To find the total kinetic energy ratio for oxygen and hydrogen at a given temperature, we first note that kinetic energy (average) of a molecule can be represented using the formula for translational kinetic energy of an ideal gas, which is given by:

$ KE_{avg} = \frac{3}{2} k_B T $

where $ k_B $ is the Boltzmann constant and $ T $ is the temperature in Kelvin.

For a given number of molecules $ N $, the total kinetic energy $ KE_{total} $ can be expressed as:

$ KE_{total} = N KE_{avg} $

However, in order to find $ N $, which is the number of molecules, we need the number of moles, since $ N = nN_A $ where $ n $ is the number of moles and $ N_A $ is Avogadro’s number. We calculate $ n $ by dividing the mass of the gas $ m $ by its molar mass $ M $. This gives:

$ n = \frac{m}{M} $

For hydrogen (H₂), the molar mass $ M_{H_2} = 2 $ g/mol and for oxygen (O₂), the molar mass $ M_{O_2} = 32 $ g/mol. Given that both gases have a mass of 8 grams each, we can calculate the moles for each:

$ n_{H_2} = \frac{8 \text{ g}}{2 \text{ g/mol}} = 4 \text{ moles} $

$ n_{O_2} = \frac{8 \text{ g}}{32 \text{ g/mol}} = 0.25 \text{ moles} $

The number of molecules $ N $ for each gas becomes:

$ N_{H_2} = n_{H_2} N_A = 4 N_A $

$ N_{O_2} = n_{O_2} N_A = 0.25 N_A $

Even though the mass of each gas is the same, the number of moles (and therefore the number of molecules) is different. However, the total kinetic energy for any ideal gas sample is still dependant on the temperature and the total number of molecules. Therefore, for each gas, substituting from the kinetic energy formula, we get:

$ KE_{total, H_2} = N_{H_2} KE_{avg} = 4 N_A \times \frac{3}{2}k_B T $

$ KE_{total, O_2} = N_{O_2} KE_{avg} = 0.25 N_A \times \frac{3}{2}k_B T $

The ratio of total kinetic energies $(H_2/O_2)$ thus becomes:

$ \frac{KE_{total, H_2}}{KE_{total, O_2}} = \frac{4 N_A \times \frac{3}{2}k_B T}{0.25 N_A \times \frac{3}{2}k_B T} = \frac{4 N_A}{0.25 N_A} = 16 $

Even though both samples are at the same temperature and have the same mass, hydrogen has more molecules contributing to its kinetic energy due to its significantly lower molecular weight compared to oxygen. Hence, under these conditions, the ratio of the total kinetic energy of hydrogen to oxygen is $ 16:1 $.