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Calculate the number of atoms there are in a sample of aluminium that weighs 80 , 94 g .

  1. n = m M = 80 , 94 26 , 98 = 3 moles
  2. Number of atoms in 3 mol aluminium = 3 × 6 , 022 × 10 23

    There are 18 , 069 × 10 23 aluminium atoms in a sample of 80 , 94 g .

Some simple calculations

  1. Calculate the number of moles in each of the following samples:
    1. 5 , 6 g of calcium
    2. 0 , 02 g of manganese
    3. 40 g of aluminium
  2. A lead sinker has a mass of 5 g .
    1. Calculate the number of moles of lead the sinker contains.
    2. How many lead atoms are in the sinker?
  3. Calculate the mass of each of the following samples:
    1. 2,5 mol magnesium
    2. 12 mol lithium
    3. 4 , 5 × 10 25 atoms of silicon

Molecules and compounds

So far, we have only discussed moles, mass and molar mass in relation to elements . But what happens if we are dealing with a molecule or some other chemical compound? Do the same concepts and rules apply? The answer is 'yes'. However, you need to remember that all your calculations will apply to the whole molecule . So, when you calculate the molar mass of a molecule, you will need to add the molar mass of each atom in that compound. Also, the number of moles will also apply to the whole molecule. For example, if you have one mole of nitric acid ( HNO 3 ), it means you have 6 , 022 × 10 23 molecules of nitric acid in the sample. This also means that there are 6 , 022 × 10 23 atoms of hydrogen, 6 , 022 × 10 23 atoms of nitrogen and ( 3 × 6 , 022 × 10 23 ) atoms of oxygen in the sample.

In a balanced chemical equation, the number that is written in front of the element or compound, shows the mole ratio in which the reactants combine to form a product. If there are no numbers in front of the element symbol, this means the number is '1'.

e.g. N 2 + 3 H 2 2 N H 3

In this reaction, 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.

Calculate the molar mass of H 2 SO 4 .

  1. Hydrogen = 1,008 g · mol - 1 ; Sulphur = = 32,07 g · mol - 1 ; Oxygen = 16 g · mol - 1

  2. M ( H 2 S O 4 ) = ( 2 × 1 , 008 ) + ( 32 , 07 ) + ( 4 × 16 ) = 98 , 09 g · mol - 1

Calculate the number of moles there are in 1 kg of MgCl 2 .

  1. n = m M
    1. Convert mass into grams
      m = 1 kg × 1 000 = 1 000 g
    2. Calculate the molar mass of MgCl 2 .
      M ( MgCl 2 ) = 24 , 31 + ( 2 × 35 , 45 ) = 95 , 21 g · mol - 1
  2. n = 1000 95 , 21 = 10 , 5 mol

    There are 10 , 5 moles of magnesium chloride in a 1 kg sample.

Barium chloride and sulphuric acid react according to the following equation to produce barium sulphate and hydrochloric acid.

BaCl 2 + H 2 SO 4 BaSO 4 + 2 HCl

If you have 2 g of BaCl 2 ...

  1. What quantity (in g) of H 2 SO 4 will you need for the reaction so that all the barium chloride is used up?
  2. What mass of HCl is produced during the reaction?
  1. n = m M = 2 208 , 24 = 0 , 0096 mol
  2. According to the balanced equation, 1 mole of BaCl 2 will react with 1 mole of H 2 SO 4 . Therefore, if 0 , 0096 mol of BaCl 2 react, then there must be the same number of moles of H 2 SO 4 that react because their mole ratio is 1:1.

  3. m = n × M = 0 , 0096 × 98 , 086 = 0 , 94 g

    (answer to 1)

  4. According to the balanced equation, 2 moles of HCl are produced for every 1 mole of the two reactants. Therefore the number of moles of HCl produced is ( 2 × 0 , 0096 ), which equals 0 , 0096 moles .

  5. m = n × M = 0 , 0192 × 35 , 73 = 0 , 69 g

    (answer to 2)

Questions & Answers

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I start with an easy one. carbon nanotubes woven into a long filament like a string
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what is system testing?
AMJAD
preparation of nanomaterial
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
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AMJAD
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
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after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
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name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
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I'm interested in Nanotube
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this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
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What makes metals better to use as wires than non-metals? (please link to bonding type)??? HELP
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Source:  OpenStax, Chemistry grade 10 [caps]. OpenStax CNX. Jun 13, 2011 Download for free at http://cnx.org/content/col11303/1.4
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