Moles and molar mass  (Page 3/4)

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:

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

and

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.

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 $\times$ 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 $\times$ 10 ${}^{23}$ atoms of hydrogen, 6.023 $\times$ 10 ${}^{23}$ atoms of nitrogen and (3 $\times$ 6.023 $\times$ 10 ${}^{23}$ ) atoms of oxygen in the sample.