Maharashtra State BoardHSC Science (Electronics) 11th
Advertisement Remove all ads

Solve the following problem: Work out the percentage composition of constituent elements in the following compound: Lead phosphate [Pb3(PO4)2] (At. mass : Pb = 207; P = 31; O = 16; K = 39; - Chemistry

Sum

Solve the following problem:

Work out the percentage composition of constituent elements in the following compound:

Lead phosphate [Pb3(PO4)2]

(At. mass: Pb = 207; P = 31; O = 16; K = 39; Cr = 52; Na = 23; N = 14)

Advertisement Remove all ads

Solution

Given: Atomic mass: Pb = 207; P = 31; O = 16; K = 39; Cr = 52; Na = 23; N = 14

To find: The percentage composition of constituent element

Formula: Percentage (by weight) = `"Mass of the element in 1 mole of compound"/"Molar mass of the compound"xx100`

Calculation: 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 = `621/811xx100` = 76.57%

Percentage of P = `621/811xx100` = 7.64%

Percentage of O = `128/811xx100` = 15.78%

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

  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

Balbharati Chemistry 11th Standard Maharashtra State Board
Chapter 2 Introduction to Analytical Chemistry
Exercises | Q 4. (E)(a) | Page 24
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×