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 $H = 1)$

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Solution

$M g (N O_{3})_{2} .6 H_{2} O$
$24 + 14 \times 2 + 16 \times 6 + 6 \times 18 = 256 g m$
% of $M g = \frac{24}{256} \times 100$
$= 9.375$ %.

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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

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

]

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

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]

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 Magnesium in magnesium nitrate crystals:[Mg(NO3)26H2O.(Mg=24, O=16 and H=1)

Mg(NO3)2.6H2O24+14×2+16×6+6×18=256 gm% of Mg=24256×100=9.375%.

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 $H = 1)$

$M g (N O_{3})_{2} .6 H_{2} O$
$24 + 14 \times 2 + 16 \times 6 + 6 \times 18 = 256 g m$
% of $M g = \frac{24}{256} \times 100$
$= 9.375$ %.