# What are the mole fractions of "water, ethanol, acetic acid," in a 25%:25%:50% mixture by mass?

Jun 14, 2017

We assume a mass of $100 \cdot g .$....and come up mole fractions of chi_(H_2O)=0.503; chi_("EtOH")=0.196; chi_("HOAc")=0.301.....

#### Explanation:

We assume a mass of $100 \cdot g .$....and work out the molar quantities on that basis........

$\text{Moles of water} = \frac{25 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} = 1.39 \cdot m o l$

$\text{Moles of ethanol} = \frac{25 \cdot g}{46.07 \cdot g \cdot m o {l}^{-} 1} = 0.543 \cdot m o l$

$\text{Moles of acetic acid} = \frac{50 \cdot g}{60.05 \cdot g \cdot m o {l}^{-} 1} = 0.833 \cdot m o l$

And thus...........

${\chi}_{{H}_{2} O} = \frac{1.39 \cdot m o l}{\left(1.39 + 0.543 + 0.833\right) \cdot m o l} = 0.503$

${\chi}_{\text{EtOH}} = \frac{0.543 \cdot m o l}{\left(1.39 + 0.543 + 0.833\right) \cdot m o l} = 0.196$

${\chi}_{\text{HOAc}} = \frac{1.39 \cdot m o l}{\left(1.39 + 0.543 + 0.833\right) \cdot m o l} = 0.301$

And we note that ${\chi}_{{H}_{2} O} + {\chi}_{\text{EtOH"+chi_"HOAc}} = 1.00$, as is required for a $\Sigma {n}_{i}$ where ${n}_{i}$ are the individual molar quantities.