Question

Analysis of 20.0 g of a compound containing only calcium and bromine indicates that 4.00 g of calcium are present. What is the empirical formula of the compound formed?
A. Ca7Br2
B. Ca3Br2
C. CaBr2
D. Ca5Br2

Answer 1

Answer:

d. $Ca_5Br_2$

Explanation:

Mass of calcium = 4.00 g

Molar mass of calcium = 40.078 g/mol

Moles of calcium = 4.00 / 40.078 moles = 0.9981 moles

Given that the compound only contains calcium and bromine. So,

Mass of bromine in the sample = Total mass - Mass of calcium

Mass of the sample = 20.0 g

Mass of bromine in sample = 20.0 - 4.00 g = 16.0 g

Molar mass of bromine = 79.904 g/mol

Moles of Bromine  = 16.0 / 79.904  = 0.2002 moles

Taking the simplest ratio for Ca and Br as:

0.9981 : 0.2002  = 5 : 1

The empirical formula is = $Ca_5Br_2$