
How many formula units make up 31.8 g of magnesium chloride $MgC{{l}_{2}}$?
Answer
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Hint:One mole of any compound or element always contain same number of units or particles, called Avogadro number. This value is $6.022\times {{10}^{23}}$ particles or formula units.
Complete step-by-step answer:Magnesium chloride is an ionic compound. The exact numbers of molecules in the empirical formula of ionic compounds are termed as formula unit. The molar mass of all the species in a compound multiplied by their amount or quantity in that compound gives us the formula unit mass.
The empirical formula of a compound tells us about the formula units. For instance, in magnesium chloride, one magnesium ion is always needed for two chloride ions. This would make the total weight of magnesium chloride involving 1 mole $Mg$ and 2 mole of $Cl$ which will make 1 mole of $MgC{{l}_{2}}$ to be:
$MgC{{l}_{2}}$= molar mass of Mg $+$molar mass of $C{{l}_{2}}$
$MgC{{l}_{2}}$= 70.906$+$24.305
$MgC{{l}_{2}}$= 95.211 g/mol
As we know 1 mole of any compound contains $6.022\times {{10}^{23}}$ formula units, so 1 mole of $MgC{{l}_{2}}$ will also contain $6.022\times {{10}^{23}}$ formula units.
Now to calculate formula units in 31.8 g of $MgC{{l}_{2}}$, we will use stoichiometric calculations and conversion factors as:
Formula units in 31.8 g $MgC{{l}_{2}}$= 31.8 g $MgC{{l}_{2}}$$\times \dfrac{1\,mol\,MgC{{l}_{2}}}{95.2\,g\,MgC{{l}_{2}}}\times \dfrac{6.022\times {{10}^{23}}\,form.unit\,MgC{{l}_{2}}}{1\,mol\,MgC{{l}_{2}}}$
Formula units in 31.8 g $MgC{{l}_{2}}$= $\dfrac{31.8\,}{95.2\,}\times 6.022\times {{10}^{23}}$ formula units
Formula units in 31.8 g $MgC{{l}_{2}}$= $2.01\times {{10}^{23}}$ formula units of $MgC{{l}_{2}}$
Hence, in 31.8 g of magnesium chloride, $MgC{{l}_{2}}$, $2.01\times {{10}^{23}}$ formula units are present.
Note:Stoichiometric calculations need to be done carefully by employing correct unitary method and conversion factors. Here the molar mass of magnesium chloride is taken for 1 mole in grams, and then the number of particles (formula units) in this 1 mole of mass is multiplied.
Complete step-by-step answer:Magnesium chloride is an ionic compound. The exact numbers of molecules in the empirical formula of ionic compounds are termed as formula unit. The molar mass of all the species in a compound multiplied by their amount or quantity in that compound gives us the formula unit mass.
The empirical formula of a compound tells us about the formula units. For instance, in magnesium chloride, one magnesium ion is always needed for two chloride ions. This would make the total weight of magnesium chloride involving 1 mole $Mg$ and 2 mole of $Cl$ which will make 1 mole of $MgC{{l}_{2}}$ to be:
$MgC{{l}_{2}}$= molar mass of Mg $+$molar mass of $C{{l}_{2}}$
$MgC{{l}_{2}}$= 70.906$+$24.305
$MgC{{l}_{2}}$= 95.211 g/mol
As we know 1 mole of any compound contains $6.022\times {{10}^{23}}$ formula units, so 1 mole of $MgC{{l}_{2}}$ will also contain $6.022\times {{10}^{23}}$ formula units.
Now to calculate formula units in 31.8 g of $MgC{{l}_{2}}$, we will use stoichiometric calculations and conversion factors as:
Formula units in 31.8 g $MgC{{l}_{2}}$= 31.8 g $MgC{{l}_{2}}$$\times \dfrac{1\,mol\,MgC{{l}_{2}}}{95.2\,g\,MgC{{l}_{2}}}\times \dfrac{6.022\times {{10}^{23}}\,form.unit\,MgC{{l}_{2}}}{1\,mol\,MgC{{l}_{2}}}$
Formula units in 31.8 g $MgC{{l}_{2}}$= $\dfrac{31.8\,}{95.2\,}\times 6.022\times {{10}^{23}}$ formula units
Formula units in 31.8 g $MgC{{l}_{2}}$= $2.01\times {{10}^{23}}$ formula units of $MgC{{l}_{2}}$
Hence, in 31.8 g of magnesium chloride, $MgC{{l}_{2}}$, $2.01\times {{10}^{23}}$ formula units are present.
Note:Stoichiometric calculations need to be done carefully by employing correct unitary method and conversion factors. Here the molar mass of magnesium chloride is taken for 1 mole in grams, and then the number of particles (formula units) in this 1 mole of mass is multiplied.
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