Calculate the percentage of platinum in ammonium chloroplatinate (NH₄)₂PtCl₆.

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Solution

Atomic weight of platinum is 195 g.
Molecular mass of (NH₄)₂PtCl₆ = [2(14 + 4) + 195 + 6(35.5)] g = 444 g
$Percentage of platinum = \frac{Mass of platinum in the compound}{Molecular mass of the compound} \times 100 Percentage of platinum = \frac{195}{444} \times 100 = 43.99 \approx 44 %$

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(N H_{4})_{2} P t C l_{6}

N = 14, H = 1, C l = 35.5, P t = 195

Calculate the percentage of platinum in ammonium chloroplatinate, (NH₄)₂PtCl₆.

(Give your answer correct to the nearest whole number).

[N= 14,H= l,Pt= 195,C1 = 35.5]

Calculate the relative molecular masses of :

(a) $C H C l_{3}$ (b) $(N H_{4})_{2} C r_{2} O_{7}$

(e) $C H_{3} C O O N a$ (f) Potassium chlorate

(g) Ammonium chloroplatinate $(N H_{4})_{2} P t C l_{6}$

2.40 g

0.8 g

P t = 195 g / m o l

39

P t = 195

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