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Samenvatting Chemisch rekenen

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goede samenvatting voor toets/examen

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  • June 3, 2021
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  • 2020/2021
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  • Secondary school
  • TSO
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Chemisch rekenen
1. Atoommassa
Aa = absolute atoommassa in gram -> zeer klein dus werken met Ar
Ar = relatieve atoommassa (kan je aflezen in periodiek ss)  verteld hoeveel x een atoom zwaarder is dan een H-atoom
of amu

 We vergelijken de Aa met de Aa van een H- atoom  die komt overeen met de amu (= atoommassa unit)
 Amu = 1,66 . 10 -27 kg
 Vb. Ar (C)  Aa van H vergelijken met de Aa van C = 12, 0 (altijd 1 cijfer na komma!)


2. Moleculemassa
Ma = absolute moleculemassa
Mr = relatieve moleculemassa = de som van alle Ar van de aanwezige atomen
 Vb. Mr (H2O) = 2 . Ar (H) + Ar (O) = 2 . 1,0 + 16,0 = 18,0


3. Stofhoeveelheid (n) met eenheid mol
Stofhoeveelheid = hoeveelheid om te kunnen spreken over de hoeveelheid v/e stof
o 1 paar = 2
o 1 dozijn = 12
o 1 mol = 6,02 . 1023 = getal van Avogrado = Na  eenheid van de stofhoeveelheid

 Bepaalt aantal deeltjes per mol
 Er zijn 6,02 . 1023 deeltjes (atomen of moleculen) in 1 mol stof
 Na = 6,02 . 1023 mol -1 (mol-1 = 1/mol = per mol)

Hoeveel weegt 1 mol?
 Als je 1 mol v/e atoom wil  Ar g afwegen  vb. 1 mol ijzer = 55,8 g (Ar (fe) = 55,9)
 Als je 1 mol v/e molecule wil  Mr g afwegen  vb. 1 H2O = 18,0 g (Mr (H2O) = Ar(H) . 2 + Ar (O)
 Extra vb. 1 mol Na = 23 g (Ar (Na) = 23,0 g)

Zijn er overal evenveel deeltjes?
Ja, 1 mol = 6,02 . 1023
Is de massa overal gelijk?
Nee, dit hangt af v/d moleculen/atomen  vb. 50 grote en 50 kleine haribo bollen snoepjes

4. Molaire massa
Molaire massa = de massa in gram van 1 mol deeltjes
 De molaire massa v/e stof is de waarde van Ar of Mr uitgedrukt in g/mol
 Vb. M(fe) = 55,8 g/mol en M(H2O) = 18,0 g/mol

m m
Dus M = n= m = n . M
n M

Om een aantal deeltjes N (atomen of moleculen) te berekenen in x mol v/e stof gebruik je deze formule:
N
N = n . NA of n =
NA

 Vb. wat is de M van 1 mol zwavel?
A: 32:1

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