In a 100 g sample of the compound, there are 63.57 g of carbon, 6 g of hydrogen, 9.267 g of nitrogen, and 21.17 g of oxygen. First, convert these masses into moles (n) using the formula n = m/M, where M is the molar mass from the periodic table.
For carbon: 63.57 g C -> 63.57 g C / 12.01 g/mol = 5.29 moles C.
For hydrogen: 6 g H -> 6 g H / 1.008 g/mol = 5.95 moles H.
For nitrogen: 9.267 g N -> 9.267 g N / 14.01 g/mol = 0.6615 moles N.
For oxygen: 21.17 g O -> 21.17 g O / 16.00 g/mol = 1.32 moles O.
Thus, the mole ratio looks like this: C 5.29 H 5.95 N 0.6615 O 1.32.
Now, divide each value by the smallest number (1.32): C 4 H 4.5 N 0.5 O 1.
To eliminate fractions, multiply all values by 2, yielding C8H9N1O2.
Now, all numbers are integers! Hence, the empirical formula is C8H9NO2.
Although the empirical formula isn't always the same as the molecular formula, in this instance, it corresponds to acetaminophen.
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N₀ signifies the quantity of C-14 atoms per kg of carbon in the original sample at time = 0 seconds, when the carbon composition matched that in today’s atmosphere. As time progresses to ts, the number of C-14 atoms per kg declines to N, due to radioactive decay. λ indicates the decay constant.
Hence, we have N = N₀e - λt, which is the equation for radioactive decay. Rearranging gives us N₀/N = e λt, or In(N₀/N) = - λt, which becomes equation 1.
The sample contains mc kg of carbon, leading to an activity measured as A/mc decay per kg. The variable r represents the initial mass of C-14 in the sample at t=0 relative to the total mass of carbon which is calculated as [(total number of C-14 atoms at t = 0) × ma] / total mass of carbon. Thus, N₀ equates to r/ma, which becomes equation 2.
The activity of the radioactive element is directly related to the atom count at the moment. The activity equation A = dN/dt = λ(N) indicates that: A = λ₁(N × mc). Rearranging provides N = A / (λmc), represented in equation 3.
By integrating equations 2 and 3, we can solve for t yielding
t = (1/λ) In(rλmc/m₀A).
