I've got a - probably rather not-so-smart question regarding chemical reactions. In many publications, the authors say that they used e.g. t-BuOH/H2O (1:4) as solvent. What exactly does 1:4 mean? Is it one part of t-BuOH of 4 parts of solvent in total, i.e. $\pu{50 mL}$ t-BuOH and $\pu{150 mL}$ of H2O to give a total volume of $\pu{200 mL}$ (4*50), or is it one part of t-BuOH and four times as much water, i.e. $\pu{50 mL}$ t-BuOH and $\pu{200 mL}$ H2O to give a total volume of $\pu{250 mL}$?

Or otherwise said, do I add both numbers to get the total amount of parts, or is the higher number already the total number of parts? In my first example, we could say, that 4 is the total number of parts. In my second example, you'd have to add both numbers to get the total number of parts, i.e. 1 + 4 = 5 parts in total, one of which is t-BuOH and four of which are H2O.

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    $\begingroup$ one part of t-BuOH and four times as much water $\endgroup$
    – MaxW
    Aug 18, 2020 at 7:37
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    $\begingroup$ In literature, when in experimental section, if it says 1:4 of solvent system but didn't verify, most of the time it means 1 part of one solvent and 4 part of the other solvent in volume units (total: 1+ 4, regardless they are additive or not). This is through specifically in chromatography. If it's not in volume, the authors would definitely verify whether it is $w/v$ or $v/w$ or $w/w$. $\endgroup$ Aug 18, 2020 at 23:49

1 Answer 1


As @MaxW already pointed out, 1:4 means that you have to use four times as much water as t-BuOH. But I also want to explain why this is, because you'll probably see this more often.

In fact, 1:4 is a ratio, so what it represents is the ratio between t-BuOH and H2O:

$$ \begin{align} \text{ratio} &= \frac{V(\ce{t-BuOH})}{V(\ce{H2O})}\\ &= \frac{50\text{ mL}}{200\text{ mL}}\\ &= \frac{1}{4} \end{align} $$

And that's why your ratio says 1:4. You could have just rearranged the equation:

$$ \begin{align} 1:4 = \frac{1}{4} = \frac{V(\ce{t-BuOH})}{V(\ce{H2O})}\\ \implies V(\ce{H2O}) = 4 \cdot V(\ce{t-BuOH}) \end{align} $$

Here, both components, water and tert-butanol, are liquids. Sometimes the implicit assumption the ratio refering to the ratio of volumes is expressed explicitly by an indication like (tBuOH-water, 1:4, v/v). Conversely, a mass ratio sometimes may be expressed by m/m.

Side note: As discussed on ChemSE and here, or in the literature (example), be aware of the possibility of a volume contraction. Thus, measure the volumes of the components independently from each other before pouring them into one container in common (e.g., for chromatography).

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    $\begingroup$ +1 for pointing out that exactly 50 ml of t-BuOH and 200 ml water will not yield exactly 250 ml of solution. $\endgroup$
    – MaxW
    Aug 18, 2020 at 17:57

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