Question

A metallurgist begins with 1,250 grams of vanadinite (Pb5(VO4)3Cl) and ends up with 135 grams of pure vanadium and 915 grams of pure lead. The percent composition of vanadium in the ore is , and the percent composition of lead is

Answer 1

1. 10.8%

2. 73.2%

Step-by-step explanation:

Answer 2

1. 10.8%

2. 73.2%

Step-by-step explanation:

Sample: 1,250 grams of vanadinite (Pb₅ (VO₄)₃Cl

Mass of Vanadium = 135 g

Mass of Pb = 915 g

Percentage of V = ?

Percentage of Pb =?

Using the formula;

The percentage composition is given by the formula;

= ( mass in grams of element / mass in grams of sample) × 100

Therefore;

The percentage of Vanadium;

= (mass of V / mass of sample) × 100 = (135 g/ 1,250 g) × 100

= 10.8 %

The percentage composition of Lead;

= (mass of Pb / mass of sample) × 100 = (915 g / 1,250 g) × 100

= 73.2%