Question

# 21. (10 marks) Imagine you react 1.25 g of germanium, Ge(s), with enough liquid bromine, Br2(C), to consume all the germanium. The reaction makes 6.75 g of a single solid product with formula Ge, Br, a) (6 marks) Based on this data only (i.e., do NOT use the information given in part (b) yet), what is the empirical formula of the product? Show full calculations. b) (2 marks) If a mass spectrometry experiment determines the product's molar mass to be 392 g/mol, what is the molecular formula of the product? Explain in a few words or with a short calculation. c) 1 mark) What is the product's name? Assume the product is molecular in nature.

Answer 1

Solution:

Part A)

Atomic mass of Ge = 72.64 g mol-1

mass of Ge = 1.25 g

Number of moles of Ge = 1.25 g / 72.64 g mol-1 = 0.0172 mol = 0.02 mol

Atomic mass of Br = 79.904 g mol-1

mass of Br = total mass of product - mass of Ge = 6.75 - 1.25 = 6.50 g

Number of moles of Br = 6.50 g / 79.904 g mol-1 = 0.0813 mol = 0.08 mol

Therefore, product is written as; Ge(0.02) Br (0.08)

Ratio of moles is calculated as;

x = 0.02 / 0.02 = 1

y = 0.08 / 0.02 = 4

Hence, empirical formula = Ge (1) Br (4) = Ge Br4

Part B)

Molar mass of Ge Br4 = 72.64 + 4 x 79.904 = 392.3 g mol-1 = 392 g mol-1

Thus, GeBr4 form molecular ion peak (M+*) at m/z =392 as:

Ge Br4 + e- = [GeBr4]+* ( molecular ion peak) + 2e-

Part C)

The product GeBr4 is named as,

Germanium tetrabromide or Germanium (IV) bromide