Question

What is the empirical formula for a compound if 300.00 g of it is known to contain 82.46224 g of molybdenum, 45.741 g of chlorine and the rest is bromine

Answer 1

MoClBr₂

Step-by-step explanation:

First we calculate the mass of bromine in the compound:

• 300.00 g - (82.46224 g + 45.741 g) = 171.79676 g

Then we calculate the number of moles of each element, using their respective molar masses:

• 82.46224 g Mo ÷ 95.95 g/mol = 0.9594 mol Mo

• 45.741 g Cl ÷ 35.45 g/mol = 1.290 mol Cl

• 171.79676 g Br ÷79.9 g/mol = 2.150 mol Br

Now we divide those numbers of moles by the lowest number among them:

• 0.9594 mol Mo / 0.9594 = 1

• 1.290 mol Cl / 0.9594 = 1.34 ≅ 1

• 2.150 mol Br / 0.9594 = 2.24 ≅ 2

Meaning the empirical formula is MoClBr₂.