0
$\begingroup$

The energy of ionization of the potassium is $\pu{6.94\cdot 10^{-19} J/atom}.$ Find the necessary energy to ionize $\pu{1 g}$ of potasium that is found on its fundamental state.

Attempt. I really didn't want to overthink this so I tried with conversion factors: $$\pu{6.94\cdot 10^{-19} J/atom} \cdot \frac{\pu{1 atom}}{\pu{1 molecule}} \cdot \frac{\pu{1 molecule}}{\pu{6.022\cdot 10^{23} moles}} \cdot \frac{\pu{1 mol} \ \text{of } \ce{K}}{\pu{39 g} \ \ce{K}}\\ \approx \pu{2.96775\cdot 10^{-44} J/g},$$ but that didn't give me the correct answer since the correct answer is $\pu{10.7 kJ/g}$. I'm clueless on what else to try. May anyone shine a light?

$\endgroup$
16
  • 3
    $\begingroup$ 6.022e23 mol of potassium is about 0.4% of the Earth mass. $\endgroup$
    – Poutnik
    Sep 19, 2021 at 17:39
  • 2
    $\begingroup$ @Acyex A symbolic algebraic expression typically implies writing an equation in generic form using symbols (not numbers or physical quantities) for all constants and variables. It's a universally recommended initial step of calculation prior to plugging in the numbers so that you tackle the problem conceptually and don't have to deal with units, dimensions and significant figures at the same time as you are figuring out the solution. $\endgroup$
    – andselisk
    Sep 19, 2021 at 17:51
  • 2
    $\begingroup$ Don't overcomplicate. You know the energy per atom. You know the mass of potassium. How many potassium atoms are in 1g? Problem solved. $\endgroup$
    – matt_black
    Sep 19, 2021 at 20:37
  • 2
    $\begingroup$ You have the correct method, but one of your ratios is flipped. $\pu{1 mol}$ is $\pu{6.022E23}$ molecules, you have that 1 molecule is $\pu{6.022E23 mol}.$ As a sanity check, even if you didn't know the right result, a gram should have more energy than a single atom, but your answer was smaller. $\endgroup$
    – Tyberius
    Sep 21, 2021 at 2:22
  • 2
    $\begingroup$ @andselisk $$\pu{6.94E-19 J} \cdot \frac{\pu{6.022E23} }{\pu{1 mol}} \cdot \frac {\pu{1 mol}}{\pu{39 g}}$$ Thanks, E vs e is the very new info for me, I was not aware it is case sensitive, nor what is the difference. $\endgroup$
    – Poutnik
    Sep 23, 2021 at 10:03

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.