
A mineral contained $MgO = 31.88\% ;Si{O_2} = 63.37\% $ and ${H_2}O = 4.75\% $. Show that the simplest formula for the mineral is ${H_2}M{g_3}S{i_4}{O_{12}}$
($H = 1;Mg = 24;Si = 28;O = 16$)
Answer
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Hint:We can assume a sample of weight $100g$ with the above compositions. Then we can find the number of moles of each compound and from that, find the mole ratio of the three compounds. This will help us to write the empirical formula of the mineral.
Formulas used: $n = \dfrac{W}{M}$
Where $n$ is the number of moles, $W$ is the given mass and $M$ is the molecular mass of the compounds respectively.
Complete step by step answer:
Let us first consider a sample of the mineral weighing $100g$ . Therefore, the mass of each compound in the mineral will be equal to the percentages, as we have chosen a sample of $100g$ .
Hence, in our sample, the mass of $MgO = 31.88g;Si{O_2} = 63.37g$ and ${H_2}O = 4.75g$ .
Next, we shall find the number of moles of each constituent. As we know number of moles:
$n = \dfrac{W}{M}$
Where $n$ is the number of moles, $W$ is the given mass and $M$ is the molecular mass of the compounds respectively.
We know the molecular mass of $MgO = 40g;Si{O_2} = 60g$ and ${H_2}O = 18g$
Hence,
number of moles of $MgO = \dfrac{{31.88}}{{40}} = 0.8moles$
number of moles of $Si{O_2} = \dfrac{{63.37}}{{60}} = 1.05moles$
number of moles of ${H_2}O = \dfrac{{4.75}}{{18}} = 0.26moles$
Now, to find the mole ratio, we will divide the moles of each constituent with the number of moles of the constituent which has least number of moles. From above, the constituent with the least number of moles is water, having $0.26moles$ . Hence, mole ratio:
$ = \dfrac{{0.8}}{{0.26}}:\dfrac{{1.05}}{{0.26}}:\dfrac{{0.26}}{{0.26}}$
On solving this, we get:
Mole ratio $ = 3:4:1$
Hence, in the simplest form, the mineral contains three molecules of $MgO$ , four molecules of $Si{O_2}$ and one molecule of ${H_2}O$
Therefore, the simplest formula for this mineral is:
${(MgO)_3}{(Si{O_2})_4}({H_2}O)$
Grouping the similar atoms, we get:
${H_2}M{g_3}S{i_4}{O_{12}}$
Hence proved.
Note:
-Note that the empirical formula is a formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.
-Empirical formulas are powerful in the fact that they can be used to predict the molecular formula of a compound just by knowing the composition of the elements present in it. The molecular formula is always a whole number multiple of the empirical formula.
-Also note that we divided the number of moles of each constituent with the smallest number of moles so as to obtain the mole ratio in the form of integers.
Formulas used: $n = \dfrac{W}{M}$
Where $n$ is the number of moles, $W$ is the given mass and $M$ is the molecular mass of the compounds respectively.
Complete step by step answer:
Let us first consider a sample of the mineral weighing $100g$ . Therefore, the mass of each compound in the mineral will be equal to the percentages, as we have chosen a sample of $100g$ .
Hence, in our sample, the mass of $MgO = 31.88g;Si{O_2} = 63.37g$ and ${H_2}O = 4.75g$ .
Next, we shall find the number of moles of each constituent. As we know number of moles:
$n = \dfrac{W}{M}$
Where $n$ is the number of moles, $W$ is the given mass and $M$ is the molecular mass of the compounds respectively.
We know the molecular mass of $MgO = 40g;Si{O_2} = 60g$ and ${H_2}O = 18g$
Hence,
number of moles of $MgO = \dfrac{{31.88}}{{40}} = 0.8moles$
number of moles of $Si{O_2} = \dfrac{{63.37}}{{60}} = 1.05moles$
number of moles of ${H_2}O = \dfrac{{4.75}}{{18}} = 0.26moles$
Now, to find the mole ratio, we will divide the moles of each constituent with the number of moles of the constituent which has least number of moles. From above, the constituent with the least number of moles is water, having $0.26moles$ . Hence, mole ratio:
$ = \dfrac{{0.8}}{{0.26}}:\dfrac{{1.05}}{{0.26}}:\dfrac{{0.26}}{{0.26}}$
On solving this, we get:
Mole ratio $ = 3:4:1$
Hence, in the simplest form, the mineral contains three molecules of $MgO$ , four molecules of $Si{O_2}$ and one molecule of ${H_2}O$
Therefore, the simplest formula for this mineral is:
${(MgO)_3}{(Si{O_2})_4}({H_2}O)$
Grouping the similar atoms, we get:
${H_2}M{g_3}S{i_4}{O_{12}}$
Hence proved.
Note:
-Note that the empirical formula is a formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.
-Empirical formulas are powerful in the fact that they can be used to predict the molecular formula of a compound just by knowing the composition of the elements present in it. The molecular formula is always a whole number multiple of the empirical formula.
-Also note that we divided the number of moles of each constituent with the smallest number of moles so as to obtain the mole ratio in the form of integers.
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