Chemical calculations Flashcards

1
Q

Relative atomic mass

A

Average mass of one atom of the element relevant to 1/12 of the mass of one atom of carbon 12

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2
Q

relative isotopic mass

A

Average mass of one atom to isotope relevant to 1/12 of the mass of one atom of carbon 12

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3
Q

Amount of substances

A

mol= mass x Mr
Mol= PV/RT
Mol= CxV/1000

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4
Q

Atom economy

A

Mr of required Product/ Mr of total reactants
x100

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5
Q

empirical formula

A

smallest whole number ratio of atoms of each element in a compound

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6
Q

ideal gas equation

A

PV=nRT
P-Pressure in pascals x1000 from Kpa
V-volume in m3 /1000 from dm
n-moles
R- gas constant 8.31 JK-1mol-1
T- temp in K +273 from celcius

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7
Q

Molar mass

A

Mass of one mole of a substance in gmol-1

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8
Q

percentage yield

A

actual/theoretical x100

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9
Q

Molecular forumula

A

Mr of molecule / empirical Mr

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10
Q

Mass spectrometer

A

Technique used to identify compounds and determine their relative molecular mass

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11
Q

ionisation 1

A

A sample of Electrons are vaporised and injected into the mass spectometer where high voltage is passed over the chamber causing electrons to be removed from the atoms leaving plus charged ions in the chamber

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12
Q

acceleration 2

A

The positively charged ions are then accelerated towards negatively charged detection plate

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13
Q

deflection 3

A

Ions are then deflected by a magnetic field into a curved path radius
Path is dependent on the charge and mass of ion

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14
Q

detection 4

A

When the positive ions hit the negatively charged detection plate they gain an electron producing a flow of charge
The greater the abundance the greater the current produced

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15
Q

why is there a vaccum

A

A vacuum is needed inside the apparatus so that ions produced in the chamber have a free run through the machine without hitting air molecules

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16
Q

relative atomic mass from mass spectrometry

A

m/z x abundance / total abundance

17
Q

Volume using ideal gas equation

A

p1xv1/t1= p2xv2/t2

18
Q

Conc of solutions

A

moles/volume
cm-dm= /1000

19
Q

Solubility

A

g/100g water = 100cm3
1. Moles=massxMr
2. 100/1000=0.100 dm3
3. Moles/0.1000(vol)

20
Q

Acid based titration calculations

A
  1. Most of the solution with known concentration
  2. Stoichiometric ratio between acid and base
  3. Calculate concentration of second solution from known volume and moles
21
Q
A