Computing the change in mass $\Delta m$ for the $\beta^{+}$ decay for $\mathrm{_{\ \ 9}^{18}F}$ For the decay: 
$$\require{mhchem}\ce{_9^18F -> _8^18O + e+ + \nu}$$
To compute $E$, I need $\Delta m$, the provided answer looks like: 
$$m_i = 18.000938~\mathrm u$$
$$m_f = 17.999159~\mathrm u + 2~(5.49 \times 10^{-4}~\mathrm u)$$
I believe $m_i$ is mass for $\ce{F}$ and $m_f$ mass for $\ce{O}$? Then where did the numbers $17.999159~\mathrm u$ and $5.49 \times 10^{-4}~\mathrm u$ come from? 
 A: The energy liberated from the reaction :
$$\require{mhchem} \ce{_9^18F -> _8^18O + e+ + \nu}$$
$E=\Delta m\times c^2=(m_f-m_i)c^2=(m_{\ce{_^18O}}+m_{\ce{_^e^+}}+\ce{m_{\nu}}-m_{\ce{_^18F}})\times c^2$
Edit:
$m_{\ce{^{18}O}}=   17.9991603 u$
$m_{\ce{e+}}=   0.000548579909 u$
$m_{\nu}= $ neglected in some cases but you can find that in wikipedia
$m_{\ce{^{18}F}}=18.0009380u$ 
A: The reaction is simply $\require{mhchem}\ce{ _9^{18}F -> _8^{18}O +  _{1}^0e + $\nu_e$}$ (electron neutrino)
The mass of electron and positron are the same. The method for computing the mass defect still remains the same. Have a look at the question on finding $\Delta m$. That number $5.49\times 10^{-4}u$ is the mass of electron. Hence, $m_f$ is not just for oxygen, it's the total mass of products. Since the neutrino's mass is so small, we can just neglect it (or atleast for homework).
I don't think there's a $2$ coming around...
Edit based on comments: The reason it more mass is because in the link (you've provided), they've used oxygen of mass number $16$ which is frequent I think so. In case of this decay, the oxygen has 2 more neutrons. Since O-16, 17 and even 18 are stable, it's not a big problem. In case of the $2$ electrons issue, strictly speaking, no two electrons are formed in this reaction. But as dmckee says, the question creator should've taken both the released electron and the electron neutrino into account. So that, both the leptons may have been assumed to have the same mass.
Here's the Wiki article on F-18 decay.
