>\n<div class=\"formule bluebox\">\u03c1<sub>liq<\/sub> = d \u00d7 \u03c1<sub>eau<\/sub><\/div>\n<\/div>\n<p>\u03c1<sub>liq<\/sub> = 0,75 \u00d7 1000 = 750 g\u00b7L\u207b\u00b9<\/p>\n\n<p class=\"red\">\u00c9tape 3 : quantit\u00e9 de mati\u00e8re<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formules :<\/p>\n<div class=\"formule\">m = \u03c1 \u00d7 V<\/div>\n<div class=\"formule\">n = m \/ M<\/div>\n<p class=\"step\">On injecte m = \u03c1 \u00d7 V dans n = m \/ M :<\/p>\n<div class=\"formule bluebox\">n = \u03c1 \u00d7 V \/ M<\/div>\n<\/div>\n<p>V = 15 mL = 15 \u00d7 10\u207b\u00b3 L<\/p>\n<p>n = (750 \u00d7 15 \u00d7 10\u207b\u00b3) \/ 114<\/p>\n<p class=\"resultat\">n = 9,8 \u00d7 10\u207b\u00b2 mol<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\" id=\"exercices\">\n<h2>V. Exercices contextualis\u00e9s<\/h2>\n\n<div class=\"exercice\">\n<h3>1. Comprim\u00e9 de vitamine C<\/h3>\n<p>Calculer la quantit\u00e9 de mati\u00e8re dans 1 comprim\u00e9 de vitamine C C\u2086H\u2088O\u2086 de 500 mg.<\/p>\n<p>Donn\u00e9es : M<sub>C<\/sub> = 12 g\u00b7mol\u207b\u00b9 ; M<sub>H<\/sub> = 1 g\u00b7mol\u207b\u00b9 ; M<sub>O<\/sub> = 16 g\u00b7mol\u207b\u00b9.<\/p>\n\n<p class=\"red\">Masse molaire :<\/p>\n<p>M<sub>C\u2086H\u2088O\u2086<\/sub> = 6 \u00d7 12 + 8 \u00d7 1 + 6 \u00d7 16 = 176 g\u00b7mol\u207b\u00b9<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n = m \/ M<\/div>\n<\/div>\n<p>m = 500 mg = 500 \u00d7 10\u207b\u00b3 g = 0,500 g<\/p>\n<p>n = 0,500 \/ 176<\/p>\n<p class=\"resultat\">n = 2,84 \u00d7 10\u207b\u00b3 mol<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>2. \u00c9thanol au laboratoire<\/h3>\n<p>Calculer la masse de C\u2082H\u2086O \u00e0 peser pour pr\u00e9lever 2,5 \u00d7 10\u207b\u00b2 mol.<\/p>\n<p>Donn\u00e9es : M<sub>C<\/sub> = 12 g\u00b7mol\u207b\u00b9 ; M<sub>H<\/sub> = 1 g\u00b7mol\u207b\u00b9 ; M<sub>O<\/sub> = 16 g\u00b7mol\u207b\u00b9.<\/p>\n<p>M<sub>C\u2082H\u2086O<\/sub> = 2 \u00d7 12 + 6 \u00d7 1 + 16 = 46 g\u00b7mol\u207b\u00b9<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule :<\/p><div class=\"formule\">n = m \/ M<\/div>\n<p class=\"step\">On isole m :<\/p><div class=\"formule bluebox\">m = n \u00d7 M<\/div>\n<\/div>\n<p>m = 2,5 \u00d7 10\u207b\u00b2 \u00d7 46<\/p>\n<p class=\"resultat\">m = 1,15 g<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>3. Acide sulfurique<\/h3>\n<p>Calculer la quantit\u00e9 de mati\u00e8re qu\u2019il y a dans 100 mL d\u2019H\u2082SO\u2084 de masse volumique \u03c1 = 1800 g\u00b7L\u207b\u00b9.<\/p>\n<p>Donn\u00e9es : M<sub>H<\/sub> = 1 g\u00b7mol\u207b\u00b9 ; M<sub>S<\/sub> = 32 g\u00b7mol\u207b\u00b9 ; M<sub>O<\/sub> = 16 g\u00b7mol\u207b\u00b9.<\/p>\n<p>M<sub>H\u2082SO\u2084<\/sub> = 2 \u00d7 1 + 32 + 4 \u00d7 16 = 98 g\u00b7mol\u207b\u00b9<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formules :<\/p>\n<div class=\"formule\">m = \u03c1 \u00d7 V<\/div>\n<div class=\"formule\">n = m \/ M<\/div>\n<p class=\"step\">On injecte :<\/p>\n<div class=\"formule bluebox\">n = \u03c1 \u00d7 V \/ M<\/div>\n<\/div>\n<p>V = 100 mL = 100 \u00d7 10\u207b\u00b3 L<\/p>\n<p>n = (1800 \u00d7 100 \u00d7 10\u207b\u00b3) \/ 98<\/p>\n<p class=\"resultat\">n = 1,83 mol<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>4. Acide \u00e9thano\u00efque dans le vinaigre<\/h3>\n<p>Calculer la quantit\u00e9 de mati\u00e8re qu\u2019il y a dans 20 mL d\u2019acide \u00e9thano\u00efque CH\u2083COOH, de densit\u00e9 d = 1,07.<\/p>\n<p>Donn\u00e9es : M<sub>C<\/sub> = 12 g\u00b7mol\u207b\u00b9 ; M<sub>H<\/sub> = 1 g\u00b7mol\u207b\u00b9 ; M<sub>O<\/sub> = 16 g\u00b7mol\u207b\u00b9 ; \u03c1<sub>eau<\/sub> = 1000 g\u00b7L\u207b\u00b9.<\/p>\n<p>CH\u2083COOH correspond \u00e0 C\u2082H\u2084O\u2082.<\/p>\n<p>M<sub>C\u2082H\u2084O\u2082<\/sub> = 2 \u00d7 12 + 4 \u00d7 1 + 2 \u00d7 16 = 60 g\u00b7mol\u207b\u00b9<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formules :<\/p>\n<div class=\"formule\">\u03c1<sub>liq<\/sub> = d \u00d7 \u03c1<sub>eau<\/sub><\/div>\n<div class=\"formule\">m = \u03c1 \u00d7 V<\/div>\n<div class=\"formule\">n = m \/ M<\/div>\n<p class=\"step\">On injecte :<\/p>\n<div class=\"formule bluebox\">n = d \u00d7 \u03c1<sub>eau<\/sub> \u00d7 V \/ M<\/div>\n<\/div>\n<p>V = 20 mL = 20 \u00d7 10\u207b\u00b3 L<\/p>\n<p>n = (1,07 \u00d7 1000 \u00d7 20 \u00d7 10\u207b\u00b3) \/ 60<\/p>\n<p class=\"resultat\">n = 0,35 mol<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\" id=\"type-bac\">\n<h2>VI. Partie type bac \/ \u00e9valuation \u2014 niveau seconde<\/h2>\n\n<div class=\"exercice\">\n<h3>Situation 1 \u2014 Pr\u00e9parer une solution de glucose pour une boisson de sport<\/h3>\n<p>\nUn pr\u00e9parateur sportif souhaite utiliser 2,0 g de glucose C\u2086H\u2081\u2082O\u2086.\n<\/p>\n<ol>\n<li>Calculer la masse molaire du glucose.<\/li>\n<li>Calculer la quantit\u00e9 de mati\u00e8re de glucose utilis\u00e9e.<\/li>\n<li>Calculer le nombre de mol\u00e9cules de glucose correspondant.<\/li>\n<\/ol>\n<p>Donn\u00e9es : M<sub>C<\/sub> = 12 g\u00b7mol\u207b\u00b9 ; M<sub>H<\/sub> = 1 g\u00b7mol\u207b\u00b9 ; M<sub>O<\/sub> = 16 g\u00b7mol\u207b\u00b9 ; N<sub>A<\/sub> = 6,02 \u00d7 10\u00b2\u00b3 mol\u207b\u00b9.<\/p>\n\n<p class=\"red\">1. Masse molaire<\/p>\n<p>M<sub>C\u2086H\u2081\u2082O\u2086<\/sub> = 6 \u00d7 12 + 12 \u00d7 1 + 6 \u00d7 16 = 180 g\u00b7mol\u207b\u00b9.<\/p>\n\n<p class=\"red\">2. Quantit\u00e9 de mati\u00e8re<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule :<\/p>\n<div class=\"formule\">n = m \/ M<\/div>\n<\/div>\n<p>n = 2,0 \/ 180<\/p>\n<p>n = 1,1 \u00d7 10\u207b\u00b2 mol<\/p>\n\n<p class=\"red\">3. Nombre de mol\u00e9cules<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule :<\/p>\n<div class=\"formule\">N = n \u00d7 N<sub>A<\/sub><\/div>\n<\/div>\n<p>N = 1,1 \u00d7 10\u207b\u00b2 \u00d7 6,02 \u00d7 10\u00b2\u00b3<\/p>\n<p class=\"resultat\">N = 6,6 \u00d7 10\u00b2\u00b9 mol\u00e9cules<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Situation 2 \u2014 Carburant pour un moteur<\/h3>\n<p>\nOn pr\u00e9l\u00e8ve 15 mL d\u2019octane C\u2088H\u2081\u2088, liquide que l\u2019on assimile \u00e0 de l\u2019essence, de densit\u00e9 d = 0,75.\n<\/p>\n<ol>\n<li>Calculer la masse molaire de l\u2019octane.<\/li>\n<li>Calculer la masse volumique de l\u2019octane.<\/li>\n<li>Calculer la quantit\u00e9 de mati\u00e8re dans les 15 mL pr\u00e9lev\u00e9s.<\/li>\n<\/ol>\n<p>Donn\u00e9es : M<sub>C<\/sub> = 12 g\u00b7mol\u207b\u00b9 ; M<sub>H<\/sub> = 1 g\u00b7mol\u207b\u00b9 ; \u03c1<sub>eau<\/sub> = 1000 g\u00b7L\u207b\u00b9.<\/p>\n\n<p class=\"red\">Correction :<\/p>\n<p>M<sub>C\u2088H\u2081\u2088<\/sub> = 8 \u00d7 12 + 18 \u00d7 1 = 114 g\u00b7mol\u207b\u00b9.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule :<\/p>\n<div class=\"formule\">\u03c1<sub>liq<\/sub> = d \u00d7 \u03c1<sub>eau<\/sub><\/div>\n<\/div>\n<p>\u03c1<sub>liq<\/sub> = 0,75 \u00d7 1000 = 750 g\u00b7L\u207b\u00b9.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formules :<\/p>\n<div class=\"formule\">m = \u03c1 \u00d7 V<\/div>\n<div class=\"formule\">n = m \/ M<\/div>\n<p class=\"step\">On injecte :<\/p>\n<div class=\"formule bluebox\">n = \u03c1 \u00d7 V \/ M<\/div>\n<\/div>\n<p>V = 15 mL = 15 \u00d7 10\u207b\u00b3 L<\/p>\n<p>n = (750 \u00d7 15 \u00d7 10\u207b\u00b3) \/ 114<\/p>\n<p class=\"resultat\">n = 9,8 \u00d7 10\u207b\u00b2 mol<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Situation 3 \u2014 Un flacon d\u2019acide<\/h3>\n<p>\nUn flacon contient de l\u2019acide sulfurique H\u2082SO\u2084 de masse volumique \u03c1 = 1800 g\u00b7L\u207b\u00b9.\nOn pr\u00e9l\u00e8ve 100 mL.\n<\/p>\n<ol>\n<li>Calculer la masse molaire de H\u2082SO\u2084.<\/li>\n<li>Calculer la masse du pr\u00e9l\u00e8vement.<\/li>\n<li>Calculer la quantit\u00e9 de mati\u00e8re.<\/li>\n<\/ol>\n<p>Donn\u00e9es : M<sub>H<\/sub> = 1 g\u00b7mol\u207b\u00b9 ; M<sub>S<\/sub> = 32 g\u00b7mol\u207b\u00b9 ; M<sub>O<\/sub> = 16 g\u00b7mol\u207b\u00b9.<\/p>\n\n<p>M<sub>H\u2082SO\u2084<\/sub> = 2 \u00d7 1 + 32 + 4 \u00d7 16 = 98 g\u00b7mol\u207b\u00b9.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule :<\/p>\n<div class=\"formule\">m = \u03c1 \u00d7 V<\/div>\n<\/div>\n<p>V = 100 \u00d7 10\u207b\u00b3 L<\/p>\n<p>m = 1800 \u00d7 100 \u00d7 10\u207b\u00b3 = 180 g.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule :<\/p>\n<div class=\"formule\">n = m \/ M<\/div>\n<\/div>\n<p>n = 180 \/ 98<\/p>\n<p class=\"resultat\">n = 1,83 mol<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\" id=\"fiche-bilan\">\n<h2>\ud83d\udccc Fiche bilan \u2014 La quantit\u00e9 de mati\u00e8re<\/h2>\n<div class=\"grid3\">\n<div class=\"card\"><h3>Mole<\/h3><p>Paquet de r\u00e9f\u00e9rence pour compter atomes, mol\u00e9cules ou ions.<\/p><\/div>\n<div class=\"card\"><h3>Avogadro<\/h3><div class=\"formule bluebox\">N<sub>A<\/sub> = 6,02 \u00d7 10\u00b2\u00b3 mol\u207b\u00b9<\/div><\/div>\n<div class=\"card\"><h3>Nombre d\u2019entit\u00e9s<\/h3><div class=\"formule\">N = n \u00d7 N<sub>A<\/sub><\/div><\/div>\n<div class=\"card\"><h3>Masse molaire atomique<\/h3><p>Masse d\u2019une mole d\u2019atomes.<\/p><div class=\"formule bluebox\">M<sub>X<\/sub> = A g\u00b7mol\u207b\u00b9<\/div><\/div>\n<div class=\"card\"><h3>Masse molaire mol\u00e9culaire<\/h3><p>On additionne les masses molaires des atomes.<\/p><\/div>\n<div class=\"card\"><h3>Masse molaire ionique<\/h3><p>On n\u00e9glige la masse des charges.<\/p><\/div>\n<div class=\"card\"><h3>Quantit\u00e9 de mati\u00e8re<\/h3><div class=\"formule\">n = m \/ M<\/div><\/div>\n<div class=\"card\"><h3>Masse<\/h3><div class=\"formule bluebox\">m = n \u00d7 M<\/div><\/div>\n<div class=\"card\"><h3>Masse volumique<\/h3><div class=\"formule\">\u03c1 = m \/ V<\/div><\/div>\n<div class=\"card\"><h3>Masse d\u2019un liquide<\/h3><div class=\"formule bluebox\">m = \u03c1 \u00d7 V<\/div><\/div>\n<div class=\"card\"><h3>Densit\u00e9<\/h3><div class=\"formule\">d = \u03c1<sub>liq<\/sub> \/ \u03c1<sub>eau<\/sub><\/div><\/div>\n<div class=\"card\"><h3>\u00c0 retenir<\/h3><p class=\"red\">V en L si \u03c1 est en g\u00b7L\u207b\u00b9 ; m en g si M est en g\u00b7mol\u207b\u00b9.<\/p><\/div>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>Carte mentale<\/h2>\n<div class=\"grid3\">\n<div class=\"mm-center\">LA QUANTIT\u00c9 DE MATI\u00c8RE<\/div>\n<div class=\"card\"><h3>Compter<\/h3><p>N = n \u00d7 N\u2090.<\/p><\/div>\n<div class=\"card\"><h3>Mole<\/h3><p>1 mol = 6,02 \u00d7 10\u00b2\u00b3 entit\u00e9s.<\/p><\/div>\n<div class=\"card\"><h3>Masse molaire<\/h3><p>Atomique, mol\u00e9culaire, ionique.<\/p><\/div>\n<div class=\"card\"><h3>Solide<\/h3><p>n = m \/ M.<\/p><\/div>\n<div class=\"card\"><h3>Liquide avec \u03c1<\/h3><p>m = \u03c1V puis n = m\/M.<\/p><\/div>\n<div class=\"card\"><h3>Liquide avec d<\/h3><p>\u03c1liq = d \u00d7 \u03c1eau puis n = \u03c1V\/M.<\/p><\/div>\n<\/div>\n<\/section>\n\n<\/div>\n\n<script>\nlet utterance;\nfunction playAudioSummary(){\n speechSynthesis.cancel();\n const text=document.getElementById(\"audioText\").innerText;\n utterance=new SpeechSynthesisUtterance(text);\n utterance.lang=\"fr-FR\";\n utterance.rate=0.90;\n speechSynthesis.speak(utterance);\n}\nfunction stopAudioSummary(){speechSynthesis.cancel();}\n<\/script>\n<\/body>\n<\/html>\n","protected":false},"excerpt":{"rendered":"<p>Seconde \u2014 La quantit\u00e9 de mati\u00e8re La quantit\u00e9 de mati\u00e8re Seconde physique-chimie \u2022 Mole \u2022 Constante d\u2019Avogadro \u2022 Masse molaire \u2022 Calculs avec m, M, V,...<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-793","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/793","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/comments?post=793"}],"version-history":[{"count":1,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/793\/revisions"}],"predecessor-version":[{"id":795,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/793\/revisions\/795"}],"wp:attachment":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/media?parent=793"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}

{"id":793,"date":"2026-05-28T04:02:07","date_gmt":"2026-05-28T02:02:07","guid":{"rendered":"https:\/\/pcwallis.malo.wf\/?page_id=793"},"modified":"2026-05-28T04:02:07","modified_gmt":"2026-05-28T02:02:07","slug":"quantite-de-matiere","status":"publish","type":"page","link":"https:\/\/pcwallis.malo.wf\/index.php\/quantite-de-matiere\/","title":{"rendered":"quantit\u00e9 de mati\u00e8re"},"content":{"rendered":"\n\n\n\n\n\nSeconde \u2014 La quantit\u00e9 de mati\u00e8re<\/title>\n<style>\n:root{\n --deep:#101827; --blue:#2563eb; --cyan:#0891b2; --teal:#14b8a6;\n --purple:#7c3aed; --pink:#db2777; --orange:#f97316; --red:#b83227;\n --green:#0f766e; --amber:#f59e0b; --ink:#1f2937; --paper:#fff; --line:#d9e2ec;\n}\n*{box-sizing:border-box} html{scroll-behavior:smooth}\nbody{\n margin:0;font-family:Arial,Helvetica,sans-serif;color:var(--ink);line-height:1.65;\n background:\n radial-gradient(circle at 10% 8%,rgba(20,184,166,.13),transparent 24%),\n radial-gradient(circle at 88% 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.start{color:var(--red);font-weight:900}.litteral .step{color:var(--deep);font-weight:800}\n.grid2{display:grid;grid-template-columns:repeat(auto-fit,minmax(320px,1fr));gap:22px;align-items:start}\n.grid3{display:grid;grid-template-columns:repeat(auto-fit,minmax(250px,1fr));gap:18px;align-items:start}\n.schema{background:#fbfdff;border:1px solid var(--line);border-radius:20px;padding:18px;text-align:center;overflow-x:auto;margin:14px 0}\nsvg{max-width:100%;height:auto}\n.btns{text-align:center;margin:18px 0 4px}\nbutton,.linkbtn{\n border:0;background:var(--blue);color:white;text-decoration:none;padding:14px 22px;border-radius:14px;\n font-size:17px;cursor:pointer;margin:6px;box-shadow:0 7px 18px rgba(37,99,235,.20);display:inline-block\n}\nbutton.stop{background:#333}.cyanbtn{background:var(--cyan)}.pinkbtn{background:var(--pink)}.greenbtn{background:#0f766e}.orangebtn{background:var(--orange)}\ntable{border-collapse:collapse;width:100%;margin:14px 0;background:white}\nth,td{border:1px solid #9aa7bd;padding:10px;text-align:center}th{background:#f0f4fb}\n.card{border:2px solid #e1e8f0;border-top:6px solid var(--teal);border-radius:18px;padding:18px;background:#fff}\n.mm-center{grid-column:1\/-1;text-align:center;background:linear-gradient(135deg,var(--deep),var(--teal));color:white;border-radius:18px;padding:18px;font-size:1.3rem;font-weight:900}\n.small{font-size:.92rem;color:#5b6770}\n@media print{button,.btns,.linkbtn{display:none}body{background:white}.section{box-shadow:none;break-inside:avoid}}\n<\/style>\n<\/head>\n<body>\n<header>\n<h1>La quantit\u00e9 de mati\u00e8re<\/h1>\n<p>Seconde physique-chimie \u2022 Mole \u2022 Constante d\u2019Avogadro \u2022 Masse molaire \u2022 Calculs avec m, M, V, \u03c1 et d<\/p>\n<\/header>\n\n<div class=\"container\">\n\n<div class=\"btns\">\n<button onclick=\"playAudioSummary()\">\ud83d\udd0a \u00c9couter le r\u00e9sum\u00e9 audio clair<\/button>\n<button class=\"stop\" onclick=\"stopAudioSummary()\">\u23f9 Arr\u00eater<\/button><br>\n<a class=\"linkbtn cyanbtn\" href=\"#fiche-bilan\">\ud83d\udccc Aller \u00e0 la fiche bilan<\/a>\n<a class=\"linkbtn pinkbtn\" href=\"#exercices\">\ud83e\uddea Exercices contextualis\u00e9s<\/a>\n<a class=\"linkbtn greenbtn\" href=\"#type-bac\">\ud83c\udf93 Partie type bac \/ \u00e9valuation<\/a>\n<a class=\"linkbtn orangebtn\" href=\"#liquides\">\ud83d\udca7 Cas des liquides<\/a>\n<\/div>\n\n<div id=\"audioText\" style=\"display:none;\">\nBienvenue dans le chapitre la quantit\u00e9 de mati\u00e8re.\nSi on veut compter les mol\u00e9cules ou les atomes dans un \u00e9chantillon, on r\u00e9alise que ces nombres sont immenses.\nOn d\u00e9finit donc un paquet de r\u00e9f\u00e9rence : la mole.\nUne mole est un paquet d\u2019atomes ou de mol\u00e9cules qui contient 6,02 fois 10 puissance 23 entit\u00e9s.\nCe nombre est la constante d\u2019Avogadro.\nLa quantit\u00e9 de mati\u00e8re se note n et s\u2019exprime en mole.\nLe nombre d\u2019entit\u00e9s N est reli\u00e9 \u00e0 la quantit\u00e9 de mati\u00e8re par la relation N \u00e9gale n fois NA.\nLa masse molaire atomique est la masse d\u2019une mole d\u2019atomes. On la note M et elle s\u2019exprime en gramme par mole.\nPour une mol\u00e9cule, on calcule la masse molaire mol\u00e9culaire en additionnant les masses molaires des atomes qui la constituent.\nPour un ion, on fait comme pour une mol\u00e9cule, car on n\u00e9glige la masse des charges \u00e9lectriques.\nPour calculer une quantit\u00e9 de mati\u00e8re \u00e0 partir d\u2019une masse, on utilise n \u00e9gale m divis\u00e9 par M.\nPour un liquide pur, si on conna\u00eet son volume et sa masse volumique, on calcule d\u2019abord la masse avec m \u00e9gale rho fois V, puis on calcule n avec n \u00e9gale m divis\u00e9 par M.\nSi on donne la densit\u00e9 d\u2019un liquide, on calcule sa masse volumique avec rho liquide \u00e9gale d fois rho eau, et rho eau vaut 1000 grammes par litre.\nFin du r\u00e9sum\u00e9.\n<\/div>\n\n<section class=\"section\">\n<h2>Objectifs du chapitre<\/h2>\n<p>\nCe cours reprend les phrases du cahier comme base : la mole, la constante d\u2019Avogadro, les masses molaires,\nle calcul d\u2019une quantit\u00e9 de mati\u00e8re \u00e0 partir d\u2019une masse, puis \u00e0 partir du volume d\u2019un liquide pur avec la masse volumique ou la densit\u00e9.\n<\/p>\n<div class=\"grid3\">\n<div class=\"card\"><h3>Comprendre<\/h3><p>Pourquoi on utilise la mole pour compter des entit\u00e9s microscopiques.<\/p><\/div>\n<div class=\"card\"><h3>Calculer<\/h3><p>Une masse molaire, une quantit\u00e9 de mati\u00e8re, une masse ou un volume.<\/p><\/div>\n<div class=\"card\"><h3>R\u00e9diger<\/h3><p>Formule du cours en rouge, transformation litt\u00e9rale, application num\u00e9rique, r\u00e9sultat soulign\u00e9.<\/p><\/div>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>\ud83d\udd34 M\u00e9thode obligatoire pour les exercices<\/h2>\n<p>\nDans chaque correction, on part de la <span class=\"red\">formule du cours encadr\u00e9e<\/span>, puis on fait le\n<span class=\"blue\">travail litt\u00e9ral<\/span>, et enfin l\u2019application num\u00e9rique avec les unit\u00e9s.\n<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Exemple : isoler la masse dans n = m \/ M<\/p>\n<div class=\"formule\">n = m \/ M<\/div>\n<p class=\"step\">On multiplie par M :<\/p>\n<div class=\"formule bluebox\">n \u00d7 M = m<\/div>\n<p class=\"step\">Donc :<\/p>\n<div class=\"formule bluebox\">m = n \u00d7 M<\/div>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>I. La mole<\/h2>\n\n<p>\nSi on veut compter les mol\u00e9cules qu\u2019il y a dans un \u00e9chantillon, on r\u00e9alise que ces nombres sont immenses.\n<\/p>\n\n<div class=\"note\">\n<p>Exemple du cahier : dans 1 g de carbone \u00b9\u00b2C, on trouve environ N = 4,98 \u00d7 10\u00b2\u00b2 atomes. Ce nombre est gigantesque.<\/p>\n<\/div>\n\n<p class=\"red\">\nOn va donc d\u00e9finir un paquet de r\u00e9f\u00e9rence : la mole.\n<\/p>\n<p>\nOn a choisi Avogadro : un paquet tr\u00e8s facile \u00e0 peser.\n<\/p>\n<p class=\"red\">\nUne mole est un paquet d\u2019atomes ou de mol\u00e9cules qui contient N<sub>A<\/sub> = 6,02 \u00d7 10\u00b2\u00b3 entit\u00e9s.\n<\/p>\n\n<div class=\"formule\">N<sub>A<\/sub> = 6,02 \u00d7 10\u00b2\u00b3 mol\u207b\u00b9<\/div>\n\n<p><span class=\"red\">N<sub>A<\/sub><\/span> : constante d\u2019Avogadro.<\/p>\n<p>\nC\u2019est le nombre d\u2019atomes qu\u2019il y a dans A = 12 g de carbone \u00b9\u00b2C.\n<\/p>\n\n<p class=\"red\">\nOn va d\u00e9finir n : quantit\u00e9 de mati\u00e8re, qui s\u2019exprimera en mole, mol.\n<\/p>\n<p>\nOn peut retrouver N, le nombre d\u2019atomes ou de mol\u00e9cules, en utilisant la relation \u00e9vidente :\n<\/p>\n\n<div class=\"formule\">N = n \u00d7 N<sub>A<\/sub><\/div>\n\n<ul>\n<li><span class=\"red\">N<\/span> : nombre d\u2019entit\u00e9s, sans unit\u00e9 ;<\/li>\n<li><span class=\"red\">n<\/span> : quantit\u00e9 de mati\u00e8re en mol ;<\/li>\n<li><span class=\"red\">N<sub>A<\/sub><\/span> : constante d\u2019Avogadro en mol\u207b\u00b9.<\/li>\n<\/ul>\n\n<div class=\"schema\">\n<svg width=\"850\" height=\"360\" viewBox=\"0 0 850 360\">\n<rect x=\"55\" y=\"45\" width=\"740\" height=\"240\" rx=\"22\" fill=\"#fbfdff\" stroke=\"#D9E2EC\"\/>\n<text x=\"90\" y=\"85\" font-size=\"20\" font-weight=\"900\" fill=\"#0f172a\">La mole : un paquet de r\u00e9f\u00e9rence<\/text>\n<circle cx=\"205\" cy=\"175\" r=\"78\" fill=\"#e0f2fe\" stroke=\"#2563EB\" stroke-width=\"4\"\/>\n<text x=\"155\" y=\"165\" fill=\"#2563EB\" font-size=\"20\" font-weight=\"900\">1 mole<\/text>\n<text x=\"130\" y=\"200\" fill=\"#2563EB\" font-size=\"17\" font-weight=\"900\">6,02 \u00d7 10\u00b2\u00b3 entit\u00e9s<\/text>\n<line x1=\"310\" y1=\"175\" x2=\"505\" y2=\"175\" stroke=\"#0f172a\" stroke-width=\"5\"\/>\n<polygon points=\"505,175 480,161 480,189\" fill=\"#0f172a\"\/>\n<text x=\"540\" y=\"150\" font-size=\"20\" font-weight=\"900\">N = n \u00d7 N\u2090<\/text>\n<text x=\"540\" y=\"190\" fill=\"#B83227\" font-size=\"17\" font-weight=\"900\">La mole sert \u00e0 compter sans \u00e9crire des nombres immenses.<\/text>\n<\/svg>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Exercice type \u2014 Calculer le nombre d\u2019entit\u00e9s<\/h3>\n<p>On pr\u00e9l\u00e8ve n = 1,0 \u00d7 10\u207b\u00b2 mol de mol\u00e9cules. Calculer le nombre de mol\u00e9cules.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">N = n \u00d7 N<sub>A<\/sub><\/div>\n<p class=\"step\">La formule est d\u00e9j\u00e0 sous la bonne forme.<\/p>\n<\/div>\n<p>N = 1,0 \u00d7 10\u207b\u00b2 \u00d7 6,02 \u00d7 10\u00b2\u00b3<\/p>\n<p class=\"resultat\">N = 6,02 \u00d7 10\u00b2\u00b9 mol\u00e9cules<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>II. Masses molaires<\/h2>\n\n<h3>1. Masse molaire atomique<\/h3>\n<p>\nOn a choisi N<sub>A<\/sub> pour que 1 mole d\u2019atome <sup>A<\/sup><sub>Z<\/sub>X p\u00e8se A grammes.\n<\/p>\n\n<p class=\"red\">\nMasse molaire atomique : c\u2019est la masse d\u2019une mole d\u2019atomes.\n<\/p>\n<div class=\"formule\">M<sub>X<\/sub> = A g\u00b7mol\u207b\u00b9<\/div>\n\n<p>Il suffit de regarder dans le tableau p\u00e9riodique la valeur de A.<\/p>\n\n<table>\n<tr><th>Atome<\/th><th>Masse molaire atomique<\/th><\/tr>\n<tr><td>\u00b9\u00b2C<\/td><td>M<sub>C<\/sub> = 12 g\u00b7mol\u207b\u00b9<\/td><\/tr>\n<tr><td>\u00b9H<\/td><td>M<sub>H<\/sub> = 1 g\u00b7mol\u207b\u00b9<\/td><\/tr>\n<tr><td>\u00b9\u2076O<\/td><td>M<sub>O<\/sub> = 16 g\u00b7mol\u207b\u00b9<\/td><\/tr>\n<tr><td>Na<\/td><td>M<sub>Na<\/sub> = 23 g\u00b7mol\u207b\u00b9<\/td><\/tr>\n<tr><td>Cl<\/td><td>M<sub>Cl<\/sub> = 35,5 g\u00b7mol\u207b\u00b9<\/td><\/tr>\n<\/table>\n\n<h3>2. Masse molaire mol\u00e9culaire<\/h3>\n<p class=\"red\">\nOn ajoute les nombres molaires des atomes qui constituent la mol\u00e9cule.\n<\/p>\n\n<div class=\"formule\">M<sub>CxHyOz<\/sub> = x \u00d7 M<sub>C<\/sub> + y \u00d7 M<sub>H<\/sub> + z \u00d7 M<sub>O<\/sub><\/div>\n\n<div class=\"exercice\">\n<h3>Exemples du cahier<\/h3>\n<p><span class=\"red\">Glucose C\u2086H\u2081\u2082O\u2086 :<\/span><\/p>\n<p>M<sub>C\u2086H\u2081\u2082O\u2086<\/sub> = 6 \u00d7 12 + 12 \u00d7 1 + 6 \u00d7 16<\/p>\n<p class=\"resultat\">M<sub>C\u2086H\u2081\u2082O\u2086<\/sub> = 180 g\u00b7mol\u207b\u00b9<\/p>\n\n<p><span class=\"red\">Chlorure de sodium NaCl :<\/span><\/p>\n<p>M<sub>NaCl<\/sub> = 1 \u00d7 23 + 1 \u00d7 35,5<\/p>\n<p class=\"resultat\">M<sub>NaCl<\/sub> = 58,5 g\u00b7mol\u207b\u00b9<\/p>\n\n<p><span class=\"red\">\u00c9thanol C\u2082H\u2086O :<\/span><\/p>\n<p>M<sub>C\u2082H\u2086O<\/sub> = 2 \u00d7 12 + 6 \u00d7 1 + 1 \u00d7 16<\/p>\n<p class=\"resultat\">M<sub>C\u2082H\u2086O<\/sub> = 46 g\u00b7mol\u207b\u00b9<\/p>\n\n<p><span class=\"red\">Eau H\u2082O :<\/span><\/p>\n<p>M<sub>H\u2082O<\/sub> = 2 \u00d7 1 + 1 \u00d7 16<\/p>\n<p class=\"resultat\">M<sub>H\u2082O<\/sub> = 18 g\u00b7mol\u207b\u00b9<\/p>\n<\/div>\n\n<h3>3. Masse molaire ionique<\/h3>\n<p class=\"red\">\nC\u2019est la masse d\u2019une mole d\u2019ion.\n<\/p>\n<p class=\"red\">\nOn n\u00e9glige la masse des charges \u00e9lectriques + ou \u2212.\n<\/p>\n<p>Donc on fait comme pour une mol\u00e9cule.<\/p>\n\n<div class=\"exercice\">\n<h3>Exemple \u2014 Ion hypochlorite ClO\u207b<\/h3>\n<p>M<sub>ClO\u207b<\/sub> = M<sub>Cl<\/sub> + M<sub>O<\/sub><\/p>\n<p>M<sub>ClO\u207b<\/sub> = 35,5 + 16<\/p>\n<p class=\"resultat\">M<sub>ClO\u207b<\/sub> = 51,5 g\u00b7mol\u207b\u00b9<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>III. Calcul d\u2019une quantit\u00e9 de mati\u00e8re<\/h2>\n\n<h3>1. \u00c0 partir de la masse m d\u2019un \u00e9chantillon<\/h3>\n<div class=\"formule\">n = m \/ M<\/div>\n<ul>\n<li><span class=\"red\">n<\/span> : quantit\u00e9 de mati\u00e8re en mol ;<\/li>\n<li><span class=\"red\">m<\/span> : masse de l\u2019\u00e9chantillon en g ;<\/li>\n<li><span class=\"red\">M<\/span> : masse molaire en g\u00b7mol\u207b\u00b9.<\/li>\n<\/ul>\n\n<div class=\"exercice\">\n<h3>Exercice type 1 \u2014 Glucose<\/h3>\n<p>Calculer la quantit\u00e9 de mati\u00e8re qu\u2019il y a dans 2,0 g de glucose C\u2086H\u2081\u2082O\u2086.<\/p>\n<p>Donn\u00e9e : M<sub>C\u2086H\u2081\u2082O\u2086<\/sub> = 180 g\u00b7mol\u207b\u00b9.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n = m \/ M<\/div>\n<p class=\"step\">La formule est d\u00e9j\u00e0 sous la bonne forme.<\/p>\n<\/div>\n<p>n = 2,0 \/ 180<\/p>\n<p class=\"resultat\">n = 0,011 mol = 1,1 \u00d7 10\u207b\u00b2 mol<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Exercice type 2 \u2014 Sel NaCl<\/h3>\n<p>Calculer la quantit\u00e9 de mati\u00e8re dans 3,0 g de NaCl.<\/p>\n<p>Donn\u00e9e : M<sub>NaCl<\/sub> = 58,5 g\u00b7mol\u207b\u00b9.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n = m \/ M<\/div>\n<\/div>\n<p>n = 3,0 \/ 58,5<\/p>\n<p class=\"resultat\">n = 0,051 mol = 5,1 \u00d7 10\u207b\u00b2 mol<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Exercice type 3 \u2014 Masse \u00e0 peser<\/h3>\n<p>Calculer la masse \u00e0 peser pour pr\u00e9lever n = 2,5 \u00d7 10\u207b\u00b2 mol d\u2019\u00e9thanol C\u2082H\u2086O.<\/p>\n<p>Donn\u00e9e : M<sub>C\u2082H\u2086O<\/sub> = 46 g\u00b7mol\u207b\u00b9.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n = m \/ M<\/div>\n<p class=\"step\">On multiplie par M :<\/p>\n<div class=\"formule bluebox\">n \u00d7 M = m<\/div>\n<p class=\"step\">Donc :<\/p>\n<div class=\"formule bluebox\">m = n \u00d7 M<\/div>\n<\/div>\n<p>m = 2,5 \u00d7 10\u207b\u00b2 \u00d7 46<\/p>\n<p class=\"resultat\">m = 1,15 g<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\" id=\"liquides\">\n<h2>IV. \u00c0 partir du volume d\u2019un liquide pur<\/h2>\n\n<h3>1. En utilisant la masse volumique \u03c1<\/h3>\n<div class=\"formule\">\u03c1 = m \/ V<\/div>\n<ul>\n<li><span class=\"red\">\u03c1<\/span> : masse volumique du liquide en g\u00b7L\u207b\u00b9 ;<\/li>\n<li><span class=\"red\">m<\/span> : masse de l\u2019\u00e9chantillon en g ;<\/li>\n<li><span class=\"red\">V<\/span> : volume de l\u2019\u00e9chantillon en L.<\/li>\n<\/ul>\n\n<p class=\"red\">\nOn va utiliser \u03c1 et V pour trouver la masse de l\u2019\u00e9chantillon m.\n<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">\u03c1 = m \/ V<\/div>\n<p class=\"step\">On multiplie par V :<\/p>\n<div class=\"formule bluebox\">\u03c1 \u00d7 V = m<\/div>\n<p class=\"step\">Donc :<\/p>\n<div class=\"formule bluebox\">m = \u03c1 \u00d7 V<\/div>\n<p class=\"step\">Ensuite on utilise :<\/p>\n<div class=\"formule\">n = m \/ M<\/div>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Exercice type \u2014 Eau<\/h3>\n<p>Calculer la quantit\u00e9 de mati\u00e8re qu\u2019il y a dans 10 mL d\u2019eau.<\/p>\n<p>Donn\u00e9es : \u03c1<sub>eau<\/sub> = 1000 g\u00b7L\u207b\u00b9 ; M<sub>H\u2082O<\/sub> = 18 g\u00b7mol\u207b\u00b9.<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule 1 :<\/p>\n<div class=\"formule\">\u03c1 = m \/ V<\/div>\n<p class=\"step\">On isole m :<\/p>\n<div class=\"formule bluebox\">m = \u03c1 \u00d7 V<\/div>\n<p class=\"start\">Formule 2 :<\/p>\n<div class=\"formule\">n = m \/ M<\/div>\n<p class=\"step\">On injecte la formule 1 dans la formule 2 :<\/p>\n<div class=\"formule bluebox\">n = \u03c1 \u00d7 V \/ M<\/div>\n<\/div>\n\n<p class=\"red\">Conversion :<\/p>\n<p>V = 10 mL = 10 \u00d7 10\u207b\u00b3 L = 1,0 \u00d7 10\u207b\u00b2 L<\/p>\n<p>Application num\u00e9rique :<\/p>\n<p>n = (1000 \u00d7 10 \u00d7 10\u207b\u00b3) \/ 18<\/p>\n<p class=\"resultat\">n = 0,55 mol<\/p>\n<\/div>\n\n<h3>2. En utilisant la densit\u00e9 d\u2019un liquide<\/h3>\n\n<div class=\"formule\">d = \u03c1<sub>liquide<\/sub> \/ \u03c1<sub>eau<\/sub><\/div>\n<ul>\n<li><span class=\"red\">d<\/span> : densit\u00e9 d\u2019un liquide, sans unit\u00e9 ;<\/li>\n<li><span class=\"red\">\u03c1<sub>liquide<\/sub><\/span> : masse volumique du liquide ;<\/li>\n<li><span class=\"red\">\u03c1<sub>eau<\/sub><\/span> : masse volumique de l\u2019eau = 1000 g\u00b7L\u207b\u00b9.<\/li>\n<\/ul>\n\n<p class=\"red\">On va utiliser d pour trouver \u03c1<sub>liquide<\/sub>.<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">d = \u03c1<sub>liquide<\/sub> \/ \u03c1<sub>eau<\/sub><\/div>\n<p class=\"step\">On multiplie par \u03c1<sub>eau<\/sub> :<\/p>\n<div class=\"formule bluebox\">d \u00d7 \u03c1<sub>eau<\/sub> = \u03c1<sub>liquide<\/sub><\/div>\n<p class=\"step\">Donc :<\/p>\n<div class=\"formule bluebox\">\u03c1<sub>liquide<\/sub> = d \u00d7 \u03c1<sub>eau<\/sub><\/div>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Exercice type \u2014 Essence<\/h3>\n<p>Calculer la quantit\u00e9 de mati\u00e8re qu\u2019il y a dans 15 mL d\u2019essence C\u2088H\u2081\u2088 de densit\u00e9 d = 0,75.<\/p>\n<p>Donn\u00e9es : M<sub>C<\/sub> = 12 g\u00b7mol\u207b\u00b9 ; M<sub>H<\/sub> = 1 g\u00b7mol\u207b\u00b9 ; \u03c1<sub>eau<\/sub> = 1000 g\u00b7L\u207b\u00b9.<\/p>\n\n<p class=\"red\">\u00c9tape 1 : masse molaire<\/p>\n<p>M<sub>C\u2088H\u2081\u2088<\/sub> = 8 \u00d7 M<sub>C<\/sub> + 18 \u00d7 M<sub>H<\/sub><\/p>\n<p>M<sub>C\u2088H\u2081\u2088<\/sub> = 8 \u00d7 12 + 18 \u00d7 1<\/p>\n<p>M<sub>C\u2088H\u2081\u2088<\/sub> = 114 g\u00b7mol\u207b\u00b9<\/p>\n\n<p class=\"red\">\u00c9tape 2 : masse volumique du liquide<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule :<\/p>\n<div class=\"formule\">d = \u03c1<sub>liq<\/sub> \/ \u03c1<sub>eau<\/sub><\/div>\n<p class=\"step\">On isole \u03c1<sub>liq<\/sub> :<\/p>\n<div class=\"formule bluebox\">\u03c1<sub>liq<\/sub> = d \u00d7 \u03c1<sub>eau<\/sub><\/div>\n<\/div>\n<p>\u03c1<sub>liq<\/sub> = 0,75 \u00d7 1000 = 750 g\u00b7L\u207b\u00b9<\/p>\n\n<p class=\"red\">\u00c9tape 3 : quantit\u00e9 de mati\u00e8re<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formules :<\/p>\n<div class=\"formule\">m = \u03c1 \u00d7 V<\/div>\n<div class=\"formule\">n = m \/ M<\/div>\n<p class=\"step\">On injecte m = \u03c1 \u00d7 V dans n = m \/ M :<\/p>\n<div class=\"formule bluebox\">n = \u03c1 \u00d7 V \/ M<\/div>\n<\/div>\n<p>V = 15 mL = 15 \u00d7 10\u207b\u00b3 L<\/p>\n<p>n = (750 \u00d7 15 \u00d7 10\u207b\u00b3) \/ 114<\/p>\n<p class=\"resultat\">n = 9,8 \u00d7 10\u207b\u00b2 mol<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\" id=\"exercices\">\n<h2>V. Exercices contextualis\u00e9s<\/h2>\n\n<div class=\"exercice\">\n<h3>1. Comprim\u00e9 de vitamine C<\/h3>\n<p>Calculer la quantit\u00e9 de mati\u00e8re dans 1 comprim\u00e9 de vitamine C C\u2086H\u2088O\u2086 de 500 mg.<\/p>\n<p>Donn\u00e9es : M<sub>C<\/sub> = 12 g\u00b7mol\u207b\u00b9 ; M<sub>H<\/sub> = 1 g\u00b7mol\u207b\u00b9 ; M<sub>O<\/sub> = 16 g\u00b7mol\u207b\u00b9.<\/p>\n\n<p class=\"red\">Masse molaire :<\/p>\n<p>M<sub>C\u2086H\u2088O\u2086<\/sub> = 6 \u00d7 12 + 8 \u00d7 1 + 6 \u00d7 16 = 176 g\u00b7mol\u207b\u00b9<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n = m \/ M<\/div>\n<\/div>\n<p>m = 500 mg = 500 \u00d7 10\u207b\u00b3 g = 0,500 g<\/p>\n<p>n = 0,500 \/ 176<\/p>\n<p class=\"resultat\">n = 2,84 \u00d7 10\u207b\u00b3 mol<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>2. \u00c9thanol au laboratoire<\/h3>\n<p>Calculer la masse de C\u2082H\u2086O \u00e0 peser pour pr\u00e9lever 2,5 \u00d7 10\u207b\u00b2 mol.<\/p>\n<p>Donn\u00e9es : M<sub>C<\/sub> = 12 g\u00b7mol\u207b\u00b9 ; M<sub>H<\/sub> = 1 g\u00b7mol\u207b\u00b9 ; M<sub>O<\/sub> = 16 g\u00b7mol\u207b\u00b9.<\/p>\n<p>M<sub>C\u2082H\u2086O<\/sub> = 2 \u00d7 12 + 6 \u00d7 1 + 16 = 46 g\u00b7mol\u207b\u00b9<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule :<\/p><div class=\"formule\">n = m \/ M<\/div>\n<p class=\"step\">On isole m :<\/p><div class=\"formule bluebox\">m = n \u00d7 M<\/div>\n<\/div>\n<p>m = 2,5 \u00d7 10\u207b\u00b2 \u00d7 46<\/p>\n<p class=\"resultat\">m = 1,15 g<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>3. Acide sulfurique<\/h3>\n<p>Calculer la quantit\u00e9 de mati\u00e8re qu\u2019il y a dans 100 mL d\u2019H\u2082SO\u2084 de masse volumique \u03c1 = 1800 g\u00b7L\u207b\u00b9.<\/p>\n<p>Donn\u00e9es : M<sub>H<\/sub> = 1 g\u00b7mol\u207b\u00b9 ; M<sub>S<\/sub> = 32 g\u00b7mol\u207b\u00b9 ; M<sub>O<\/sub> = 16 g\u00b7mol\u207b\u00b9.<\/p>\n<p>M<sub>H\u2082SO\u2084<\/sub> = 2 \u00d7 1 + 32 + 4 \u00d7 16 = 98 g\u00b7mol\u207b\u00b9<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formules :<\/p>\n<div class=\"formule\">m = \u03c1 \u00d7 V<\/div>\n<div class=\"formule\">n = m \/ M<\/div>\n<p class=\"step\">On injecte :<\/p>\n<div class=\"formule bluebox\">n = \u03c1 \u00d7 V \/ M<\/div>\n<\/div>\n<p>V = 100 mL = 100 \u00d7 10\u207b\u00b3 L<\/p>\n<p>n = (1800 \u00d7 100 \u00d7 10\u207b\u00b3) \/ 98<\/p>\n<p class=\"resultat\">n = 1,83 mol<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>4. Acide \u00e9thano\u00efque dans le vinaigre<\/h3>\n<p>Calculer la quantit\u00e9 de mati\u00e8re qu\u2019il y a dans 20 mL d\u2019acide \u00e9thano\u00efque CH\u2083COOH, de densit\u00e9 d = 1,07.<\/p>\n<p>Donn\u00e9es : M<sub>C<\/sub> = 12 g\u00b7mol\u207b\u00b9 ; M<sub>H<\/sub> = 1 g\u00b7mol\u207b\u00b9 ; M<sub>O<\/sub> = 16 g\u00b7mol\u207b\u00b9 ; \u03c1<sub>eau<\/sub> = 1000 g\u00b7L\u207b\u00b9.<\/p>\n<p>CH\u2083COOH correspond \u00e0 C\u2082H\u2084O\u2082.<\/p>\n<p>M<sub>C\u2082H\u2084O\u2082<\/sub> = 2 \u00d7 12 + 4 \u00d7 1 + 2 \u00d7 16 = 60 g\u00b7mol\u207b\u00b9<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formules :<\/p>\n<div class=\"formule\">\u03c1<sub>liq<\/sub> = d \u00d7 \u03c1<sub>eau<\/sub><\/div>\n<div class=\"formule\">m = \u03c1 \u00d7 V<\/div>\n<div class=\"formule\">n = m \/ M<\/div>\n<p class=\"step\">On injecte :<\/p>\n<div class=\"formule bluebox\">n = d \u00d7 \u03c1<sub>eau<\/sub> \u00d7 V \/ M<\/div>\n<\/div>\n<p>V = 20 mL = 20 \u00d7 10\u207b\u00b3 L<\/p>\n<p>n = (1,07 \u00d7 1000 \u00d7 20 \u00d7 10\u207b\u00b3) \/ 60<\/p>\n<p class=\"resultat\">n = 0,35 mol<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\" id=\"type-bac\">\n<h2>VI. Partie type bac \/ \u00e9valuation \u2014 niveau seconde<\/h2>\n\n<div class=\"exercice\">\n<h3>Situation 1 \u2014 Pr\u00e9parer une solution de glucose pour une boisson de sport<\/h3>\n<p>\nUn pr\u00e9parateur sportif souhaite utiliser 2,0 g de glucose C\u2086H\u2081\u2082O\u2086.\n<\/p>\n<ol>\n<li>Calculer la masse molaire du glucose.<\/li>\n<li>Calculer la quantit\u00e9 de mati\u00e8re de glucose utilis\u00e9e.<\/li>\n<li>Calculer le nombre de mol\u00e9cules de glucose correspondant.<\/li>\n<\/ol>\n<p>Donn\u00e9es : M<sub>C<\/sub> = 12 g\u00b7mol\u207b\u00b9 ; M<sub>H<\/sub> = 1 g\u00b7mol\u207b\u00b9 ; M<sub>O<\/sub> = 16 g\u00b7mol\u207b\u00b9 ; N<sub>A<\/sub> = 6,02 \u00d7 10\u00b2\u00b3 mol\u207b\u00b9.<\/p>\n\n<p class=\"red\">1. Masse molaire<\/p>\n<p>M<sub>C\u2086H\u2081\u2082O\u2086<\/sub> = 6 \u00d7 12 + 12 \u00d7 1 + 6 \u00d7 16 = 180 g\u00b7mol\u207b\u00b9.<\/p>\n\n<p class=\"red\">2. Quantit\u00e9 de mati\u00e8re<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule :<\/p>\n<div class=\"formule\">n = m \/ M<\/div>\n<\/div>\n<p>n = 2,0 \/ 180<\/p>\n<p>n = 1,1 \u00d7 10\u207b\u00b2 mol<\/p>\n\n<p class=\"red\">3. Nombre de mol\u00e9cules<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule :<\/p>\n<div class=\"formule\">N = n \u00d7 N<sub>A<\/sub><\/div>\n<\/div>\n<p>N = 1,1 \u00d7 10\u207b\u00b2 \u00d7 6,02 \u00d7 10\u00b2\u00b3<\/p>\n<p class=\"resultat\">N = 6,6 \u00d7 10\u00b2\u00b9 mol\u00e9cules<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Situation 2 \u2014 Carburant pour un moteur<\/h3>\n<p>\nOn pr\u00e9l\u00e8ve 15 mL d\u2019octane C\u2088H\u2081\u2088, liquide que l\u2019on assimile \u00e0 de l\u2019essence, de densit\u00e9 d = 0,75.\n<\/p>\n<ol>\n<li>Calculer la masse molaire de l\u2019octane.<\/li>\n<li>Calculer la masse volumique de l\u2019octane.<\/li>\n<li>Calculer la quantit\u00e9 de mati\u00e8re dans les 15 mL pr\u00e9lev\u00e9s.<\/li>\n<\/ol>\n<p>Donn\u00e9es : M<sub>C<\/sub> = 12 g\u00b7mol\u207b\u00b9 ; M<sub>H<\/sub> = 1 g\u00b7mol\u207b\u00b9 ; \u03c1<sub>eau<\/sub> = 1000 g\u00b7L\u207b\u00b9.<\/p>\n\n<p class=\"red\">Correction :<\/p>\n<p>M<sub>C\u2088H\u2081\u2088<\/sub> = 8 \u00d7 12 + 18 \u00d7 1 = 114 g\u00b7mol\u207b\u00b9.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule :<\/p>\n<div class=\"formule\">\u03c1<sub>liq<\/sub> = d \u00d7 \u03c1<sub>eau<\/sub><\/div>\n<\/div>\n<p>\u03c1<sub>liq<\/sub> = 0,75 \u00d7 1000 = 750 g\u00b7L\u207b\u00b9.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formules :<\/p>\n<div class=\"formule\">m = \u03c1 \u00d7 V<\/div>\n<div class=\"formule\">n = m \/ M<\/div>\n<p class=\"step\">On injecte :<\/p>\n<div class=\"formule bluebox\">n = \u03c1 \u00d7 V \/ M<\/div>\n<\/div>\n<p>V = 15 mL = 15 \u00d7 10\u207b\u00b3 L<\/p>\n<p>n = (750 \u00d7 15 \u00d7 10\u207b\u00b3) \/ 114<\/p>\n<p class=\"resultat\">n = 9,8 \u00d7 10\u207b\u00b2 mol<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Situation 3 \u2014 Un flacon d\u2019acide<\/h3>\n<p>\nUn flacon contient de l\u2019acide sulfurique H\u2082SO\u2084 de masse volumique \u03c1 = 1800 g\u00b7L\u207b\u00b9.\nOn pr\u00e9l\u00e8ve 100 mL.\n<\/p>\n<ol>\n<li>Calculer la masse molaire de H\u2082SO\u2084.<\/li>\n<li>Calculer la masse du pr\u00e9l\u00e8vement.<\/li>\n<li>Calculer la quantit\u00e9 de mati\u00e8re.<\/li>\n<\/ol>\n<p>Donn\u00e9es : M<sub>H<\/sub> = 1 g\u00b7mol\u207b\u00b9 ; M<sub>S<\/sub> = 32 g\u00b7mol\u207b\u00b9 ; M<sub>O<\/sub> = 16 g\u00b7mol\u207b\u00b9.<\/p>\n\n<p>M<sub>H\u2082SO\u2084<\/sub> = 2 \u00d7 1 + 32 + 4 \u00d7 16 = 98 g\u00b7mol\u207b\u00b9.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule :<\/p>\n<div class=\"formule\">m = \u03c1 \u00d7 V<\/div>\n<\/div>\n<p>V = 100 \u00d7 10\u207b\u00b3 L<\/p>\n<p>m = 1800 \u00d7 100 \u00d7 10\u207b\u00b3 = 180 g.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule :<\/p>\n<div class=\"formule\">n = m \/ M<\/div>\n<\/div>\n<p>n = 180 \/ 98<\/p>\n<p class=\"resultat\">n = 1,83 mol<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\" id=\"fiche-bilan\">\n<h2>\ud83d\udccc Fiche bilan \u2014 La quantit\u00e9 de mati\u00e8re<\/h2>\n<div class=\"grid3\">\n<div class=\"card\"><h3>Mole<\/h3><p>Paquet de r\u00e9f\u00e9rence pour compter atomes, mol\u00e9cules ou ions.<\/p><\/div>\n<div class=\"card\"><h3>Avogadro<\/h3><div class=\"formule bluebox\">N<sub>A<\/sub> = 6,02 \u00d7 10\u00b2\u00b3 mol\u207b\u00b9<\/div><\/div>\n<div class=\"card\"><h3>Nombre d\u2019entit\u00e9s<\/h3><div class=\"formule\">N = n \u00d7 N<sub>A<\/sub><\/div><\/div>\n<div class=\"card\"><h3>Masse molaire atomique<\/h3><p>Masse d\u2019une mole d\u2019atomes.<\/p><div class=\"formule bluebox\">M<sub>X<\/sub> = A g\u00b7mol\u207b\u00b9<\/div><\/div>\n<div class=\"card\"><h3>Masse molaire mol\u00e9culaire<\/h3><p>On additionne les masses molaires des atomes.<\/p><\/div>\n<div class=\"card\"><h3>Masse molaire ionique<\/h3><p>On n\u00e9glige la masse des charges.<\/p><\/div>\n<div class=\"card\"><h3>Quantit\u00e9 de mati\u00e8re<\/h3><div class=\"formule\">n = m \/ M<\/div><\/div>\n<div class=\"card\"><h3>Masse<\/h3><div class=\"formule bluebox\">m = n \u00d7 M<\/div><\/div>\n<div class=\"card\"><h3>Masse volumique<\/h3><div class=\"formule\">\u03c1 = m \/ V<\/div><\/div>\n<div class=\"card\"><h3>Masse d\u2019un liquide<\/h3><div class=\"formule bluebox\">m = \u03c1 \u00d7 V<\/div><\/div>\n<div class=\"card\"><h3>Densit\u00e9<\/h3><div class=\"formule\">d = \u03c1<sub>liq<\/sub> \/ \u03c1<sub>eau<\/sub><\/div><\/div>\n<div class=\"card\"><h3>\u00c0 retenir<\/h3><p class=\"red\">V en L si \u03c1 est en g\u00b7L\u207b\u00b9 ; m en g si M est en g\u00b7mol\u207b\u00b9.<\/p><\/div>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>Carte mentale<\/h2>\n<div class=\"grid3\">\n<div class=\"mm-center\">LA QUANTIT\u00c9 DE MATI\u00c8RE<\/div>\n<div class=\"card\"><h3>Compter<\/h3><p>N = n \u00d7 N\u2090.<\/p><\/div>\n<div class=\"card\"><h3>Mole<\/h3><p>1 mol = 6,02 \u00d7 10\u00b2\u00b3 entit\u00e9s.<\/p><\/div>\n<div class=\"card\"><h3>Masse molaire<\/h3><p>Atomique, mol\u00e9culaire, ionique.<\/p><\/div>\n<div class=\"card\"><h3>Solide<\/h3><p>n = m \/ M.<\/p><\/div>\n<div class=\"card\"><h3>Liquide avec \u03c1<\/h3><p>m = \u03c1V puis n = m\/M.<\/p><\/div>\n<div class=\"card\"><h3>Liquide avec d<\/h3><p>\u03c1liq = d \u00d7 \u03c1eau puis n = \u03c1V\/M.<\/p><\/div>\n<\/div>\n<\/section>\n\n<\/div>\n\n<script>\nlet utterance;\nfunction playAudioSummary(){\n speechSynthesis.cancel();\n const text=document.getElementById(\"audioText\").innerText;\n utterance=new SpeechSynthesisUtterance(text);\n utterance.lang=\"fr-FR\";\n utterance.rate=0.90;\n speechSynthesis.speak(utterance);\n}\nfunction stopAudioSummary(){speechSynthesis.cancel();}\n<\/script>\n<\/body>\n<\/html>\n","protected":false},"excerpt":{"rendered":"<p>Seconde \u2014 La quantit\u00e9 de mati\u00e8re La quantit\u00e9 de mati\u00e8re Seconde physique-chimie \u2022 Mole \u2022 Constante d\u2019Avogadro \u2022 Masse molaire \u2022 Calculs avec m, M, V,...<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-793","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/793","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/comments?post=793"}],"version-history":[{"count":1,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/793\/revisions"}],"predecessor-version":[{"id":795,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/793\/revisions\/795"}],"wp:attachment":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/media?parent=793"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}