Find the total percentage of oxygen in magnesium nitrate crystal Mg(NO₃)₂·6H₂O.
[H = 1, N = 14, O = 16, Mg = 24]

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Solution

Mass of oxygen present in magnesium nitrate crystal = [(6 $\times$ 16) + (6 $\times$ 16)] g = 192 g
Molecular mass of Mg(NO₃)₂·6H₂O = [24 + 2 (14+48) + 6 (18)] g = 256 g
$Percentage of O = \frac{Mass of O in compound}{Molecular mass of Mg ({NO}_{3})_{2} . 6 H_{2} O} \times 100 = \frac{192}{256} \times 100 = 75 %$

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Q. What is the total percentage of oxygen in magnesium nitrate crystals

(M g (N O_{3})_{2} .6 H_{2} O)

[

O = 16, N = 14, H = 1, M g = 24

]

Q. The total percentage of magnesium present in magnesium nitrate crystals

M g (N O_{3})_{2} .6 H_{2} O

is:
[

M g = 24; N = 14; O = 16; H = 1

Find the percentage of oxygen in magnesium nitrate crystals, $M g (N O_{3})_{2} .6 H_{2} O$ [Mg = 24, N = 14, O = 16, H = 1]

Q. Find the total percentage of Magnesium in magnesium nitrate crystals:

[M g (N O_{3})_{2} 6 H_{2} O .

(M g = 24, O = 16

and

H = 1)

Find the percentage of
(a) oxygen in magnesium nitrate crystals $[M g (N O_{3})_{2} 6 H_{2} O]$ .
(b) boron in $N a_{2} B_{4} O_{7} . 10 H_{2} O . [H = 1, B = 11, O = 16, N a = 23] .$
(c) phosphorus in the fertilizer superphosphate $C a (H_{2} P O_{4})_{2}$

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Find the total percentage of oxygen in magnesium nitrate crystal Mg(NO3)2·6H2O. [H = 1, N = 14, O = 16, Mg = 24]

Mass of oxygen present in magnesium nitrate crystal = [(6×16) + (6×16)] g = 192 g Molecular mass of Mg(NO3)2·6H2O = [24 + 2 (14+48) + 6 (18)] g = 256 g Percentage of O = Mass of O in compoundMolecular mass of Mg(NO3)2.6H2O×100=192256×100=75%

Find the total percentage of oxygen in magnesium nitrate crystal Mg(NO₃)₂·6H₂O.
[H = 1, N = 14, O = 16, Mg = 24]