Find the percentage of

(a) oxygen in magnesium nitrate crystals [Mg(NO₃)₂.6H₂O].

(b) boron in Na₂B₄O₇.10H₂O. H=1,B=11,O=16,Na=23.

(c) phosphorus in the fertilizer superphosphate Ca(H₂PO₄)₂

Relative molecular mass of [Mg (NO₃)₂.6H₂O]

= 24 + 2[14 + 3(16)] + 12(1) + 6(16)
= 24 + 2[14 + 48] + 12 + 96
= 24 + 28 + 96 + 12 + 96
= 256 g

Molar mass of oxygen in [Mg (NO₃)₂.6H₂O] = 96 + 96 = 192 g

Since, 256 g of [Mg(NO₃)₂.6H₂O] contains 192 g of oxygen

∴ 100 g of [Mg(NO₃)₂.6H₂O] contains

$\dfrac{192}{256}$ x 100 = 75% of oxygen

(b) Relative molecular mass of Na₂B₄O₇.10H₂O
= 2(23) + 4(11) + 7(16) + 20(1) + 10(16)
= 46 + 44 + 112 + 20 + 160
= 382 g

Molar mass of boron in Na₂B₄O₇.10H₂O = 44 g

382 g of Na₂B₄O₇.10H₂O contains 44 g of boron

∴ 100 g Na₂B₄O₇.10H₂O contains $\dfrac{44}{382}$ x 100 = 11.5%

(c) Relative molecular mass of Ca(H₂PO₄)₂
= 40 + 4(1) + 2(31) + 8(16)
= 40 + 4 + 62 + 128
= 234 g

Molar mass of phosphorus in Ca(H₂PO₄)₂ = 62 g

234g of Ca(H₂PO₄)₂ contains 62 g of phosphorus

∴ 100 g of Ca(H₂PO₄)₂ contains $\dfrac{62}{234}$ x 100 = 26.5%