Two pure samples of atoms,labeled A and B,contain titanium (Ti)atoms and carbon (C)atoms,respectively.Each sample contains the same number of atoms.What is the ratio of the mass of sample B to that of sample A,mB/mA? Note the following atomic masses: C = 12 u;Ti = 48 u.

Question

Quizplus · Accepted Answer

Since each sample contains the same number of atoms, we only need to consider the mass of one atom of each element in order to calculate the mass ratio of the two samples.

The atomic mass of carbon is 12 u, so the mass of one carbon atom is 12 u.

The atomic mass of titanium is 48 u, so the mass of one titanium atom is 48 u.

Therefore, the ratio of the mass of sample B (containing carbon atoms) to that of sample A (containing titanium atoms) is:

mB/mA = [(mass of one carbon atom) x (number of atoms in sample B)] / [(mass of one titanium atom) x (number of atoms in sample A)]

mB/mA = (12 x 1) / (48 x 1)

mB/mA = 0.25

Thus, the answer is B) 0.25.

Two Pure Samples of Atoms,labeled a and B,contain Titanium (Ti)atoms