Calculate the percent composition of Mg(NO3)2.

Considering the definition of percent composition, the percent composition of Mg(NO₃)₂ is:

Percent Composition is a measure of the amount of mass that an element occupies in a compound. In other words, the percentage composition indicates the percentage by mass of each element that is part of a compound.

Knowing the formula of a chemical compound, it is possible to know the percentage composition with which each element that forms said compound is present in it.

First, the molecular weight of the compound is obtained. The molecular weight is obtained by multiplying the atomic weight by the number of atoms there are of an element.

In this case, being:

So, the molar mass of Mg(NO₃)₂ is calculated as:

Molar mass of Mg(NO₃)₂ = 24 + [14+(16×3)]×2

Molar mass of Mg(NO₃)₂ = 24 + [14+48]×2

Molar mass of Mg(NO₃)₂ = 24 + 62×2

Molar mass of Mg(NO₃)₂ = 24 + 124

Molar mass of Mg(NO₃)₂ = 148

Then divide the atomic weight of each of the atoms of the compound by the molecular weight of the entire compound, and multiply by 100 to obtain the percentage value.

In this case, :

So, the percent composition of Mg(NO₃)₂ is:

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%M = 16.2%

%N = 18.9%

%O = 64.9%

Explanation:

Mg(NO3)2

Atomic mass of Magnesium M = 24

Atomic mass of Nitrogen N=14

Atomic mass of Oxygen O=16

Molar mass of Mg(NO3)2 = 24 + (14+[16*3})*2

Mg(NO3)2 = 24 + (14+48)*2

Mg(NO3)2 = 24 + 124

Mg(NO3)2 = 148

Percentage composition;

Magnesium M= (24/148) × 100

Magnesium M=0.162 × 100 =16.2%

Oxygen O=(16*6/148) × 100

Oxygen O= 0.649 × 100 = 64.9%

Nitrogen N= (14*2/148) × 100

Nitrogen N=0.189 × 100=18.9%