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Given : 

Atomic mass : 

Pb = 207; 

P = 31; 

O = 16; 

K = 39; 

Cr = 52; 

Na = 23; 

N = 14 

To find : 

The percentage composition of constituent elements 

Formula :

Percentage (by weight) = \(\frac{Mass\,of\,the\,element\,in\,mole\,of\,compound}{Molar\,mass\,of\,the\,compound}\times 100\)

Calculation :

a. Lead phosphate [Pb₃(PO₄)₂]

Molar mass of Pb₃(PO₄)₂ = 3 × (207) + 2 × (31) + 8 × (16) 

= 621 + 62 + 128 = 811 g mol^-1

Percentage of Pb = \(\frac{621}{811}\) \(\times 100\)

= 76.57%

Percentage of P = \(\frac{621}{811}\) \(\times 100\)

= 7.64%

Percentage of O = \(\frac{128}{811}\) \(\times 100\)

= 15.78%

b. Potassium dichromate (K₂Cr₂O₇)

Molar mass of K₂Cr₂O₇ = 2 × (39) + 2 × (52) + 7 × (16) 

= 78 + 104 + 112 

= 294 g mol^-1

Percentage of K = \(\frac{78}{294}\) \(\times 100\)

= 26.53%

Percentage of Cr =  \(\frac{104}{294}\) \(\times 100\)

= 35.37%

Percentage of O = \(\frac{112}{294}\) \(\times 100\)

= 38.10%

c. Macrocosmic salt – Sodium ammonium hydrogen phosphate, tetrahydrate (NaNH₄HPO₄ .4H₂O)

Molar mass of NaNH₄HPO₄.4H₂O = 1 × (23) + 1 × (14) + 1 × (31) + 13 × (1) + 8 × (16) 

= 23 + 14 + 31 + 13 + 128 = 209 g mol^-1

Percentage of Na = \(\frac{23}{209}\) \(\times 100\)

= 11.00%

Percentage of N = \(\frac{14}{209}\) \(\times 100\)

= 6.70%

Percentage of P = \(\frac{31}{209}\) \(\times 100\)

= 14.83%

Percentage of H = \(\frac{13}{209}\) \(\times 100\)

= 6.22%

Percentage of O = \(\frac{128}{209}\) \(\times 100\)

= 61.24%

∴ i. Mass percentage of Pb, P and O in lead phosphate [Pb₃(PO₄)₂] are 76.57%, 7.64% and 15.78% respectively.

ii. Mass percentage of K, Cr and O in potassium dichromate (K₂Cr₂O₇) are 26.53%, 35.37% and 38.10% respectively.

iii. Mass percentage of Na, N, P, H and O in NaNH₄HPO₄.4H₂O are 11.00%, 6.70%, 14.83%, 6.22% and 61.24% respectively.

1. White ppt. insoluble in excess ammonium hydroxide.

2.  Reddish brown precipitate is formed which in insoluble in excess of NaOH.