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You have a sample that contains 5 moles of zinc.

  1. What is the mass of the zinc in the sample?
  2. How many atoms of zinc are in the sample?
  1. Molar mass of zinc is 65.38 g.mol - 1 , meaning that 1 mole of zinc has a mass of 65.38 g.

  2. If 1 mole of zinc has a mass of 65.38 g, then 5 moles of zinc has a mass of:

    65.38 g x 5 mol = 326.9 g (answer to a)

  3. 5 × 6 . 023 × 10 23 = 30 . 115 × 10 23

    (answer to b)

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Moles and molar mass

  1. Give the molar mass of each of the following elements:
    1. hydrogen
    2. nitrogen
    3. bromine
  2. Calculate the number of moles in each of the following samples:
    1. 21.62 g of boron (B)
    2. 54.94 g of manganese (Mn)
    3. 100.3 g of mercury (Hg)
    4. 50 g of barium (Ba)
    5. 40 g of lead (Pb)

An equation to calculate moles and mass in chemical reactions

The calculations that have been used so far, can be made much simpler by using the following equation:

n ( number of moles ) = m ( mass of substance in g ) M ( molar mass of substance in g · mol - 1 )
Remember that when you use the equation n = m M , the mass is always in grams (g) and molar mass is in grams per mol ( g · mol - 1 ).

The equation can also be used to calculate mass and molar mass, using the following equations:

m = n × M

and

M = m n

The following diagram may help to remember the relationship between these three variables. You need to imagine that the horizontal line is like a 'division' sign and that the vertical line is like a 'multiplication' sign. So, for example, if you want to calculate 'M', then the remaining two letters in the triangle are 'm' and 'n' and 'm' is above 'n' with a division sign between them. In your calculation then, 'm' will be the numerator and 'n' will be the denominator.

Calculate the number of moles of copper there are in a sample that weighs 127 g.

  1. n = m M
  2. n = 127 63 . 55 = 2

    There are 2 moles of copper in the sample.

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You are given a 5 mol sample of sodium. What mass of sodium is in the sample?

  1. m = n × M
  2. M N a = 22.99 g.mol - 1

    Therefore,

    m = 5 × 22 . 99 = 114 . 95 g

    The sample of sodium has a mass of 114.95 g.

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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 m o l e s
  2. Number of atoms in 3 mol aluminium = 3 × 6.023 × 10 23

    There are 18.069 × 10 23 aluminium atoms in a sample of 80.94 g.

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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 silica

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 6.023 x 10 23 molecules of nitric acid in the sample. This also means that there are 6.023 × 10 23 atoms of hydrogen, 6.023 × 10 23 atoms of nitrogen and (3 × 6.023 × 10 23 ) atoms of oxygen in the sample.

Questions & Answers

during a snooker competition ,a 200g ball A m moving with velocity va collide head on with a identical ball B that was at rest.A after the collision ball A remains at rest wile ball B moves on with a velocity of 4m/s? With what speed was ball a moving before the collision
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Source:  OpenStax, Siyavula textbooks: grade 11 physical science. OpenStax CNX. Jul 29, 2011 Download for free at http://cnx.org/content/col11241/1.2
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