Question

In an experiment to determine the empirical formula of magnesium chloride, 0.50 g of magnesium ribbon was taken in a beaker and a few drops of HCl was added until all the Mg reacted with the acid liberating the hydrogen gas. The reaction completion was noted by observing that the bubbling stopped and no solid Mg was seen in the beaker. The beaker was heated to remove water and the dried magnesium chloride was weighed and found to be 1.99 g. What is the empirical formula for magnesium chloride based on this experiment

Answer 1

The the empirical formula for magnesium chloride based on this experiment will be
`MgCl_(2)`

Step-by-step explanation:

Given that,

Mass of Mg = 0.50 g

Mass of magnesium chloride found = 1.99 g

Let the formula of magnesium chloride be
`Mg_(x)Cl_(y)`

We know that,

Molar mass of Mg= 24 g/mol

Molar mass of magnesium chloride = (24x+35.5y) g/mol

We need to calculate the moles of Mg

Using formula of moles

`Number\ of\ moles=(mass)/(molar\ mass\ of\ Mg)`

Put the value into the formula

`Number\ of\ moles=(0.50)/(24)`

`Number\ of\ moles=0.020\ mole`

We need to calculate the mole of magnesium chloride

Using formula of moles

`Number\ of\ moles=(mass)/(molar\ mass\ of\ magnesium\ chloride)`

Put the value into the formula

`Number\ of\ moles=(1.99)/(24x+35.5y)\ mole`

The reaction will be,

`Mg+HCl\Rightarrow Mg_(x)Cl_(y)+H_(2)`

We need to calculate the value of x and y

Using number of moles of Mg in reactant and product

Moles of Mg atom in reactant=Moles of Mg atom in product

`(0.50)/(24)=x*(1.99)/(24x+35.5y)`

`0.020(24x+35.5y)=x*1.99`

`0.48x+0.71y=1.99x`

`0.71y=(1.99-0.48)x`

`(x)/(y)=(0.71)/(1.99-0.48)`

`(x)/(y)=0.47\approx0.5`

`(x)/(y)=(1)/(2)`

Hence, The the empirical formula for magnesium chloride based on this experiment will be
`MgCl_(2)`